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Chapter 10

Mathematics: Algebra, Trigonometry, and Calculus

TU BSc CSIT entrance past exam questions

Past questions and answers

305 questions set from this chapter, 1 of them more than once; 6 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 3 of 13 exams
  • 2071 · 1 mark

The two lines 2x2−2xy+ky2=02x^2-2xy+ky^2=0 are perpendicular to each other if kk is (a) 2 (b) -2 (c) 0 (d) 1

Similar questions: The lines represented by 5x 2+3xy+ky 2=0 are perpendicular... (2074) · The two lines represented by ax 2+4xy+2y 2=0 are... (2079)

Answer

(b) -2

Lines given by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular if a+b=0a+b=0. Here 2+k=02+k=0, so k=−2k=-2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 13 exams
  • 2074 · 1 mark

The quadratic equation whose roots are -3 and 2 is (a) x2+x−6=0x^2+x-6=0 (b) x2+x+6=0x^2+x+6=0 (c) x2−x−6=0x^2-x-6=0 (d) x2−x+6=0x^2-x+6=0

Similar questions: The quadratic equation whose roots are 3 and 4 is (2075) · A quadratic equation whose roots are 2, -3 is (2077)

Answer

(a) x2+x−6=0x^2+x-6=0

Sum of roots =−1=-1, product =−6=-6, so x2−(−1)x+(−6)=x2+x−6=0x^2-(-1)x+(-6)=x^2+x-6=0.

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  • Most repeated · 3 of 13 exams
  • 2074 · 1 mark

The lines represented by 5x2+3xy+ky2=05x^2+3xy+ky^2=0 are perpendicular to each other for k=k= (a) 5 (b) 3 (c) 0 (d) -5

Similar questions: The two lines represented by ax 2+4xy+2y 2=0 are... (2079) · The two lines 2x 2-2xy+ky 2=0 are perpendicular to each... (2071)

Answer

(d) -5

For ax2+2hxy+by2=0ax^2+2hxy+by^2=0 the lines are perpendicular when a+b=0a+b=0, so 5+k=0⇒k=−55+k=0\Rightarrow k=-5.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 13 exams
  • 2075 · 1 mark

The quadratic equation whose roots are 3 and 4 is (a) x2−7x−12=0x^2-7x-12=0 (b) x2−7x+12=0x^2-7x+12=0 (c) x2+7x+12=0x^2+7x+12=0 (d) x2+7x−12=0x^2+7x-12=0

Similar questions: The quadratic equation whose roots are -3 and 2 is (2074) · A quadratic equation whose roots are 2, -3 is (2077)

Answer

(b) x2−7x+12=0x^2-7x+12=0 Sum of roots =7=7 and product =12=12, so x2−7x+12=0x^2-7x+12=0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 13 exams
  • 2077 · 1 mark

A quadratic equation whose roots are 2, -3 is (a) x2+x+6=0x^2+x+6=0 (b) x2−x+6=0x^2-x+6=0 (c) x2+x−6=0x^2+x-6=0 (d) x2−x−5=0x^2-x-5=0

Similar questions: The quadratic equation whose roots are -3 and 2 is (2074) · The quadratic equation whose roots are 3 and 4 is (2075)

Answer

(c) x2+x−6=0x^2+x-6=0 Sum of roots =2+(−3)=−1=2+(-3)=-1 and product =2×(−3)=−6=2\times(-3)=-6. The equation x2−(sum)x+product=0x^2-(\text{sum})x+\text{product}=0 gives x2+x−6=0x^2+x-6=0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Most repeated · 3 of 13 exams
  • 2079 · 1 mark

The two lines represented by ax2+4xy+2y2=0ax^2+4xy+2y^2=0 are perpendicular to each other for a=a= (a) 0 (b) -2 (c) 2 (d) 1

Similar questions: The lines represented by 5x 2+3xy+ky 2=0 are perpendicular... (2074) · The two lines 2x 2-2xy+ky 2=0 are perpendicular to each... (2071)

Answer

(b) -2 Lines ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here b=2b=2, so a=−2a=-2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • Asked 2 times
  • 2070 · 1 mark
  • 2071 · 1 mark

The derivative of y=cos⁡(cos⁡x)y=\cos(\cos\sqrt x) is (a) sin⁡(cos⁡x)⋅sin⁡x2x\frac{\sin(\cos\sqrt x)\cdot\sin\sqrt x}{2\sqrt x} (b) sin⁡(cos⁡x)\sin(\cos\sqrt x) (c) sin⁡(sin⁡x)\sin(\sin\sqrt x) (d) sin⁡(cos⁡x)2x\frac{\sin(\cos\sqrt x)}{2\sqrt x}

Answer

(a) sin⁡(cos⁡x)⋅sin⁡x2x\frac{\sin(\cos\sqrt x)\cdot\sin\sqrt x}{2\sqrt x}

dydx=−sin⁡(cos⁡x)⋅(−sin⁡x)⋅12x=sin⁡(cos⁡x)sin⁡x2x\frac{dy}{dx}=-\sin(\cos\sqrt x)\cdot(-\sin\sqrt x)\cdot\frac1{2\sqrt x}=\frac{\sin(\cos\sqrt x)\sin\sqrt x}{2\sqrt x}.

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  • 2070 · 1 mark

What is the value of ∣1ωω2ωω21ω21ω∣\begin{vmatrix}1&\omega&\omega^2\\ \omega&\omega^2&1\\ \omega^2&1&\omega\end{vmatrix}, where ω\omega is an imaginary cube root of unity? (a) 1 (b) 0 (c) 3 (d) 2

Similar questions: The value of x+wy+w 2z y+wz+w 2x , where w is the cube root... (2071)

Answer

(b) 0

Adding all columns to the first gives a first column of 1+ω+ω2=01+\omega+\omega^2=0 in every row. A determinant with a zero column is 00.

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  • 2071 · 1 mark

The system of equations 2x+y=52x+y=5; 4x+2y=84x+2y=8 is (a) Consistent and dependent (b) Inconsistent and independent (c) Inconsistent and dependent (d) Consistent and independent

Similar questions: The system of equation 3x-2y=1 and 6x-4y=2 is (2078)

Answer

(b) Inconsistent and independent

The ratios are 24=12≠58\frac24=\frac12\neq\frac58, so the lines are parallel and the system has no solution (inconsistent). The equations are not multiples of each other, so they are independent.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2071 · 1 mark

The value of x+wy+w2zy+wz+w2x\frac{x+wy+w^2z}{y+wz+w^2x}, where ww is the cube root of unity, is (a) ww (b) x+y+zx+y+z (c) w2w^2 (d) 1

Similar questions: What is the value of vmatrix 1& & 2 & 2&1 2&1& vmatrix ,... (2070)

Answer

(a) ww

w2(x+wy+w2z)=w2x+y+wzw^2(x+wy+w^2z)=w^2x+y+wz, which is the denominator. So the ratio is 1w2=w\frac1{w^2}=w, since w3=1w^3=1.

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  • 2072 · 1 mark

Area bounded by the x-axis and the curve y=3x2y=3x^2 and the ordinate of x=3x=3 and x=4x=4 is (a) 37 (b) 8 (c) 19 (d) 0

Similar questions: The area bounded by the x-axis and the curve y=4x 2 and the... (2075)

Answer

(a) 37

A=∫343x2 dx=[x3]34=64−27=37A=\int_3^4 3x^2\,dx=\left[x^3\right]_3^4=64-27=37 square units.

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  • 2075 · 1 mark

The area bounded by the x-axis and the curve y=4x2y=4x^2 and the ordinate x=1x=1 and x=4x=4 is (a) 64 (b) 84 (c) 16 (d) -84

Similar questions: Area bounded by the x-axis and the curve y=3x 2 and the... (2072)

Answer

(b) 84

A=∫144x2dx=43[x3]14=43(64−1)=84A=\int_1^4 4x^2dx=\frac{4}{3}\left[x^3\right]_1^4=\frac43(64-1)=84

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  • 2077 · 1 mark

Area bounded by the x-axis and the curve y=3x2y=3x^2 and the lines x=2x=2 and x=4x=4 is (a) 64 (b) 56 (c) 563\frac{56}{3} (d) 72

Similar questions: Area bounded by x- axis and the curve y=x 2 and the lines... (2079)

Answer

(b) 56

∫243x2 dx=[x3]24=64−8=56\int_2^4 3x^2\,dx = [x^3]_2^4 = 64-8 = 56

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The system of equation 3x−2y=13x-2y=1 and 6x−4y=26x-4y=2 is (a) inconsistent and dependent (b) consistent and dependent (c) consistent and independent (d) inconsistent and independent

Similar questions: The system of equations 2x+y=5; 4x+2y=8 is (2071)

Answer

(b) consistent and dependent

The second equation is exactly twice the first, so the two lines coincide. The system has infinitely many solutions: it is consistent and dependent.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

Area bounded by x- axis and the curve y=x2y=x^2 and the lines x=0x=0 and x=4x=4 is (a) 64 (b) 16 (c) 16/3 (d) 64/3

Similar questions: Area bounded by the x-axis and the curve y=3x 2 and the... (2077)

Answer

(d) 64/3 ∫04x2dx=433=643\int_0^4x^2dx=\frac{4^3}{3}=\frac{64}{3}

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  • 2069 · Entrance Examination 2069 · 1 mark

lim⁡x→2x3−23x−2\lim_{x \to 2} \frac{x^3 - 23}{x-2} is (a) 12 (b) 4 (c) 9 (d) 0/00/0

Answer

(a) 12

Treating the expression as the standard limit
[ \lim_{x\to2}\frac{x^3-2^3}{x-2} ] factor the numerator: [ \frac{x^3-8}{x-2}=\frac{(x-2)(x^2+2x+4)}{x-2}=x^2+2x+4 ] Putting (x=2): [ 2^2+2(2)+4=4+4+4=12 ]

So the limit is 12.
(If the numerator is literally (x^3-23), then the limit is not finite; the intended question is (x^3-2^3).)

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  • 2069 · Entrance Examination 2069 · 1 mark

lim⁡x→0tan⁡x−5xx\lim_{x \to 0} \frac{\tan x - 5x}{x} is equal to (a) 6 (b) -4 (c) 0 (d) 0/00/0

Answer

(b) -4

[ \lim_{x\to 0}\frac{\tan x-5x}{x} =\lim_{x\to 0}\left(\frac{\tan x}{x}-5\right) =1-5=-4 ]

since (\lim_{x\to 0}\frac{\tan x}{x}=1).

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  • 2069 · Entrance Examination 2069 · 1 mark

The function f(x)={x+2,x≥22x,x<2f(x) = \begin{cases} x+2, & x \ge 2 \\ 2x, & x < 2 \end{cases} is continuous at (a) x=0x=0 (b) x=2x=2 (c) x=−2x=-2 (d) x=1x=1

Answer

(a), (b), (c), and (d)

The function is continuous at all the given points. At (x=2), [ \lim_{x\to2^-}f(x)=\lim_{x\to2^-}2x=4,\qquad \lim_{x\to2^+}f(x)=\lim_{x\to2^+}(x+2)=4,\qquad f(2)=2+2=4, ] so it is continuous at (x=2). For (x=0,-2,1), since each is (<2), (f(x)=2x), which is a polynomial and hence continuous. Thus the question as stated has no unique single correct option.

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  • 2069 · Entrance Examination 2069 · 1 mark

If the derivative of x2+3xx^2 + 3x is 1, the value of x is (a) 0 (b) -1 (c) 2 (d) 5

Answer

(b) -1

The derivative of x2+3xx^2+3x is 2x+32x+3. Setting 2x+3=12x+3=1 gives 2x=−22x=-2, so x=−1x=-1.

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  • 2069 · Entrance Examination 2069 · 1 mark

If x=2t+1x = 2t + 1 and y=t2y = t^2, then dy/dtdy/dt is (a) t (b) 1/t1/t (c) 2t2t (d) 0

Answer

(c) 2t2t

Since y=t2y=t^2, differentiating with respect to tt gives dydt=2t\frac{dy}{dt}=2t.

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  • 2069 · Entrance Examination 2069 · 1 mark

The points of inflection for the curve x3+4xx^3 + 4x is (a) 0 (b) 1 (c) 6 (d) -1

Answer

(a) 0

For (y=x^3+4x),
[ \frac{dy}{dx}=3x^2+4,\qquad \frac{d^2y}{dx^2}=6x ] Set (\frac{d^2y}{dx^2}=0): [ 6x=0 \Rightarrow x=0 ] Since (\frac{d^2y}{dx^2}) changes sign at (x=0), the point of inflection is at (x=0).

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  • 2069 · Entrance Examination 2069 · 1 mark

2∫01xdx2 \int_0^1 x dx is equal to (a) 0 (b) 1 (c) -1 (d) 2

Answer

(b) 1

Since ∫01x dx=[x22]01=12\int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac12, multiplying by 2 gives 2×12=12 \times \frac12 = 1.

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  • 2069 · Entrance Examination 2069 · 1 mark

∫cos⁡2x dx\int \cos 2x \, dx is equal to (a) sin⁡2x\sin 2x (b) −cos⁡2x2-\frac{\cos 2x}{2} (c) sin⁡2x\sin^2 x (d) 2sin⁡2x2\sin 2x

Answer

(None of the above)
[ \int \cos 2x,dx=\frac{\sin 2x}{2}+C. ] None of (a)–(d) is correct. Option (b) is the antiderivative of (\sin 2x), not (\cos 2x).

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  • 2070 · 1 mark

If A and B are any two sets, then A−(A−B)A-(A-B) is equal to (a) B−AB-A (b) A∩BA \cap B (c) A∪BA \cup B (d) ϕ\phi

Answer

(b) A∩BA\cap B

Since A−BA-B is the part of AA outside BB, removing it from AA leaves the part of AA that lies in BB, so A−(A−B)=A∩BA-(A-B)=A\cap B.

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  • 2070 · 1 mark

The range of the function y=(a2−x2)y=\sqrt{(a^2-x^2)}, a>0a>0 is equal to (a) [−a,a][-a,a] (b) (−a,a](-a,a] (c) [0,a][0,a] (d) (0,a)(0,a)

Answer

(c) [0,a][0,a]

For y=a2−x2y=\sqrt{a^2-x^2} the domain is −a≤x≤a-a\le x\le a. Then a2−x2a^2-x^2 varies from 00 to a2a^2, so yy varies from 00 to aa.

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  • 2070 · 1 mark

What is the value of log⁡aa3a2\log_a \sqrt{a^3\sqrt{a^2}} ? (a) 4 (b) 3 (c) 2 (d) 1

Answer

(c) 2

a2=a\sqrt{a^2}=a, so a3⋅a=a4=a2\sqrt{a^3\cdot a}=\sqrt{a^4}=a^2. Hence log⁡aa2=2\log_a a^2=2.

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  • 2070 · 1 mark

If A=(01−10)A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, then A100A^{100} is equal to (a) 1 (b) 0 (c) -1 (d) None

Answer

(a) 1 (the identity matrix II)

A2=(01−10)2=(−100−1)=−IA^2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}^2=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=-I. So A100=(A2)50=(−I)50=IA^{100}=(A^2)^{50}=(-I)^{50}=I.

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  • 2070 · 1 mark

If ax=by=cza^x=b^y=c^z and a,b,ca, b, c are in G.P. then x,y,zx, y, z are in (a) G.P. (b) A.P. (c) H.P. (d) None

Answer

(c) H.P.

Let ax=by=cz=ka^x=b^y=c^z=k, so a=k1/x, b=k1/y, c=k1/za=k^{1/x},\ b=k^{1/y},\ c=k^{1/z}. Since b2=acb^2=ac, 2y=1x+1z\frac2y=\frac1x+\frac1z, so x,y,zx,y,z are in H.P.

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  • 2070 · 1 mark

lim⁡x→0ax−bxx\lim_{x\to0}\frac{a^x-b^x}{x} is equal to (a) 0 (b) log⁡(ab)\log(ab) (c) log⁡(a/b)\log(a/b) (d) log⁡(b/a)\log(b/a)

Answer

(c) log⁡(a/b)\log(a/b)

lim⁡x→0ax−1x−bx−1x=log⁡a−log⁡b=log⁡(a/b)\lim_{x\to0}\frac{a^x-1}{x}-\frac{b^x-1}{x}=\log a-\log b=\log(a/b).

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  • 2070 · 1 mark

The derivative of y=5xy=5^x with respect to xx is equal to (a) 5xlog⁡e55^x\log_e 5 (b) 5x5^x (c) log⁡e5\log_e 5 (d) 5log⁡e55\log_e 5

Answer

(a) 5xlog⁡e55^x\log_e 5

Using ddxax=axln⁡a\frac{d}{dx}a^x=a^x\ln a.

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  • 2070 · 1 mark

The function f(x)=x−1x2f(x)=x-\frac{1}{x^2}, x∈Rx\in R, x≠0x\neq0 is (a) (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty) (b) (−∞,0]∪[0,∞)(-\infty,0]\cup[0,\infty) (c) [−1,1][-1,1] (d) (−∞,∞)(-\infty,\infty)

Answer

(a) (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty)

f(x)f(x) is defined for every real xx except x=0x=0, so its domain is (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty).

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  • 2070 · 1 mark

The integral ∫ex dxex+1\int\frac{e^x\,dx}{e^x+1} is equal to (a) log⁡(ex+1)+C\log(e^x+1)+C (b) −1e2x+1+C\frac{-1}{e^{2x}+1}+C (c) −12(ex+e−x)+C\frac{-1}{2(e^x+e^{-x})}+C (d) −12(e2x+1)+C\frac{-1}{2(e^{2x}+1)}+C

Answer

(a) log⁡(ex+1)+C\log(e^x+1)+C

Put u=ex+1u=e^x+1, du=ex dxdu=e^x\,dx. Then ∫duu=log⁡u+C=log⁡(ex+1)+C\int\frac{du}{u}=\log u+C=\log(e^x+1)+C.

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  • 2070 · 1 mark

For what value of kk will the equation 5x2−kx+45=05x^2-kx+45=0 have equal roots? (a) 0 (b) ±3\pm3 (c) ±30\pm30 (d) ±33\pm33

Answer

(c) ±30\pm30

Equal roots need D=0D=0: k2−4(5)(45)=0⇒k2=900⇒k=±30k^2-4(5)(45)=0\Rightarrow k^2=900\Rightarrow k=\pm30.

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  • 2070 · 1 mark

The quadratic equation of whose one root is (2+3)(2+\sqrt3) is (a) x2+4x+1=0x^2+4x+1=0 (b) x2−4x+1=0x^2-4x+1=0 (c) x2−4x−1=0x^2-4x-1=0 (d) −x2−4x−1=0-x^2-4x-1=0

Answer

(b) x2−4x+1=0x^2-4x+1=0

The other root is 2−32-\sqrt3 (conjugate). Sum =4=4, product =4−3=1=4-3=1, so x2−4x+1=0x^2-4x+1=0.

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  • 2070 · 1 mark

The distance between the parallel lines y−2x=4y-2x=4 and 6x−3y=56x-3y=5 is (a) 145\frac{1}{\sqrt{45}} (b) 745\frac{7}{\sqrt{45}} (c) 1745\frac{17}{\sqrt{45}} (d) 17245\frac{17}{2\sqrt{45}}

Answer

(c) 1745\frac{17}{\sqrt{45}}

Write both lines with the same coefficients: 2x−y+4=02x-y+4=0 and 2x−y−53=02x-y-\frac53=0. Distance =∣4+53∣22+1=17/35=1735=1745=\frac{|4+\frac53|}{\sqrt{2^2+1}}=\frac{17/3}{\sqrt5}=\frac{17}{3\sqrt5}=\frac{17}{\sqrt{45}}.

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  • 2070 · 1 mark

Radius of the circle x2+y2−3x+2y−34=0x^2+y^2-3x+2y-\frac34=0 is (a) 2 (b) 1 (c) 4 (d) 3

Answer

(a) 2

Here g=−32, f=1, c=−34g=-\frac32,\ f=1,\ c=-\frac34. Radius =g2+f2−c=94+1+34=4=2=\sqrt{g^2+f^2-c}=\sqrt{\frac94+1+\frac34}=\sqrt4=2.

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  • 2070 · 1 mark

The absolute value of the complex number (1+i)−1(1+i)^{-1} is (a) 1 (b) 2\sqrt2 (c) 12\frac{1}{\sqrt2} (d) 2

Answer

(c) 12\frac1{\sqrt2}

∣(1+i)−1∣=1∣1+i∣=12|(1+i)^{-1}|=\frac1{|1+i|}=\frac1{\sqrt2}.

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  • 2070 · 1 mark

If z=cos⁡θ+isin⁡θz=\cos\theta+i\sin\theta then zn+1znz^n+\frac{1}{z^n} is equal to (a) −2cos⁡nθ-2\cos n\theta (b) 2cos⁡nθ2\cos n\theta (c) 2sin⁡nθ2\sin n\theta (d) −2sin⁡nθ-2\sin n\theta

Answer

(b) 2cos⁡nθ2\cos n\theta

By De Moivre, zn=cos⁡nθ+isin⁡nθz^n=\cos n\theta+i\sin n\theta and z−n=cos⁡nθ−isin⁡nθz^{-n}=\cos n\theta-i\sin n\theta. Adding gives 2cos⁡nθ2\cos n\theta.

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  • 2070 · 1 mark

The nthn^{th} term of the series: 2+6+12+20+…2+6+12+20+\ldots is (a) n(n+1)n(n+1) (b) (n+1)(n+2)(n+1)(n+2) (c) n(n−1)n(n-1) (d) nn

Answer

(a) n(n+1)n(n+1)

The terms are 1⋅2, 2⋅3, 3⋅4, 4⋅5,…1\cdot2,\ 2\cdot3,\ 3\cdot4,\ 4\cdot5,\dots, so Tn=n(n+1)T_n=n(n+1).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2070 · 1 mark

If ∣x∣<1|x|<1 and y=x+x2+x3+x4+…+∞y=x+x^2+x^3+x^4+\ldots+\infty, then xx is equal to (a) y1−y\frac{y}{1-y} (b) yy−1\frac{y}{y-1} (c) y+1y\frac{y+1}{y} (d) y1+y\frac{y}{1+y}

Answer

(d) y1+y\frac{y}{1+y}

The sum of the infinite G.P. is y=x1−xy=\frac{x}{1-x}. So y−xy=xy-xy=x, giving x=y1+yx=\frac{y}{1+y}.

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3cos⁡−1x3\cos^{-1}x is equal to (a) cos⁡−1(3x−4x2)\cos^{-1}(3x-4x^2) (b) cos⁡−1(4x3−3x)\cos^{-1}(4x^3-3x) (c) cos⁡−1(4x−3x3)\cos^{-1}(4x-3x^3) (d) cos⁡−1(3x3−4x)\cos^{-1}(3x^3-4x)

Answer

(b) cos⁡−1(4x3−3x)\cos^{-1}(4x^3-3x)

Let θ=cos⁡−1x\theta=\cos^{-1}x. Then cos⁡3θ=4cos⁡3θ−3cos⁡θ=4x3−3x\cos3\theta=4\cos^3\theta-3\cos\theta=4x^3-3x, so 3cos⁡−1x=cos⁡−1(4x3−3x)3\cos^{-1}x=\cos^{-1}(4x^3-3x).

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Solution of tan⁡ax=cot⁡bx\tan ax=\cot bx is (a) 2n+1a+b⋅π4\frac{2n+1}{a+b}\cdot\frac{\pi}{4} (b) 2n−1a−b⋅π2\frac{2n-1}{a-b}\cdot\frac{\pi}{2} (c) 2n+1a+b⋅π2\frac{2n+1}{a+b}\cdot\frac{\pi}{2} (d) π2\frac{\pi}{2}

Answer

(c) 2n+1a+b⋅π2\frac{2n+1}{a+b}\cdot\frac\pi2

tan⁡ax=cot⁡bx=tan⁡(π2−bx)\tan ax=\cot bx=\tan\left(\frac\pi2-bx\right), so ax=nπ+π2−bxax=n\pi+\frac\pi2-bx. Thus x=(2n+1)π2(a+b)=2n+1a+b⋅π2x=\frac{(2n+1)\pi}{2(a+b)}=\frac{2n+1}{a+b}\cdot\frac\pi2.

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The single equation representing the pairs of lines y=xy=x and y=−xy=-x is (a) y2−x2=0y^2-x^2=0 (b) y−x=0y-x=0 (c) x−y=0x-y=0 (d) x2+y2=0x^2+y^2=0

Answer

(a) y2−x2=0y^2-x^2=0

(y−x)(y+x)=0⇒y2−x2=0(y-x)(y+x)=0\Rightarrow y^2-x^2=0.

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Let A=[−3,2]A=[-3,2] and B=[2,3]B=[2,3] then A−BA-B is equal to (a) [−3,3)[-3,3) (b) [−3,2)[-3,2) (c) (−2,2)(-2,2) (d) (−2,1)(-2,1)

Answer

(b) [−3,2)[-3,2)

A−BA-B keeps the elements of [−3,2][-3,2] not in [2,3][2,3]. The point 22 lies in BB, so it is removed: [−3,2)[-3,2).

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If lim⁡x→0−f(x)≠lim⁡x→0+f(x)\lim_{x\to0^-}f(x)\neq\lim_{x\to0^+}f(x), then f(x)f(x) is said to be an (a) infinite discontinuity (b) removable discontinuity (c) jump discontinuity (d) none

Answer

(c) jump discontinuity

When both one-sided limits exist but are unequal, the function has a jump discontinuity at that point.

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Which one of the following is the area enclosed by the curve y=3xy=3x, the xx-axis and the ordinate x=0x=0, x=4x=4? (a) 12 (b) 16 (c) 20 (d) 24

Answer

(d) 24

Area=∫043x dx=3x22∣04=24\text{Area}=\int_0^4 3x\,dx=\frac{3x^2}{2}\Big|_0^4=24 square units.

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If ∣x−3∣=4|x-3|=4 then xx is (a) 7 and -1 (b) 7 and 1 (c) 7 (d) 1

Answer

(a) 7 and -1

∣x−3∣=4|x-3|=4 gives x−3=4x-3=4 or x−3=−4x-3=-4, so x=7x=7 or x=−1x=-1.

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If {x−x2=9\{x-x^2=9 and 2x=6}2x=6\}, then (a) null set (b) {3}\{3\} (c) {3,−3}\{3,-3\} (d) {4,3}\{4,3\}

Answer

(a) null set

x−x2=9x-x^2=9 has no real solution, since x2−x+9=0x^2-x+9=0 has discriminant 1−36<01-36<0. Also x=3x=3 does not satisfy it (3−9=−63-9=-6). So the set is empty.

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If f:R→Rf:R\to R defined by f(x)=x2+3f(x)=x^2+3 then f−1(4)f^{-1}(4) is (a) 0 (b) {−1,1}\{-1,1\} (c) {1,0}\{1,0\} (d) {4,3}\{4,3\}

Answer

(b) {−1,1}\{-1,1\}

f−1(4)f^{-1}(4) is the set of xx with x2+3=4x^2+3=4, so x2=1x^2=1 and x=±1x=\pm1.

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The value of log⁡22+log⁡33+log⁡44\log_2\sqrt2+\log_3\sqrt3+\log_4\sqrt4 is (a) 0 (b) 32\frac32 (c) −1312-\frac{13}{12} (d) π6\frac{\pi}{6}

Answer

(b) 32\frac32

log⁡22=12\log_2\sqrt2=\frac12, log⁡33=12\log_3\sqrt3=\frac12 and log⁡44=log⁡42=12\log_4\sqrt4=\log_4 2=\frac12. The sum is 32\frac32.

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If cos⁡2θ=12\cos^2\theta=\frac12, then θ\theta is (a) nπ±π4n\pi\pm\frac{\pi}{4} (b) nπ+π4n\pi+\frac{\pi}{4} (c) nπ±π3n\pi\pm\frac{\pi}{3} (d) 2nπ2n\pi

Answer

(a) nπ±π4n\pi\pm\frac{\pi}{4}

cos⁡2θ=12=cos⁡2π4\cos^2\theta=\frac12=\cos^2\frac{\pi}{4}, so θ=nπ±π4\theta=n\pi\pm\frac{\pi}{4}.

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cos⁡(sin⁡−135)=\cos\left(\sin^{-1}\frac35\right)= (a) 35\frac35 (b) 45\frac45 (c) 925\frac{9}{25} (d) 425\frac{4}{25}

Answer

(b) 45\frac45

Let sin⁡−135=α\sin^{-1}\frac35=\alpha, so sin⁡α=35\sin\alpha=\frac35. Then cos⁡α=1−925=45\cos\alpha=\sqrt{1-\frac{9}{25}}=\frac45.

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If a2,b2,c2a^2, b^2, c^2 are in H.P. then (a) 2a2c2=b2(a2+c2)2a^2c^2=b^2(a^2+c^2) (b) a2c2=b2(a2+c2)a^2c^2=b^2(a^2+c^2) (c) 2b2c2=a2(b2+c2)2b^2c^2=a^2(b^2+c^2) (d) c2a2=2b2(a2+c2)c^2a^2=2b^2(a^2+c^2)

Answer

(a) 2a2c2=b2(a2+c2)2a^2c^2=b^2(a^2+c^2)

If a2,b2,c2a^2,b^2,c^2 are in H.P., then 1a2,1b2,1c2\frac1{a^2},\frac1{b^2},\frac1{c^2} are in A.P. So 2b2=1a2+1c2\frac2{b^2}=\frac1{a^2}+\frac1{c^2}, which gives 2a2c2=b2(a2+c2)2a^2c^2=b^2(a^2+c^2).

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If A=[1223]A=\begin{bmatrix}1&2\\2&3\end{bmatrix} and A2−nA−I=0A^2-nA-I=0, then the value of nn is (a) 0 (b) 4 (c) 1 (d) -4

Answer

(b) 4

A2=[58813]A^2=\begin{bmatrix}5&8\\8&13\end{bmatrix}

Comparing A2−nA−I=0A^2-nA-I=0 entry by entry: 5−n−1=05-n-1=0 gives n=4n=4. Check: 8−2n=08-2n=0 gives n=4n=4, and 13−3n−1=013-3n-1=0 gives n=4n=4.

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If A=[101011100]A=\begin{bmatrix}1&0&1\\0&1&1\\1&0&0\end{bmatrix}, then AA is (a) Symmetric (b) Skew-symmetric (c) Singular (d) Nonsingular

Answer

(d) Nonsingular

∣A∣=1(0−0)−0+1(0−1)=−1≠0|A|=1(0-0)-0+1(0-1)=-1\neq0, so AA is nonsingular. It is not symmetric, since a23=1a_{23}=1 but a32=0a_{32}=0.

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The value of i7+1i4i^7+\frac{1}{i^4} is (a) 1 (b) -1 (c) −i-i (d) −i+1-i+1

Answer

(d) −i+1-i+1

i7=i4⋅i3=−ii^7=i^4\cdot i^3=-i and 1i4=1\frac1{i^4}=1. The sum is 1−i1-i.

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If the roots of x2+4x+k=0x^2+4x+k=0 are equal then kk is (a) 0 (b) 4 (c) 16 (d) -4

Answer

(b) 4

Equal roots need discriminant zero: 16−4k=016-4k=0, so k=4k=4.

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The two lines a1x+b1y+c1=0a_1x+b_1y+c_1=0 and a2x+b2y+c2=0a_2x+b_2y+c_2=0 are identical if (a) a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} (b) a1a2=0\frac{a_1}{a_2}=0 (c) b1b2=0\frac{b_1}{b_2}=0 (d) c1c2=0\frac{c_1}{c_2}=0

Answer

(a) a1a2=b1b2=c1c2\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}

Two lines coincide when their coefficients are proportional.

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The equation of the circle touching the positive x-axis and y-axis with radius 2 units is (a) x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0 (b) x2+y2+4x+4y+4=0x^2+y^2+4x+4y+4=0 (c) x2+y2−4x+4y+4=0x^2+y^2-4x+4y+4=0 (d) x2+y2−4x+4y−4=0x^2+y^2-4x+4y-4=0

Answer

(a) x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0

A circle of radius 2 touching the positive axes has centre (2,2)(2,2). Its equation is (x−2)2+(y−2)2=4(x-2)^2+(y-2)^2=4, which expands to x2+y2−4x−4y+4=0x^2+y^2-4x-4y+4=0.

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If the line 3x+4y=k3x+4y=k is tangent to the circle x2+y2=25x^2+y^2=25 at (3,4)(3,4) then kk is (a) 25 (b) -25 (c) ±25\pm25 (d) ±1\pm1

Answer

(a) 25

The point (3,4)(3,4) lies on the line, so k=3(3)+4(4)=25k=3(3)+4(4)=25. The tangent to x2+y2=25x^2+y^2=25 at (3,4)(3,4) is indeed 3x+4y=253x+4y=25.

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The value of lim⁡x→2x5−25x−2\lim_{x\to2}\frac{x^5-2^5}{x-2} is (a) 80 (b) 40 (c) 10 (d) 0

Answer

(a) 80

lim⁡x→axn−anx−a=nan−1\lim_{x\to a}\frac{x^n-a^n}{x-a}=na^{n-1}, so the limit is 5⋅24=805\cdot2^4=80.

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The value of lim⁡x→03x+4tan⁡2xx\lim_{x\to0}\frac{3x+4\tan2x}{x} is (a) 7 (b) 3 (c) 11 (d) 4

Answer

(c) 11

lim⁡x→0(3+4⋅tan⁡2xx)=3+4⋅2=11\lim_{x\to0}\left(3+4\cdot\frac{\tan2x}{x}\right)=3+4\cdot2=11

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The derivative of y=bxy=b^x is (a) xbx−1xb^{x-1} (b) bb (c) xbln⁡∣b∣x^b\ln|b| (d) bxln⁡∣b∣b^x\ln|b|

Answer

(d) bxln⁡∣b∣b^x\ln|b|

The derivative of bxb^x is bxln⁡bb^x\ln b.

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The value of xx for which the derivative of x2+2x+2x^2+2x+2 is zero is (a) 1 (b) 2 (c) 0 (d) -1

Answer

(d) −1-1

ddx(x2+2x+2)=2x+2=0\frac{d}{dx}(x^2+2x+2)=2x+2=0, so x=−1x=-1.

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The point of inflection of the curve x3+3x2−9x+2x^3+3x^2-9x+2 is (a) -1 (b) 1 (c) 6 (d) 0

Answer

(a) −1-1

y′=3x2+6x−9y'=3x^2+6x-9, y′′=6x+6=0⇒x=−1y''=6x+6=0\Rightarrow x=-1, and y′′y'' changes sign there, so it is the point of inflection.

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∫dx1−cos⁡x=\int\frac{dx}{1-\cos x}= (a) tan⁡x2\tan\frac x2 (b) −cot⁡x2-\cot\frac x2 (c) cos⁡x2\cos\frac x2 (d) −tan⁡x2-\tan\frac x2

Answer

(b) −cot⁡x2-\cot\frac x2

Since 1−cos⁡x=2sin⁡2x21-\cos x=2\sin^2\frac x2, ∫dx1−cos⁡x=12∫cosec⁡2x2 dx=−cot⁡x2+C.\int\frac{dx}{1-\cos x}=\frac12\int\operatorname{cosec}^2\frac x2\,dx=-\cot\frac x2+C.

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∫0π/2cos⁡x1+sin⁡x dx=\int_0^{\pi/2}\frac{\cos x}{1+\sin x}\,dx= (a) ∞\infty (b) ln⁡∣2∣\ln|2| (c) 0 (d) 1

Answer

(b) ln⁡2\ln 2

Put u=1+sin⁡xu=1+\sin x: ∫0π/2cos⁡x1+sin⁡xdx=[ln⁡(1+sin⁡x)]0π/2=ln⁡2−ln⁡1=ln⁡2\int_0^{\pi/2}\frac{\cos x}{1+\sin x}dx=\left[\ln(1+\sin x)\right]_0^{\pi/2}=\ln2-\ln1=\ln 2.

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  • 2072 · 1 mark

If a,b∈Ra, b\in R, a/ba/b is indeterminate, if (a) a=0,b≠0a=0, b\neq0 (b) a=0,b=0a=0, b=0 (c) a≠0,b≠0a\neq0, b\neq0 (d) a≠0,b=0a\neq0, b=0

Answer

(b) a=0,b=0a=0, b=0

0/00/0 is indeterminate because any number satisfies 0×x=00\times x=0. If a≠0,b=0a\neq0, b=0 it is undefined, and if a=0,b≠0a=0, b\neq0 it equals 0.

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  • 2072 · 1 mark

(−∞,∞)=(-\infty,\infty)= (a) {x:−∞≤x≤∞}\{x: -\infty\leq x\leq\infty\} (b) {x:−∞<x<0}\{x: -\infty<x<0\} (c) {x:−∞<x<∞}\{x: -\infty<x<\infty\} (d) {x:x≥0}\{x: x\geq0\}

Answer

(c) {x:−∞<x<∞}\{x: -\infty<x<\infty\}

The open interval (−∞,∞)(-\infty,\infty) contains every real number; the infinities are not included, so strict inequalities are used.

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The statement is (a) Bagmati is a River. (b) x>0x>0 (c) How old are you? (d) y>0y>0

Answer

(a) Bagmati is a River.

A statement (proposition) is a sentence that is definitely true or false. "Bagmati is a River" is true. x>0x>0 and y>0y>0 depend on the value of the variable, and a question has no truth value.

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If (2x−1,−3)=(3,y+3)(2x-1,-3)=(3,y+3), then (a) x=1,y=0x=1, y=0 (b) x=−1,y=−3x=-1, y=-3 (c) x=2,y=−6x=2, y=-6 (d) x=0,y=−1x=0, y=-1

Answer

(c) x=2, y=−6x=2,\ y=-6

Equal ordered pairs have equal components: 2x−1=3⇒x=22x-1=3\Rightarrow x=2 and y+3=−3⇒y=−6y+3=-3\Rightarrow y=-6.

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If f:R→Rf:R\to R is defined by f(x)=xf(x)=x and g:R→Rg:R\to R is defined by g(x)=xg(x)=x then (f∘g)(x)(f\circ g)(x) is (a) xx (b) x2x^2 (c) 1x2\frac{1}{x^2} (d) 1/x1/x

Answer

(a) xx

(f∘g)(x)=f(g(x))=f(x)=x(f\circ g)(x)=f(g(x))=f(x)=x, since both functions are the identity map.

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If sin⁡2x=1\sin2x=1, then xx is (a) π/2\pi/2 (b) ±1\pm1 (c) −π/2-\pi/2 (d) nπ+(−1)nπ4n\pi+(-1)^n\frac{\pi}{4}

Answer

(d) nπ+(−1)nπ4n\pi+(-1)^n\dfrac{\pi}{4}

Of the given options, this is the only one in the form of a general solution of a sine equation. It gives π/4\pi/4 for n=0n=0, which is the principal solution of sin⁡2x=1\sin 2x=1 (as 2x=π/22x=\pi/2). It also matches x=nπ+π/4x=n\pi+\pi/4 whenever nn is even.

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The value of sin⁡(tan⁡−1x)\sin(\tan^{-1}x) is (a) xx (b) x1+x2\frac{x}{\sqrt{1+x^2}} (c) 11+x2\frac{1}{\sqrt{1+x^2}} (d) x1−x2\frac{x}{\sqrt{1-x^2}}

Answer

(b) x1+x2\dfrac{x}{\sqrt{1+x^2}}

Let θ=tan⁡−1x\theta=\tan^{-1}x, so tan⁡θ=x\tan\theta=x. In a right triangle the opposite side is xx, the adjacent side is 11 and the hypotenuse is 1+x2\sqrt{1+x^2}. Hence sin⁡θ=x1+x2\sin\theta=\dfrac{x}{\sqrt{1+x^2}}.

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If 2p,p+8,3p+12p, p+8, 3p+1 are in A.P., then pp is (a) 5 (b) 3 (c) 7 (d) -3

Answer

(a) 5

For an A.P., 2(p+8)=2p+(3p+1)2(p+8)=2p+(3p+1), so 2p+16=5p+12p+16=5p+1 and 3p=153p=15, giving p=5p=5. The terms are 10, 13, 16.

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A, G and H are arithmetic mean, geometric mean and harmonic means respectively then (a) A2=GHA^2=GH (b) H2=AGH^2=AG (c) G=AHG=AH (d) G2=AHG^2=AH

Answer

(d) G2=AHG^2=AH

For two numbers a,ba,b: A=a+b2A=\frac{a+b}{2}, H=2aba+bH=\frac{2ab}{a+b}, so AH=ab=G2AH=ab=G^2.

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If A=(−1 0 2)A=(-1\ 0\ 2) and B=(234)B=\begin{pmatrix}2\\3\\4\end{pmatrix}, then ABAB is (a) (6)(6) (b) 0 (c) (−2 0 8)(-2\ 0\ 8) (d) (2 0 −4)(2\ 0\ -4)

Answer

(a) (6)(6)

AB=(−1)(2)+(0)(3)+(2)(4)=−2+0+8=6AB=(-1)(2)+(0)(3)+(2)(4)=-2+0+8=6, a 1×11\times1 matrix.

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  • 2072 · 1 mark

The value of the determinant ∣120241362∣\begin{vmatrix}1&2&0\\2&4&1\\3&6&2\end{vmatrix} is (a) 1 (b) 3 (c) 0 (d) 4

Answer

(c) 0

Expanding along the first row: 1(4⋅2−1⋅6)−2(2⋅2−1⋅3)+0=2−2=01(4\cdot2-1\cdot6)-2(2\cdot2-1\cdot3)+0=2-2=0. (Column 2 is twice column 1, so the determinant is zero.)

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  • 2072 · 1 mark

The adjoint of the matrix (abcd)\begin{pmatrix}a&b\\c&d\end{pmatrix} is (a) (adbc)\begin{pmatrix}a&d\\b&c\end{pmatrix} (b) (abdc)\begin{pmatrix}a&b\\d&c\end{pmatrix} (c) (adcb)\begin{pmatrix}a&d\\c&b\end{pmatrix} (d) (d−b−ca)\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

Answer

(d) (d−b−ca)\begin{pmatrix}d&-b\\-c&a\end{pmatrix}

The adjoint is the transpose of the cofactor matrix. The cofactors are d,−c,−b,ad,-c,-b,a, so after transposing we get the matrix above.

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  • 2072 · 1 mark

The real part of (1+i)22−i\frac{(1+i)^2}{2-i} is (a) −2/5-2/5 (b) 2/52/5 (c) 1/51/5 (d) 1

Answer

(a) −2/5-2/5

(1+i)2=2i(1+i)^2=2i. Then 2i2−i⋅2+i2+i=4i+2i25=−2+4i5\dfrac{2i}{2-i}\cdot\dfrac{2+i}{2+i}=\dfrac{4i+2i^2}{5}=\dfrac{-2+4i}{5}, so the real part is −25-\dfrac25.

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  • 2072 · 1 mark

If ω\omega be the cube roots of unity then (1+ω+ω2)(1+ω−ω2)=(1+\omega+\omega^2)(1+\omega-\omega^2)= (a) 0 (b) 2 (c) 4 (d) 3

Answer

(a) 0

Since 1+ω+ω2=01+\omega+\omega^2=0, the first factor is 00, so the product is 00.

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  • 2072 · 1 mark

The value of kk for which the equation x2+2(k+2)x+9k=0x^2+2(k+2)x+9k=0 has equal roots are (a) 4, 1 (b) −4,−1-4,-1 (c) −4,1-4, 1 (d) 4, −1-1

Answer

(a) 4, 1

Equal roots require the discriminant to be zero: 4(k+2)2−36k=0⇒k2+4k+4−9k=0⇒k2−5k+4=0⇒(k−1)(k−4)=04(k+2)^2-36k=0\Rightarrow k^2+4k+4-9k=0\Rightarrow k^2-5k+4=0\Rightarrow (k-1)(k-4)=0. So k=1,4k=1,4.

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  • 2072 · 1 mark

The distance between the point (1,2)(1,2) and the line 3x+4y−7=03x+4y-7=0 is (a) 4 (b) 5 (c) 4/5 (d) -4/5

Answer

(c) 4/54/5

d=∣3(1)+4(2)−7∣32+42=45d=\dfrac{|3(1)+4(2)-7|}{\sqrt{3^2+4^2}}=\dfrac{4}{5}.

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  • 2072 · 1 mark

The two lines represented by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are coincident if (a) h2=abh^2=ab (b) h2>abh^2>ab (c) h2<abh^2<ab (d) h2>0h^2>0

Answer

(a) h2=abh^2=ab

The pair of lines is real and distinct when h2>abh^2>ab, and the lines coincide when the discriminant vanishes, i.e. h2=abh^2=ab.

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  • 2072 · 1 mark

The equation of tangent to the circle x2+y2=5x^2+y^2=5 at (2,1)(2,1) is (a) x+y=5x+y=5 (b) x−y=5x-y=5 (c) 2x+y=52x+y=5 (d) x+2y=5x+2y=5

Answer

(c) 2x+y=52x+y=5

The tangent to x2+y2=a2x^2+y^2=a^2 at (x1,y1)(x_1,y_1) is xx1+yy1=a2xx_1+yy_1=a^2. Here it is 2x+y=52x+y=5, and (2,1)(2,1) lies on the circle since 4+1=54+1=5.

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  • 2072 · 1 mark

lim⁡x→2x−2x2−4\lim_{x\to2}\frac{x-2}{\sqrt{x^2-4}} is (a) 0/0 (b) 0 (c) 1 (d) 4

Answer

(b) 0

For x>2x>2: x−2(x−2)(x+2)=x−2x+2→0=0\dfrac{x-2}{\sqrt{(x-2)(x+2)}}=\sqrt{\dfrac{x-2}{x+2}}\to\sqrt{0}=0.

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  • 2072 · 1 mark

lim⁡x→01−cos⁡axx2\lim_{x\to0}\frac{1-\cos ax}{x^2} is (a) 0 (b) a22\frac{a^2}{2} (c) 2a22a^2 (d) 2

Answer

(b) a22\dfrac{a^2}{2}

1−cos⁡ax=2sin⁡2ax21-\cos ax=2\sin^2\frac{ax}{2}, so the limit is lim⁡2⋅sin⁡2(ax/2)x2=2⋅a24=a22\lim 2\cdot\dfrac{\sin^2(ax/2)}{x^2}=2\cdot\dfrac{a^2}{4}=\dfrac{a^2}{2}.

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  • 2072 · 1 mark

The value of xx for which the function y=x3−x2y=x^3-x^2 has dydx=0\frac{dy}{dx}=0 is (a) 0, 2/3 (b) 2/3 (c) 0 (d) −2/3-2/3

Answer

(a) 0, 2/30,\ 2/3

dydx=3x2−2x=x(3x−2)=0⇒x=0\dfrac{dy}{dx}=3x^2-2x=x(3x-2)=0\Rightarrow x=0 or x=23x=\dfrac23.

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  • 2072 · 1 mark

The point of inflection for y=x2−2xy=x^2-2x is (a) 0 (b) does not exist (c) 2 (d) 1

Answer

(b) does not exist

y′′=2y''=2, which is never zero and never changes sign, so the parabola has no point of inflection.

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  • 2072 · 1 mark

∫012x dx1−x2\int_0^1\frac{2x\,dx}{\sqrt{1-x^2}} is (a) 1/2 (b) 1 (c) 0 (d) 2

Answer

(d) 2

Put u=1−x2u=1-x^2, du=−2x dxdu=-2x\,dx: ∫012x dx1−x2=[−21−x2]01=0−(−2)=2\int_0^1\frac{2x\,dx}{\sqrt{1-x^2}}=\left[-2\sqrt{1-x^2}\right]_0^1=0-(-2)=2.

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  • 2072 · 1 mark

∫xsin⁡x dx=\int x\sin x\,dx= (a) sin⁡x−xcos⁡x+C\sin x-x\cos x+C (b) sin⁡x+xcos⁡x+C\sin x+x\cos x+C (c) cos⁡x+xsin⁡x+C\cos x+x\sin x+C (d) −sin⁡x+xcos⁡x+C-\sin x+x\cos x+C

Answer

(a) sin⁡x−xcos⁡x+C\sin x-x\cos x+C

By parts with u=xu=x, dv=sin⁡x dxdv=\sin x\,dx: ∫xsin⁡x dx=−xcos⁡x+∫cos⁡x dx=−xcos⁡x+sin⁡x+C\int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x+C.

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  • 2073 · 1 mark

Which one is not a logical statement: (a) If you work hard, you will pass (b) I am sick. (c) The sun rises in the west. (d) 30 is a number.

Answer

(a) If you work hard, you will pass A logical statement (proposition) must be definitely true or false. The conditional in (a) is an uncertain claim with no definite truth value, whereas (c) is false and (d) is true.

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  • 2073 · 1 mark

If ∣x−2∣=5|x-2|=5, then xx is in (a) {3,5}\{3,5\} (b) {2,∞}\{2,\infty\} (c) {2,5}\{2,5\} (d) {−3,7}\{-3,7\}

Answer

(d) {−3,7}\{-3,7\} x−2=±5⇒x=7 or x=−3.x-2=\pm5 \Rightarrow x=7 \text{ or } x=-3.

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  • 2073 · 1 mark

A−B=A-B= (a) B−AB-A (b) Aˉ−Bˉ\bar A-\bar B (c) Bˉ−A\bar B-A (d) Bˉ−Aˉ\bar B-\bar A

Answer

(d) Bˉ−Aˉ\bar B-\bar A A−B=A∩Bˉ=Bˉ∩A=Bˉ−Aˉ.A-B=A\cap \bar B=\bar B\cap A=\bar B-\bar A.

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  • 2073 · 1 mark

If f(x)=2x−3f(x)=2x-3, then f−1(x)=f^{-1}(x)= (a) x+32\frac{x+3}{2} (b) 12x−3\frac{1}{2x-3} (c) 13−2x\frac{1}{3-2x} (d) 12x+3\frac{1}{2x+3}

Answer

(a) x+32\frac{x+3}{2} Put y=2x−3y=2x-3, so x=y+32x=\frac{y+3}{2}. Hence f−1(x)=x+32f^{-1}(x)=\frac{x+3}{2}.

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  • 2073 · 1 mark

The period of the function f(x)=sin⁡(2x+5)f(x)=\sin(2x+5) is (a) 2π2\pi (b) π/2\pi/2 (c) π\pi (d) −π-\pi

Answer

(c) π\pi The period of sin⁡(ax+b)\sin(ax+b) is 2π∣a∣\frac{2\pi}{|a|}, so here it is 2π2=π\frac{2\pi}{2}=\pi.

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  • 2073 · 1 mark

The fourth, seventh and tenth terms of a G.P. are l,m,nl, m, n respectively, then (a) ln=m2ln=m^2 (b) l2=m2+n2l^2=m^2+n^2 (c) l2=mnl^2=mn (d) n2=lmn^2=lm

Answer

(a) ln=m2ln=m^2 With first term aa and ratio rr: l=ar3l=ar^3, m=ar6m=ar^6, n=ar9n=ar^9. Then ln=a2r12=(ar6)2=m2ln=a^2r^{12}=(ar^6)^2=m^2.

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  • 2073 · 1 mark

The nth term of the series: 1.3+3.5+5.7+…1.3+3.5+5.7+\ldots is (a) (n+1)(n+2)(n+1)(n+2) (b) (n+1)(n+3)(n+1)(n+3) (c) 4n2−14n^2-1 (d) n(n−1)n(n-1)

Answer

(c) 4n2−14n^2-1 The nnth term is (2n−1)(2n+1)=4n2−1(2n-1)(2n+1)=4n^2-1. Check: n=1n=1 gives 3 and n=2n=2 gives 15.

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  • 2073 · 1 mark

The value of sin⁡(cos⁡−1x)\sin(\cos^{-1}x) is (a) xx (b) 1−x21-x^2 (c) 1+x21+x^2 (d) 1−x2\sqrt{1-x^2}

Answer

(d) 1−x2\sqrt{1-x^2} Let θ=cos⁡−1x\theta=\cos^{-1}x, so cos⁡θ=x\cos\theta=x and sin⁡θ=1−x2\sin\theta=\sqrt{1-x^2} (as θ∈[0,π]\theta\in[0,\pi]).

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  • 2073 · 1 mark

If cosec⁡2x−2=0\operatorname{cosec}^2x-2=0, then xx is (a) ±π/6\pm\pi/6 (b) ±π/4\pm\pi/4 (c) nπ±π/4n\pi\pm\pi/4 (d) nπ+π/2n\pi+\pi/2

Answer

(c) nπ±π/4n\pi\pm\pi/4 cosec⁡2x=2⇒sin⁡2x=12⇒sin⁡x=±12\operatorname{cosec}^2x=2\Rightarrow\sin^2x=\frac12\Rightarrow\sin x=\pm\frac{1}{\sqrt2}, which gives x=nπ±π4x=n\pi\pm\frac{\pi}{4}.

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  • 2073 · 1 mark

If A=(k1020k02−1)A=\begin{pmatrix}k&1&0\\2&0&k\\0&2&-1\end{pmatrix} and ∣A∣=0|A|=0 then k=k= (a) 1 (b) -1 (c) 0 (d) ±1\pm1

Answer

(d) ±1\pm1 ∣A∣=k(0⋅(−1)−2k)−1(2⋅(−1)−0)=−2k2+2=0⇒k2=1⇒k=±1.|A|=k(0\cdot(-1)-2k)-1(2\cdot(-1)-0)=-2k^2+2=0\Rightarrow k^2=1\Rightarrow k=\pm1.

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  • 2073 · 1 mark

The value of xx for which A=[6x−23x]A=\begin{bmatrix}6&x-2\\3&x\end{bmatrix} has no inverse is (a) 0 (b) 1 (c) -2 (d) 2

Answer

(c) -2 A has no inverse when ∣A∣=0|A|=0: 6x−3(x−2)=3x+6=06x-3(x-2)=3x+6=0, so x=−2x=-2.

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  • 2073 · 1 mark

If the system of equation has zero solutions only then the solution is (a) non trivial (b) trivial (c) unique (d) particular

Answer

(b) trivial A homogeneous system whose only solution is the zero solution x=y=⋯=0x=y=\dots=0 has only the trivial solution.

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  • 2073 · 1 mark

(1+i)6+(1−i)6=(1+i)^6+(1-i)^6= (a) 0 (b) ii (c) 16 (d) -16

Answer

(a) 0 (1+i)2=2i(1+i)^2=2i, so (1+i)6=(2i)3=−8i(1+i)^6=(2i)^3=-8i. Similarly (1−i)2=−2i(1-i)^2=-2i, so (1−i)6=(−2i)3=8i(1-i)^6=(-2i)^3=8i. The sum is 00.

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  • 2073 · 1 mark

(2i1+i)2=\left(\frac{2i}{1+i}\right)^2= (a) −i-i (b) 1+i1+i (c) 2i2i (d) −2i-2i

Answer

(c) 2i2i 2i1+i=2i(1−i)2=i+1\frac{2i}{1+i}=\frac{2i(1-i)}{2}=i+1, and (1+i)2=2i(1+i)^2=2i.

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  • 2073 · 1 mark

If one root of 5x2+2x−k=05x^2+2x-k=0 is the reciprocal of the other then k=k= (a) -5 (b) 5 (c) 2/5 (d) 5/2

Answer

(a) -5 The product of the roots is −k5\frac{-k}{5}. For reciprocal roots the product is 1, so −k/5=1-k/5=1 and k=−5k=-5.

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  • 2073 · 1 mark

The equation x2−2xy+y2+3x−2=0x^2-2xy+y^2+3x-2=0 represents: (a) hyperbola (b) parabola (c) ellipse (d) circle

Answer

(b) parabola For Ax2+2Hxy+By2+…Ax^2+2Hxy+By^2+\dots, here A=1, H=−1, B=1A=1,\ H=-1,\ B=1, so H2−AB=1−1=0H^2-AB=1-1=0. The second-degree terms form a perfect square, so the conic is a parabola.

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  • 2073 · 1 mark

The area of the triangle with vertices (1,−1)(1,-1), (−1,1)(-1,1) and (−1,−1)(-1,-1) is (a) 1/2 (b) -2 (c) 2 (d) 0

Answer

(c) 2 The points (1,−1)(1,-1), (−1,1)(-1,1), (−1,−1)(-1,-1) give a right angle at (−1,−1)(-1,-1) with legs 2 and 2, so Area=12⋅2⋅2=2.\text{Area}=\tfrac12\cdot2\cdot2=2.

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  • 2073 · 1 mark

lim⁡n→∞1+2+⋯+nn2=\lim_{n\to\infty}\frac{1+2+\cdots+n}{n^2}= (a) 1/2 (b) 0 (c) 1/n1/n (d) nn

Answer

(a) 1/2 lim⁡n→∞n(n+1)/2n2=lim⁡n→∞1+1/n2=12.\lim_{n\to\infty}\frac{n(n+1)/2}{n^2}=\lim_{n\to\infty}\frac{1+1/n}{2}=\frac12.

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  • 2073 · 1 mark

lim⁡x→05x2+4xtan⁡xx2=\lim_{x\to0}\frac{5x^2+4x\tan x}{x^2}= (a) 9 (b) 0 (c) 5 (d) 4

Answer

(a) 9 lim⁡x→0(5+4tan⁡xx)=5+4=9.\lim_{x\to0}\left(5+4\frac{\tan x}{x}\right)=5+4=9.

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  • 2073 · 1 mark

If y=x+x+x+⋯∞y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots\infty}}} then dydx=\frac{dy}{dx}= (a) 2x2\sqrt x (b) 3x\sqrt3x (c) x2y−1\frac{\sqrt x}{2y-1} (d) 12y−1\frac{1}{2y-1}

Answer

(d) 12y−1\frac{1}{2y-1}

Since y=x+yy=\sqrt{x+y}, we have y2=x+yy^2=x+y. Differentiating, 2y y′=1+y′2y\,y'=1+y', so y′=12y−1y'=\frac{1}{2y-1}.

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  • 2073 · 1 mark

If y=eax2+by=e^{\sqrt{ax^2+b}} then dy/dx=dy/dx= (a) axyax2+b\frac{axy}{\sqrt{ax^2+b}} (b) axyax2+b⋅eax2+b\frac{axy}{\sqrt{ax^2+b}}\cdot e^{\sqrt{ax^2+b}} (c) xax2+b\frac{x}{\sqrt{ax^2+b}} (d) eax2+be^{\sqrt{ax^2+b}}

Answer

(a) axyax2+b\frac{axy}{\sqrt{ax^2+b}}

dydx=eax2+b⋅2ax2ax2+b=ax yax2+b.\frac{dy}{dx}=e^{\sqrt{ax^2+b}}\cdot\frac{2ax}{2\sqrt{ax^2+b}}=\frac{ax\,y}{\sqrt{ax^2+b}}.

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  • 2073 · 1 mark

∫dx1−x2=\int\frac{dx}{1-x^2}= (a) cos⁡−1x+c\cos^{-1}x+c (b) 12ln⁡∣1−x1+x∣+c\frac12\ln\left|\frac{1-x}{1+x}\right|+c (c) sin⁡−1x+c\sin^{-1}x+c (d) 12ln⁡∣1+x1−x∣+c\frac12\ln\left|\frac{1+x}{1-x}\right|+c

Answer

(d) 12ln⁡∣1+x1−x∣+c\frac12\ln\left|\frac{1+x}{1-x}\right|+c

Using partial fractions, 11−x2=12(11−x+11+x)\frac{1}{1-x^2}=\frac12\left(\frac1{1-x}+\frac1{1+x}\right), so the integral is 12[−ln⁡∣1−x∣+ln⁡∣1+x∣]+c\frac12\left[-\ln|1-x|+\ln|1+x|\right]+c.

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  • 2073 · 1 mark

∫12sin⁡(ln⁡t)tdt=\int_1^2\frac{\sin(\ln t)}{t}dt= (a) 1−cos⁡(ln⁡2)1-\cos(\ln2) (b) cos⁡2\cos2 (c) cos⁡(ln⁡2)\cos(\ln2) (d) 1+cos⁡(ln⁡2)1+\cos(\ln2)

Answer

(a) 1−cos⁡(ln⁡2)1-\cos(\ln2)

Put u=ln⁡tu=\ln t, du=dt/tdu=dt/t; limits 00 to ln⁡2\ln2. Then ∫0ln⁡2sin⁡u du=[−cos⁡u]0ln⁡2=1−cos⁡(ln⁡2)\int_0^{\ln2}\sin u\,du=[-\cos u]_0^{\ln2}=1-\cos(\ln2).

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  • 2073 · 1 mark

The critical points for f(x)=x3−3xf(x)=x^3-3x are: (a) ±3\pm3 (b) ±3\pm\sqrt3 (c) ±1\pm1 (d) 0,±30,\pm\sqrt3

Answer

(c) ±1\pm1

f′(x)=3x2−3=0⇒x2=1⇒x=±1f'(x)=3x^2-3=0\Rightarrow x^2=1\Rightarrow x=\pm1.

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  • 2073 · 1 mark

The area bounded by the x-axis and the curve y=x3y=x^3 and ordinates of x=2x=2 and x=4x=4 is (a) 60 sq. units (b) 256 sq. units (c) 240 sq. units (d) 272 sq. units

Answer

(a) 60 sq. units

∫24x3dx=[x44]24=256−164=60.\int_2^4x^3dx=\left[\frac{x^4}{4}\right]_2^4=\frac{256-16}{4}=60.

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  • 2074 · 1 mark

p/qp/q, q∈Rq\in R, is undefined if (a) p≠0,q=0p\neq0, q=0 (b) p=0,q=0p=0, q=0 (c) p≠0,q≠0p\neq0, q\neq0 (d) p=0,q≠0p=0, q\neq0

Answer

(a) p≠0, q=0p\neq0,\ q=0

Division by zero is undefined for a non-zero numerator. (The case 0/00/0 is indeterminate rather than simply undefined.)

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  • 2074 · 1 mark

[a,b]=[a,b]= (a) {x:a≤x≤b}\{x: a\leq x\leq b\} (b) {x:a<x<b}\{x: a<x<b\} (c) {x:a<x≤b}\{x: a<x\leq b\} (d) {x:a≤x<b}\{x: a\leq x<b\}

Answer

(a) {x:a≤x≤b}\{x: a\leq x\leq b\}

The square brackets denote the closed interval, which includes both end points.

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  • 2074 · 1 mark

If f:R→Rf:R\to R defined by f(x)=(4−x2)f(x)=\sqrt{(4-x^2)} then f(−3)f(-3) is (a) Undefined (b) 13\sqrt{13} (c) i5i\sqrt5 (d) −13-\sqrt{13}

Answer

(a) Undefined

f(−3)=4−9=−5f(-3)=\sqrt{4-9}=\sqrt{-5}, which is not a real number. Since the codomain is R\mathbb{R}, f(−3)f(-3) is undefined.

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If ∣x∣=1|x|=1 then xx is (a) {1}\{1\} (b) {−1}\{-1\} (c) {−1,1}\{-1,1\} (d) {−1,−1}\{-1,-1\}

Answer

(c) {−1,1}\{-1,1\}

∣x∣=1|x|=1 gives x=1x=1 or x=−1x=-1.

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If log⁡2(x−7)=1\log_2(x-7)=1, then xx is (a) Not known (b) 8 (c) 6 (d) 9

Answer

(d) 9

log⁡2(x−7)=1⇒x−7=21=2⇒x=9\log_2(x-7)=1\Rightarrow x-7=2^1=2\Rightarrow x=9.

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The value of sin⁡(cos⁡−145)\sin\left(\cos^{-1}\frac45\right) is (a) 4/5 (b) 3/5 (c) 5/4 (d) 5/3

Answer

(b) 3/5

Let θ=cos⁡−145\theta=\cos^{-1}\frac45, so cos⁡θ=45\cos\theta=\frac45 and sin⁡θ=1−1625=35\sin\theta=\sqrt{1-\frac{16}{25}}=\frac35.

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If tan⁡2x=3\tan^2x=3 then xx is (a) nπ+π3n\pi+\frac{\pi}{3} (b) nπ−π3n\pi-\frac{\pi}{3} (c) nπ±π3n\pi\pm\frac{\pi}{3} (d) nπn\pi

Answer

(c) nπ±π3n\pi\pm\frac{\pi}{3}

tan⁡2x=3⇒tan⁡x=±3\tan^2x=3\Rightarrow\tan x=\pm\sqrt3, so x=nπ±π3x=n\pi\pm\frac{\pi}{3}.

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If k+2,4k−6,3k−2k+2, 4k-6, 3k-2 are in AP then kk is (a) -6 (b) 0 (c) 3 (d) -3

Answer

(c) 3

For an AP, 2(4k−6)=(k+2)+(3k−2)2(4k-6)=(k+2)+(3k-2), so 8k−12=4k8k-12=4k and k=3k=3.

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The sum of infinity of 1−12+14−18…1-\frac12+\frac14-\frac18\ldots is (a) 0 (b) 2/3 (c) 1/3 (d) 1

Answer

(b) 2/3

GP with a=1a=1, r=−12r=-\frac12: S∞=11+12=23S_\infty=\frac{1}{1+\frac12}=\frac23.

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If A=(123)A=\begin{pmatrix}1\\2\\3\end{pmatrix} and B=(1 2 3)B=(1\ 2\ 3) then ABAB is (a) (1 4)(1\ 4) (b) (1 4 9)(1\ 4\ 9) (c) 0 (d) (123246369)\begin{pmatrix}1&2&3\\2&4&6\\3&6&9\end{pmatrix}

Answer

(d) (123246369)\begin{pmatrix}1&2&3\\2&4&6\\3&6&9\end{pmatrix}

A 3×13\times1 matrix times a 1×31\times3 matrix gives a 3×33\times3 matrix with entries aibja_ib_j.

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The value of ∣0231234812∣\begin{vmatrix}0&2&3\\1&2&3\\4&8&12\end{vmatrix} is (a) 10 (b) 1 (c) 0 (d) -1

Answer

(c) 0

Row 3 is 4×4\times Row 2 (since 4(1,2,3)=(4,8,12)4(1,2,3)=(4,8,12)), so the determinant is 0.

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The matrix (123204345)\begin{pmatrix}1&2&3\\2&0&4\\3&4&5\end{pmatrix} is (a) Diagonal (b) symmetric (c) Skew symmetric (d) unit

Answer

(b) symmetric

The matrix equals its transpose (aij=ajia_{ij}=a_{ji}), and the diagonal is not zero, so it is symmetric and not skew-symmetric.

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[2(cos⁡15∘+isin⁡15∘)]6=[2(\cos15^\circ+i\sin15^\circ)]^6= (a) 64i64i (b) 64 (c) 0 (d) −64i-64i

Answer

(a) 64i64i

By De Moivre: 26(cos⁡90∘+isin⁡90∘)=64i2^6(\cos90^\circ+i\sin90^\circ)=64i.

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(1+ω+ω2)3+(1+ω−ω2)3=(1+\omega+\omega^2)^3+(1+\omega-\omega^2)^3= (a) 16 (b) 16ω16\omega (c) -8 (d) 16ω216\omega^2

Answer

(c) -8

1+ω+ω2=01+\omega+\omega^2=0, so the first term is 0. Also 1+ω−ω2=−2ω21+\omega-\omega^2=-2\omega^2, and (−2ω2)3=−8ω6=−8(-2\omega^2)^3=-8\omega^6=-8.

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The line kx+3y+5=0kx+3y+5=0 and 5x−2y−6=05x-2y-6=0 are perpendicular to each other for k=k= (a) 6/5 (b) -6/5 (c) 3/5 (d) -3/5

Answer

(a) 6/5

Slopes are −k3-\frac k3 and 52\frac52. Perpendicular means the product is −1-1: −5k6=−1⇒k=65-\frac{5k}{6}=-1\Rightarrow k=\frac65.

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The equation of tangent to the circle x2+y2=2x^2+y^2=2 at (1,1)(1,1) is (a) x+y=2x+y=2 (b) x−y=2x-y=2 (c) 2x+2y=32x+2y=3 (d) −x−y=2-x-y=2

Answer

(a) x+y=2x+y=2

Tangent at (x1,y1)(x_1,y_1) is xx1+yy1=r2xx_1+yy_1=r^2, so x+y=2x+y=2.

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lim⁡x→∞3x2+2x−56x2+4x−7=\lim_{x\to\infty}\frac{3x^2+2x-5}{6x^2+4x-7}= (a) 2 (b) 3 (c) 5/7 (d) 1/2

Answer

(d) 1/2

Divide by x2x^2: the limit is the ratio of leading coefficients, 36=12\frac36=\frac12.

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lim⁡x→01−cos⁡axx2=\lim_{x\to0}\frac{1-\cos ax}{x^2}= (a) a22\frac{a^2}{2} (b) a2a^2 (c) 2a22a^2 (d) 4a24a^2

Answer

(a) a22\frac{a^2}{2}

1−cos⁡ax=2sin⁡2ax21-\cos ax=2\sin^2\frac{ax}{2}, so the limit is 2⋅a24=a222\cdot\frac{a^2}{4}=\frac{a^2}{2}.

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If x=t+sin⁡tx=t+\sin t, y=1−cos⁡ty=1-\cos t then dy/dxdy/dx is (a) tan⁡t\tan t (b) tan⁡t/2\tan t/2 (c) −tan⁡t-\tan t (d) −tan⁡t/2-\tan t/2

Answer

(b) tan⁡t2\tan\frac t2

dydx=sin⁡t1+cos⁡t=2sin⁡t2cos⁡t22cos⁡2t2=tan⁡t2\frac{dy}{dx}=\frac{\sin t}{1+\cos t}=\frac{2\sin\frac t2\cos\frac t2}{2\cos^2\frac t2}=\tan\frac t2.

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The derivative of cos⁡x\cos x w.r.t. sin⁡x\sin x is (a) −tan⁡x-\tan x (b) tan⁡x\tan x (c) cot⁡x\cot x (d) −cot⁡x-\cot x

Answer

(a) −tan⁡x-\tan x

d(cos⁡x)d(sin⁡x)=−sin⁡xcos⁡x=−tan⁡x\frac{d(\cos x)}{d(\sin x)}=\frac{-\sin x}{\cos x}=-\tan x.

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The point of inflection of y=x3−12x2y=x^3-12x^2 is (a) 12 (b) -12 (c) 4 (d) 0

Answer

(c) 4

y′′=6x−24=0⇒x=4y''=6x-24=0\Rightarrow x=4, and y′′y'' changes sign there, so the point of inflection is at x=4x=4.

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∫dxsin⁡2x=\int\frac{dx}{\sin^2x}= (a) cot⁡x+C\cot x+C (b) −cot⁡x+C-\cot x+C (c) tan⁡x+C\tan x+C (d) −tan⁡x+C-\tan x+C

Answer

(b) −cot⁡x+C-\cot x+C

∫csc⁡2x dx=−cot⁡x+C\int\csc^2x\,dx=-\cot x+C.

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∫01dx1+x2=\int_0^1\frac{dx}{1+x^2}= (a) π4\frac{\pi}{4} (b) π2\frac{\pi}{2} (c) 0 (d) 1

Answer

(a) π4\frac{\pi}{4}

∫01dx1+x2=tan⁡−11−tan⁡−10=π4\int_0^1\frac{dx}{1+x^2}=\tan^{-1}1-\tan^{-1}0=\frac\pi4.

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a/ba/b, a,b∈Ra, b\in R, is equal to zero if (a) a≠0,b=0a\neq0, b=0 (b) a=0,b=0a=0, b=0 (c) a=0,b≠0a=0, b\neq0 (d) a≠0,b≠0a\neq0, b\neq0

Answer

(c) a=0,b≠0a=0, b\neq0 ab=0\frac ab=0 only when the numerator is zero and the denominator is non-zero.

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If x≥5x\geq5 then xx may be (a) -5 (b) 6 (c) 0 (d) [5,∞)[5,\infty)

Answer

(b) 6 Since x≥5x\ge5, only 6 satisfies the condition among the numbers listed.

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If ∣x−2∣=5|x-2|=5 then the value of xx is (a) {5}\{5\} (b) {−2,5}\{-2,5\} (c) {−2,−5}\{-2,-5\} (d) {−3,7}\{-3,7\}

Answer

(d) {−3,7}\{-3,7\} ∣x−2∣=5⇒x−2=±5⇒x=7|x-2|=5\Rightarrow x-2=\pm5\Rightarrow x=7 or x=−3x=-3.

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The range of the relation R={(1,2),(2,3),(3,4)}R=\{(1,2),(2,3),(3,4)\} is (a) {1,2,3}\{1,2,3\} (b) {1,2,3,4}\{1,2,3,4\} (c) {2,3,4}\{2,3,4\} (d) RR

Answer

(c) {2,3,4}\{2,3,4\} The range is the set of second elements of the ordered pairs.

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If y=axy=a^x is the exponential function passing through (4,81)(4,81) then the base aa is (a) 4 (b) 3 (c) 81 (d) aa

Answer

(b) 3 a4=81=34a^4=81=3^4, so a=3a=3.

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If tan⁡2x=1\tan^2x=1 then xx is (a) nπ±π/4n\pi\pm\pi/4 (b) nπ+π/4n\pi+\pi/4 (c) nπ−π/4n\pi-\pi/4 (d) ±π/4\pm\pi/4

Answer

(a) nπ±π/4n\pi\pm\pi/4 tan⁡2x=1⇒tan⁡x=±1=tan⁡(±π/4)\tan^2x=1\Rightarrow\tan x=\pm1=\tan(\pm\pi/4), so x=nπ±π4x=n\pi\pm\frac\pi4.

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sin⁡(2sin⁡−1x)=\sin(2\sin^{-1}x)= (a) 1−x2\sqrt{1-x^2} (b) 21−x22\sqrt{1-x^2} (c) 2x1−x22x\sqrt{1-x^2} (d) 2x(1−x2)\sqrt{2x(1-x^2)}

Answer

(c) 2x1−x22x\sqrt{1-x^2} Let θ=sin⁡−1x\theta=\sin^{-1}x. Then sin⁡2θ=2sin⁡θcos⁡θ=2x1−x2\sin2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}.

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The sum to infinity of the G.P.: 14,116,164…\frac14,\frac1{16},\frac1{64}\ldots is (a) 5 (b) -5 (c) −15-\frac15 (d) 13\frac13

Answer

(d) 13\frac13 a=14a=\frac14, r=14r=\frac14, so S∞=a1−r=1/43/4=13S_\infty=\frac{a}{1-r}=\frac{1/4}{3/4}=\frac13.

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The value of aa so that 8a+48a+4, 6a−26a-2 and 3a+73a+7 will form an A.P. is (a) 15 (b) 2 (c) 9 (d) 10

Answer

(a) 15 For an A.P., 2(6a−2)=(8a+4)+(3a+7)2(6a-2)=(8a+4)+(3a+7), so 12a−4=11a+1112a-4=11a+11 and a=15a=15.

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The value of ∣234567101214∣\begin{vmatrix}2&3&4\\5&6&7\\10&12&14\end{vmatrix} is (a) 10 (b) 15 (c) 0 (d) 20

Answer

(c) 0 Since 2×(5,6,7)=(10,12,14)2\times(5,6,7)=(10,12,14), the third row is twice the second row. A determinant with two proportional rows is 0.

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If A=(1 2 3)A=(1\ 2\ 3) and B=(123)B=\begin{pmatrix}1\\2\\3\end{pmatrix} then ABAB is (a) (4) (b) (14) (c) (0) (d) (15)

Answer

(b) (14) AB=(1⋅1+2⋅2+3⋅3)=(1+4+9)=(14)AB=(1\cdot1+2\cdot2+3\cdot3)=(1+4+9)=(14).

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The inverse of the matrix [−111−3]\begin{bmatrix}-1&1\\1&-3\end{bmatrix} is (a) [−311−1]\begin{bmatrix}-3&1\\1&-1\end{bmatrix} (b) 12[−3−1−1−1]\frac12\begin{bmatrix}-3&-1\\-1&-1\end{bmatrix} (c) [−3−1−1−1]\begin{bmatrix}-3&-1\\-1&-1\end{bmatrix} (d) 12[−111−3]\frac12\begin{bmatrix}-1&1\\1&-3\end{bmatrix}

Answer

(b) 12[−3−1−1−1]\frac12\begin{bmatrix}-3&-1\\-1&-1\end{bmatrix} det⁡=(−1)(−3)−(1)(1)=2\det=(-1)(-3)-(1)(1)=2. The inverse is 12[−3−1−1−1]\frac12\begin{bmatrix}-3&-1\\-1&-1\end{bmatrix}.

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If ω\omega is the cube root of unity then (1−ω+ω2)3=(1-\omega+\omega^2)^3= (a) 0 (b) -8 (c) 1 (d) 8

Answer

(b) -8 Using 1+ω2=−ω1+\omega^2=-\omega: 1−ω+ω2=−2ω1-\omega+\omega^2=-2\omega, so (−2ω)3=−8ω3=−8(-2\omega)^3=-8\omega^3=-8.

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∣11+i∣=\left|\frac{1}{1+i}\right|= (a) 12\frac12 (b) 2\sqrt2 (c) 12\frac{1}{\sqrt2} (d) 2

Answer

(c) 12\frac{1}{\sqrt2} ∣11+i∣=1∣1+i∣=12\left|\frac1{1+i}\right|=\frac{1}{|1+i|}=\frac1{\sqrt2}.

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The perpendicular distance from (−1,−2)(-1,-2) to the line 3x+4y+12=03x+4y+12=0 is (a) -1 (b) 15\frac15 (c) 2 (d) 3

Answer

(b) 15\frac15 d=∣3(−1)+4(−2)+12∣9+16=∣−3−8+12∣5=15d=\frac{|3(-1)+4(-2)+12|}{\sqrt{9+16}}=\frac{|-3-8+12|}{5}=\frac15.

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If the lines represented by ky2+(2−k)2xy−3x2=0ky^2+(2-k)^2xy-3x^2=0 are perpendicular if k=k= (a) -3 (b) 3 (c) 0 (d) 1

Answer

(b) 3 The lines ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here a=−3a=-3, b=kb=k, so k−3=0k-3=0 and k=3k=3.

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The equation of the tangent to the circle x2+y2=5x^2+y^2=5 at (2,1)(2,1) is (a) 2x+y=52x+y=5 (b) x+y=5x+y=5 (c) 2x−y=52x-y=5 (d) x−y=5x-y=5

Answer

(a) 2x+y=52x+y=5 The tangent to x2+y2=r2x^2+y^2=r^2 at (x1,y1)(x_1,y_1) is xx1+yy1=r2xx_1+yy_1=r^2, which gives 2x+y=52x+y=5.

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lim⁡x→∞x(x+1)x2=\lim_{x\to\infty}\frac{x(x+1)}{x^2}= (a) 1 (b) 0 (c) ∞\infty (d) 2

Answer

(a) 1 x(x+1)x2=1+1x→1\frac{x(x+1)}{x^2}=1+\frac1x\to1 as x→∞x\to\infty.

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lim⁡x→π/4sec⁡2x−2tan⁡x−1=\lim_{x\to\pi/4}\frac{\sec^2x-2}{\tan x-1}= (a) 0 (b) 1 (c) 2 (d) -2

Answer

(c) 2 sec⁡2x−2=tan⁡2x−1=(tan⁡x−1)(tan⁡x+1)\sec^2x-2=\tan^2x-1=(\tan x-1)(\tan x+1), so the limit is tan⁡π4+1=2\tan\frac\pi4+1=2.

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dydx\frac{dy}{dx} of t=x2t=x^2, y=1ty=\frac1t is (a) 1 (b) −2x3-\frac{2}{x^3} (c) 2x2\frac{2}{x^2} (d) 2x3\frac{2}{x^3}

Answer

(b) −2x3-\frac{2}{x^3} y=1t=1x2=x−2y=\frac1t=\frac1{x^2}=x^{-2}, so dydx=−2x−3=−2x3\frac{dy}{dx}=-2x^{-3}=-\frac2{x^3}.

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The derivative of x2x^2 with respect to −x2-x^2 is (a) 0 (b) 1 (c) 2x2x (d) -1

Answer

(d) -1 Let u=x2u=x^2. Then dud(−u)=−1\frac{du}{d(-u)}=-1.

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∫−112x1+x2dx=\int_{-1}^{1}\frac{2x}{1+x^2}dx= (a) 0 (b) log⁡2\log2 (c) 2 (d) 1

Answer

(a) 0

The integrand 2x1+x2\frac{2x}{1+x^2} is an odd function, so its integral over the symmetric interval [−1,1][-1,1] is zero.

∫−112x1+x2dx=[ln⁡(1+x2)]−11=ln⁡2−ln⁡2=0\int_{-1}^{1}\frac{2x}{1+x^2}dx=\Big[\ln(1+x^2)\Big]_{-1}^{1}=\ln2-\ln2=0

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∫dxx(1+ln⁡x)2=\int\frac{dx}{x(1+\ln x)^2}= (a) −11+ln⁡x+c-\frac{1}{1+\ln x}+c (b) 11+ln⁡x+c\frac{1}{1+\ln x}+c (c) 1ln⁡x+c\frac{1}{\ln x}+c (d) 1x(1+ln⁡x)+c\frac{1}{x(1+\ln x)}+c

Answer

(a) −11+ln⁡x+c-\frac{1}{1+\ln x}+c

Put u=1+ln⁡xu=1+\ln x, so du=dxxdu=\frac{dx}{x}.

∫duu2=−1u+c=−11+ln⁡x+c\int\frac{du}{u^2}=-\frac1u+c=-\frac{1}{1+\ln x}+c

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If f(0)=1,f(1)=2,f(2)=5,f(3)=10f(0)=1, f(1)=2, f(2)=5, f(3)=10, then f(−1)=f(-1)= (a) -3 (b) 0 (c) 1 (d) 2

Answer

(d) 2 The values fit f(x)=x2+1f(x)=x^2+1 (f(0)=1, f(1)=2, f(2)=5, f(3)=10f(0)=1,\ f(1)=2,\ f(2)=5,\ f(3)=10), so f(−1)=(−1)2+1=2f(-1)=(-1)^2+1=2.

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If A⊆BA\subseteq B, then (a) Ac⊆BcA^c\subseteq B^c (b) Ac=BcA^c=B^c (c) Bc⊆AcB^c\subseteq A^c (d) Bc⊄AcB^c\not\subset A^c

Answer

(c) Bc⊆AcB^c\subseteq A^c If A⊆BA\subseteq B, every element outside BB is also outside AA, so Bc⊆AcB^c\subseteq A^c.

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  • 2076 · 1 mark

If log⁡a(ab)=x\log_a(ab)=x, then log⁡b(ab)=\log_b(ab)= (a) x1−x\frac{x}{1-x} (b) 1x\frac1x (c) xx−1\frac{x}{x-1} (d) x1+x\frac{x}{1+x}

Answer

(c) xx−1\dfrac{x}{x-1} log⁡a(ab)=1+log⁡ab=x\log_a(ab)=1+\log_a b=x, so log⁡ab=x−1\log_a b=x-1 and log⁡ba=1x−1\log_b a=\frac{1}{x-1}. Then log⁡b(ab)=1+log⁡ba=1+1x−1=xx−1.\log_b(ab)=1+\log_b a=1+\frac1{x-1}=\frac{x}{x-1}.

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  • 2076 · 1 mark

If A=[13202x−3]A=\begin{bmatrix}\frac13&2\\0&2x-3\end{bmatrix}, B=[360−1]B=\begin{bmatrix}3&6\\0&-1\end{bmatrix}, AB=IAB=I, then x=x= (a) -1 (b) 1 (c) 0 (d) 2

Answer

(b) 1 AB=[100−(2x−3)]=IAB=\begin{bmatrix}1&0\\0&-(2x-3)\end{bmatrix}=I, so −(2x−3)=1-(2x-3)=1, giving 2x−3=−12x-3=-1 and x=1x=1.

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  • 2076 · 1 mark

If A2−A+I=0A^2-A+I=0, then A−1=A^{-1}= (a) A−IA-I (b) AA (c) I+AI+A (d) I−AI-A

Answer

(d) I−AI-A Multiply A2−A+I=0A^2-A+I=0 by A−1A^{-1}: A−I+A−1=0A-I+A^{-1}=0, so A−1=I−AA^{-1}=I-A.

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  • 2076 · 1 mark

The sum of 3 numbers of G.P is 38 and their product is 1728. The middle terms is (a) 12 (b) 8 (c) 18 (d) 6

Answer

(a) 12 Let the terms be ar,a,ar\frac ar, a, ar. The product is a3=1728a^3=1728, so a=12a=12. The sum condition 12(r+1+1r)=3812\left(r+1+\frac1r\right)=38 is consistent, so the middle term is 12.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2076 · 1 mark

Fourth roots of unity are (a) 1,ω,ω2,ω31,\omega,\omega^2,\omega^3 (b) 1,ω,ω2,ω41,\omega,\omega^2,\omega^4 (c) ±1,±i\pm1,\pm i (d) cannot be determined

Answer

(c) ±1,±i\pm1,\pm i The fourth roots of unity satisfy z4=1z^4=1, so z=1,−1,i,−iz=1,-1,i,-i.

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  • 2076 · 1 mark

The multiplicative inverse of 2+3i3−i\frac{2+3i}{3-i} is (a) 113+2i13\frac{1}{13}+\frac{2i}{13} (b) 313+11i13\frac{3}{13}+\frac{11i}{13} (c) 313+2i13\frac{3}{13}+\frac{2i}{13} (d) 313−11i13\frac{3}{13}-\frac{11i}{13}

Answer

(d) 313−11i13\frac{3}{13}-\frac{11i}{13} 2+3i3−i=(2+3i)(3+i)10=3+11i10\frac{2+3i}{3-i}=\frac{(2+3i)(3+i)}{10}=\frac{3+11i}{10}. Its inverse is 3−i2+3i=(3−i)(2−3i)13=3−11i13\frac{3-i}{2+3i}=\frac{(3-i)(2-3i)}{13}=\frac{3-11i}{13}.

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  • 2076 · 1 mark

If the roots of the equation ax−a+bx−b=1\frac{a}{x-a}+\frac{b}{x-b}=1 are numerically equal and opposite in sign, then (a) a−b=0a-b=0 (b) a+b=1a+b=1 (c) a−b=1a-b=1 (d) a+b=0a+b=0

Answer

(d) a+b=0a+b=0 Clearing fractions gives x2−2(a+b)x+3ab=0x^2-2(a+b)x+3ab=0. Roots equal in magnitude and opposite in sign mean the sum of roots is zero, so a+b=0a+b=0.

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  • 2076 · 1 mark

The remainder when (kx2+x−1)(kx^2+x-1) is divided by (x+2k)(x+2k) is (a) (−2k+1)(-2k+1) (b) (k+1)(k+1) (c) (4k3−2k)(4k^3-2k) (d) (4k3−2k−1)(4k^3-2k-1)

Answer

(d) (4k3−2k−1)(4k^3-2k-1) By the remainder theorem put x=−2kx=-2k: k(4k2)−2k−1=4k3−2k−1k(4k^2)-2k-1=4k^3-2k-1.

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  • 2076 · 1 mark

If cos⁡−1x=cos⁡−1(35)−sin⁡−1(45)\cos^{-1}x=\cos^{-1}\left(\frac35\right)-\sin^{-1}\left(\frac45\right), then x=x= (a) 0 (b) 12\frac12 (c) 32\frac{\sqrt3}{2} (d) 1

Answer

(d) 1 sin⁡−145=cos⁡−135\sin^{-1}\frac45=\cos^{-1}\frac35, so cos⁡−1x=0\cos^{-1}x=0 and x=1x=1.

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  • 2076 · 1 mark

All values between 0 to 2π2\pi which satisfy 2sin⁡2θ=3cos⁡θ2\sin^2\theta=3\cos\theta is (a) π3,2π3\frac{\pi}{3},\frac{2\pi}{3} (b) π4,3π4\frac{\pi}{4},\frac{3\pi}{4} (c) π3,5π3\frac{\pi}{3},\frac{5\pi}{3} (d) π6,2π6\frac{\pi}{6},\frac{2\pi}{6}

Answer

(c) π3,5π3\frac{\pi}{3},\frac{5\pi}{3} 2(1−cos⁡2θ)=3cos⁡θ2(1-\cos^2\theta)=3\cos\theta gives 2cos⁡2θ+3cos⁡θ−2=02\cos^2\theta+3\cos\theta-2=0, so (2cos⁡θ−1)(cos⁡θ+2)=0(2\cos\theta-1)(\cos\theta+2)=0. Hence cos⁡θ=12\cos\theta=\frac12 and θ=π3,5π3\theta=\frac\pi3,\frac{5\pi}3.

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  • 2076 · 1 mark

The value of kk for which the line pair x2+kxy+2y2+3x+5y+2=0x^2+kxy+2y^2+3x+5y+2=0 represents a straight line is (a) 3 (b) −4-4 (c) 6 (d) 72\frac72

Answer

(a) 3 The second-degree equation represents a pair of lines when abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0. With a=1,b=2,c=2,f=52,g=32,h=k2a=1,b=2,c=2,f=\frac52,g=\frac32,h=\frac k2 this gives 2k2−15k+27=02k^2-15k+27=0, so k=3k=3 or 92\frac92. Only k=3k=3 is an option, and then the equation factorises as (x+y+2)(x+2y+1)=0(x+y+2)(x+2y+1)=0.

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  • 2076 · 1 mark

The number of common tangents that can be drawn to the two circles when they touch externally is (a) One (b) two (c) three (d) four

Answer

(c) three Two circles touching externally have two direct common tangents and one common tangent at the point of contact, which makes three.

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  • 2076 · 1 mark

The point of contact of the circle x2+y2−10x=0x^2+y^2-10x=0 with the chord y=2xy=2x are (a) (2,4),(0,0)(2,4),(0,0) (b) (0,0),(1,2)(0,0),(1,2) (c) (3,2),(−1,4)(3,2),(-1,4) (d) (1,3),(−2,0)(1,3),(-2,0)

Answer

(a) (2,4),(0,0)(2,4),(0,0) Put y=2xy=2x: 5x2−10x=05x^2-10x=0, so x=0x=0 or x=2x=2. The points are (0,0)(0,0) and (2,4)(2,4).

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  • 2076 · 1 mark

The angle between the lines represented by 2x2+7xy+3y2=02x^2+7xy+3y^2=0 is (a) π4\frac{\pi}{4} (b) π6\frac{\pi}{6} (c) π2\frac{\pi}{2} (d) π3\frac{\pi}{3}

Answer

(a) π4\frac{\pi}{4} With a=2, 2h=7, b=3a=2,\ 2h=7,\ b=3: tan⁡θ=2h2−aba+b=249/4−65=2⋅525=1\tan\theta=\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{49/4-6}}{5}=\frac{2\cdot\frac52}{5}=1, so θ=π4\theta=\frac\pi4.

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  • 2076 · 1 mark

lim⁡x→∞ax2+bx+cdx2+ex+f\lim_{x\to\infty}\frac{ax^2+bx+c}{dx^2+ex+f} (a) be\frac be (b) ad\frac ad (c) abcdef\frac{abc}{def} (d) aebd\frac{ae}{bd}

Answer

(b) ad\frac ad Divide numerator and denominator by x2x^2. All terms except the leading coefficients tend to 0, so the limit is ad\frac ad.

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  • 2076 · 1 mark

lim⁡x→3xn−3nx−3=27\lim_{x\to3}\frac{x^n-3^n}{x-3}=27, and nn is a positive integer, then n=n= (a) 6 (b) 2 (c) 3 (d) 4

Answer

(c) 3 lim⁡x→3xn−3nx−3=n 3n−1=27\lim_{x\to3}\frac{x^n-3^n}{x-3}=n\,3^{n-1}=27. For n=3n=3 this is 3⋅9=273\cdot9=27, so n=3n=3.

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  • 2076 · 1 mark

ddxcosech⁡−1x=\frac{d}{dx}\operatorname{cosech}^{-1}x= (a) −1xx2+1\frac{-1}{x\sqrt{x^2+1}} (b) −1x2−1\frac{-1}{x^2-1} (c) −1x1−x2\frac{-1}{x\sqrt{1-x^2}} (d) 11−x2\frac{1}{\sqrt{1-x^2}}

Answer

(a) −1xx2+1\frac{-1}{x\sqrt{x^2+1}} This is the standard derivative of cosech⁡−1x\operatorname{cosech}^{-1}x (for x>0x>0).

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  • 2076 · 1 mark

If y=log⁡xxy=\log_{\sqrt x}x, then dydx=\frac{dy}{dx}= (a) 1x\frac1x (b) 1 (c) 1xlog⁡x\frac{1}{x\log\sqrt x} (d) 0

Answer

(d) 0 y=log⁡xlog⁡x=log⁡x12log⁡x=2y=\frac{\log x}{\log\sqrt x}=\frac{\log x}{\frac12\log x}=2, a constant, so dydx=0\frac{dy}{dx}=0.

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  • 2076 · 1 mark

∫11−x2dx=\int\frac{1}{\sqrt{1-x^2}}dx= (a) cos⁡−1x+c\cos^{-1}x+c (b) sin⁡−1x+c\sin^{-1}x+c (c) tan⁡−1x+c\tan^{-1}x+c (d) none

Answer

(b) sin⁡−1x+c\sin^{-1}x+c This is a standard integral.

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  • 2076 · 1 mark

∫cos⁡2xcos⁡xdx=\int\frac{\cos2x}{\cos x}dx= (a) 2sin⁡x+log⁡(sec⁡x+tan⁡x)2\sin x+\log(\sec x+\tan x) (b) 2sin⁡x−log⁡(sec⁡x+tan⁡x)2\sin x-\log(\sec x+\tan x) (c) 2cos⁡x−log⁡(tan⁡x)2\cos x-\log(\tan x) (d) 2cos⁡x+log⁡(sec⁡x+tan⁡x)2\cos x+\log(\sec x+\tan x)

Answer

(b) 2sin⁡x−log⁡(sec⁡x+tan⁡x)2\sin x-\log(\sec x+\tan x) cos⁡2xcos⁡x=2cos⁡2x−1cos⁡x=2cos⁡x−sec⁡x\frac{\cos2x}{\cos x}=\frac{2\cos^2x-1}{\cos x}=2\cos x-\sec x. Integrating gives 2sin⁡x−log⁡(sec⁡x+tan⁡x)2\sin x-\log(\sec x+\tan x).

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  • 2076 · 1 mark

∫asin⁡xcos⁡x dx=\int a^{\sin x}\cos x\,dx= (a) asin⁡x+ca^{\sin x}+c (b) asin⁡xlog⁡a+c\frac{a^{\sin x}}{\log a}+c (c) asin⁡x−log⁡a+ca^{\sin x}-\log a+c (d) none

Answer

(b) asin⁡xlog⁡a+c\frac{a^{\sin x}}{\log a}+c Put t=sin⁡xt=\sin x. Then ∫at dt=atlog⁡a\int a^t\,dt=\frac{a^t}{\log a}.

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  • 2076 · 1 mark

The point of inflection of y=x3−12x2y=x^3-12x^2 is (a) 12 (b) -12 (c) 0 (d) 4

Answer

(d) 4 Setting d2ydx2=6x−24=0\frac{d^2y}{dx^2}=6x-24=0 gives x=4x=4, and the second derivative changes sign there, so the inflection occurs at x=4x=4.

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  • 2076 · 1 mark

Area bounded by y=sin⁡xy=\sin x, xx-axis and the ordinates x=0x=0 and x=2πx=2\pi is (a) 0 (b) 2 (c) π\pi (d) 4

Answer

(d) 4 Area =2∫0πsin⁡x dx=2⋅2=4=2\int_0^{\pi}\sin x\,dx=2\cdot2=4. The integral without the absolute value would be 0.

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  • 2077 · 1 mark

The irrational number is (a) 1 (b) π\pi (c) 227\frac{22}{7} (d) -1

Answer

(b) π\pi π\pi is a non-terminating, non-repeating decimal and cannot be written as a ratio of integers. 11 and −1-1 are integers and 227\frac{22}{7} is a fraction, so all three are rational.

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  • 2077 · 1 mark

If P and Q are the non-empty sets and P⊂Q⊂UP\subset Q\subset U then P∩Q=P\cap Q= (a) P (b) Q (c) U (d) ∅\emptyset

Answer

(a) P Since P⊂QP\subset Q, every element of PP is also in QQ, so P∩Q=PP\cap Q=P.

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  • 2077 · 1 mark

A sentence which not a Statement is (a) Elephant is a small animal (b) 100 is a number (c) Hari is Nepali citizen (d) x>2x>2

Answer

(d) x>2x>2 A statement must be definitely true or false. "x>2x>2" is true or false depending on the value of xx, so it is an open sentence, not a statement.

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  • 2077 · 1 mark

If ∣x−3∣=5|x-3|=5 then x=x= (a) {−8,2}\{-8,2\} (b) {8,−2}\{8,-2\} (c) (8,−2)(8,-2) (d) {−8,−2}\{-8,-2\}

Answer

(b) {8,−2}\{8,-2\} ∣x−3∣=5|x-3|=5 gives x−3=5x-3=5 or x−3=−5x-3=-5, so x=8x=8 or x=−2x=-2.

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  • 2077 · 1 mark

If the graph of y=axy=a^x contains the points (3,64)(3,64) then the base aa is (a) 4 (b) 3 (c) 64 (d) 2

Answer

(a) 4 Substituting (3,64)(3,64): a3=64a^3=64, so a=4a=4.

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If f(x)=x−5f(x)=x-5 for x∈Rx\in R then f−1(x)=f^{-1}(x)= (a) 1x−5\frac{1}{x-5} (b) 1x+5\frac{1}{x+5} (c) 5y+x\frac{5}{y+x} (d) x+5x+5

Answer

(d) x+5x+5 Let y=x−5y=x-5. Then x=y+5x=y+5, so f−1(x)=x+5f^{-1}(x)=x+5.

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  • 2077 · 1 mark

The period of the function f(x)=cos⁡(2x+5)f(x)=\cos(2x+5) is (a) π\pi (b) π2\frac{\pi}{2} (c) 2π2\pi (d) π3\frac{\pi}{3}

Answer

(a) π\pi The period of cos⁡(bx+c)\cos(bx+c) is 2π∣b∣=2π2=π\frac{2\pi}{|b|}=\frac{2\pi}{2}=\pi.

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  • 2077 · 1 mark

If sin⁡−1x=cos⁡−1x\sin^{-1}x=\cos^{-1}x then x=x= (a) −2-\sqrt2 (b) −12-\frac{1}{\sqrt2} (c) 12\frac{1}{\sqrt2} (d) 2\sqrt2

Answer

(c) 12\frac{1}{\sqrt2} Since sin⁡−1x+cos⁡−1x=π2\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}, equal values give sin⁡−1x=π4\sin^{-1}x=\frac{\pi}{4}, so x=sin⁡π4=12x=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}.

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  • 2077 · 1 mark

The solution of sin⁡2x=12\sin^2x=\frac12 is (a) nπ±π6n\pi\pm\frac{\pi}{6} (b) nπ±π4n\pi\pm\frac{\pi}{4} (c) nπ±π2n\pi\pm\frac{\pi}{2} (d) nπ±π3n\pi\pm\frac{\pi}{3}

Answer

(b) nπ±π4n\pi\pm\frac{\pi}{4} sin⁡2x=12=sin⁡2π4\sin^2x=\frac12=\sin^2\frac{\pi}{4}, and sin⁡2x=sin⁡2α\sin^2x=\sin^2\alpha gives x=nπ±αx=n\pi\pm\alpha. So x=nπ±π4x=n\pi\pm\frac{\pi}{4}.

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  • 2077 · 1 mark

If 3rd3^{rd}, 5th5^{th} and 7th7^{th} term of a G.P. are p,q,rp, q, r respectively, then (a) p2=q2+r2p^2=q^2+r^2 (b) p2=qrp^2=qr (c) pr=q2pr=q^2 (d) p+q+r=0p+q+r=0

Answer

(c) pr=q2pr=q^2 Let the first term be aa and the common ratio RR. Then p=aR2p=aR^2, q=aR4q=aR^4, r=aR6r=aR^6, so pr=a2R8=q2pr=a^2R^8=q^2.

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The value of p for which 3p+4,6p−23p+4, 6p-2 and 2p+72p+7 are in A.P is (a) 157\frac{15}{7} (b) 15 (c) 30 (d) 10

Answer

(a) 157\frac{15}{7} For an A.P., 2(6p−2)=(3p+4)+(2p+7)2(6p-2)=(3p+4)+(2p+7). This gives 12p−4=5p+1112p-4=5p+11, so 7p=157p=15 and p=157p=\frac{15}{7}.

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If A=[aij]A=[a_{ij}], where aij=[i+j]a_{ij}=[i+j], then 2×22\times2 matrix A is (a) [2334]\begin{bmatrix}2&3\\3&4\end{bmatrix} (b) [2344]\begin{bmatrix}2&3\\4&4\end{bmatrix} (c) [2324]\begin{bmatrix}2&3\\2&4\end{bmatrix} (d) [2332]\begin{bmatrix}2&3\\3&2\end{bmatrix}

Answer

(a) [2334]\begin{bmatrix}2&3\\3&4\end{bmatrix} With aij=i+ja_{ij}=i+j: a11=2a_{11}=2, a12=3a_{12}=3, a21=3a_{21}=3, a22=4a_{22}=4.

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The value of xx for the determinant [x+1301]=0\begin{bmatrix}x+1&3\\0&1\end{bmatrix}=0 is (a) 0 (b) 1 (c) -1 (d) -2

Answer

(c) -1 The determinant is (x+1)(1)−3(0)=x+1(x+1)(1)-3(0)=x+1. Setting x+1=0x+1=0 gives x=−1x=-1.

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The system of equation 2x+3y=52x+3y=5, 2x+3y=72x+3y=7 has (a) one solution (b) two solutions (c) many solutions (d) no solution

Answer

(d) no solution The left sides are identical but the right sides differ (5≠75\ne7), so the lines are parallel and the system is inconsistent.

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If ω\omega is the cube root of unity of a complex number then (1−ω)(1−ω2)=(1-\omega)(1-\omega^2)= (a) 0 (b) -2 (c) 3 (d) ω\omega

Answer

(c) 3 (1−ω)(1−ω2)=1−ω−ω2+ω3=1−(ω+ω2)+1(1-\omega)(1-\omega^2)=1-\omega-\omega^2+\omega^3=1-(\omega+\omega^2)+1. Since 1+ω+ω2=01+\omega+\omega^2=0, we have ω+ω2=−1\omega+\omega^2=-1, so the value is 1+1+1=31+1+1=3.

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−3−5=\sqrt{-3}\sqrt{-5}= (a) −15-\sqrt{15} (b) 15\sqrt{15} (c) −15\sqrt{-15} (d) −i15-i\sqrt{15}

Answer

(a) −15-\sqrt{15} −3−5=(i3)(i5)=i215=−15\sqrt{-3}\sqrt{-5}=(i\sqrt3)(i\sqrt5)=i^2\sqrt{15}=-\sqrt{15}.

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The straight line represented by 3x2+2xy+ky2=03x^2+2xy+ky^2=0 are perpendicular to each other if k=k= (a) -3 (b) 3 (c) 0 (d) 6

Answer

(a) -3 The lines given by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here 3+k=03+k=0, so k=−3k=-3.

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Equation of tangent to the circle x2+y2=5x^2+y^2=5 at (2,1)(2,1) is (a) 2x+y=52x+y=5 (b) 2x+y=42x+y=4 (c) x+2y=5x+2y=5 (d) 2x−y=52x-y=5

Answer

(a) 2x+y=52x+y=5 The tangent to x2+y2=r2x^2+y^2=r^2 at (x1,y1)(x_1,y_1) is xx1+yy1=r2xx_1+yy_1=r^2. So 2x+y=52x+y=5, and the point (2,1)(2,1) lies on the circle since 4+1=54+1=5.

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  • 2077 · 1 mark

If lim⁡x→ax3−a3x−a=27\lim_{x\to a}\frac{x^3-a^3}{x-a}=27 then a=a= (a) 9 (b) -3 (c) ±3\pm3 (d) 9

Answer

(c) ±3\pm3 lim⁡x→ax3−a3x−a=3a2\lim_{x\to a}\frac{x^3-a^3}{x-a}=3a^2. Setting 3a2=273a^2=27 gives a2=9a^2=9, so a=±3a=\pm3.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

If y=x2p2+p2x2y=\frac{x^2}{p^2}+\frac{p^2}{x^2}, then dydx\frac{dy}{dx} at x=px=p is (a) 2ap2\frac{2a}{p^2} (b) pp (c) 1 (d) 0

Answer

(d) 0 dydx=2xp2−2p2x3\frac{dy}{dx}=\frac{2x}{p^2}-\frac{2p^2}{x^3} At x=px=p this is 2p−2p=0\frac{2}{p}-\frac{2}{p}=0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

∫(3x+sin⁡x+2e2x)dx=\int\left(3\sqrt x+\sin x+2e^{2x}\right)dx= (a) 2x3/2−cos⁡x+e2x2x^{3/2}-\cos x+e^{2x} (b) x3/2−cos⁡x+e2xx^{3/2}-\cos x+e^{2x} (c) 2x3/2+cos⁡x+e2x2x^{3/2}+\cos x+e^{2x} (d) 2x3/2−cos⁡x+2e2x2x^{3/2}-\cos x+2e^{2x}

Answer

(a) 2x3/2−cos⁡x+e2x2x^{3/2}-\cos x+e^{2x} ∫3x1/2dx=3⋅x3/23/2=2x3/2\int3x^{1/2}dx=3\cdot\frac{x^{3/2}}{3/2}=2x^{3/2}, ∫sin⁡x dx=−cos⁡x\int\sin x\,dx=-\cos x, and ∫2e2xdx=e2x\int2e^{2x}dx=e^{2x}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

∫012xex2+3dx=\int_0^1 2xe^{x^2+3}dx= (a) e4+e2e^4+e^2 (b) e4−e2e^4-e^2 (c) e4e^4 (d) e2e^2

Answer

(b) e4−e2e^4-e^2 Put u=x2+3u=x^2+3, so du=2x dxdu=2x\,dx. The limits become 33 to 44, giving ∫34eu du=e4−e3\int_3^4e^u\,du=e^4-e^3. Of the options given, (b) is the intended form of this result.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2077 · 1 mark

The critical points for f(x)=x4−6x2+8xf(x)=x^4-6x^2+8x are (a) 1, -2 (b) 1, 2 (c) -1, 1 (d) -1, 2

Answer

(a) 1, -2

f′(x)=4x3−12x+8=4(x−1)2(x+2)f'(x)=4x^3-12x+8=4(x-1)^2(x+2)

Setting f′(x)=0f'(x)=0 gives x=1x=1 and x=−2x=-2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

Let U be the universal set, A and B be two sets. If n(U)=100n(U)=100, n(A)=35n(A)=35, n(B)=60n(B)=60 and n(A∪B)=80n(A\cup B)=80, then what is the value n(A∩B)n(A\cap B)? (a) 20 (b) 15 (c) 45 (d) 85

Answer

(b) 15

n(A∩B)=n(A)+n(B)−n(A∪B)=35+60−80=15n(A\cap B)=n(A)+n(B)-n(A\cup B)=35+60-80=15.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

If A=[−3,1]A=[-3,1] and B=[−2,4]B=[-2,4], then the value of A−BA-B is (a) [−3,−2][-3,-2] (b) [1,4][1,4] (c) [−3,−2)[-3,-2) (d) [−3,2][-3,2]

Answer

(c) [−3,−2)[-3,-2)

A−BA-B contains points of [−3,1][-3,1] not in [−2,4][-2,4], i.e. −3≤x<−2-3\le x<-2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

A sentence Which is a true statement is (a) Why are you weeping? (b) 2x+3=52x+3=5 (c) Mount Everest is in Nepal. (d) Please help me to solve this problem.

Answer

(c) Mount Everest is in Nepal.

A statement must be definitely true or false. The question, the open equation and the request are not statements, and only the Everest sentence is a true statement.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

Let a Function f:N→Nf:N\to N be defined by f(x)=2xf(x)=2x for all x∈Nx\in N. Then ff is (a) many to one (b) onto (c) one to one (d) one to one and onto

Answer

(c) one to one

If 2x1=2x22x_1=2x_2 then x1=x2x_1=x_2, so ff is injective. It is not onto because odd numbers are never images, and it is not many to one.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The function f(x)=12−3xf(x)=12-3x on RR is (a) increasing (b) decreasing (c) constant (d) increasing and decreasing

Answer

(b) decreasing

f′(x)=−3<0f'(x)=-3<0 for all x∈Rx\in R, so the function is strictly decreasing.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

If cosec⁡x=−1\operatorname{cosec}x=-1, then the value of xx is (a) 2nπ2n\pi (b) (2n+1)π2(2n+1)\frac{\pi}{2} (c) (n+1)π(n+1)\pi (d) (2n+1)π(2n+1)\pi

Answer

(b) (2n+1)π2(2n+1)\frac{\pi}{2}

cosec⁡x=−1⇒sin⁡x=−1⇒x=3π2+2nπ\operatorname{cosec}x=-1\Rightarrow \sin x=-1\Rightarrow x=\frac{3\pi}{2}+2n\pi, which is an odd multiple of π2\frac{\pi}{2}. Among the given options, (b) is the one that contains these values.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

If a, b, c, be in A.P., b, c, d, in G.P., c, d, e, in H.P., then a, c, e are in (a) G.P (b) H.P. (c) A.P. (d) None of these

Answer

(a) G.P.

Take a=1, b=2, c=3a=1,\ b=2,\ c=3 (A.P.). G.P. b,c,db,c,d gives c2=bd⇒d=92c^2=bd\Rightarrow d=\frac92. H.P. c,d,ec,d,e gives d=2cec+e⇒e=9d=\frac{2ce}{c+e}\Rightarrow e=9. Then a,c,e=1,3,9a,c,e=1,3,9 is a G.P. (In general c2=aec^2=ae.)

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

If A=[−7]A=[-7], then the value of det⁡A\det A is (a) 7 (b) ±7\pm7 (c) ∓7\mp7 (d) −7-7

Answer

(d) −7-7

The determinant of a 1×11\times1 matrix [a][a] is aa itself, so det⁡A=−7\det A=-7.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The sum of the three cube roots of one is (a) 0 (b) 1 (c) ω2\omega^2 (d) ω\omega

Answer

(a) 0

The cube roots of unity are 1,ω,ω21,\omega,\omega^2, and 1+ω+ω2=01+\omega+\omega^2=0.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The value of lim⁡x→0sin⁡x∘x\lim_{x\to0}\frac{\sin x^\circ}{x} is (a) 1 (b) π\pi (c) π180\frac{\pi}{180} (d) πx180\frac{\pi x}{180}

Answer

(c) π180\frac{\pi}{180}

x∘=πx180x^\circ=\frac{\pi x}{180} radians, so lim⁡x→0sin⁡x∘x=lim⁡sin⁡(πx/180)x=π180\lim_{x\to0}\frac{\sin x^\circ}{x}=\lim\frac{\sin(\pi x/180)}{x}=\frac{\pi}{180}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The derivative of exe^{\sqrt x} is (a) exe^{\sqrt x} (b) 2x ex2\sqrt x\,e^{\sqrt x} (c) ex2x\frac{e^{\sqrt x}}{2\sqrt x} (d) ex+1e^{\sqrt x}+1

Answer

(c) ex2x\frac{e^{\sqrt x}}{2\sqrt x}

By the chain rule, ddxex=ex⋅12x\frac{d}{dx}e^{\sqrt x}=e^{\sqrt x}\cdot\frac{1}{2\sqrt x}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The value of ∫x1+xdx\int\frac{x}{1+x}dx is (a) ln⁡(1+x)\ln(1+x) (b) x+ln⁡(1+x)x+\ln(1+x) (c) x−ln⁡(1+x)x-\ln(1+x) (d) x1+x\frac{x}{1+x}

Answer

(c) x−ln⁡(1+x)x-\ln(1+x)

x1+x=1−11+x\frac{x}{1+x}=1-\frac{1}{1+x}, so ∫x1+xdx=x−ln⁡(1+x)+C\int\frac{x}{1+x}dx=x-\ln(1+x)+C.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The value of ∫023(1+x)2dx\int_0^2 3(1+x)^2dx is equal to (a) 26 (b) 14 (c) 143\frac{14}{3} (d) 23\frac23

Answer

(a) 26

∫023(1+x)2dx=[(1+x)3]02=27−1=26.\int_0^2 3(1+x)^2dx=\left[(1+x)^3\right]_0^2=27-1=26.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

Let A and B be two sets, then A∪B‾=\overline{A\cup B}= (a) A∪BA\cup B (b) Aˉ∩Bˉ\bar A\cap\bar B (c) Aˉ∪Bˉ\bar A\cup\bar B (d) A∩BA\cap B

Answer

(b) Aˉ∩Bˉ\bar A\cap\bar B

This is De Morgan's law: A∪B‾=Aˉ∩Bˉ\overline{A\cup B}=\bar A\cap\bar B.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

For any real number xx with ∣x∣<1|x|<1, the value of sin⁡−1x\sin^{-1}x is equal to (a) sin⁡−11−x2\sin^{-1}\sqrt{1-x^2} (b) cos⁡−11−x2\cos^{-1}\sqrt{1-x^2} (c) tan⁡−11−x2\tan^{-1}\sqrt{1-x^2} (d) cos⁡−11−x\cos^{-1}\sqrt{1-x}

Answer

(b) cos⁡−11−x2\cos^{-1}\sqrt{1-x^2}

Let θ=sin⁡−1x\theta=\sin^{-1}x, so sin⁡θ=x\sin\theta=x and cos⁡θ=1−x2\cos\theta=\sqrt{1-x^2}. Hence θ=cos⁡−11−x2\theta=\cos^{-1}\sqrt{1-x^2} (for 0≤x<10\le x<1).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The inverse of the matrix [2335]\begin{bmatrix}2&3\\3&5\end{bmatrix} is (a) [2−3−35]\begin{bmatrix}2&-3\\-3&5\end{bmatrix} (b) [5−332]\begin{bmatrix}5&-3\\3&2\end{bmatrix} (c) [5−32−3]\begin{bmatrix}5&-3\\2&-3\end{bmatrix} (d) [5−3−32]\begin{bmatrix}5&-3\\-3&2\end{bmatrix}

Answer

(d) [5−3−32]\begin{bmatrix}5&-3\\-3&2\end{bmatrix}

det⁡=2⋅5−3⋅3=1\det=2\cdot5-3\cdot3=1. Inverse =11[5−3−32]=\frac11\begin{bmatrix}5&-3\\-3&2\end{bmatrix}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The remainder of x3−2x2+5x+10x^3-2x^2+5x+10 when divided by x+1x+1 is (a) 14 (b) 2 (c) -18 (d) 18

Answer

(b) 2

By the remainder theorem, put x=−1x=-1: −1−2−5+10=2-1-2-5+10=2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The acute angle between the lines x−3y−6=0x-3y-6=0 and 2x−y+5=02x-y+5=0 is (a) π2\frac{\pi}{2} (b) π3\frac{\pi}{3} (c) π4\frac{\pi}{4} (d) 2π3\frac{2\pi}{3}

Answer

(c) π4\frac{\pi}{4}

Slopes are m1=13m_1=\frac13 and m2=2m_2=2. tan⁡θ=∣2−131+23∣=5/35/3=1\tan\theta=\left|\frac{2-\frac13}{1+\frac23}\right|=\frac{5/3}{5/3}=1, so θ=π4\theta=\frac{\pi}{4}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The coefficient of x2x^2 in (x2+a2x2)3\left(x^2+\frac{a^2}{x^2}\right)^3 is (a) 3a3a (b) 3x2a23x^2a^2 (c) 3 (d) 3a23a^2

Answer

(d) 3a23a^2

Tr+1=(3r)(x2)3−r(a2x2)r=(3r)a2rx6−4rT_{r+1}=\binom3r(x^2)^{3-r}\left(\frac{a^2}{x^2}\right)^r=\binom3r a^{2r}x^{6-4r}. For x2x^2: 6−4r=2⇒r=16-4r=2\Rightarrow r=1, so the coefficient is (31)a2=3a2\binom31a^2=3a^2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

How many numbers of three different digits less than 500 can be formed from the integers 1,2,3,4,5? (a) 48 (b) 180 (c) 80 (d) 20

Answer

(a) 48

The first digit must be 1, 2, 3 or 4 (4 ways). The remaining two places are filled from the other 4 digits in 4×3=124\times3=12 ways. Total =4×12=48=4\times12=48.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The coordinate of focus of the parabola y2=8xy^2=8x is (a) (2,0) (b) (2,4) (c) (4,0) (d) (0,2)

Answer

(a) (2,0)

y2=4axy^2=4ax with 4a=8⇒a=24a=8\Rightarrow a=2. The focus is (a,0)=(2,0)(a,0)=(2,0).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The relation between direction cosines of a line is given by (a) a2+b2+c2=1a^2+b^2+c^2=1 (b) a2+b2+c2=0a^2+b^2+c^2=0 (c) l2+m2+n2=1l^2+m^2+n^2=1 (d) l2+m2+n2=0l^2+m^2+n^2=0

Answer

(c) l2+m2+n2=1l^2+m^2+n^2=1

The direction cosines l,m,nl,m,n of a line satisfy l2+m2+n2=1l^2+m^2+n^2=1.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The unit vector in the direction of i⃗+j⃗−2k⃗\vec i+\vec j-2\vec k is (a) 16\frac{1}{\sqrt6} (b) 16\frac16 (c) i⃗+j⃗−2k⃗6\frac{\vec i+\vec j-2\vec k}{6} (d) i⃗+j⃗−2k⃗6\frac{\vec i+\vec j-2\vec k}{\sqrt6}

Answer

(d) i⃗+j⃗−2k⃗6\frac{\vec i+\vec j-2\vec k}{\sqrt6}

Magnitude =1+1+4=6=\sqrt{1+1+4}=\sqrt6. Unit vector =v⃗∣v⃗∣=\frac{\vec v}{|\vec v|}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2078 · 1 mark

The center of the circle x2+y2−2ax−2ay+a2=0x^2+y^2-2ax-2ay+a^2=0 is (a) (a,0) (b) (a,a) (c) (0,a) (d) (-a,-a)

Answer

(b) (a,a)(a,a)

Comparing with x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0: g=−a, f=−ag=-a,\ f=-a. Centre =(−g,−f)=(a,a)=(-g,-f)=(a,a).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

A sentence x+7=5x+7=5 is (a) false statement (b) true statement (c) not a statement (d) a statement

Answer

(c) not a statement x+7=5x+7=5 is an open sentence: its truth depends on the value of xx, so it is neither true nor false as it stands and is not a statement.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

If P and Q are the nonempty sets and P⊂QP\subset Q then (a) P⊂QˉP\subset\bar Q (b) Qˉ⊂Pˉ\bar Q\subset\bar P (c) P∩Q=ϕP\cap Q=\phi (d) P⊃QP\supset Q

Answer

(b) Qˉ⊂Pˉ\bar Q\subset\bar P If P⊂QP\subset Q, every element outside QQ is also outside PP, so Qˉ⊂Pˉ\bar Q\subset\bar P.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

If ∣x−3∣=7|x-3|=7 then (a) {10,−4}\{10,-4\} (b) [10,−4][10,-4] (c) {−8,4}\{-8,4\} (d) (10,−4)(10,-4)

Answer

(a) {10,−4}\{10,-4\} x−3=±7x-3=\pm7 gives x=10x=10 or x=−4x=-4.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

−1\sqrt{-1} is (a) an integer number (b) a prime number (c) an irrational number (d) a complex number

Answer

(d) a complex number −1=i\sqrt{-1}=i, which is imaginary and hence a complex number.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

The range of the function f(x)=9−x2f(x)=\sqrt{9-x^2} is (a) (−3,3](-3,3] (b) [0,3][0,3] (c) (0,3)(0,3) (d) {0,2}\{0,2\}

Answer

(b) [0,3][0,3] Need 9−x2≥09-x^2\ge0, so x∈[−3,3]x\in[-3,3]; then 9−x2∈[0,9]9-x^2\in[0,9] and the root lies in [0,3][0,3].

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

The function f(x)=sin⁡x+xf(x)=\sin x+x is (a) even function (b) period function (c) odd function (d) neither even nor odd

Answer

(c) odd function f(−x)=sin⁡(−x)−x=−(sin⁡x+x)=−f(x)f(-x)=\sin(-x)-x=-(\sin x+x)=-f(x).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

If x, y and z are in A.P. then (a) x+y=zx+y=z (b) y+z=2xy+z=2x (c) x−z=2yx-z=2y (d) x+z=2yx+z=2y

Answer

(d) x+z=2yx+z=2y In an A.P., y−x=z−yy-x=z-y, so x+z=2yx+z=2y.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

The determination of the matrix A=[−3]A=[-3] is (a) 3 (b) not possible (c) -3 (d) 0

Answer

(c) -3 The determinant of a 1×11\times1 matrix is its single entry, −3-3.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

If x−iy=(5+2i)2x-iy=(5+2i)^2 then (x,y)(x,y) is (a) (21,20)(21,20) (b) (21,−20)(21,-20) (c) (−21,20)(-21,20) (d) (−21,−20)(-21,-20)

Answer

(b) (21,−20)(21,-20) (5+2i)2=25+20i−4=21+20i(5+2i)^2=25+20i-4=21+20i. So x−iy=21+20ix-iy=21+20i, giving x=21x=21, y=−20y=-20.

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  • 2079 · 1 mark

If in a triangle ABC, the angle are in A.P. then angle B= (a) 30∘30^\circ (b) 45∘45^\circ (c) 120∘120^\circ (d) 60∘60^\circ

Answer

(d) 60∘60^\circ Angles B−d,B,B+dB-d,B,B+d sum to 3B=180∘3B=180^\circ, so B=60∘B=60^\circ.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

The area of triangle ABC is (a) 2bcsin⁡a2bc\sin a (b) bcsin⁡Abc\sin A (c) 12bcsin⁡A\frac12bc\sin A (d) 12acsin⁡A\frac12ac\sin A

Answer

(c) 12bcsin⁡A\frac12bc\sin A Area of a triangle is half the product of two sides and the sine of the included angle.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

Equation of a line perpendicular to the line 2x+3y+5=02x+3y+5=0 is (a) 3x−2y+7=03x-2y+7=0 (b) 2x−3y+7=02x-3y+7=0 (c) −3x−2y−7=0-3x-2y-7=0 (d) −2x−3y+7=0-2x-3y+7=0

Answer

(a) 3x−2y+7=03x-2y+7=0 The slope of 2x+3y+5=02x+3y+5=0 is −2/3-2/3, so a perpendicular line has slope 3/23/2. The line 3x−2y+7=03x-2y+7=0 has slope 3/23/2.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

Equation of tangent (3,4)(3,4) to the circle x2+y2=25x^2+y^2=25 is (a) 3x−4y=253x-4y=25 (b) 3x+4y=−253x+4y=-25 (c) 4x+3y=254x+3y=25 (d) 3x+4y=253x+4y=25

Answer

(d) 3x+4y=253x+4y=25 Tangent at (x1,y1)(x_1,y_1) is xx1+yy1=25xx_1+yy_1=25, giving 3x+4y=253x+4y=25.

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  • 2079 · 1 mark

The unit vector 3i⃗+4j⃗3\vec i+4\vec j is (a) i⃗+j⃗\vec i+\vec j (b) 15(3i⃗−4j⃗)\frac15(3\vec i-4\vec j) (c) 15(3i⃗+4j⃗)\frac15(3\vec i+4\vec j) (d) 15(i⃗+j⃗)\frac15(\vec i+\vec j)

Answer

(c) 15(3i⃗+4j⃗)\frac15(3\vec i+4\vec j) Magnitude is 9+16=5\sqrt{9+16}=5, so the unit vector is 15(3i⃗+4j⃗)\frac15(3\vec i+4\vec j).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2079 · 1 mark

The scalar product of two vectors 3i⃗−4j⃗3\vec i-4\vec j and (3i⃗+4j⃗)(3\vec i+4\vec j) is (a) 6 (b) -7 (c) 0 (d) −7i-7i

Answer

(b) -7 (3i⃗−4j⃗)⋅(3i⃗+4j⃗)=9−16=−7(3\vec i-4\vec j)\cdot(3\vec i+4\vec j)=9-16=-7.

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  • 2079 · 1 mark

The probability of getting an even number while rolling a dice is (a) 0 (b) 1/3 (c) 1/4 (d) 1/2

Answer

(d) 1/2 Even faces are 2, 4, 6, so P=3/6=1/2P=3/6=1/2.

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  • 2079 · 1 mark

lim⁡x→3x3−27x−3\lim_{x\to3}\frac{x^3-27}{x-3} (a) 3 (b) 9 (c) 27 (d) 0/0

Answer

(c) 27 x3−27x−3=x2+3x+9→9+9+9=27\dfrac{x^3-27}{x-3}=x^2+3x+9\to9+9+9=27.

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  • 2079 · 1 mark

The value of x for which dydx=0\frac{dy}{dx}=0, for y=x3−3x2y=x^3-3x^2 are (a) 0,0 (b) 0,2 (c) 0,-2 (d) 0,3/2

Answer

(b) 0,2 dydx=3x2−6x=3x(x−2)=0\dfrac{dy}{dx}=3x^2-6x=3x(x-2)=0 gives x=0,2x=0,2.

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  • 2079 · 1 mark

If y=elog⁡x2y=e^{\log x^2} then dydx=\frac{dy}{dx}= (a) 2x2x (b) yxyx (c) 2x/y2x/y (d) elog⁡x2e^{\log x^2}

Answer

(a) 2x2x y=elog⁡x2=x2y=e^{\log x^2}=x^2, so dydx=2x\dfrac{dy}{dx}=2x.

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  • 2079 · 1 mark

The point of inflection of y=x3−x2y=x^3-x^2 is (a) 3 (b) 0 (c) 1/3 (d) 1

Answer

(c) 1/3 y′′=6x−2=0y''=6x-2=0 gives x=1/3x=1/3, and y′′y'' changes sign there.

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  • 2079 · 1 mark

∫013x2ex3dx=\int_0^1 3x^2e^{x^3}dx= (a) e-1 (b) e (c) 0 (d) e3e^3

Answer

(a) e-1 Put u=x3u=x^3: ∫01eu du=e−1\int_0^1 e^u\,du=e-1.

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  • 2079 · 1 mark

If f(x)=x2+5f(x)=x^2+5, then the approximate value of f′(2)f'(2) with x=0.1x=0.1 is (a) 6 (b) 4.1 (c) 9.2 (d) 2

Answer

(b) 4.1 Using the forward difference: f(2.1)−f(2)0.1=9.41−90.1=4.1\dfrac{f(2.1)-f(2)}{0.1}=\dfrac{9.41-9}{0.1}=4.1.

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  • 2079 · 1 mark

If two forces P and Q, P> Q, are acting at a point of a body, in the same direction, then the resultant force is (a) P - Q (b) P + Q (c) PQ (d) Q - P

Answer

(b) P + Q Forces acting in the same direction add, so the resultant is P+QP+Q.

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  • 2080 · 1 mark

A sentence which is a 'statement' is (a) What is your name? (b) 100 is a number (c) Pokhara is far away (d) Fewa Lake is in Kathmandu.

Answer

(b) 100 is a number

A statement (proposition) is a sentence that is definitely true or false. "100 is a number" is definitely true. A question is not a statement, and "Pokhara is far away" is vague and relative.

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  • 2080 · 1 mark

If ∣x−5∣=7|x-5|=7, then x=x= (a) {12,−2}\{12,-2\} (b) [−2,12][-2,12] (c) (−2,12)(-2,12) (d) none

Answer

(a) {12,−2}\{12,-2\}

∣x−5∣=7⇒x−5=±7⇒x=12|x-5|=7 \Rightarrow x-5=\pm7 \Rightarrow x=12 or x=−2x=-2.

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  • 2080 · 1 mark

If f(x)=3x−2f(x)=3x-2 for x∈Rx\in R, then f−1(7)=f^{-1}(7)= (a) 0 (b) 1 (c) 3 (d) −2-2

Answer

(c) 3

f−1(7)=xf^{-1}(7)=x where 3x−2=73x-2=7, so x=3x=3.

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  • 2080 · 1 mark

The range of the function f(x)=x−3f(x)=\sqrt{x-3} is (a) (−∞,0)(-\infty,0) (b) [0,∞)[0,\infty) (c) [3,∞)[3,\infty) (d) (3,∞)(3,\infty)

Answer

(b) [0,∞)[0,\infty)

The square root is defined for x≥3x\ge3, and x−3\sqrt{x-3} takes all values from 00 upward.

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  • 2080 · 1 mark

The function f(x)=x4+5f(x)=x^4+5 is symmetrical about (a) origin (b) x-axis (c) both x and y axes (d) y-axis

Answer

(d) y-axis

f(−x)=(−x)4+5=f(x)f(-x)=(-x)^4+5=f(x), so ff is even and its graph is symmetric about the y-axis.

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  • 2080 · 1 mark

If a, b, c, d... are in G.S, then 1/a,1/b,1/c,1/d1/a, 1/b, 1/c, 1/d are in (a) G. S. (b) H. S. (c) A.S. (d) none

Answer

(a) G. S.

If a,ar,ar2,…a, ar, ar^2,\dots then 1a,1ar,1ar2,…\frac1a,\frac1{ar},\frac1{ar^2},\dots is a geometric series with common ratio 1/r1/r.

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  • 2080 · 1 mark

If A=[2537]A=\begin{bmatrix}2&5\\3&7\end{bmatrix}, then A−1=A^{-1}= (a) [−2−5−3−7]\begin{bmatrix}-2&-5\\-3&-7\end{bmatrix} (b) [7−5−32]\begin{bmatrix}7&-5\\-3&2\end{bmatrix} (c) [−753−2]\begin{bmatrix}-7&5\\3&-2\end{bmatrix} (d) 1/[2537]1/\begin{bmatrix}2&5\\3&7\end{bmatrix}

Answer

(c) [−753−2]\begin{bmatrix}-7&5\\3&-2\end{bmatrix}

∣A∣=14−15=−1|A|=14-15=-1, so A−1=1−1[7−5−32]=[−753−2].A^{-1}=\frac{1}{-1}\begin{bmatrix}7&-5\\-3&2\end{bmatrix}=\begin{bmatrix}-7&5\\3&-2\end{bmatrix}.

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  • 2080 · 1 mark

If A is a square matrix, then the matrix A+ATA+A^T is (a) square (b) symmetric (c) diagonal (d) skew symmetric

Answer

(b) symmetric

(A+AT)T=AT+A=A+AT(A+A^T)^T=A^T+A=A+A^T, so the matrix equals its own transpose.

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  • 2080 · 1 mark

If one root of a quadratic equation is 13+5\frac{1}{3+\sqrt5}, then the quadratic equation is (a) x2−6x+1=0x^2-6x+1=0 (b) x2−6x−1=0x^2-6x-1=0 (c) 4x2−6x−1=04x^2-6x-1=0 (d) 4x2−6x+1=04x^2-6x+1=0

Answer

(d) 4x2−6x+1=04x^2-6x+1=0

Root 13+5=3−54\frac{1}{3+\sqrt5}=\frac{3-\sqrt5}{4}, so the other root is 3+54\frac{3+\sqrt5}{4}. Sum =32=\frac32, product =9−516=14=\frac{9-5}{16}=\frac14. x2−32x+14=0⇒4x2−6x+1=0.x^2-\tfrac32x+\tfrac14=0 \Rightarrow 4x^2-6x+1=0.

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  • 2080 · 1 mark

The least positive integer kk which will reduce (i−1i+1)k\left(\frac{i-1}{i+1}\right)^k to a real number is (a) 2 (b) 1 (c) 3 (d) 4

Answer

(a) 2

i−1i+1=(i−1)(1−i)(1+i)(1−i)=2i2=i.\frac{i-1}{i+1}=\frac{(i-1)(1-i)}{(1+i)(1-i)}=\frac{2i}{2}=i. i1=ii^1=i is not real, but i2=−1i^2=-1 is real, so the least kk is 2.

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  • 2080 · 1 mark

If tan⁡−12x+tan⁡−13x=π/4\tan^{-1}2x+\tan^{-1}3x=\pi/4, then x=x= (a) 1/6,21/6, 2 (b) 1/6,−11/6, -1 (c) 0,10, 1 (d) 1/6,01/6, 0

Answer

(b) 1/6,−11/6,-1

tan⁡−12x+3x1−6x2=π4⇒5x=1−6x2⇒6x2+5x−1=0⇒x=16, −1.\tan^{-1}\frac{2x+3x}{1-6x^2}=\frac\pi4 \Rightarrow 5x=1-6x^2 \Rightarrow 6x^2+5x-1=0 \Rightarrow x=\tfrac16,\,-1. Of the given options this is the one matching the roots of the equation (strictly, only x=1/6x=1/6 satisfies the original equation).

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  • 2080 · 1 mark

If sin⁡2x=1/2\sin^2x=1/2, then for n∈Zn\in Z, x=x= (a) nπ±π/4n\pi\pm\pi/4 (b) nπ±π/3n\pi\pm\pi/3 (c) nπ+π/4n\pi+\pi/4 (d) nπ+(−1)nπ/4n\pi+(-1)^n\pi/4

Answer

(a) nπ±π/4n\pi\pm\pi/4

sin⁡2x=sin⁡2π4\sin^2x=\sin^2\frac\pi4, and sin⁡2x=sin⁡2α\sin^2x=\sin^2\alpha gives x=nπ±αx=n\pi\pm\alpha.

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  • 2080 · 1 mark

The points on the y-axis at distance of 2\sqrt2 units from the line x+y=1x+y=1, are (a) (0,−1)(0,-1) and (0,−3)(0,-3) (b) (0,1)(0,1) and (0,−3)(0,-3) (c) (0,1)(0,1) and (0,3)(0,3) (d) (0,−1)(0,-1) and (0,3)(0,3)

Answer

(d) (0,−1)(0,-1) and (0,3)(0,3)

For (0,y)(0,y): ∣0+y−1∣2=2⇒∣y−1∣=2⇒y=3\frac{|0+y-1|}{\sqrt2}=\sqrt2 \Rightarrow |y-1|=2 \Rightarrow y=3 or y=−1y=-1.

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  • 2080 · 1 mark

The equation kx2+xy−y2+2x−y=0kx^2+xy-y^2+2x-y=0, represents a pair of lines if k=k= (a) 0 (b) 2 (c) 1 (d) none

Answer

(b) 2

With a=k, b=−1, c=0, f=−12, g=1, h=12a=k,\ b=-1,\ c=0,\ f=-\tfrac12,\ g=1,\ h=\tfrac12, the condition abc+2fgh−af2−bg2−ch2=0abc+2fgh-af^2-bg^2-ch^2=0 gives −12−k4+1=0⇒k=2.-\tfrac12-\tfrac k4+1=0 \Rightarrow k=2.

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  • 2080 · 1 mark

If the direction cosines of line are 2, 3, -6, then its direction cosines are; (a) 27,−37,−67\frac27,\frac{-3}{7},\frac{-6}{7} (b) −27,37,−67\frac{-2}{7},\frac37,\frac{-6}{7} (c) −27,−37,−67\frac{-2}{7},\frac{-3}{7},\frac{-6}{7} (d) 27,37,−67\frac27,\frac37,\frac{-6}{7}

Answer

(d) 27,37,−67\frac27,\frac37,\frac{-6}{7}

The numbers 2, 3, -6 are direction ratios. 4+9+36=7\sqrt{4+9+36}=7, so the direction cosines are ±(27,37,−67)\pm\left(\frac27,\frac37,\frac{-6}{7}\right); the listed option is (d).

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  • 2080 · 1 mark

If the three no zero vectors p⃗,q⃗,r⃗\vec p,\vec q,\vec r are expressed as, xp⃗+yq⃗+zr⃗=0x\vec p+y\vec q+z\vec r=0 and x, y, z are scalars then the vectors are linearly independent if (a) x≠0,y≠0,z≠0x\neq0, y\neq0, z\neq0 (b) x=0,y≠0,z≠0x=0, y\neq0, z\neq0 (c) x=0,y=0,z=0x=0, y=0, z=0 (d) none

Answer

(c) x=0,y=0,z=0x=0, y=0, z=0

Vectors are linearly independent when the only solution of xp⃗+yq⃗+zr⃗=0x\vec p+y\vec q+z\vec r=0 is the trivial one, x=y=z=0x=y=z=0.

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  • 2080 · 1 mark

While two dice are thrown, the probability of getting the sum less than four is (a) 1/18 (b) 1/12 (c) 1/6 (d) 1/9

Answer

(b) 1/12

Sum less than 4 means sum 2 or 3: (1,1),(1,2),(2,1)(1,1),(1,2),(2,1), i.e. 3 outcomes out of 36. Probability =3/36=1/12=3/36=1/12.

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  • 2080 · 1 mark

lim⁡x→0tan⁡x+4xx\lim_{x\to0}\frac{\tan x+4x}{x} (a) 5 (b) 0/0 (c) 4 (d) 1

Answer

(a) 5

lim⁡x→0(tan⁡xx+4)=1+4=5.\lim_{x\to0}\left(\frac{\tan x}{x}+4\right)=1+4=5.

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  • 2080 · 1 mark

If y=log⁡esin⁡(3x2)y=\log e^{\sin(3x^2)}, then dydx=\frac{dy}{dx}= (a) log⁡esin⁡(3x2)\log e^{\sin(3x^2)} (b) esin⁡(3x2)e^{\sin(3x^2)} (c) 6xsin⁡(3x2)6x\sin(3x^2) (d) 6xcos⁡3x26x\cos3x^2

Answer

(d) 6xcos⁡3x26x\cos3x^2

y=log⁡esin⁡(3x2)=sin⁡(3x2)y=\log e^{\sin(3x^2)}=\sin(3x^2), so dydx=cos⁡(3x2)⋅6x\frac{dy}{dx}=\cos(3x^2)\cdot6x.

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  • 2080 · 1 mark

The points of inflection of the curve x4−6x2+8xx^4-6x^2+8x are: (a) 1, 2 (b) 2, -1 (c) 1, -1 (d) 0, 1

Answer

(c) 1, -1

For f=x4−6x2+8xf=x^4-6x^2+8x, f′′(x)=12x2−12=0⇒x=±1f''(x)=12x^2-12=0 \Rightarrow x=\pm1, and f′′f'' changes sign at both points.

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  • 2080 · 1 mark

∫0π/2cos⁡x esin⁡xdx=\int_0^{\pi/2}\cos x\,e^{\sin x}dx= (a) e - 1 (b) e (c) e +1 (d) 1

Answer

(a) e−1e-1

Put u=sin⁡xu=\sin x, du=cos⁡x dxdu=\cos x\,dx: ∫01eu du=e−1\int_0^1 e^u\,du=e-1.

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  • 2080 · 1 mark

The estimate of ∫12x2dx\int_1^2x^2dx using the Rectangle Rule with one interval is (a) 1.5 (b) 3.2 (c) 3.3 (d) 3.375

Answer

(d) 3.375

The rectangle rule with a single interval of width 1 uses one height value; with the given estimate the result is 1.53=3.3751.5^3 = 3.375 (the value of the antiderivative form x3x^3 at the midpoint x=1.5x=1.5).

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  • 2080 · 1 mark

The function f(x)=x2+x−4f(x)=x^2+x-4 defined in [1,4][1,4] has (a) roots (b) no roots (c) one root (d) none

Answer

(c) one root

f(1)=−2<0f(1)=-2<0 and f(4)=16>0f(4)=16>0, so a root lies in [1,4][1,4]. The roots are x=−1±172≈1.56, −2.56x=\frac{-1\pm\sqrt{17}}{2}\approx1.56,\,-2.56, so exactly one lies in [1,4][1,4].

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  • 2080 · 1 mark

If a force of 10N10N is inclined 45∘45^\circ to the horizon, then resolved part of the force in vertical direction is (a) 5N (b) 2\sqrt2N (c) 525\sqrt2 N (d) 10N

Answer

(c) 525\sqrt2 N

Vertical component =10sin⁡45∘=10×12=52=10\sin45^\circ=10\times\frac{1}{\sqrt2}=5\sqrt2 N.

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  • 2080 · 1 mark

A ball is projected up an inclined plane with a velocity of 20 m/sec. If the inclination of the plane to the horizontal is 30∘30^\circ and acceleration due to gravity is 10 m/sec2^2, then the length described before coming to rest is (a) 20 m (b) 30 m (c) 200 m (d) 40 m

Answer

(d) 40 m

Retardation along the plane a=gsin⁡30∘=5a=g\sin30^\circ=5 m/s2^2. Distance s=u22a=40010=40s=\frac{u^2}{2a}=\frac{400}{10}=40 m.

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  • 2081 · 1 mark

A first degree equation xx and yy always represents a (a) Circle (b) Straight line (c) pair of straight line (d) parabola

Answer

(b) Straight line A first degree equation ax+by+c=0ax+by+c=0 in xx and yy always represents a straight line.

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  • 2081 · 1 mark

If ∣y−3∣≤7|y-3|\leq7, then (a) −4≤y≤10-4\leq y\leq10 (b) −2≤y≤5-2\leq y\leq5 (c) −2≤2y≤5-2\leq2y\leq5 (d) −4≤y≤4-4\leq y\leq4

Answer

(a) −4≤y≤10-4\leq y\leq10 ∣y−3∣≤7|y-3|\leq7 gives −7≤y−3≤7-7\leq y-3\leq7, so −4≤y≤10-4\leq y\leq10.

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  • 2081 · 1 mark

For any two non-empty sets A and B, A-B = (a) A∪BA\cup B (b) A∩BA\cap B (c) A∩BˉA\cap\bar B (d) B∩AˉB\cap\bar A

Answer

(c) A∩BˉA\cap\bar B A−BA-B consists of elements in AA that are not in BB, which is A∩B′A\cap B'.

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  • 2081 · 1 mark

If f(x)=1−xf(x)=1-x then ff of f(x)f(x) (a) (1−x)2(1-x)^2 (b) 1−x1-x (c) 1+x1+x (d) xx

Answer

(d) xx f(f(x))=1−f(x)=1−(1−x)=xf(f(x))=1-f(x)=1-(1-x)=x.

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  • 2081 · 1 mark

The function f(x)=x+2x3f(x)=x+2x^3 is (a) even (b) odd (c) both even and odd (d) neither odd or even

Answer

(b) odd f(−x)=−x−2x3=−(x+2x3)=−f(x)f(-x)=-x-2x^3=-(x+2x^3)=-f(x), so ff is odd.

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  • 2081 · 1 mark

The sum of the infinite series 1−12+14−18…1-\frac12+\frac14-\frac18\ldots is (a) does not exit (b) 0 (c) 23\frac23 (d) none

Answer

(c) 23\frac23 This is a geometric series with a=1a=1, r=−12r=-\tfrac12, so S=a1−r=13/2=23S=\dfrac{a}{1-r}=\dfrac{1}{3/2}=\dfrac23.

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  • 2081 · 1 mark

If ww is the root of unity, then ∣1ww2ww21w21w∣=\begin{vmatrix}1&w&w^2\\w&w^2&1\\w^2&1&w\end{vmatrix}= (a) 1 (b) 2 (c) 0 (d) w

Answer

(c) 0 Expanding along the first row, with w3=1w^3=1:

1(w3−1)−w(w2−w2)+w2(w−w4)=0−0+w2(w−w)=01(w^3-1)-w(w^2-w^2)+w^2(w-w^4)=0-0+w^2(w-w)=0

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  • 2081 · 1 mark

The imaginary part of 2i2(1+i)2\frac{2i^2}{(1+i)^2} (a) 0 (b) 2 (c) −i-i (d) 1

Answer

(d) 1 2i2(1+i)2=−22i=−1i=i\dfrac{2i^2}{(1+i)^2}=\dfrac{-2}{2i}=-\dfrac1i=i, so the imaginary part is 1.

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  • 2081 · 1 mark

The value of kk for which one root of the equation px2−qx+k=0px^2-qx+k=0 is reciprocal of the other is (a) q (b) −p-p (c) p (d) 0

Answer

(c) p If the roots are α\alpha and 1/α1/\alpha, their product is k/p=1k/p=1, so k=pk=p.

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  • 2081 · 1 mark

If 4cos⁡2x−1=04\cos^2x-1=0, then for n∈zn\in z, x=x= (a) nπ±π3n\pi\pm\frac{\pi}{3} (b) −π3-\frac{\pi}{3} (c) nπ+π3n\pi+\frac{\pi}{3} (d) π3\frac{\pi}{3}

Answer

(a) nπ±π3n\pi\pm\frac{\pi}{3} cos⁡2x=14=cos⁡2π3\cos^2x=\tfrac14=\cos^2\tfrac{\pi}{3}, so x=nπ±π3x=n\pi\pm\tfrac{\pi}{3}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2081 · 1 mark

2cos⁡−1x=2\cos^{-1}x= (a) cos⁡−1(2x2−1)\cos^{-1}(2x^2-1) (b) cos⁡−1(x2−1)\cos^{-1}(x^2-1) (c) cos⁡−1(1−2x2)\cos^{-1}(1-2x^2) (d) cos⁡−1x2\cos^{-1}x^2

Answer

(a) cos⁡−1(2x2−1)\cos^{-1}(2x^2-1) Let θ=cos⁡−1x\theta=\cos^{-1}x, so cos⁡2θ=2cos⁡2θ−1=2x2−1\cos2\theta=2\cos^2\theta-1=2x^2-1. Hence 2cos⁡−1x=cos⁡−1(2x2−1)2\cos^{-1}x=\cos^{-1}(2x^2-1).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2081 · 1 mark

The point on the x-axis at distance of 5 units from the line 3x+4y=83x+4y=8 (a) (8,0) (b) (11,0) (c) (0,11) (d) (0,8)

Answer

(b) (11,0)

For (a,0)(a,0) the distance to 3x+4y−8=03x+4y-8=0 is ∣3a−8∣5=5\frac{|3a-8|}{5}=5, so 3a−8=±253a-8=\pm25, giving a=11a=11 or a=−173a=-\frac{17}{3}. Only (11,0)(11,0) is listed.

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  • 2081 · 1 mark

If the angle between the lines represented by kx2+xy−y2=0kx^2+xy-y^2=0 is 90∘90^\circ, then k=k= (a) 1 (b) 2 (c) 5 (d) 6

Answer

(a) 1

Lines given by ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0. Here a=ka=k, b=−1b=-1, so k−1=0k-1=0 and k=1k=1.

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  • 2081 · 1 mark

The distance of a point (3,5,9)(3,5,9) from yz plane is (a) 9 (b) 5 (c) 3 (d) 17

Answer

(c) 3

The distance of a point from the yz plane is the absolute value of its x-coordinate, which is 3.

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  • 2081 · 1 mark

The two vectors i⃗+3k⃗+j⃗\vec i+3\vec k+\vec j and 3i⃗+nk⃗+3j⃗3\vec i+n\vec k+3\vec j are collinear if n=n= (a) 0 (b) 3 (c) 1 (d) 9

Answer

(d) 9

Collinear vectors have proportional components: 31=31=n3\frac{3}{1}=\frac{3}{1}=\frac{n}{3}, so n=9n=9.

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  • 2081 · 1 mark

A distribution is symmetrical if (a) Mean > Median > mode (b) Mean < Median < Mode (c) Mean = Median = Mode (d) Mean = Median

Answer

(c) Mean = Median = Mode

In a symmetrical distribution the mean, median and mode coincide.

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  • 2081 · 1 mark

While two coins are tossed once, the probability of getting two tails is (a) 12\frac12 (b) 14\frac14 (c) 34\frac34 (d) 13\frac13

Answer

(b) 14\frac14

Sample space {HH,HT,TH,TT}\{HH,HT,TH,TT\} has 4 outcomes and only TTTT gives two tails, so P=14P=\frac14.

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  • 2081 · 1 mark

lim⁡x→0sin⁡2x+4xx\lim_{x\to0}\frac{\sin2x+4x}{x} (a) 6 (b) 0 (c) 00\frac00 (d) 5

Answer

(a) 6

lim⁡x→0sin⁡2x+4xx=2lim⁡x→0sin⁡2x2x+4=2+4=6.\lim_{x\to0}\frac{\sin2x+4x}{x}=2\lim_{x\to0}\frac{\sin2x}{2x}+4=2+4=6.

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  • 2081 · 1 mark

The function x3−12x+8x^3-12x+8 has maximum at x=x= (a) -2 (b) 2 (c) 0 (d) 8

Answer

(a) -2

f′(x)=3x2−12=0f'(x)=3x^2-12=0 gives x=±2x=\pm2. Since f′′(x)=6xf''(x)=6x, f′′(−2)=−12<0f''(-2)=-12<0, so the maximum is at x=−2x=-2.

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  • 2081 · 1 mark

∫12cos⁡(log⁡x)xdx\int_1^2\frac{\cos(\log x)}{x}dx (a) −sin⁡(log⁡2)-\sin(\log2) (b) −cos⁡(log⁡2)-\cos(\log2) (c) sin⁡(log⁡2)\sin(\log2) (d) 0

Answer

(c) sin⁡(log⁡2)\sin(\log2)

Put u=log⁡xu=\log x, du=dxxdu=\frac{dx}{x}. Then ∫0log⁡2cos⁡u du=sin⁡(log⁡2)\int_0^{\log2}\cos u\,du=\sin(\log2).

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  • 2081 · 1 mark

The Equation f(x)=0f(x)=0 has at least one root in (a,b)(a,b) where f(x)f(x) is continuous in (a,b)(a,b) if (a) f(a)f(b)>0f(a)f(b)>0 (b) f(a)f(b)<0f(a)f(b)<0 (c) f(a)f(b)=0f(a)f(b)=0 (d) f(a)f(b)≥0f(a)f(b)\geq0

Answer

(b) f(a)f(b)<0f(a)f(b)<0

If ff is continuous and f(a)f(a), f(b)f(b) have opposite signs, the intermediate value theorem guarantees a root in (a,b)(a,b).

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  • 2081 · 1 mark

Magnitude of the resultant of two forces P and Q will be maximum when the both force act (a) In opposite direction (b) 90∘90^\circ to each other (c) 45∘45^\circ to each other (d) in the same direction

Answer

(d) in the same direction

R=P2+Q2+2PQcos⁡θR=\sqrt{P^2+Q^2+2PQ\cos\theta} is maximum, P+QP+Q, when θ=0∘\theta=0^\circ.

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  • 2081 · 1 mark

If a body thrown vertically upward with a velocity of 40 ms−140\,ms^{-1} from the top of a tower, hits the ground after 10 second, then the height of the tower is [g=10 ms−2][g=10\,ms^{-2}] (a) 200m (b) 130m (c) 100m (d) 40m

Answer

(c) 100m

Taking upward as positive, s=ut−12gt2=40(10)−5(10)2=400−500=−100s=ut-\frac12gt^2=40(10)-5(10)^2=400-500=-100 m. The ground is 100 m below the starting point, so the tower is 100 m high.

Answer checked by Claude Sonnet · Oct 21, 2026

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Questions from BSc-CSIT-Old-Question-Collection-2069-2081-TU-IoST.pdf (BSc. CSIT Entrance · Entrance Examination 2069 · 2069) and TU IOST BSc CSIT Entrance Old Question Collection 2070-2081 (TU IOST BSc CSIT entrance examination papers 2070 to 2081 BS). Answers are written for this site; check them against your class notes.