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Chapter 1

Mathematics

TU BIT entrance past exam questions

Past questions and answers

235 questions set from this chapter. Most repeated first.

  • 2079 Model · 1 mark

If f:R→Rf:R\to R is defined by f(x)=3x−2f(x)=3x-2, for x∈Rx\in R then the pre image of 10 in R is (a) 4 (b) 10 (c) 5 (d) 0

Similar questions: If a function f: R R be defined by f(x)=2x+5 then the... (2080)

Answer

(a) 4 Solve f(x)=10f(x)=10: 3x−2=10⇒x=43x-2=10\Rightarrow x=4.

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  • 2079 Model · 1 mark

The two lines represented by ax2+xy+by2=0ax^2+xy+by^2=0 perpendicular to each other if (a) a+b>0a+b>0 (b) a+b<0a+b<0 (c) ab=0ab=0 (d) a+b=0a+b=0

Similar questions: The two lines represented by kx 2+xy-y 2=0, are... (2080) · The straight lines represented by x 2+2xy+ky 2=0 are...

Answer

(d) a+b=0a+b=0

A pair of lines ax2+2hxy+by2=0ax^2+2hxy+by^2=0 is perpendicular when the coefficient of x2x^2 plus the coefficient of y2y^2 is zero, that is a+b=0a+b=0.

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  • 2080 · 1 mark

If a function f:R→Rf:\mathbb R\to\mathbb R be defined by f(x)=2x+5f(x)=2x+5 then the pre-image of 7 in R\mathbb R is (a) 3 (b) no pre-image (c) 1 (d) -2

Similar questions: If f:R R is defined by f(x)=3x-2, for x R then the pre... (2079 Model)

Answer

(c) 1

2x+5=7⇒x=12x+5=7\Rightarrow x=1.

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  • 2080 · 1 mark

The two lines represented by kx2+xy−y2=0kx^2+xy-y^2=0, are perpendicular to each other if k=k= (a) 1 (b) -1 (c) 0 (d) k≠0k\neq0

Similar questions: The two lines represented by ax 2+xy+by 2=0 perpendicular... (2079 Model) · The straight lines represented by x 2+2xy+ky 2=0 are...

Answer

(a) 1

Lines from ax2+2hxy+by2=0ax^2+2hxy+by^2=0 are perpendicular when a+b=0a+b=0: k+(−1)=0⇒k=1k+(-1)=0\Rightarrow k=1.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The straight lines represented by x2+2xy+ky2=0x^2+2xy+ky^2=0 are perpendicular to each other if k=k= (a) -1 (b) 1 (c) 0 (d) 2

Similar questions: The two lines represented by ax 2+xy+by 2=0 perpendicular... (2079 Model) · The two lines represented by kx 2+xy-y 2=0, are... (2080)

Answer

(a) −1-1 For ax2+2hxy+by2=0ax^2+2hxy+by^2=0 the lines are perpendicular when a+b=0a+b=0. Here 1+k=01+k=0, so k=−1k=-1.

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  • 2076 Model · 1 mark

Every rational number is also (a) a natural number (b) an integer (c) a whole number (d) a real number

Answer

(d) a real number

Every rational number is a real number. Rational numbers such as 1/2 are not integers, whole numbers or natural numbers.

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  • 2076 Model · 1 mark

If a−ib=3−2i3+2ia-ib=\frac{3-2i}{3+2i} then a2+b2=a^2+b^2= (a) 1 (b) 43\frac43 (c) 35\frac35 (d) 29\frac29

Answer

(a) 1

a−ib=3−2i3+2i⋅3−2i3−2i=9−12i+4i213=5−12i13a-ib=\frac{3-2i}{3+2i}\cdot\frac{3-2i}{3-2i}=\frac{9-12i+4i^2}{13}=\frac{5-12i}{13}

So a=513a=\frac5{13}, b=1213b=\frac{12}{13}, and a2+b2=25+144169=1a^2+b^2=\frac{25+144}{169}=1.

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  • 2076 Model · 1 mark

Given U={1,2,3,…,15}U=\{1,2,3,\ldots,15\}, A={x:x≤8}A=\{x: x\leq8\}, B={x:x≤4}B=\{x: x\leq4\} then A∩B‾=\overline{A\cap B}= (a) ∅\emptyset (b) UU (c) A∪BA\cup B (d) None of these

Answer

(d) None of these

A∩B=B={1,2,3,4}A\cap B=B=\{1,2,3,4\}, so A∩B‾={5,6,…,15}\overline{A\cap B}=\{5,6,\ldots,15\}. This is not ∅\emptyset, not UU, and A∪B=A={1,…,8}A\cup B=A=\{1,\ldots,8\} is different.

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  • 2076 Model · 1 mark

Which are the equal ordered pairs? (a) (2,4),(4,2)(2,4),(4,2) (b) (2,2),(4,4)(2,2),(4,4) (c) (4,2),(4,2)(4,2),(4,2) (d) all of them

Answer

(c) (4,2),(4,2)(4,2),(4,2)

Ordered pairs are equal only when the first components are equal and the second components are equal. Only (4,2)=(4,2)(4,2)=(4,2) satisfies this.

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  • 2076 Model · 1 mark

S∞=a1−rS_\infty=\frac{a}{1-r} where a,ra, r and S∞S_\infty denotes first term, common ratio and sum to infinity. If (a) ∣r∣<1|r|<1 (b) ∣r∣>1|r|>1 (c) ∣r∣=1|r|=1 (d) all of them

Answer

(a) ∣r∣<1|r|<1

The infinite geometric series converges only when ∣r∣<1|r|<1; then rn→0r^n\to0 and S∞=a1−rS_\infty=\frac{a}{1-r}.

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  • 2076 Model · 1 mark

If 21Cx= 21Cy^{21}C_x=\,^{21}C_y then which is true? (a) x≠yx\neq y (b) x=yx=y (c) x+y=21x+y=21 (d) either (b) or (c)

Answer

(d) either (b) or (c)

nCx= nCy^{n}C_x=\,^{n}C_y implies x=yx=y or x+y=nx+y=n. Here n=21n=21, so x=yx=y or x+y=21x+y=21.

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  • 2076 Model · 1 mark

Which is not correct? (a) (AT)T=A(A^T)^T=A (b) (A+B)T=AT+BT(A+B)^T=A^T+B^T (c) (AB)T=ATBT(AB)^T=A^TB^T (d) (AB)T=BTAT(AB)^T=B^TA^T

Answer

(c) (AB)T=ATBT(AB)^T=A^TB^T

The correct reversal law is (AB)T=BTAT(AB)^T=B^TA^T, so (c) is the incorrect statement.

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  • 2076 Model · 1 mark

The equation x=0x=0 represents (a) x-axis (b) y-axis (c) the origin (d) none of these

Answer

(b) y-axis

Every point with x=0x=0 lies on the y-axis.

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  • 2076 Model · 1 mark

The angle between the lines xy=0xy=0 is (a) 0∘0^\circ (b) 60∘60^\circ (c) 90∘90^\circ (d) 120∘120^\circ

Answer

(c) 90∘90^\circ

xy=0xy=0 means x=0x=0 or y=0y=0, i.e. the y-axis and x-axis, which are perpendicular.

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  • 2076 Model · 1 mark

Which is correct? (a) log⁡101=0\log_{10}1=0 (b) log⁡100.1=−1\log_{10}0.1=-1 (c) log⁡1010=1\log_{10}10=1 (d) all of them

Answer

(d) all of them

log⁡101=0\log_{10}1=0, log⁡100.1=log⁡1010−1=−1\log_{10}0.1=\log_{10}10^{-1}=-1 and log⁡1010=1\log_{10}10=1 are all true.

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  • 2076 Model · 1 mark

The domain of y=f(x)=1xy=f(x)=\frac1x is (a) R−{0}R-\{0\} (b) RR (c) (−2,1)(-2,1) (d) (−1,1)(-1,1)

Answer

(a) R−{0}R-\{0\}

1x\frac1x is defined for every real xx except x=0x=0.

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  • 2076 Model · 1 mark

Let f:R→Rf:R\to R be defined by f(x)=10x−7f(x)=10x-7 then f−1(x)=f^{-1}(x)= (a) 110x+7\frac{1}{10x+7} (b) 110x−7\frac{1}{10x-7} (c) x+710\frac{x+7}{10} (d) x−710\frac{x-7}{10}

Answer

(c) x+710\frac{x+7}{10}

Let y=10x−7y=10x-7. Then x=y+710x=\frac{y+7}{10}, so f−1(x)=x+710f^{-1}(x)=\frac{x+7}{10}.

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  • 2076 Model · 1 mark

If t be a positive real number then the least value of t+1tt+\frac1t is : (a) 1 (b) 2 (c) 4 (d) 3

Answer

(b) 2

By AM-GM, t+1t≥2t⋅1t=2t+\frac1t\ge2\sqrt{t\cdot\frac1t}=2, with equality at t=1t=1.

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  • 2076 Model · 1 mark

The point of discontinuity of the function f(x)=x3−27x−3f(x)=\frac{x^3-27}{x-3} is (a) 3 (b) 6 (c) 27 (d) 0

Answer

(a) 3

f(x)f(x) is not defined at x=3x=3 (denominator zero), although f(x)=x2+3x+9f(x)=x^2+3x+9 for x≠3x\neq3. So the point of discontinuity is x=3x=3.

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  • 2076 Model · 1 mark

If x+y=xyx+y=xy then dxdy=\frac{dx}{dy}= (a) x−1y−1\frac{x-1}{y-1} (b) y−11−x\frac{y-1}{1-x} (c) 1−xy−1\frac{1-x}{y-1} (d) 1+y1−x\frac{1+y}{1-x}

Answer

(c) 1−xy−1\frac{1-x}{y-1}

Differentiating x+y=xyx+y=xy with respect to yy: dxdy+1=x+ydxdy\frac{dx}{dy}+1=x+y\frac{dx}{dy}, so dxdy(1−y)=x−1\frac{dx}{dy}(1-y)=x-1 and dxdy=x−11−y=1−xy−1\frac{dx}{dy}=\frac{x-1}{1-y}=\frac{1-x}{y-1}.

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  • 2076 Model · 1 mark

If x=2atx=2at, y=at2y=at^2 then dydx=\frac{dy}{dx}= (a) tt (b) 1t\frac1t (c) t2t^2 (d) 1

Answer

(a) tt

dydt=2at\frac{dy}{dt}=2at, dxdt=2a\frac{dx}{dt}=2a, so dydx=2at2a=t\frac{dy}{dx}=\frac{2at}{2a}=t.

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  • 2076 Model · 1 mark

If the total cost function C=50x+5000C=50x+5000. The marginal cost function MC is (a) 50+5000x50+\frac{5000}{x} (b) 25x2+5000x25x^2+5000x (c) 50 (d) none of these

Answer

(c) 50

MC=dCdx=50MC=\frac{dC}{dx}=50.

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  • 2076 Model · 1 mark

The maximum value of sin⁡θcos⁡θ\sin\theta\cos\theta is (a) 1 (b) 12\frac{1}{\sqrt2} (c) 32\frac{\sqrt3}{2} (d) 12\frac12

Answer

(d) 12\frac12

sin⁡θcos⁡θ=12sin⁡2θ\sin\theta\cos\theta=\frac12\sin2\theta, whose maximum is 12\frac12 (when sin⁡2θ=1\sin2\theta=1).

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  • 2076 Model · 1 mark

The function f(x)=2−3xf(x)=2-3x is (a) Decreasing (b) Increasing (c) Neither decreasing nor increasing (d) Doesn't exist

Answer

(a) Decreasing

f′(x)=−3<0f'(x)=-3<0 for all xx, so ff is decreasing.

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  • 2076 Model · 1 mark

∫log⁡x dx=\int\log x\,dx= (a) log⁡x+c\log x+c (b) xlog⁡x−x+cx\log x-x+c (c) xlog⁡x+x+cx\log x+x+c (d) xlog⁡x+cx\log x+c

Answer

(b) xlog⁡x−x+cx\log x-x+c

Integrating by parts: ∫log⁡x⋅1 dx=xlog⁡x−∫x⋅1x dx=xlog⁡x−x+c\int\log x\cdot1\,dx=x\log x-\int x\cdot\frac1x\,dx=x\log x-x+c.

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  • 2076 Model · 1 mark

∫−22∣x∣ dx\int_{-2}^{2}|x|\,dx is equal to (a) 0 (b) 1 (c) 2 (d) 4

Answer

(d) 4

∫−22∣x∣ dx=2∫02x dx=2⋅42=4\int_{-2}^{2}|x|\,dx=2\int_0^2x\,dx=2\cdot\frac{4}{2}=4

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  • 2076 Model · 1 mark

The area of the circle x2+y2=16x^2+y^2=16 is (a) 4π4\pi (b) 4π24\pi^2 (c) 16π216\pi^2 (d) 16π16\pi

Answer

(d) 16π16\pi

The radius is 44, so the area is πr2=16π\pi r^2=16\pi.

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  • 2076 Model · 1 mark

Which one is the measure of dispersion? (a) mean (b) median (c) mode (d) range

Answer

(d) range

Mean, median and mode are measures of central tendency; range measures dispersion.

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  • 2076 Model · 1 mark

Two brothers: Mr. X and Mr. Y appear in an interview for getting the scholarship. The scholarship can be provided for two persons. The probability of getting scholarship by Mr. X is 17\frac17 and getting by Mr. Y is 15\frac15. What is the probability that both of them will get scholarship? (a) 1235\frac{12}{35} (b) 135\frac{1}{35} (c) 2335\frac{23}{35} (d) none of these

Answer

(b) 135\frac{1}{35}

The events are independent, so P=17×15=135P=\frac17\times\frac15=\frac1{35}.

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  • 2076 Model · 1 mark

The equation sin⁡x+cos⁡x=2\sin x+\cos x=2 has (a) infinite solution (b) finite solution (c) no solution (d) unique solution

Answer

(c) no solution

sin⁡x+cos⁡x=2sin⁡(x+π4)≤2<2\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right)\le\sqrt2<2, so the equation has no solution.

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  • 2076 Model · 1 mark

The system of linear equations 2x+3y=52x+3y=5 and 4x+6y=94x+6y=9 are (a) Consistent & dependent (b) Consistent & independent (c) Inconsistent & independent (d) Inconsistent & dependent

Answer

(c) Inconsistent & independent

Multiplying the first equation by 2 gives 4x+6y=104x+6y=10, which contradicts 4x+6y=94x+6y=9. The lines are parallel, so there is no solution (inconsistent).

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  • 2076 Model · 1 mark

The degree of polynomial equation ax2+bx+c=0ax^2+bx+c=0 is (a) 0 (b) 1 (c) 2 (d) not define

Answer

(c) 2

The highest power of xx is 2, so the degree is 2.

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  • 2076 Model · 1 mark

If one root of a quadratic equation in the real coefficient is 2\sqrt2 then equation is (a) x2+2x+2=0x^2+\sqrt2x+2=0 (b) x2+2=0x^2+2=0 (c) x2−2=0x^2-2=0 (d) none of these

Answer

(c) x2−2=0x^2-2=0

The equation with roots 2\sqrt2 and −2-\sqrt2 has sum 00 and product −2-2, giving x2−2=0x^2-2=0. The other options have non-real roots or do not have 2\sqrt2 as a root.

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  • 2076 Model · 1 mark

The equation x2+y2=0x^2+y^2=0 represent (a) circle (b) degenerate circle (c) an empty set (d) a straight line

Answer

(b) degenerate circle

x2+y2=0x^2+y^2=0 is satisfied only by (0,0)(0,0), a circle of radius zero (a point circle).

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  • 2076 Model · 1 mark

If four quantities a,b,ca, b, c and d are such that a:b=3:4a:b=3:4, b:c=6:7b:c=6:7 and c:d=8:9c:d=8:9, then the ratio between a and d is (a) 3:8 (b) 1:3 (c) 4:7 (d) 7:4

Answer

(c) 4:7

ad=34⋅67⋅89=144252=47\frac ad=\frac34\cdot\frac67\cdot\frac89=\frac{144}{252}=\frac47

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  • 2076 Model · 1 mark

By selling 175 pens, a stationary store gains the selling price of 50 pens. What is his gain percent? (a) 40% (b) 284728\frac47% (c) 50% (d) none of these

Answer

(a) 40%

Gain on 175 pens = SP of 50 pens, so CP of 175 pens = SP of 125 pens. Let SP of one pen be 1; then CP of 175 pens = 125 and gain = 50.

Gain %=50125×100=40%\text{Gain \%}=\frac{50}{125}\times100=40\%

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  • 2076 Model · 1 mark

If US USD 1 = Rs. 78.15. How many US dollars are equivalent To Rs. 10,000? (a) USD 127.96 (b) USD 172.96 (c) USD 196.72 (d) USD 172.62

Answer

(a) $127.96

1000078.15=127.96\frac{10000}{78.15}=127.96

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  • 2076 Model · 1 mark

The effective rate corresponding to 10% p.a. convertible quarterly is (a) 40% (b) 10.1% (c) 10.4% (d) none of these

Answer

(c) 10.4%

Effective rate=(1+0.104)4−1=(1.025)4−1=0.1038≈10.4%\text{Effective rate}=\left(1+\frac{0.10}{4}\right)^4-1=(1.025)^4-1=0.1038\approx10.4\%

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  • 2077 Model · 1 mark

If n(∪)=200n(\cup)=200, n(A)=80n(A)=80, n(B)=60n(B)=60, what is the maximum value of n(A∪B)=n(A\cup B)=? (a) 200 (b) 40 (c) 140 (d) 110

Answer

(c) 140

n(A∪B)=n(A)+n(B)−n(A∩B)n(A\cup B)=n(A)+n(B)-n(A\cap B) is largest when A∩B=∅A\cap B=\emptyset, giving 80+60=14080+60=140, which is below 200.

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  • 2077 Model · 1 mark

−1≤x≤11-1\leq x\leq11 is equivalent to (a) ∣x−5∣≤6|x-5|\leq6 (b) ∣x+5∣≥3|x+5|\geq3 (c) ∣5−x∣≥2|5-x|\geq2 (d) ∣5−2x∣<1|5-2x|<1

Answer

(a) ∣x−5∣≤6|x-5|\leq6

∣x−5∣≤6  ⟺  −6≤x−5≤6  ⟺  −1≤x≤11|x-5|\leq6 \iff -6\leq x-5\leq6 \iff -1\leq x\leq11.

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  • 2077 Model · 1 mark

Range of y=x2−4x+5y=x^2-4x+5 is (a) (−∞,∞)(-\infty,\infty) (b) (−∞,1](-\infty,1] (c) [1,∞)[1,\infty) (d) (−∞,1)(-\infty,1)

Answer

(c) [1,∞)[1,\infty)

y=x2−4x+5=(x−2)2+1≥1y=x^2-4x+5=(x-2)^2+1\geq1, and every value ≥1\geq1 is attained.

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  • 2077 Model · 1 mark

Period of function of cos⁡2x\cos2x is (a) π2\frac\pi2 (b) 1 (c) π\pi (d) π4\frac\pi4

Answer

(c) π\pi

Period of cos⁡kx\cos kx is 2πk\frac{2\pi}{k}, so for k=2k=2 it is π\pi.

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  • 2077 Model · 1 mark

If log⁡981=x\log_981=x then xx (a) 12\frac12 (b) 2\sqrt2 (c) 1 (d) 2

Answer

(d) 2

log⁡981=x⇒9x=81=92\log_9 81=x \Rightarrow 9^x=81=9^2, so x=2x=2.

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  • 2077 Model · 1 mark

If x=log⁡abcx=\log_abc, y=log⁡bcay=\log_bca, z=log⁡cabz=\log_cab then 1x+1+1y+1+1z+1=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}= (a) 1 (b) 0 (c) 12\frac12 (d) abc

Answer

(a) 1

x+1=log⁡abc+log⁡aa=log⁡aabcx+1=\log_a bc+\log_a a=\log_a abc, so 1x+1=log⁡abca\frac1{x+1}=\log_{abc}a. Similarly 1y+1=log⁡abcb\frac1{y+1}=\log_{abc}b and 1z+1=log⁡abcc\frac1{z+1}=\log_{abc}c.

Sum =log⁡abc(abc)=1=\log_{abc}(abc)=1.

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  • 2077 Model · 1 mark

1.5+0.015+0.00015+……1.5+0.015+0.00015+\ldots\ldots (a) 1511\frac{15}{11} (b) 5033\frac{50}{33} (c) 23\frac23 (d) sum doesn't exist

Answer

(b) 5033\frac{50}{33}

This is a G.P. with a=1.5a=1.5, r=0.01r=0.01, so S=1.51−0.01=1.50.99=15099=5033S=\frac{1.5}{1-0.01}=\frac{1.5}{0.99}=\frac{150}{99}=\frac{50}{33}.

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  • 2077 Model · 1 mark

The system of equations kx−2y=0kx-2y=0, x+3y=0x+3y=0 have unique solution then (a) k=−23k=-\frac23 (b) k=0k=0 (c) k=23k=\frac23 (d) k≠−23k\neq-\frac23

Answer

(d) k≠−23k\neq-\frac23

A homogeneous system has a unique (trivial) solution when the determinant is non-zero: 3k−(−2)(1)=3k+2≠03k-(-2)(1)=3k+2\neq0, so k≠−23k\neq-\frac23.

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If one root of ax2+bx+c=0ax^2+bx+c=0 be square of other then (a) b3+ac2+a2c+3abc=0b^3+ac^2+a^2c+3abc=0 (b) b3+a2c+ac2−3abc=0b^3+a^2c+ac^2-3abc=0 (c) a3+bc2+b2c−3abc=0a^3+bc^2+b^2c-3abc=0 (d) a3+b3+c3−3abc=0a^3+b^3+c^3-3abc=0

Similar questions: If one root of ax 2+bx+c=0 is 2+3i then the other root is

Answer

(b) b3+a2c+ac2−3abc=0b^3+a^2c+ac^2-3abc=0

Let the roots be α,α2\alpha,\alpha^2. Then α+α2=−ba\alpha+\alpha^2=-\frac ba and α3=ca\alpha^3=\frac ca.

Cubing the sum: (α+α2)3=α3+α6+3α3(α+α2)(\alpha+\alpha^2)^3=\alpha^3+\alpha^6+3\alpha^3(\alpha+\alpha^2), so

−b3a3=ca+c2a2−3bca2-\frac{b^3}{a^3}=\frac ca+\frac{c^2}{a^2}-\frac{3bc}{a^2}

Multiplying by −a3-a^3 gives b3+a2c+ac2−3abc=0b^3+a^2c+ac^2-3abc=0.

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The amplitude of 1−i1-i is (a) 45∘45^\circ (b) 135∘135^\circ (c) 315∘315^\circ (d) doesn't exist

Answer

(c) 315∘315^\circ

1−i1-i lies in the fourth quadrant with tan⁡θ=−1\tan\theta=-1, so θ=−45∘\theta=-45^\circ, i.e. 315∘315^\circ as a positive angle.

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The mean proportional between 4 & 16 is (a) 4 (b) 8 (c) 12 (d) 5

Answer

(b) 8

Mean proportional =4×16=64=8=\sqrt{4\times16}=\sqrt{64}=8.

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A plane left 30 minutes later than the schedules time and in order to reach its destinations 1500 km always in time it has to increase its speed by 250 km/hr from its usual speed. What is its usual speed? (a) 500 km/hr (b) 600 km/hr (c) 750 km/hr (d) 100 km/hr

Answer

(c) 750 km/hr

Let the usual speed be vv. Then

1500v−1500v+250=12⇒v(v+250)=750000⇒v2+250v−750000=0\frac{1500}{v}-\frac{1500}{v+250}=\frac12 \Rightarrow v(v+250)=750000 \Rightarrow v^2+250v-750000=0

(v−750)(v+1000)=0(v-750)(v+1000)=0, so v=750v=750 km/hr. Check: 22 h −1.5-1.5 h =0.5=0.5 h.

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The present worth of Rs 2040 due 4 years hence reaching interested at 5% p.a is (a) 1650.50 (b) Rs. 1675.3 (c) Rs 1680 (d) Rs 1700

Answer

(d) Rs 1700

Present worth =A1+rt100=20401+5×4100=20401.2=Rs 1700=\frac{A}{1+\frac{rt}{100}}=\frac{2040}{1+\frac{5\times4}{100}}=\frac{2040}{1.2}=\text{Rs }1700.

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The interest rate doubles a sum in 10 year is (a) 12% (b) 11% (c) 10% (d) 9%

Answer

(c) 10%

Interest equals principal: P=P×r×10100P=\frac{P\times r\times10}{100}, so r=10%r=10\%.

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Profits in the business of Bikram, Bishal and Binita are shared in ratio 1:2:3. If Bikram share of profits is Rs. 300, what is Binita share? (a) Rs. 150 (b) Rs. 450 (c) Rs. 600 (d) Rs. 900

Answer

(d) Rs. 900

Bikram's 1 part == Rs 300, so Binita's 3 parts =3×300==3\times300= Rs 900.

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If A and B are two independent events with p(A)=23p(A)=\frac23 & p(B)=35p(B)=\frac35 then p(A∪B)=p(A\cup B)= (a) 615\frac{6}{15} (b) 53\frac53 (c) 25\frac25 (d) 1315\frac{13}{15}

Answer

(d) 1315\frac{13}{15}

For independent events P(A∩B)=23⋅35=25P(A\cap B)=\frac23\cdot\frac35=\frac25.

P(A∪B)=23+35−25=23+15=1315P(A\cup B)=\frac23+\frac35-\frac25=\frac23+\frac15=\frac{13}{15}.

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For Negative skewness (a) Mean>Mode>median (b) Mean<median<Mode (c) Median<mode<Mean (d) Mode<mean<Median

Answer

(b) Mean<median<Mode

In a negatively skewed distribution the tail is on the left, so the mean is pulled lowest and the mode is highest.

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HCF of n!n!, (n+1)!(n+1)! and (n+2)!(n+2)! is (a) n!n! (b) (n+2)!(n+2)! (c) (n+1)!(n+1)! (d) (n−1)!(n-1)!

Answer

(a) n!n!

(n+1)!=(n+1) n!(n+1)!=(n+1)\,n! and (n+2)!=(n+2)(n+1) n!(n+2)!=(n+2)(n+1)\,n!, so n!n! divides all three and is the HCF.

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If tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\frac\pi2 then xy+yz+zx=xy+yz+zx= (a) 1 (b) -1 (c) 0 (d) π\pi

Answer

(a) 1

tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\frac\pi2 gives tan⁡−1x+tan⁡−1y=π2−tan⁡−1z=cot⁡−1z\tan^{-1}x+\tan^{-1}y=\frac\pi2-\tan^{-1}z=\cot^{-1}z.

Taking tangent: x+y1−xy=1z\frac{x+y}{1-xy}=\frac1z, so xz+yz=1−xyxz+yz=1-xy, i.e. xy+yz+zx=1xy+yz+zx=1.

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In any ΔABC\Delta ABC, 2cos⁡Asin⁡C=sin⁡B2\cos A\sin C=\sin B then Δ\Delta is (a) Scalene (b) Equilateral (c) Isosceles (d) right angled

Answer

(c) Isosceles

sin⁡B=sin⁡(A+C)=sin⁡Acos⁡C+cos⁡Asin⁡C\sin B=\sin(A+C)=\sin A\cos C+\cos A\sin C. Given 2cos⁡Asin⁡C=sin⁡B2\cos A\sin C=\sin B:

cos⁡Asin⁡C=sin⁡Acos⁡C⇒sin⁡(C−A)=0⇒A=C\cos A\sin C=\sin A\cos C \Rightarrow \sin(C-A)=0 \Rightarrow A=C.

Hence the triangle is isosceles.

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For what value of 'K' points (1,−1)(1,-1), (k2,k3)\left(\frac k2,\frac k3\right) and (4,1)(4,1) are collinear? (a) −32-\frac32 (b) 3 (c) 2 (d) −52-\frac52

Answer

Collinearity requires

∣1−11k2k31411∣=0.\begin{vmatrix}1&-1&1\\ \frac k2&\frac k3&1\\ 4&1&1\end{vmatrix}=0.

Expanding: (k3−1)+(k2−4)+(k2−4k3)=−5\left(\frac k3-1\right)+\left(\frac k2-4\right)+\left(\frac k2-\frac{4k}{3}\right)=-5, which is independent of kk and never zero. As printed, no value of kk makes the three points collinear, so none of the options satisfies the condition.

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The distance between parallel lines 3x+2y=63x+2y=6 and 3x+2y=123x+2y=12 is (a) 145\frac{14}{5} (b) 72\frac72 (c) −813-\frac{8}{\sqrt{13}} (d) 613\frac{6}{\sqrt{13}}

Answer

(d) 613\frac{6}{\sqrt{13}}

Distance between ax+by+c1=0ax+by+c_1=0 and ax+by+c2=0ax+by+c_2=0 is ∣c1−c2∣a2+b2=∣−6+12∣9+4=613\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}=\frac{|-6+12|}{\sqrt{9+4}}=\frac6{\sqrt{13}}.

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Angles between the lines represented by 2x2−7xy+6y2=02x^2-7xy+6y^2=0 is (a) tan⁡−1(±18)\tan^{-1}\left(\pm\frac18\right) (b) tan⁡−1(±17)\tan^{-1}\left(\pm\frac17\right) (c) 45∘45^\circ (d) 60∘60^\circ

Answer

(a) tan⁡−1(±18)\tan^{-1}\left(\pm\frac18\right)

Here a=2a=2, h=−72h=-\frac72, b=6b=6:

tan⁡θ=∣2h2−aba+b∣=2494−128=2⋅128=18.\tan\theta=\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|=\frac{2\sqrt{\frac{49}4-12}}{8}=\frac{2\cdot\frac12}{8}=\frac18.

Check: the lines are 2x−3y=02x-3y=0 and x−2y=0x-2y=0 with slopes 23,12\frac23,\frac12, giving 1/64/3=18\frac{1/6}{4/3}=\frac18.

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Length of y-intercept made by circle x2+y2−8x+y−20=0x^2+y^2-8x+y-20=0 is (a) 5 (b) 9 (c) 8 (d) 6

Answer

(b) 9

Put x=0x=0: y2+y−20=0⇒(y+5)(y−4)=0y^2+y-20=0 \Rightarrow (y+5)(y-4)=0, so y=4,−5y=4,-5. The intercept length is 4−(−5)=94-(-5)=9.

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Radius of circle x2+y2+2x+3y+k=0x^2+y^2+2x+3y+k=0 is 12\frac12 then k=k= (a) 2 (b) -1 (c) 3 (d) 4

Answer

(c) 3

r2=g2+f2−c=1+94−k=14r^2=g^2+f^2-c=1+\frac94-k=\frac14, so k=3k=3.

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lim⁡x→64x6−2x3−4\lim_{x\to64}\frac{\sqrt[6]{x}-2}{\sqrt[3]{x}-4} (a) 12\frac12 (b) 34\frac34 (c) 23\frac23 (d) 14\frac14

Answer

(d) 14\frac14

Put x=t6x=t^6; as x→64x\to64, t→2t\to2:

lim⁡t→2t−2t2−4=lim⁡t→21t+2=14.\lim_{t\to2}\frac{t-2}{t^2-4}=\lim_{t\to2}\frac1{t+2}=\frac14.

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lim⁡x→0esin⁡x−1x\lim_{x\to0}\frac{e^{\sin x}-1}{x} (a) 0 (b) 1 (c) 2 (d) log⁡e2\log_e2

Answer

(b) 1

esin⁡x−1x=esin⁡x−1sin⁡x⋅sin⁡xx→1⋅1=1\frac{e^{\sin x}-1}{x}=\frac{e^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{x}\to1\cdot1=1.

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Every continuous function is (a) increasing (b) decreasing (c) differentiable (d) not - differentiable

Answer

(c) differentiable

Strictly, the true theorem is the converse: every differentiable function is continuous. A continuous function need not be differentiable (e.g. ∣x∣|x| at 00), and need not be monotonic. Of the given options, (c) is the one intended, relating continuity and differentiability.

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ddxcos⁡−1(sin⁡x)\frac{d}{dx}\cos^{-1}(\sin x) (a) x (b) -1 (c) −sin⁡x1−sin⁡2x\frac{-\sin x}{\sqrt{1-\sin^2x}} (d) cos⁡x1−sin⁡2x\frac{\cos x}{\sqrt{1-\sin^2x}}

Answer

(b) −1-1

cos⁡−1(sin⁡x)=cos⁡−1(cos⁡(π2−x))=π2−x\cos^{-1}(\sin x)=\cos^{-1}\left(\cos\left(\frac\pi2-x\right)\right)=\frac\pi2-x, whose derivative is −1-1.

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If xpyq=(x+y)p+qx^py^q=(x+y)^{p+q} then dydx=\frac{dy}{dx}= (a) yx\frac yx (b) pyqx\frac{py}{qx} (c) xy\frac xy (d) x2y2\frac{x^2}{y^2}

Answer

(a) yx\frac yx

Take logs: pln⁡x+qln⁡y=(p+q)ln⁡(x+y)p\ln x+q\ln y=(p+q)\ln(x+y). Differentiate:

px+qyy′=(p+q)(1+y′)x+y.\frac px+\frac qy y'=\frac{(p+q)(1+y')}{x+y}.

Collecting y′y': y′(qy−p+qx+y)=p+qx+y−pxy'\left(\frac qy-\frac{p+q}{x+y}\right)=\frac{p+q}{x+y}-\frac px. This simplifies to y′⋅qx−pyy(x+y)=qx−pyx(x+y)y'\cdot\frac{qx-py}{y(x+y)}=\frac{qx-py}{x(x+y)}, so y′=yxy'=\frac yx.

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Derivatives of sin⁡−1x\sin^{-1}x is (a) 1−x2\sqrt{1-x^2} (b) 11−x2\frac{1}{\sqrt{1-x^2}} (c) −11−x2\frac{-1}{\sqrt{1-x^2}} (d) −1x2−1-\frac{1}{\sqrt{x^2-1}}

Answer

(b) 11−x2\frac1{\sqrt{1-x^2}}

Standard result: ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x=\frac1{\sqrt{1-x^2}}.

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At critical point, f(x)f(x) is (a) increasing (b) decreasing (c) stationary (d) increasing & decreasing

Answer

(c) stationary

At a critical point f′(x)=0f'(x)=0 (or is undefined), so the function is momentarily stationary there, neither increasing nor decreasing.

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∫cot⁡x dx=\int\cot x\,dx= (a) ln⁡(sec⁡x)+c\ln(\sec x)+c (b) ln⁡(cosec⁡x)+c\ln(\operatorname{cosec}x)+c (c) ln⁡(sin⁡x)+c\ln(\sin x)+c (d) −cosec⁡2x+c-\operatorname{cosec}^2x+c

Answer

(c) ln⁡(sin⁡x)+c\ln(\sin x)+c

∫cot⁡x dx=∫cos⁡xsin⁡xdx=ln⁡∣sin⁡x∣+c\int\cot x\,dx=\int\frac{\cos x}{\sin x}dx=\ln|\sin x|+c.

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∫ex−1ex+1dx\int\frac{e^x-1}{e^x+1}dx (a) ex−log⁡(ex+1)e^x-\log(e^x+1) (b) ex+e−xe^x+e^{-x} (c) ex−log⁡(e−x+1)e^x-\log(e^{-x}+1) (d) 2log⁡(ex2+e−x2)2\log\left(e^{\frac x2}+e^{-\frac x2}\right)

Answer

(d) 2log⁡(ex/2+e−x/2)2\log\left(e^{x/2}+e^{-x/2}\right)

ex−1ex+1=ex/2−e−x/2ex/2+e−x/2.\frac{e^x-1}{e^x+1}=\frac{e^{x/2}-e^{-x/2}}{e^{x/2}+e^{-x/2}}.

The numerator is the derivative of the denominator times 22 (ddx(ex/2+e−x/2)=12(ex/2−e−x/2)\frac{d}{dx}(e^{x/2}+e^{-x/2})=\frac12(e^{x/2}-e^{-x/2})), so the integral is 2log⁡(ex/2+e−x/2)+c2\log\left(e^{x/2}+e^{-x/2}\right)+c.

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∫log⁡x dx\int\log x\,dx (a) 1x\frac1x (b) xlog⁡x−xx\log x-x (c) exe^x (d) exx\frac{e^x}{x}

Answer

(b) xlog⁡x−xx\log x-x

By parts with u=log⁡xu=\log x, dv=dxdv=dx: ∫log⁡x dx=xlog⁡x−∫1 dx=xlog⁡x−x+c\int\log x\,dx=x\log x-\int1\,dx=x\log x-x+c.

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Area of region bounded by curve y2=4axy^2=4ax and x2=4byx^2=4by is (a) 4ab3\frac{4ab}{3} (b) 8ab3\frac{8ab}{3} (c) 16ab3\frac{16ab}{3} (d) 2ab3\frac{2ab}{3}

Answer

(c) 16ab3\frac{16ab}{3}

The curves meet at x=0x=0 and x=4a1/3b2/3x=4a^{1/3}b^{2/3} (so x3=64ab2x^3=64ab^2). Area =∫(2ax−x24b)dx=\int\left(2\sqrt{a}\sqrt{x}-\frac{x^2}{4b}\right)dx between these limits:

A=43a x3/2−x312b=43a⋅8a b−64ab212b=32ab3−16ab3=16ab3.A=\frac43\sqrt a\,x^{3/2}-\frac{x^3}{12b}=\frac43\sqrt a\cdot8\sqrt a\,b-\frac{64ab^2}{12b}=\frac{32ab}{3}-\frac{16ab}{3}=\frac{16ab}{3}.

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If the ratio of cost price and selling price of an article is 2:3 then (a) Profit % = 50% (b) loss % = 30% (c) Neither profit or loss (d) Profit % = 1003\frac{100}{3}%

Answer

(a) Profit % = 50%

Let CP = 2k and SP = 3k. Then profit = k, and

Profit %=k2k×100=50%.\text{Profit \%}=\frac{k}{2k}\times100=50\%.

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lim⁡n→∞∑n2n3=\lim_{n\to\infty}\frac{\sum n^2}{n^3}= (a) 0 (b) ∞\infty (c) 2 (d) 13\frac13

Answer

(d) 13\frac13

Since ∑n2=n(n+1)(2n+1)6\sum n^2=\frac{n(n+1)(2n+1)}{6},

lim⁡n→∞n(n+1)(2n+1)6n3=lim⁡n→∞(1+1/n)(2+1/n)6=26=13.\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}\frac{(1+1/n)(2+1/n)}{6}=\frac26=\frac13.

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lim⁡x→1(x−1)tan⁡πx2\lim_{x\to1}(x-1)\tan\frac{\pi x}{2} (a) −2π-\frac2\pi (b) π2\frac\pi2 (c) −12-\frac12 (d) −1π-\frac1\pi

Answer

(a) −2π-\frac{2}{\pi}

Put x=1+tx=1+t with t→0t\to0. Then tan⁡π(1+t)2=−cot⁡πt2\tan\frac{\pi(1+t)}{2}=-\cot\frac{\pi t}{2}, so

lim⁡t→0(−tcot⁡πt2)=−lim⁡t→0ttan⁡(πt/2)=−2π.\lim_{t\to0}\left(-t\cot\frac{\pi t}{2}\right)=-\lim_{t\to0}\frac{t}{\tan(\pi t/2)}=-\frac{2}{\pi}.

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If y=1+cos⁡2x1−cos⁡2xy=\sqrt{\frac{1+\cos2x}{1-\cos2x}} then dydx\frac{dy}{dx} at x=π2x=\frac\pi2 is (a) 0 (b) 1 (c) not defined (d) -1

Answer

(c) not defined

Since 1+cos⁡2x=2cos⁡2x1+\cos2x=2\cos^2x and 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x, we get y=∣cot⁡x∣y=|\cot x|.

For xx just below π2\frac\pi2, y=cot⁡xy=\cot x and y=−csc2xto−1y=-csc^2xto-1.

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Derivative of ex2+a2e^{\sqrt{x^2+a^2}} w.r.t x2+a2\sqrt{x^2+a^2} (a) e−x2+a2e^{-\sqrt{x^2+a^2}} (b) ex2+a22x\frac{e^{\sqrt{x^2+a^2}}}{2x} (c) xex2+a22x2+a2\frac{xe^{\sqrt{x^2+a^2}}}{2\sqrt{x^2+a^2}} (d) ex2+a2e^{\sqrt{x^2+a^2}}

Answer

(d) ex2+a2e^{\sqrt{x^2+a^2}}

Let u=x2+a2u=\sqrt{x^2+a^2}. Then ddueu=eu=ex2+a2\frac{d}{du}e^u=e^u=e^{\sqrt{x^2+a^2}}.

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If a+b+c=0a+b+c=0 in quadratic equation ax2+bx+c=0ax^2+bx+c=0 then roots of quadratic equation is (a) 1 and -2 (b) 1 and ca\frac ca (c) −ba-\frac ba and ca\frac ca (d) 0 and 1

Answer

(b) 1 and ca\frac{c}{a}

If a+b+c=0a+b+c=0, then x=1x=1 satisfies the equation. The product of the roots is ca\frac ca, so the other root is ca\frac ca.

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If (m+n)th(m+n)^{th} term of a G.P is p and (m−n)th(m-n)^{th} term is q then mthm^{th} term is (a) pq\sqrt{pq} (b) p+q2\frac{p+q}{2} (c) (p+q)\sqrt{(p+q)} (d) pqpq

Answer

(a) pq\sqrt{pq}

Let the first term be aa and the common ratio rr. Then arm+n−1=par^{m+n-1}=p and arm−n−1=qar^{m-n-1}=q.

pq=a2r2m−2=(arm−1)2pq=a^2r^{2m-2}=(ar^{m-1})^2

The mthm^{th} term is arm−1=pqar^{m-1}=\sqrt{pq}.

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  • 2078 Model · 1 mark

(a+b−c)x2+(c+a−b)x+(b+c−a)=0(a+b-c)x^2+(c+a-b)x+(b+c-a)=0 and a+b+c=0a+b+c=0 then roots quadratic equation, if a, b and c are rational (a) rational (b) irrational (c) imaginary (d) Real and equal

Answer

(a) rational

The sum of the coefficients is (a+b−c)+(c+a−b)+(b+c−a)=a+b+c=0(a+b-c)+(c+a-b)+(b+c-a)=a+b+c=0, so x=1x=1 is a root. The product of the roots is b+c−aa+b−c\frac{b+c-a}{a+b-c}, so the other root is b+c−aa+b−c\frac{b+c-a}{a+b-c}. This is rational when a,b,ca,b,c are rational, so both roots are rational.

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  • 2078 Model · 1 mark

The quadratic equation whose roots are 35\frac35 times the roots if quadratic equation x2−8x+7=0x^2-8x+7=0 (a) 25x2−30x+7=025x^2-30x+7=0 (b) 25x2−120x+63=025x^2-120x+63=0 (c) 25x2+120x−63=025x^2+120x-63=0 (d) none of these

Answer

(b) 25x2−120x+63=025x^2-120x+63=0

The roots of x2−8x+7=0x^2-8x+7=0 are 1 and 7. The new roots are 35\frac35 and 215\frac{21}{5}.

Sum =245=\frac{24}{5} and product =6325=\frac{63}{25}, so the equation is

x2−245x+6325=0  ⇒  25x2−120x+63=0.x^2-\frac{24}{5}x+\frac{63}{25}=0\;\Rightarrow\;25x^2-120x+63=0.

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  • 2078 Model · 1 mark

Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}, sum of geometric progression is true if (a) ∣r∣>1|r|>1 (b) ∣r∣<1|r|<1 (c) ∣r∣=1|r|=1 (d) ∣rn∣≤1|r^n|\leq1

Answer

(b) ∣r∣<1|r|<1

The formula Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r} is the standard form used when ∣r∣<1|r|<1. It is also valid algebraically for any r≠1r\ne1. For ∣r∣<1|r|<1 the sum also converges to a1−r\frac{a}{1-r} as n→∞n\to\infty.

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  • 2078 Model · 1 mark

Square roots of (3+4i)(3+4i) are (a) ±(2+i)\pm(2+i) (b) ±(3+i)\pm(3+i) (c) ±(2−i)\pm(2-i) (d) ±(3−i)\pm(3-i)

Answer

(a) ±(2+i)\pm(2+i)

(2+i)2=4+4i+i2=3+4i(2+i)^2=4+4i+i^2=3+4i, so the square roots of 3+4i3+4i are ±(2+i)\pm(2+i).

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  • 2078 Model · 1 mark

(1+i)100+(1−i)100=(1+i)^{100}+(1-i)^{100}= (a) 0 (b) -1 (c) 1 (d) i

Answer

None of the options is correct. The value is −251-2^{51}.

(1+i)2=2i(1+i)^2=2i and (1−i)2=−2i(1-i)^2=-2i, so

(1+i)100=(2i)50=250i50=−250,(1−i)100=(−2i)50=250i50=−250.(1+i)^{100}=(2i)^{50}=2^{50}i^{50}=-2^{50},\qquad (1-i)^{100}=(-2i)^{50}=2^{50}i^{50}=-2^{50}.

Hence the sum is −251-2^{51}.

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  • 2078 Model · 1 mark

The point of inflection of quadratic function is (a) always (1,0) (b) always (0,0) (c) does not exist (d) (−∞,0)(-\infty,0)

Answer

(c) does not exist

A quadratic function ax2+bx+cax^2+bx+c has f′′(x)=2a≠0f''(x)=2a\ne0 everywhere. Its concavity never changes, so it has no point of inflection.

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  • 2078 Model · 1 mark

The area of triangle with vertices (1,2),(2,3)(1,2),(2,3) and (3,4)(3,4) is (a) 0 sq. unit (b) 1 sq. unit (c) 2 sq. unit (d) 3 sq. unit

Answer

(a) 0 sq. unit

Area=12∣1(3−4)+2(4−2)+3(2−3)∣=12∣−1+4−3∣=0\text{Area}=\tfrac12|1(3-4)+2(4-2)+3(2-3)|=\tfrac12|-1+4-3|=0

The points all lie on the line y=x+1y=x+1, so they are collinear.

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  • 2078 Model · 1 mark

The polar coordinate of point (−1,−3)(-1,-\sqrt3) is (a) (π3,2)\left(\frac\pi3,2\right) (b) (2,π3)\left(2,\frac\pi3\right) (c) (2,4π3)\left(2,\frac{4\pi}{3}\right) (d) (4π3,2)\left(\frac{4\pi}{3},2\right)

Answer

(c) (2,4π3)\left(2,\frac{4\pi}{3}\right)

r=1+3=2r=\sqrt{1+3}=2. The point (−1,−3)(-1,-\sqrt3) lies in the third quadrant, where tan⁡θ=3\tan\theta=\sqrt3, so θ=π+π3=4π3\theta=\pi+\frac\pi3=\frac{4\pi}{3}. In the form (r,θ)(r,\theta) this is (2,4π3)\left(2,\frac{4\pi}{3}\right).

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  • 2078 Model · 1 mark

If the pair of lines represented by ax2+2hxy+by2+2gx+2fy+c=0ax^2+2hxy+by^2+2gx+2fy+c=0 intersect at x−axisx-axis then (a) abc+2fgh=bg2+ch2abc+2fgh=bg^2+ch^2 (b) abc+2fgh=0abc+2fgh=0 (c) abc+2fgh=af2+ch2abc+2fgh=af^2+ch^2 (d) 2fgh=bg2+ch22fgh=bg^2+ch^2

Answer

(c) abc+2fgh=af2+ch2abc+2fgh=af^2+ch^2

Let the lines meet at (x0,0)(x_0,0) on the x-axis. Putting y=0y=0 gives ax2+2gx+c=0ax^2+2gx+c=0, which must have equal roots, so g2=acg^2=ac and hence abc=bg2abc=bg^2. The pair condition is

abc+2fgh=af2+bg2+ch2.abc+2fgh=af^2+bg^2+ch^2.

Substituting abc=bg2abc=bg^2 gives 2fgh=af2+ch22fgh=af^2+ch^2, which is the relation of the form given in option (c).

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  • 2078 Model · 1 mark

∫−11sin⁡7x dx=\int_{-1}^{1}\sin^7x\,dx= (a) 6 (b) -6 (c) 0 (d) -1

Answer

(c) 0

sin⁡7x\sin^7x is an odd function, and the interval [−1,1][-1,1] is symmetric about 0, so

∫−11sin⁡7x dx=0.\int_{-1}^{1}\sin^7x\,dx=0.

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  • 2078 Model · 1 mark

∣x−a∣=|x-a|= (a) (a−x)(a-x) if x>ax>a (b) ±(x−a)\pm(x-a) if x≥ax\geq a (c) (x−a)(x-a) if x<ax<a (d) (a−x)(a-x) if x<ax<a

Answer

(d) (a−x)(a-x) if x<ax<a

By definition ∣x−a∣=x−a|x-a|=x-a for x≥ax\ge a and ∣x−a∣=−(x−a)=a−x|x-a|=-(x-a)=a-x for x<ax<a.

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  • 2078 Model · 1 mark

∫xx+1dx=\int\frac{x}{x+1}dx= (a) x−log⁡x+cx-\log x+c (b) x−log⁡(1+x)+cx-\log(1+x)+c (c) x+log⁡x+cx+\log x+c (d) x+log⁡(1+x)+cx+\log(1+x)+c

Answer

(b) x−log⁡(1+x)+cx-\log(1+x)+c

∫xx+1dx=∫(1−1x+1)dx=x−log⁡(1+x)+c.\int\frac{x}{x+1}dx=\int\left(1-\frac{1}{x+1}\right)dx=x-\log(1+x)+c.

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  • 2078 Model · 1 mark

Which of the following statements is are true (a) All scalar matrix are diagonal matrix (b) All diagonal matric are scalar matric but not vice-versa (c) if aij=−aji ∀ ia_{ij}=-a_{ji}\ \forall\ i and jj and aij=0a_{ij}=0 for i=ji=j then matrix is called skew -symmetric matrix (d) both a and c

Answer

(d) both a and c

A scalar matrix is a diagonal matrix with all diagonal entries equal, so (a) is true. (b) is false, because not every diagonal matrix is scalar (the converse is the true statement). (c) is the correct definition of a skew-symmetric matrix. So (a) and (c) are true.

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  • 2078 Model · 1 mark

Find the inverse of the function f(x)=3x4+2f(x)=3x^4+2 (a) x−23\sqrt{\frac{x-2}{3}} (b) x−234\sqrt[4]{\frac{x-2}{3}} (c) x−23\frac{x-2}{3} (d) 1+x43\frac{1+x^4}{3}

Answer

(b) x−234\sqrt[4]{\dfrac{x-2}{3}}

Let y=3x4+2y=3x^4+2. Then x4=y−23x^4=\frac{y-2}{3}, so x=y−234x=\sqrt[4]{\frac{y-2}{3}} (taking the principal root, i.e. x≥0x\ge0). Interchanging variables gives f−1(x)=x−234f^{-1}(x)=\sqrt[4]{\frac{x-2}{3}}.

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  • 2078 Model · 1 mark

Domain of the function f(x)=x2−4x−2f(x)=\frac{x^2-4}{x-2} is (a) R−{2}R-\{2\} (b) RR (c) R−{4}R-\{4\} (d) [0,∞][0,\infty]

Answer

(a) R−{2}R-\{2\}

The function is undefined only where the denominator is zero, that is at x=2x=2. So the domain is R−{2}R-\{2\}.

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  • 2078 Model · 1 mark

log⁡222222=\log_2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt2}}}}= (a) 3132\frac{31}{32} (b) 2312^{31} (c) 6364\frac{63}{64} (d) 2642^{64}

Answer

(a) 3132\frac{31}{32}

The nested expression equals 21/2+1/4+1/8+1/16+1/32=231/322^{1/2+1/4+1/8+1/16+1/32}=2^{31/32}, so

log⁡2231/32=3132.\log_2 2^{31/32}=\frac{31}{32}.

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  • 2078 Model · 1 mark

∣123456789∣=\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}= (a) 0 (b) 1 (c) -1 (d) 2

Answer

(a) 0

∣123456789∣=1(45−48)−2(36−42)+3(32−35)=−3+12−9=0.\begin{vmatrix}1&2&3\\4&5&6\\7&8&9\end{vmatrix}=1(45-48)-2(36-42)+3(32-35)=-3+12-9=0.

(Equivalently, R3−R2=R2−R1R_3-R_2=R_2-R_1.)

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  • 2078 Model · 1 mark

tan⁡15∘=\tan15^\circ= (a) 12+3\frac{1}{2+\sqrt3} (b) 2−32-\sqrt3 (c) 12−3\frac{1}{2-\sqrt3} (d) 2+32+\sqrt3

Answer

(b) 2−32-\sqrt3

tan⁡15∘=tan⁡(45∘−30∘)=1−131+13=3−13+1=(3−1)22=2−3.\tan15^\circ=\tan(45^\circ-30^\circ)=\frac{1-\frac1{\sqrt3}}{1+\frac1{\sqrt3}}=\frac{\sqrt3-1}{\sqrt3+1}=\frac{(\sqrt3-1)^2}{2}=2-\sqrt3.

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  • 2078 Model · 1 mark

a⃗×b⃗=∣a⃗∣∣b⃗∣sin⁡θ η^\vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat\eta, for any non-zero vectors a⃗\vec a and b⃗\vec b (a) η^\hat\eta is the unit vector along a⃗\vec a (b) η^\hat\eta is the unit vector along b⃗\vec b (c) η^\hat\eta is the unit vector along a⃗×b⃗\vec a\times\vec b (d) η^\hat\eta is the unit vector along a⃗\vec a and b⃗\vec b

Answer

(c) η^\hat\eta is the unit vector along a⃗×b⃗\vec a\times\vec b. By definition the cross product has magnitude ∣a⃗∣∣b⃗∣sin⁡θ|\vec a||\vec b|\sin\theta and direction given by the unit vector η^\hat\eta perpendicular to both a⃗\vec a and b⃗\vec b (right-hand rule), i.e. along a⃗×b⃗\vec a\times\vec b.

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  • 2078 Model · 1 mark

If A={x:x∈R,2x2−1=−1}A=\{x: x\in R, \sqrt{2x^2-1}=-1\} then A is (a) Finite set with elements ±1\pm1 (b) null set (c) infinite set (d) data insufficient

Answer

(b) null set. Since 2x2−1≥0\sqrt{2x^2-1}\ge 0 for every real xx for which it is defined, it can never equal −1-1. So no real xx satisfies the condition and A=∅A=\varnothing.

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  • 2078 Model · 1 mark

sin⁡−11−x21+x2+cos⁡−12x1+x2=\sin^{-1}\frac{1-x^2}{1+x^2}+\cos^{-1}\frac{2x}{1+x^2}= (a) π2−tan⁡−1x\frac\pi2-\tan^{-1}x (b) π2−2tan⁡−1x\frac\pi2-2\tan^{-1}x (c) π−2tan⁡−1x\pi-2\tan^{-1}x (d) none of these

Answer

(d) none of these. Put x=tan⁡θx=\tan\theta. Then 1−x21+x2=cos⁡2θ\frac{1-x^2}{1+x^2}=\cos2\theta and 2x1+x2=sin⁡2θ\frac{2x}{1+x^2}=\sin2\theta, so for 0≤θ≤π/40\le\theta\le\pi/4 the sum is (π2−2θ)+(π2−2θ)=π−4tan⁡−1x(\frac\pi2-2\theta)+(\frac\pi2-2\theta)=\pi-4\tan^{-1}x. Check: at x=1x=1 the sum is 0+0=00+0=0, which matches π−4⋅π4=0\pi-4\cdot\frac\pi4=0 but not options (a), (b) or (c).

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  • 2078 Model · 1 mark

cos⁡x=\cos x= (a) 2cos⁡2x2−12\cos^2\frac x2-1 (b) 1−2sin⁡2x21-2\sin^2\frac x2 (c) 1−tan⁡2x21+tan⁡2x2\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2} (d) All of these

Answer

(d) All of these. cos⁡x=2cos⁡2x2−1=1−2sin⁡2x2=1−tan⁡2x21+tan⁡2x2\cos x=2\cos^2\frac x2-1=1-2\sin^2\frac x2=\dfrac{1-\tan^2\frac x2}{1+\tan^2\frac x2} are all standard half-angle forms.

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  • 2078 Model · 1 mark

Non- intersecting lines are (a) consistent and dependent (b) Inconsistent and independent (c) Inconsistent and dependent (d) consistent and independent

Answer

(b) Inconsistent and independent. Non-intersecting (parallel) lines have no common point, so the system has no solution (inconsistent), and the equations are not multiples of each other (independent).

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  • 2078 Model · 1 mark

If the price of petrol in Nepal is increased by 25%. How much percent must a Saral Shikshya student reduce his consumption of petrol so as not to increase his expenditure on petrol? (a) 20% (b) 25% (c) 30% (d) 35%

Answer

(a) 20% Let old price =100=100 and consumption =1=1. New price =125=125. To keep expenditure the same, new consumption =100125=0.8=\frac{100}{125}=0.8, a reduction of 25125×100=20%\frac{25}{125}\times100=20\%.

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  • 2078 Model · 1 mark

If A earns 331333\frac13% more than B, how much percent does B earn less than A? (a) 23% (b) 24% (c) 20% (d) 25%

Answer

(d) 25% Let B earn 100100. Then A earns 13313=4003133\frac13=\frac{400}{3}. B earns less than A by 400/3−100400/3×100=100/3400/3×100=25%\dfrac{400/3-100}{400/3}\times100=\dfrac{100/3}{400/3}\times100=25\%.

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  • 2078 Model · 1 mark

If a2+b2+c2−ab−bc−ca=0a^2+b^2+c^2-ab-bc-ca=0 then a:b:ca:b:c is (a) 1:1:2 (b) 1:1:1 (c) 1:2:1 (d) 2:1:1

Answer

(b) 1:1:1 a2+b2+c2−ab−bc−ca=12[(a−b)2+(b−c)2+(c−a)2]=0a^2+b^2+c^2-ab-bc-ca=\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]=0 forces a=b=ca=b=c, so a:b:c=1:1:1a:b:c=1:1:1.

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  • 2078 Model · 1 mark

A certain number of people were supposed to complete work in 20 days. The work, however, took 32 days since 9 people were absent throughout. How many people were supposed to be working originally? (a) 36 (b) 38 (c) 34 (d) 30

Answer

No option matches; the correct number is 24. Let nn be the original number of people. Total work is constant: 20n=32(n−9)20n=32(n-9), so 12n=28812n=288 and n=24n=24. (None of 36, 38, 34, 30 satisfies this.)

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  • 2078 Model · 1 mark

The speed of a boat while traveling downstream is 32km/hr, whereas when travelling upstream it is 28 km/hr, what is the speed of a boat in still water? (a) 30km/hr (b) 2km/hr (c) 3km/hr (d) 34km/hr

Answer

(a) 30 km/hr Speed in still water =32+282=30=\dfrac{32+28}{2}=30 km/hr.

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  • 2078 Model · 1 mark

The surface area of a cube is 150 cm2150\,cm^2. Its volume is (a) 64 cm264\,cm^2 (b) 125 cm3125\,cm^3 (c) 150 cm3150\,cm^3 (d) 216 cm3216\,cm^3

Answer

(b) 125 cm3125\,cm^3 6a2=150⇒a2=25⇒a=56a^2=150\Rightarrow a^2=25\Rightarrow a=5 cm, so V=a3=125 cm3V=a^3=125\,cm^3.

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  • 2078 Model · 1 mark

If the supply function Ps=160+2Q2P_s=160+2Q^2 and the demand function is Pd=240−3Q2P_d=240-3Q^2. The market equilibrium price is (a) 192 (b) 194 (c) 190 (d) 180

Answer

(a) 192 At equilibrium Ps=PdP_s=P_d: 160+2Q2=240−3Q2⇒5Q2=80⇒Q2=16160+2Q^2=240-3Q^2\Rightarrow5Q^2=80\Rightarrow Q^2=16. Then P=160+2(16)=192P=160+2(16)=192.

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  • 2078 Model · 1 mark

Probability of an event E(x)E(x) is (a) 1≤ε(x)1\leq\varepsilon(x) (b) 0≤p(ε)≤10\leq p(\varepsilon)\leq1 (c) 0≤p(ε)0\leq p(\varepsilon) (d) −1≤p(ε)≤1-1\leq p(\varepsilon)\leq1

Answer

(b) 0≤p(ε)≤10\leq p(\varepsilon)\leq1 The probability of any event always lies between 0 (impossible) and 1 (certain).

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  • 2079 Model · 1 mark

Which is not a rational number? (a) 1 (b) ee (c) 3 (d) -1

Answer

(b) ee e≈2.71828…e\approx2.71828\ldots is irrational, while 1, 3 and −1-1 are rational.

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  • 2079 Model · 1 mark

If A and B are the nonempty set and A⊂B⊂UA\subset B\subset U then A∩B=A\cap B= (a) A (b) B (c) U (d) A-B

Answer

(a) A Since A⊂BA\subset B, every element of AA is in BB, so A∩B=AA\cap B=A.

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If ∣x−1∣≤5|x-1|\leq5 then x=x= (a) {−4,6}\{-4,6\} (b) [−4,6][-4,6] (c) U (d) [−4,−6][-4,-6]

Answer

(b) [−4,6][-4,6] ∣x−1∣≤5⇒−5≤x−1≤5⇒−4≤x≤6|x-1|\le5\Rightarrow-5\le x-1\le5\Rightarrow-4\le x\le6.

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  • 2079 Model · 1 mark

If log⁡ax=y\log_ax=y then x=x= (a) y (b) a (c) aya^y (d) 1/y1/y

Answer

(c) aya^y By definition, log⁡ax=y  ⟺  x=ay\log_ax=y\iff x=a^y.

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  • 2079 Model · 1 mark

The domain of the function f(x)=16−x2f(x)=\sqrt{16-x^2} is (a) (−4,4](-4,4] (b) [−4,4][-4,4] (c) (0,4)(0,4) (d) {0,5}\{0,5\}

Answer

(b) [−4,4][-4,4] We need 16−x2≥0⇒x2≤16⇒−4≤x≤416-x^2\ge0\Rightarrow x^2\le16\Rightarrow-4\le x\le4.

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A sentence x+5=4x+5=4 is (a) false statement (b) true statement (c) not a statement (d) a statement

Answer

(c) not a statement x+5=4x+5=4 is an open sentence: it is true for x=−1x=-1 and false otherwise, so it has no definite truth value until xx is specified.

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  • 2079 Model · 1 mark

The function f(x)=cos⁡xf(x)=\cos x is (a) an even function (b) period function (c) an odd function (d) neither even or odd

Answer

(a) an even function cos⁡(−x)=cos⁡x\cos(-x)=\cos x for all xx, so cos⁡x\cos x is even. (It is also periodic, but the defining property asked here is evenness.)

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  • 2079 Model · 1 mark

If p, q and r are in G.P, then (a) p+r=2qp+r=2q (b) q2<prq^2<pr (c) q2<prq^2<pr (d) q2=prq^2=pr

Answer

(d) q2=prq^2=pr In a G.P., qp=rq⇒q2=pr\dfrac qp=\dfrac rq\Rightarrow q^2=pr.

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If k+3k+3, 4k+74k+7 and 3k−13k-1 are in AP then the sequence is (a) 7, 8, 9 (b) 5, 7, 9 (c) 7, 9, 11 (d) 1, 3, 5

Answer

(c) 7, 9, 11

For an AP the middle term is the mean of the other two: 2(4k−7)=(k+3)+(3k−1)2(4k-7)=(k+3)+(3k-1), so 8k−14=4k+28k-14=4k+2 and k=4k=4. The terms are 7,9,117, 9, 11 with common difference 2.

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If [065+k0]\begin{bmatrix}0&6\\5+k&0\end{bmatrix} is a skew symmetric matrix then k=k= (a) -11 (b) -5 (c) 0 (d) -6

Answer

(a) −11-11

A skew-symmetric matrix satisfies aij=−ajia_{ij}=-a_{ji}, so 6=−(5+k)6=-(5+k), giving k=−11k=-11.

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The value of x for the determinant ∣x+3011∣=0\begin{vmatrix}x+3&0\\1&1\end{vmatrix}=0 is (a) 3 (b) -3 (c) 1 (d) 0

Answer

(b) −3-3

∣x+3011∣=(x+3)(1)−0=x+3=0\begin{vmatrix}x+3&0\\1&1\end{vmatrix}=(x+3)(1)-0=x+3=0, so x=−3x=-3.

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The value of −8−2\sqrt{-8}\sqrt{-2} is (a) 16 (b) 4 (c) -4 (d) 4i4i

Answer

(c) −4-4

−8−2=(22 i)(2 i)=4i2=−4\sqrt{-8}\sqrt{-2}=(2\sqrt2\,i)(\sqrt2\,i)=4i^2=-4.

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The modulus of 1+i1−i\frac{1+i}{1-i} is (a) 1 (b) -2 (c) 2\sqrt2 (d) 1/21/\sqrt2

Answer

(a) 1

∣1+i1−i∣=∣1+i∣∣1−i∣=22=1\left|\frac{1+i}{1-i}\right|=\frac{|1+i|}{|1-i|}=\frac{\sqrt2}{\sqrt2}=1.

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In a triangle ABC, if a=23a=2\sqrt3, b=2b=2, c=4c=4 then A=A= (a) 15∘15^\circ (b) 30∘30^\circ (c) 45∘45^\circ (d) 60∘60^\circ

Answer

(d) 60∘60^\circ

cos⁡A=b2+c2−a22bc=4+16−1216=12\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{4+16-12}{16}=\frac12, so A=60∘A=60^\circ.

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The perimeter 2s of the triangle ABC is (a) a+b+ca+b+c (b) (a+b+c)/R(a+b+c)/R (c) 2a+2b+2c2a+2b+2c (d) A+b+CA+b+C

Answer

(a) a+b+ca+b+c

The perimeter is the sum of the three sides, 2s=a+b+c2s=a+b+c.

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Area of the triangle ABC is (a) abc (b) abc/R (c) abc/4R (d) 4R/abc

Answer

(c) abc4R\dfrac{abc}{4R}

Since Δ=12bcsin⁡A\Delta=\frac12 bc\sin A and sin⁡A=a2R\sin A=\frac{a}{2R}, we get Δ=abc4R\Delta=\frac{abc}{4R}.

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Equation of a line perpendicular to the line x−3y+7=0x-3y+7=0 is (a) 3x+y+10=03x+y+10=0 (b) 2x−3y=102x-3y=10 (c) −3x−y−10=0-3x-y-10=0 (d) x+3y=10x+3y=10

Answer

(a) 3x+y+10=03x+y+10=0

The line x−3y+7=0x-3y+7=0 has slope 13\frac13, so a perpendicular line has slope −3-3. The line 3x+y+10=03x+y+10=0 has slope −3-3.

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Equation of tangent to (0,1)(0,1) to the circle x2+y2=1x^2+y^2=1 is (a) x=1x=1 (b) y=1y=1 (c) x=0x=0 (d) y=−1y=-1

Answer

(b) y=1y=1

The tangent at (x1,y1)(x_1,y_1) to x2+y2=1x^2+y^2=1 is xx1+yy1=1xx_1+yy_1=1. At (0,1)(0,1) this gives y=1y=1.

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The focus of the parabola y2=4xy^2=4x is (a) (0,1)(0,1) (b) (1,0)(1,0) (c) (0,0)(0,0) (d) (−1,0)(-1,0)

Answer

(b) (1,0)(1,0)

Comparing y2=4xy^2=4x with y2=4axy^2=4ax gives a=1a=1, so the focus is (a,0)=(1,0)(a,0)=(1,0).

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If lim⁡x→3xn−3nx−3=27\lim_{x\to3}\frac{x^n-3^n}{x-3}=27, then n=n= (a) 3 (b) 4 (c) 9 (d) 0

Answer

(a) 3

The limit equals n⋅3n−1n\cdot3^{n-1}. For n=3n=3 this is 3⋅9=273\cdot9=27.

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lim⁡x→0sin⁡3xx=\lim_{x\to0}\frac{\sin3x}{x}= (a) 1 (b) 0 (c) 0/0 (d) 3

Answer

(d) 3

lim⁡x→0sin⁡3xx=3lim⁡x→0sin⁡3x3x=3⋅1=3\lim_{x\to0}\frac{\sin3x}{x}=3\lim_{x\to0}\frac{\sin3x}{3x}=3\cdot1=3.

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The function f(x)={x2−xx−1x≠1kx=1f(x)=\begin{cases}\frac{x^2-x}{x-1}&x\neq1\\k&x=1\end{cases} is continuous at x=1x=1 for k=k= (a) 0 (b) 2 (c) 1 (d) 0/0

Answer

(c) 1

For x≠1x\neq1, x2−xx−1=x\frac{x^2-x}{x-1}=x, so lim⁡x→1f(x)=1\lim_{x\to1}f(x)=1. Continuity requires k=1k=1.

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If y=eax2+by=e^{ax^2+b} then dydx=\frac{dy}{dx}= (a) ax2+bax^2+b (b) 2axy2axy (c) 2ax2ax (d) eax2+be^{ax^2+b}

Answer

(b) 2axy2axy

dydx=eax2+b⋅2ax=2ax y\frac{dy}{dx}=e^{ax^2+b}\cdot2ax=2ax\,y.

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If x2+y2=25x^2+y^2=25 then dydx\frac{dy}{dx} at (0,5)(0,5) is (a) 0 (b) undefined (c) x/yx/y (d) −x/y-x/y

Answer

(a) 0

Differentiating, 2x+2ydydx=02x+2y\frac{dy}{dx}=0, so dydx=−xy\frac{dy}{dx}=-\frac xy. At (0,5)(0,5) this is 00.

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The stationary point of the function f(x)=x2−6x+3f(x)=x^2-6x+3 is x=x= (a) 0 (b) 3 (c) -3 (d) 6

Answer

(b) 3

f′(x)=2x−6=0f'(x)=2x-6=0 gives x=3x=3.

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∫elog⁡ex dx=\int e^{\log_ex}\,dx= (a) log⁡ex+c\log_ex+c (b) x2+cx^2+c (c) x22+c\frac{x^2}{2}+c (d) elog⁡ex+ce^{\log_ex}+c

Answer

(c) x22+c\dfrac{x^2}{2}+c

Since elog⁡ex=xe^{\log_e x}=x, ∫x dx=x22+c\int x\,dx=\frac{x^2}{2}+c.

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∫012xx2+2dx=\int_0^1\frac{2x}{x^2+2}dx= (a) log⁡e23\log_e\frac23 (b) log⁡e32\log_e\frac32 (c) 0 (d) 3

Answer

(b) log⁡e32\log_e\dfrac32

∫012xx2+2dx=[log⁡e(x2+2)]01=log⁡e3−log⁡e2=log⁡e32\int_0^1\frac{2x}{x^2+2}dx=\left[\log_e(x^2+2)\right]_0^1=\log_e3-\log_e2=\log_e\frac32.

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∫1/cos⁡2x dx=\int1/\cos^2x\,dx= (a) cot⁡x+c\cot x+c (b) cos⁡x+c\cos x+c (c) sec⁡2x+c\sec^2x+c (d) tan⁡x+c\tan x+c

Answer

(d) tan⁡x+c\tan x+c

1cos⁡2x=sec⁡2x\frac1{\cos^2x}=\sec^2x and ∫sec⁡2x dx=tan⁡x+c\int\sec^2x\,dx=\tan x+c.

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Area bounded by the x-axis and the curve y=x+2y=x+2 and the lines x=0x=0 and x=2x=2 is (a) -6 sq. units (b) 2 sq. units (c) 4 sq. units (d) 6 sq. units

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

Answer

(d) 6 sq. units

A=∫02(x+2)dx=[x22+2x]02=2+4=6A=\int_0^2(x+2)dx=\left[\frac{x^2}{2}+2x\right]_0^2=2+4=6.

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The unit vector of i⃗+j⃗\vec i+\vec j is (a) i⃗2+i⃗2\frac{\vec i}{\sqrt2}+\frac{\vec i}{\sqrt2} (b) i⃗2−i⃗2\frac{\vec i}{\sqrt2}-\frac{\vec i}{\sqrt2} (c) i⃗−j⃗\vec i-\vec j (d) 1

Answer

(a) i⃗2+j⃗2\dfrac{\vec i}{\sqrt2}+\dfrac{\vec j}{\sqrt2} (the option as printed, with the second term written as i⃗\vec i)

∣i⃗+j⃗∣=2|\vec i+\vec j|=\sqrt2, so the unit vector is i⃗+j⃗2\frac{\vec i+\vec j}{\sqrt2}. Option (a) is the only one of this form.

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The direction of vector v⃗=ai⃗+bj⃗\vec v=a\vec i+b\vec j is (a) tan⁡−1ab\tan^{-1}\frac ab (b) sin⁡−1ab\sin^{-1}\frac ab (c) tan⁡−1ba\tan^{-1}\frac ba (d) cos⁡−1ab\cos^{-1}\frac ab

Answer

(c) tan⁡−1ba\tan^{-1}\dfrac ba

The direction angle θ\theta of ai⃗+bj⃗a\vec i+b\vec j satisfies tan⁡θ=ba\tan\theta=\frac ba.

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The probability of selecting a vowel from the English alphabet is (a) 1/26 (b) 0 (c) 1/5 (d) 5/26

Answer

(d) 526\dfrac5{26}

There are 5 vowels among the 26 letters.

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The standard deviation of the data: 5, 10, 15, 20 and 25 is (a) 525\sqrt2 (b) 5 (c) 2\sqrt2 (d) 10

Answer

(a) 525\sqrt2

The mean is 1515. The squared deviations are 100,25,0,25,100100, 25, 0, 25, 100, which sum to 250250. The variance is 2505=50\frac{250}{5}=50, so the standard deviation is 50=52\sqrt{50}=5\sqrt2.

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The solution set of 3x+5>2x+13x+5>2x+1 is (a) x>4x>4 (b) x>−4x>-4 (c) x<−4x<-4 (d) x≥−4x\geq-4

Answer

(b) x>−4x>-4

3x+5>2x+13x+5>2x+1 gives x>−4x>-4.

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If f(x)=2x+1f(x)=2x+1 then the approximate value of f′(2)f'(2) with h=0.1h=0.1 is (a) 5.1 (b) 0 (c) 2.0 (d) 2.1

Answer

(c) 2.0

f′(2)≈f(2.1)−f(2)0.1=5.2−50.1=2.0f'(2)\approx\frac{f(2.1)-f(2)}{0.1}=\frac{5.2-5}{0.1}=2.0.

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  • 2079 Model · 1 mark

In a manufacturing company the production cost of 50 items is Rs 5000 in one day and Rs. 12000 to produce 150 items in one day. If x represents the number of items and y represents the cost, the model equation is (a) y=70x−1500y=70x-1500 (b) y=70x−5000y=70x-5000 (c) y=−70x−1500y=-70x-1500 (d) y=70x+1500y=70x+1500

Answer

(d) y=70x+1500y=70x+1500

Slope =12000−5000150−50=70=\frac{12000-5000}{150-50}=70. Using (50,5000)(50, 5000): 5000=70(50)+b5000=70(50)+b gives b=1500b=1500, so y=70x+1500y=70x+1500.

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A company finds the variable cost of Rs .5 per unit output and fixed cost of Rs 25, where Q is the number of outputs, then the total cost function, T is (a) T=5Q+25T=5Q+25 (b) T=25Q+5T=25Q+5 (c) T=5Q−25T=5Q-25 (d) T=25Q−5T=25Q-5

Answer

(a) T=5Q+25T=5Q+25

Total cost = variable cost + fixed cost = 5Q+255Q+25.

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If a ball is projected vertically upwards with a velocity of 24.9 m/s, then the maximum height described by the ball is (a) 44 m (b) 14.7 m (c) 78.4 m (d) 44.1 m

Answer

(d) 44.1 m

Maximum height H=u22gH=\frac{u^2}{2g} with g=9.8g=9.8 m/s2^2. The option 44.1 m corresponds to u=29.4u=29.4 m/s, since 29.4219.6=44.1\frac{29.4^2}{19.6}=44.1. It is the intended answer.

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  • 2080 · 1 mark

In the followings, the statement is (a) give me a book (b) x−2=10x-2=10 (c) Rara lake is in Pokhara (d) y>0y>0

Answer

(c) Rara lake is in Pokhara

A statement (proposition) is a sentence that is definitely true or false. This one is a (false) statement. "Give me a book" is a command, and x−2=10x-2=10 and y>0y>0 are open sentences whose truth depends on the variable.

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If S={x:x∈R and x2+4=0}S=\{x: x\in\mathbb R\text{ and }x^2+4=0\} then the set S is (a) singleton set (b) x−2=10x-2=10 (c) S={−2i,2i}S=\{-2i,2i\} (d) S=∅S=\emptyset

Answer

(d) S=∅S=\emptyset

x2+4=0x^2+4=0 gives x=±2ix=\pm2i, which are not real. So no real xx satisfies it and SS is the empty set.

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  • 2080 · 1 mark

The set is a collection of (a) long rivers of Nepal (b) longest rivers of Nepal (c) tall plants in a garden (d) nice books

Answer

(b) longest rivers of Nepal

A set must be well defined. "Longest rivers" can be decided exactly, whereas "long", "tall" and "nice" are vague.

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S={x:x≥−2,x∈R}S=\{x: x\geq-2, x\in\mathbb R\} is represented by (a) [−2,∞)[-2,\infty) (b) (−2,∞)(-2,\infty) (c) (∞,−2](\infty,-2] (d) [−2,∞][-2,\infty]

Answer

(a) [−2,∞)[-2,\infty)

x≥−2x\ge-2 includes −2-2 (closed bracket), and ∞\infty is never included.

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∣x−5∣≤3|x-5|\leq3 has values of x=x= (a) (2,8)(2,8) (b) [2,8][2,8] (c) {−2,8}\{-2,8\} (d) {2,8}\{2,8\}

Answer

(b) [2,8][2,8]

∣x−5∣≤3⇒−3≤x−5≤3⇒2≤x≤8|x-5|\le3\Rightarrow -3\le x-5\le3\Rightarrow 2\le x\le8.

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  • 2080 · 1 mark

The ordered pair (a,b)(a,b) for (2a−1,−3)=(1,b+3)(2a-1,-3)=(1,b+3) is (a) (1,−6)(1,-6) (b) [1,−6][1,-6] (c) {1,−6}\{1,-6\} (d) [1,6][1,6]

Answer

(a) (1,−6)(1,-6)

2a−1=1⇒a=12a-1=1\Rightarrow a=1 and b+3=−3⇒b=−6b+3=-3\Rightarrow b=-6.

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The domain of the function f:R→Rf:\mathbb R\to\mathbb R defined by f(x)=1−x2f(x)=\sqrt{1-x^2} is (a) [0,1][0,1] (b) {1,−1}\{1,-1\} (c) [0,∞)[0,\infty) (d) [−1,1][-1,1]

Answer

(d) [−1,1][-1,1]

We need 1−x2≥0⇒x2≤1⇒−1≤x≤11-x^2\ge0\Rightarrow x^2\le1\Rightarrow -1\le x\le1.

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The base of an exponential function y=axy=a^x passing through (2,4)(2,4) is (a) 4 (b) 2 (c) 8 (d) 1

Answer

(b) 2

a2=4⇒a=2a^2=4\Rightarrow a=2 (the base must be positive).

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  • 2080 · 1 mark

32log⁡3x=3^{2\log_3x}= (a) 2x2x (b) xx (c) x2x^2 (d) x/2x/2

Answer

(c) x2x^2

32log⁡3x=3log⁡3x2=x23^{2\log_3x}=3^{\log_3x^2}=x^2.

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  • 2080 · 1 mark

The function f(x)=x2−2f(x)=x^2-2 is (a) an even function (b) an odd number (c) not a function (d) neither odd nor even

Answer

(a) an even function

f(−x)=(−x)2−2=x2−2=f(x)f(-x)=(-x)^2-2=x^2-2=f(x).

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  • 2080 · 1 mark

A number which can be expressed in the form p/qp/q; p,qp,q are integers, q≠0q\neq0 is called (a) a rational number (b) an integer number (c) an irrational number (d) a prime number

Answer

(a) a rational number

This is the definition of a rational number.

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The sum of an infinite geometric sequence a,ak,ak2,…a, ak, ak^2,\ldots exists if (a) k>1k>1 (b) k>−1k>-1 (c) −1≤k≤1-1\leq k\leq1 (d) −1<k<1-1<k<1

Answer

(d) −1<k<1-1<k<1

The sum a1−k\dfrac{a}{1-k} exists only when the common ratio satisfies ∣k∣<1|k|<1.

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  • 2080 · 1 mark

If 2k+12k+1, 3k+23k+2 and 5k+15k+1 are in A.P., then k=k= (a) 2 (b) 6 (c) -8 (d) 0

Answer

(a) 2

For an A.P., 2(3k+2)=(2k+1)+(5k+1)2(3k+2)=(2k+1)+(5k+1), so 6k+4=7k+26k+4=7k+2 and k=2k=2.

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  • 2080 · 1 mark

The matrix [01−10]\begin{bmatrix}0&1\\-1&0\end{bmatrix} is (a) symmetric (b) skew-symmetric (c) unit (d) none

Answer

(b) skew-symmetric

The transpose is [0−110]=−A\begin{bmatrix}0&-1\\1&0\end{bmatrix}=-A, so AT=−AA^T=-A.

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  • 2080 · 1 mark

If A=[−532−1]A=\begin{bmatrix}-5&3\\2&-1\end{bmatrix} then A−1=A^{-1}= (a) [53−2−1]\begin{bmatrix}5&3\\-2&-1\end{bmatrix} (b) [−5−3−21]\begin{bmatrix}-5&-3\\-2&1\end{bmatrix} (c) [5321]\begin{bmatrix}5&3\\2&1\end{bmatrix} (d) [13−2−5]\begin{bmatrix}1&3\\-2&-5\end{bmatrix}

Answer

(d) [13−2−5]\begin{bmatrix}1&3\\-2&-5\end{bmatrix} (nearest option)

∣A∣=(−5)(−1)−(3)(2)=−1|A|=(-5)(-1)-(3)(2)=-1, so A−1=1−1[−1−3−2−5]=[1325]A^{-1}=\dfrac{1}{-1}\begin{bmatrix}-1&-3\\-2&-5\end{bmatrix}=\begin{bmatrix}1&3\\2&5\end{bmatrix}. Option (d) is the closest of the given options (only the sign of the second-row entries differs).

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  • 2080 · 1 mark

If ∣2a+113a2∣=0\begin{vmatrix}2a+1&1\\3a&2\end{vmatrix}=0, then a=a= (a) 0 (b) 2 (c) -2 (d) 1

Answer

(c) −2-2

2(2a+1)−3a=a+2=0⇒a=−22(2a+1)-3a=a+2=0\Rightarrow a=-2.

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  • 2080 · 1 mark

A square matrix A is said to be singular if (a) ∣A∣>0|A|>0 (b) ∣A∣<0|A|<0 (c) ∣A∣≠0|A|\neq0 (d) ∣A∣=0|A|=0

Answer

(d) ∣A∣=0|A|=0

A square matrix is singular when its determinant is zero (no inverse exists).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

If the roots of the equation px2+qx+r=0px^2+qx+r=0 are equal in magnitude but opposite in sign, then (a) p=0p=0 (b) q=0q=0 (c) r=0r=0 (d) p=qp=q

Answer

(b) q=0q=0

Sum of roots =−q/p=0=-q/p=0, so q=0q=0.

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  • 2080 · 1 mark

If the discriminant of the equation x2+kx+4=0x^2+kx+4=0 is zero then k=k= (a) ±4\pm4 (b) −4-4 (c) 4 (d) 0

Answer

(a) ±4\pm4

D=k2−16=0⇒k=±4D=k^2-16=0\Rightarrow k=\pm4.

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  • 2080 · 1 mark

If ii is the imaginary unit, then 1−i1+i=\frac{1-i}{1+i}= (a) 1 (b) −i-i (c) ii (d) −1-1

Answer

(b) −i-i

1−i1+i=(1−i)22=−2i2=−i\dfrac{1-i}{1+i}=\dfrac{(1-i)^2}{2}=\dfrac{-2i}{2}=-i.

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  • 2080 · 1 mark

The square roots of a complex number 5+12i5+12i is (a) ±(3+2i)\pm(3+2i) (b) ±(2+3i)\pm(2+3i) (c) ±(3−2i)\pm(3-2i) (d) ±(2−3i)\pm(2-3i)

Answer

(a) ±(3+2i)\pm(3+2i)

(3+2i)2=9−4+12i=5+12i(3+2i)^2=9-4+12i=5+12i.

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  • 2080 · 1 mark

sin⁡(cos⁡−1x)=\sin(\cos^{-1}x)= (a) 1+x2\sqrt{1+x^2} (b) sin⁡θ\sin\theta (c) 1−x2\sqrt{1-x^2} (d) x1−x2x\sqrt{1-x^2}

Answer

(c) 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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  • 2080 · 1 mark

The general solution of 4sin⁡2x=14\sin^2x=1, for n∈Zn\in Z is (a) x=nπ±π6x=n\pi\pm\frac\pi6 (b) x=nπ±π4x=n\pi\pm\frac\pi4 (c) x=nπ±π3x=n\pi\pm\frac\pi3 (d) x=2nπ±π6x=2n\pi\pm\frac\pi6

Answer

(a) x=nπ±π6x=n\pi\pm\dfrac{\pi}{6}

sin⁡2x=14=sin⁡2π6\sin^2x=\tfrac14=\sin^2\dfrac{\pi}{6}, 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 slope of the line px+qy+c=0px+qy+c=0, is (a) p (b) q (c) p/q (d) -p/q

Answer

(d) −p/q-p/q

y=−pqx−cqy=-\dfrac{p}{q}x-\dfrac{c}{q}, so the slope is −p/q-p/q.

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  • 2080 · 1 mark

If the length of the perpendicular from the point (k,2)(k,2) to 3x−4y+6=03x-4y+6=0 is 2, then k=k= (a) 8 (b) 6 (c) 4 (d) 5

Answer

(c) 4

∣3k−8+6∣9+16=2⇒∣3k−2∣=10\dfrac{|3k-8+6|}{\sqrt{9+16}}=2\Rightarrow|3k-2|=10, so k=4k=4 or k=−83k=-\tfrac83. Only k=4k=4 is among the options.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

The homogeneous equation is (a) ax2+bxy+c=0ax^2+bxy+c=0 (b) ax2+xy+by2=0ax^2+xy+by^2=0 (c) ax2+bxy+by2+c=0ax^2+bxy+by^2+c=0 (d) x2+by2+y3=0x^2+by^2+y^3=0

Answer

(b) ax2+xy+by2=0ax^2+xy+by^2=0

A homogeneous equation has every term of the same degree (here 2); the others contain a constant or a cubic term.

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  • 2080 · 1 mark

The distance of the point (1,2,3)(1,2,3) from origin is (a) 1 (b) 3 (c) 14\sqrt{14} (d) 2

Answer

(c) 14\sqrt{14}

12+22+32=14\sqrt{1^2+2^2+3^2}=\sqrt{14}.

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  • 2080 · 1 mark

The direction ratios of the line AB joining the points A(4,5,0)A(4,5,0) and B(2,3,4)B(2,3,4) are (a) −2,2,−4-2,2,-4 (b) 2,2,−42,2,-4 (c) 2,2,42,2,4 (d) −2,−2,−4-2,-2,-4

Answer

(b) 2,2,−42,2,-4

Direction ratios are the differences of coordinates: A−B=(4−2, 5−3, 0−4)=(2,2,−4)A-B=(4-2,\,5-3,\,0-4)=(2,2,-4). This is the only option proportional to the vector joining the two points (B−A=(−2,−2,4)B-A=(-2,-2,4) is its negative, and is not offered).

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  • 2080 · 1 mark

The direction of the vector ai+bja\mathbf i+b\mathbf j is (a) tan⁡−1(−ab)\tan^{-1}\left(-\frac ab\right) (b) tan⁡−1(ba)\tan^{-1}\left(\frac ba\right) (c) tan⁡−1(−ba)\tan^{-1}\left(-\frac ba\right) (d) none

Answer

(b) tan⁡−1(ba)\tan^{-1}\left(\frac ba\right)

The vector ai+bja\mathbf i+b\mathbf j makes angle θ\theta with the x-axis where tan⁡θ=ba\tan\theta=\frac ba.

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  • 2080 · 1 mark

The unit vector along xi−yjx\mathbf i-y\mathbf j is (a) xi−yjx2−y2\frac{x\mathbf i-y\mathbf j}{\sqrt{x^2-y^2}} (b) xi−yjx2y2\frac{x\mathbf i-y\mathbf j}{\sqrt{x^2y^2}} (c) xi−yjx2+y2\frac{x\mathbf i-y\mathbf j}{x^2+y^2} (d) xi−yjx2+y2\frac{x\mathbf i-y\mathbf j}{\sqrt{x^2+y^2}}

Answer

(d) xi−yjx2+y2\frac{x\mathbf i-y\mathbf j}{\sqrt{x^2+y^2}}

A unit vector is the vector divided by its magnitude, and ∣xi−yj∣=x2+y2|x\mathbf i-y\mathbf j|=\sqrt{x^2+y^2}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

The standard deviation of 3, 6, 9, 12, 15 is (a) 323\sqrt2 (b) 12 (c) 232\sqrt3 (d) 2

Answer

(a) 323\sqrt2

Mean =455=9=\frac{45}{5}=9. Squared deviations: 36+9+0+9+36=9036+9+0+9+36=90.

σ=905=18=32\sigma=\sqrt{\frac{90}{5}}=\sqrt{18}=3\sqrt2

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

The probability of getting an ace of club or a king of spade from a pack of 52 cards is (a) 1/26 (b) 1/13 (c) 1/52 (d) 1/2

Answer

(a) 126\frac1{26}

The events are mutually exclusive, so P=152+152=252=126P=\frac1{52}+\frac1{52}=\frac2{52}=\frac1{26}.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

If lim⁡x→ax4−a4x−a=256\lim_{x\to a}\frac{x^4-a^4}{x-a}=256, then a=a= (a) 5 (b) 3 (c) 4 (d) 6

Answer

(c) 4

lim⁡x→ax4−a4x−a=4a3=256\lim_{x\to a}\frac{x^4-a^4}{x-a}=4a^3=256, so a3=64a^3=64 and a=4a=4.

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  • 2080 · 1 mark

lim⁡x→0tan⁡3xx=\lim_{x\to0}\frac{\tan3x}{x}= (a) 3 (b) 0/0 (c) 0 (d) xx

Answer

(a) 3

lim⁡x→0tan⁡3xx=3lim⁡x→0tan⁡3x3x=3⋅1=3\lim_{x\to0}\frac{\tan3x}{x}=3\lim_{x\to0}\frac{\tan3x}{3x}=3\cdot1=3.

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  • 2080 · 1 mark

A function f(x)=2x+1f(x)=2x+1 is continuous at x=1x=1, if (a) lim⁡x→1f(x)=3\lim_{x\to1}f(x)=3 (b) lim⁡x→1f(x)\lim_{x\to1}f(x) exists (c) lim⁡x→1f(x)=f(1)\lim_{x\to1}f(x)=f(1) (d) none

Answer

(c) lim⁡x→1f(x)=f(1)\lim_{x\to1}f(x)=f(1)

A function is continuous at a point when the limit exists and equals the function value there. Here both equal 3, but the defining condition is (c).

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  • 2080 · 1 mark

If y=esin⁡x2y=e^{\sin x^2}, then dydx=\frac{dy}{dx}= (a) esin⁡x2e^{\sin x^2} (b) 2xycos⁡x22xy\cos x^2 (c) 2xesin⁡x22xe^{\sin x^2} (d) ycos⁡x2y\cos x^2

Answer

(b) 2xycos⁡x22xy\cos x^2

By the chain rule, dydx=esin⁡x2⋅cos⁡x2⋅2x=2xycos⁡x2\frac{dy}{dx}=e^{\sin x^2}\cdot\cos x^2\cdot2x=2xy\cos x^2.

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  • 2080 · 1 mark

If y=x3−x2y=x^3-x^2, then dydx=0\frac{dy}{dx}=0, x=x= (a) 0,−2/30,-2/3 (b) 0,3/20,3/2 (c) 0,30,3 (d) 0,2/30,2/3

Answer

(d) 0,230,\frac23

dydx=3x2−2x=x(3x−2)=0\frac{dy}{dx}=3x^2-2x=x(3x-2)=0, so x=0x=0 or x=23x=\frac23.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2080 · 1 mark

The function f(x)=2x+5f(x)=2x+5, on RR is (a) decreasing (b) stationary (c) increasing (d) none

Answer

(c) increasing

f′(x)=2>0f'(x)=2>0 for all real xx, so ff is increasing on RR.

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  • 2080 · 1 mark

∫dxax+b=\int\frac{dx}{ax+b}= (a) 1alog⁡∣ax+b∣+c\frac1a\log|ax+b|+c (b) alog⁡∣ax+b∣+ca\log|ax+b|+c (c) ∣log⁡(ax+b)∣+c|\log(ax+b)|+c (d) log⁡∣ax+b∣+c\log|ax+b|+c

Answer

(a) 1alog⁡∣ax+b∣+c\frac1a\log|ax+b|+c

Put u=ax+bu=ax+b, so du=a dxdu=a\,dx and the integral becomes 1a∫duu\frac1a\int\frac{du}{u}.

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  • 2080 · 1 mark

∫dx2cos⁡2xsin⁡2x=\int\frac{dx}{2\cos^2x\sin^2x}= (a) 2cot⁡2x+c2\cot2x+c (b) −2cot⁡2x+c-2\cot2x+c (c) cot⁡2x+c\cot2x+c (d) −cot⁡2x+c-\cot2x+c

Answer

(d) −cot⁡2x+c-\cot2x+c

Since sin⁡22x=4sin⁡2xcos⁡2x\sin^22x=4\sin^2x\cos^2x, we have 12sin⁡2xcos⁡2x=2sin⁡22x=2csc⁡22x\frac1{2\sin^2x\cos^2x}=\frac{2}{\sin^22x}=2\csc^22x.

∫2csc⁡22x dx=2⋅(−cot⁡2x2)+c=−cot⁡2x+c\int2\csc^22x\,dx=2\cdot\left(-\frac{\cot2x}{2}\right)+c=-\cot2x+c

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  • 2080 · 1 mark

∫13sin⁡(ln⁡t)tdt=\int_1^3\frac{\sin(\ln t)}{t}dt= (a) cos⁡(ln⁡3)\cos(\ln3) (b) −cos⁡(ln⁡3)-\cos(\ln3) (c) 1−cos⁡(ln⁡3)1-\cos(\ln3) (d) 1−sin⁡(ln⁡3)1-\sin(\ln3)

Answer

(c) 1−cos⁡(ln⁡3)1-\cos(\ln3)

Put u=ln⁡tu=\ln t, so du=dttdu=\frac{dt}{t}. The limits become 00 to ln⁡3\ln3.

∫0ln⁡3sin⁡u du=[−cos⁡u]0ln⁡3=1−cos⁡(ln⁡3)\int_0^{\ln3}\sin u\,du=\left[-\cos u\right]_0^{\ln3}=1-\cos(\ln3)

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  • 2080 · 1 mark

The area under the curve y=x2y=x^2 bounded by the x axis and the ordinates x=1x=1 and x=4x=4 is (a) 21 (b) 64/3 (c) 64 (d) 7

Answer

(a) 21

∫14x2dx=[x33]14=64−13=21\int_1^4x^2dx=\left[\frac{x^3}{3}\right]_1^4=\frac{64-1}{3}=21

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  • 2080 · 1 mark

The general formula for Newton-Raphson method is (a) xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} (b) xn+1=xn−f′(xn)f(xn)x_{n+1}=x_n-\frac{f'(x_n)}{f(x_n)} (c) xn+1=xn+f′(xn)f(xn)x_{n+1}=x_n+\frac{f'(x_n)}{f(x_n)} (d) xn+1=xn+1−f′(xn)f(xn)x_{n+1}=x_{n+1}-\frac{f'(x_n)}{f(x_n)}

Answer

(a) xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

This is the Newton-Raphson iteration, obtained by taking the x-intercept of the tangent at xnx_n.

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  • 2080 · 1 mark

The approximate value of ∫02x2dx\int_0^2x^2dx using midpoint rule with one interval is (a) 4 (b) 8 (c) 2 (d) 0

Answer

(c) 2

With one interval [0,2][0,2], the midpoint is x=1x=1. The approximation is h f(1)=2×12=2h\,f(1)=2\times1^2=2.

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  • 2080 · 1 mark

If three equal forces P, Q and R are acting at a point of rigid body are in equilibrium, then the angle between any two forces is (a) 120∘120^\circ (b) 30∘30^\circ (c) 60∘60^\circ (d) none

Answer

(a) 120∘120^\circ

Three equal forces in equilibrium form an equilateral triangle when drawn head to tail. The exterior angle of that triangle, which is the angle between the forces, is 120∘120^\circ. Check: P2+P2+2P2cos⁡θ=P2P^2+P^2+2P^2\cos\theta=P^2 gives cos⁡θ=−12\cos\theta=-\frac12.

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  • 2080 · 1 mark

A force equal to 4N4N acting upwards is resolved into forces, one being horizontal and equal to 3N3N. The magnitude of the other remaining force is (a) 7N7N (b) 1N1N (c) 2N2N (d) 5N5N

Answer

(d) 5N5N

The 4 N force is the resultant, so the vertical component of the other force is 4 N and its horizontal component is 3 N (equal and opposite to the given one, which has no vertical part). Magnitude =32+42=5 N=\sqrt{3^2+4^2}=5\,N.

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  • 2080 · 1 mark

If the acceleration of a particle, starting from rest is 5m/sec25m/sec^2, then the velocity at the end of 3 seconds is (a) 8m/sec28m/sec^2 (b) 15m/sec215m/sec^2 (c) 5m/sec25m/sec^2 (d) none

Answer

(b) 15 (m/sec)

v=u+at=0+5×3=15v=u+at=0+5\times3=15 m/sec. The unit of velocity is m/sec, not m/sec2^2, but 15 is the value.

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  • 2080 · 1 mark

A stone is thrown vertically upwards with a velocity of 30m/sec from the top of a tower and hits the ground after 10 secs, then the height of the tower is, (a) 100 m100\,m (b) 300 m300\,m (c) 200 m200\,m (d) 50 m50\,m

Answer

(c) 200 m200\,m

Take upward as positive, u=30u=30 m/s, g=10g=10 m/s2^2 and displacement −h-h after t=10t=10 s:

−h=30(10)−12(10)(10)2=300−500=−200-h=30(10)-\frac12(10)(10)^2=300-500=-200

So h=200h=200 m.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If one root of ax2+bx+c=0ax^2+bx+c=0 is 2+3i2+3i then the other root is (a) 4+6i4+6i (b) 2−3i2-3i (c) 2i−32i-3 (d) 2i+32i+3

Similar questions: If one root of ax 2+bx+c=0 be square of other then (2077 Model)

Answer

(b) 2−3i2-3i For a quadratic with real coefficients, complex roots occur in conjugate pairs, so the other root is 2−3i2-3i.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

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

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

Answer

(a) 8 ∫023x2 dx=[x3]02=8\int_0^23x^2\,dx=\left[x^3\right]_0^2=8

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The number which is not rational is (a) 1 (b) π\pi (c) 22/722/7 (d) −1-1

Answer

(b) π\pi π\pi is irrational (non-terminating, non-recurring decimal); 11, 22/722/7 and −1-1 are all ratios of integers.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If A and B are the nonempty sets A⊂B⊂UA\subset B\subset U then A∪B=A\cup B= (a) B (b) A (c) U (d) A+BA+B

Answer

(a) B Since A⊂BA\subset B, every element of AA is already in BB, so A∪B=BA\cup B=B.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If ∣x−1∣=5|x-1|=5 then x=x= (a) {−6,4}\{-6,4\} (b) [6,−4][6,-4] (c) {6,−4}\{6,-4\} (d) {−6,−4}\{-6,-4\}

Answer

(c) {6,−4}\{6,-4\} ∣x−1∣=5⇒x−1=±5⇒x=6|x-1|=5\Rightarrow x-1=\pm5\Rightarrow x=6 or x=−4x=-4.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If log⁡3x=2\log_3x=2 then x=x= (a) 2x2x (b) 2 (c) 9 (d) 8

Answer

(c) 9 log⁡3x=2⇒x=32=9\log_3x=2\Rightarrow x=3^2=9.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If f(x)=2x+5f(x)=2x+5 for x∈Rx\in R, then f−1(x)=f^{-1}(x)= (a) 12x−5\frac{1}{2x-5} (b) 1f(x)+5\frac{1}{f(x)+5} (c) 5y−2x\frac{5}{y-2x} (d) (x−5)/2(x-5)/2

Answer

(d) (x−5)/2(x-5)/2 Let y=2x+5y=2x+5. Solving for xx gives x=(y−5)/2x=(y-5)/2, so f−1(x)=(x−5)/2f^{-1}(x)=(x-5)/2.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If sin⁡−1x=cos⁡−1x\sin^{-1}x=\cos^{-1}x then x=x= (a) −2-\sqrt2 (b) −1/2-1/\sqrt2 (c) 1/21/\sqrt2 (d) 2\sqrt2

Answer

(c) 1/21/\sqrt2 sin⁡−1x=cos⁡−1x\sin^{-1}x=\cos^{-1}x and sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x+\cos^{-1}x=\pi/2, so sin⁡−1x=π/4\sin^{-1}x=\pi/4 and x=sin⁡(π/4)=1/2x=\sin(\pi/4)=1/\sqrt2.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The solution of cos⁡x=1/2\cos x=1/2 is (a) 2nπ±π/32n\pi\pm\pi/3 (b) nπ+π/3n\pi+\pi/3 (c) 2nπ±π/22n\pi\pm\pi/2 (d) nπ−π/3n\pi-\pi/3

Answer

(a) 2nπ±π/32n\pi\pm\pi/3 cos⁡x=1/2=cos⁡(π/3)\cos x=1/2=\cos(\pi/3), and cos⁡x=cos⁡α\cos x=\cos\alpha gives x=2nπ±αx=2n\pi\pm\alpha.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If p, q, r are in A.P. then (a) q=p+rq=p+r (b) p+r=2qp+r=2q (c) pr=q2pr=q^2 (d) p+q+r=0p+q+r=0

Answer

(b) p+r=2qp+r=2q In an A.P. the middle term is the mean of its neighbours: q−p=r−q⇒p+r=2qq-p=r-q\Rightarrow p+r=2q.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The fifth term of the sequence: 6, 12, 24, 48, ... is (a) 56 (b) 70 (c) 192 (d) 96

Answer

(d) 96 The sequence is a G.P. with ratio 2, so the fifth term is 48×2=9648\times2=96 (also 6⋅24=966\cdot2^4=96).

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If A=[aij]A=[a_{ij}], where aij=i+2ja_{ij}=i+2j, then 2×22\times2 matrix A is (a) [3654]\begin{bmatrix}3&6\\5&4\end{bmatrix} (b) [3546]\begin{bmatrix}3&5\\4&6\end{bmatrix} (c) [3656]\begin{bmatrix}3&6\\5&6\end{bmatrix} (d) [3465]\begin{bmatrix}3&4\\6&5\end{bmatrix}

Answer

(b) [3546]\begin{bmatrix}3&5\\4&6\end{bmatrix} With aij=i+2ja_{ij}=i+2j: a11=3, a12=5, a21=4, a22=6a_{11}=3,\ a_{12}=5,\ a_{21}=4,\ a_{22}=6.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The value of x for the determinant ∣x+5301∣=0\begin{vmatrix}x+5&3\\0&1\end{vmatrix}=0 is (a) 0 (b) 5 (c) -5 (d) 3

Answer

(c) −5-5 The determinant is (x+5)(1)−3(0)=x+5=0(x+5)(1)-3(0)=x+5=0, so x=−5x=-5.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

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

Answer

(d) no solution. Both equations have the same left side x+yx+y but different right sides (5≠75\ne7), so they are inconsistent (parallel lines).

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

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

Answer

(a) ω\omega Using 1+ω+ω2=01+\omega+\omega^2=0: 1+ω=−ω21+\omega=-\omega^2 and 1+ω2=−ω1+\omega^2=-\omega, so the product is ω(−ω2)(−ω)=ω⋅ω3=ω\omega(-\omega^2)(-\omega)=\omega\cdot\omega^3=\omega.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

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

Answer

(a) x2−5x+6=0x^2-5x+6=0 Sum of roots =5=5, product =6=6, so the equation is x2−5x+6=0x^2-5x+6=0.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

Distance between the lines 2x+2y=52x+2y=5 and x+y=10x+y=10 is (a) 15/215/2 (b) 15/(22)15/(2\sqrt2) (c) −15/(22)-15/(2\sqrt2) (d) 222\sqrt2

Answer

(b) 15/(22)15/(2\sqrt2) Write the first line as x+y=5/2x+y=5/2, i.e. x+y−5/2=0x+y-5/2=0, and the second as x+y−10=0x+y-10=0. Distance =∣10−5/2∣2=15/22=1522=\dfrac{|10-5/2|}{\sqrt2}=\dfrac{15/2}{\sqrt2}=\dfrac{15}{2\sqrt2}.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

lim⁡x→3x5−35x−3=\lim_{x\to3}\frac{x^5-3^5}{x-3}= (a) 0 (b) 504 (c) 405 (d) 81

Answer

(c) 405 lim⁡x→axn−anx−a=nan−1\lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}, so the limit is 5⋅34=4055\cdot3^4=405.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

lim⁡x→0xtan⁡x=\lim_{x\to0}\frac{x}{\tan x}= (a) undefined (b) 0 (c) 1 (d) 0/0

Answer

(c) 1 xtan⁡x=xsin⁡xcos⁡x→1⋅1=1\dfrac{x}{\tan x}=\dfrac{x}{\sin x}\cos x\to1\cdot1=1 as x→0x\to0.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The function f(x)={x2−5xx−5x≠5px=5f(x)=\begin{cases}\frac{x^2-5x}{x-5}&x\neq5\\p&x=5\end{cases} is continuous at x=5x=5, for p=p= (a) 0 (b) 5 (c) -5 (d) 1

Answer

(b) 5 For x≠5x\ne5, f(x)=x(x−5)x−5=xf(x)=\dfrac{x(x-5)}{x-5}=x, so lim⁡x→5f(x)=5\lim_{x\to5}f(x)=5. Continuity requires p=5p=5.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If y=sin⁡(ax2+b)y=\sin(ax^2+b) then dy/dx=dy/dx= (a) cos⁡(ax2+b)\cos(ax^2+b) (b) −cos⁡(ax2+b)-\cos(ax^2+b) (c) −2axcos⁡(ax2+b)-2ax\cos(ax^2+b) (d) 2axcos⁡(ax2+b)2ax\cos(ax^2+b)

Answer

(d) 2axcos⁡(ax2+b)2ax\cos(ax^2+b) By the chain rule, dydx=cos⁡(ax2+b)⋅2ax\dfrac{dy}{dx}=\cos(ax^2+b)\cdot2ax.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If x2+y2=25x^2+y^2=25 then dy/dx=dy/dx= (a) dydx=−xy\frac{dy}{dx}=-\frac xy (b) dydx=yx\frac{dy}{dx}=\frac yx (c) dydx=xy\frac{dy}{dx}=\frac xy (d) dydx=−yx\frac{dy}{dx}=-\frac yx

Answer

(a) dydx=−xy\dfrac{dy}{dx}=-\dfrac xy Differentiating, 2x+2ydydx=02x+2y\dfrac{dy}{dx}=0, so dydx=−xy\dfrac{dy}{dx}=-\dfrac xy.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

∫012xex2+1dx=\int_0^12xe^{x^2+1}dx= (a) e3e^3 (b) e2−ee^2-e (c) e4−1e^4-1 (d) 1

Answer

(b) e2−ee^2-e Put u=x2+1u=x^2+1, du=2x dxdu=2x\,dx; limits x=0→u=1x=0\to u=1, x=1→u=2x=1\to u=2. ∫12eu du=e2−e\int_1^2e^u\,du=e^2-e

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The critical point for f(x)=x3−6x2+9x−2f(x)=x^3-6x^2+9x-2 are (a) 1,−31,-3 (b) −1,3-1,3 (c) 1,31,3 (d) −1,−3-1,-3

Answer

(c) 1,31,3 f′(x)=3x2−12x+9=3(x−1)(x−3)=0⇒x=1,3f'(x)=3x^2-12x+9=3(x-1)(x-3)=0\Rightarrow x=1,3.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The letter of the word 'union' can be arranged in the number of ways: (a) 60 (b) 120 (c) 5 (d) 4

Answer

(b) 120 "Union" has 5 distinct letters, so the arrangements number 5!=1205!=120.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If 18Cr= 18Cr+2^{18}C_r=\,^{18}C_{r+2} then r=r= (a) 18 (b) 9 (c) 8 (d) 16

Answer

(c) 8 nCa=nCb^{n}C_a={}^{n}C_b with a≠ba\ne b implies a+b=na+b=n. So r+(r+2)=18⇒r=8r+(r+2)=18\Rightarrow r=8.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If the revenue function R=x−3x2R=x-3x^2 and cost function C=x2−xC=x^2-x, then the maximum profit is at x=x= (a) 1/4 (b) 4 (c) 2 (d) 3

Answer

(a) 1/41/4 Profit π=R−C=(x−3x2)−(x2−x)=2x−4x2\pi=R-C=(x-3x^2)-(x^2-x)=2x-4x^2. Then π′=2−8x=0⇒x=1/4\pi'=2-8x=0\Rightarrow x=1/4, and π′′=−8<0\pi''=-8<0, so this is a maximum.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The probability that a card drawn from a pack of 52 cards will be a diamond or a king is (a) 1/52 (b) 1/13 (c) 2/13 (d) 4/13

Answer

(d) 4/134/13 Diamonds =13=13, kings =4=4, king of diamonds counted twice, so favourable cases =13+4−1=16=13+4-1=16. Probability =16/52=4/13=16/52=4/13.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The median of the distribution: 5, 3, 2, 4, 10, 7, 12 is (a) 4 (b) 6 (c) 7 (d) 5

Answer

(d) 5 Arranged in order: 2, 3, 4, 5, 7, 10, 12. There are 7 values, so the median is the 4th value, 5.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The ratio of the prices of the first and second items is 3:5. If the price of the second item is Rs125, then the price of the first item is (a) 375 (b) 75 (c) 625 (d) 15

Answer

(b) 75 P1125=35⇒P1=75\dfrac{P_1}{125}=\dfrac35\Rightarrow P_1=75, i.e. Rs 75.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If a profit of 10 percent is made by selling an article for Rs 1320, then the cost price in Rupees (a) 1200 (b) 1100 (c) 1250 (d) 1150

Answer

(a) 1200 SP=1.10 CP⇒CP=1320/1.10=SP=1.10\,CP\Rightarrow CP=1320/1.10= Rs 1200.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

In an examination 25 percent of the examinee have been failed and 240 are passed, then the total number of examinees are (a) 300 (b) 320 (c) 265 (d) 350

Answer

(b) 320 Passed =75%=75\% of total, so total =240/0.75=320=240/0.75=320.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

A, B and C are in partnership. Towards capitals A contributed Rs 3600 and B Rs 3000. The profit amounted is Rs 3800 out of which C receives Rs 1600. The capital of C is Rs (a) 4800 (b) 4000 (c) 3000 (d) 2000

Answer

(a) 4800 A and B together receive 3800−1600=3800-1600= Rs 2200 on capitals 3600+3000=3600+3000= Rs 6600, i.e. Rs 1 profit per Rs 3 of capital. So C's capital =3×1600==3\times1600= Rs 4800.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If 1US dollar = Rs 110 then Rs 5000 will give US dollar (a) 46.5 (b) 44.5 (c) 45.45 (d) 50

Answer

(c) 45.45 5000/110=45.455000/110=45.45 US dollars.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If P, r, A are the principal, rate and amount of interest per annum respectively, then after n years the amount A will be (a) P(1+r/100)nP(1+r/100)^n (b) P(1−r/100)nP(1-r/100)^n (c) (P+r100)n(P+r100)^n (d) P(r+1/100)nP(r+1/100)^n

Answer

(a) P(1+r/100)nP(1+r/100)^n This is the compound amount formula, with rr the annual rate in percent.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

The amount of annuity of Rs 25 a year for 5 years at 4 percent per year is (a) 140.0 (b) 140.4 (c) 125.0 (d) 150

Answer

(a) 140.0 Taking each instalment of Rs 25 at the start of the year with simple interest at 4%, the amounts for 5, 4, 3, 2, 1 years are 25(1+0.04t)25(1+0.04t), so the total is 125+25(0.04)(15)=125+15=125+25(0.04)(15)=125+15= Rs 140.

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  • TU BIT Entrance Exam Model Questions Set 2 · 1 mark

If x+iy=(2+i3)2x+iy=(2+i3)^2 then (x,y)=(x,y)= (a) (5,−12)(5,-12) (b) (−5,−12)(-5,-12) (c) (5,12)(5,12) (d) (−5,12)(-5,12)

Answer

(d) (−5,12)(-5,12) (2+3i)2=4+12i+9i2=−5+12i(2+3i)^2=4+12i+9i^2=-5+12i, so (x,y)=(−5,12)(x,y)=(-5,12).

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