Algebra

Given: $f(x)=\frac{2x+5}{3},\qquad g(x)=\frac{x+5}{2}$

First find $f^{-1}(x)$: $y=\frac{2x+5}{3}$ $3y=2x+5 \Rightarrow x=\frac{3y-5}{2}$ So, $f^{-1}(x)=\frac{3x-5}{2}$

Now, $g\circ g(x)=g(g(x))$ $g(x)=\frac{x+5}{2}$

Given, $g\circ g(x)=f^{-1}(x)$ $\frac{x+15}{4}=\frac{3x-5}{2}$ $x+15=2(3x-5)$ $x+15=6x-10$ $25=5x$ $x=5$

$\boxed{x=5}$

For a geometric series with first term $a$, common ratio $r$ ($r>1$) and $n$ terms,

$S_n=\frac{a(r^n-1)}{r-1}$

(When $r<1$, the equivalent form $S_n=\dfrac{a(1-r^n)}{1-r}$ is used.)

By the factor theorem, since $y-2$ is a factor of $p(y)$, we have $p(2)=0$.

$p(2)=2^3-2a(2)^2+a(2)-2$ $=8-8a+2a-2=6-6a$

So, $6-6a=0 \Rightarrow a=1$

$\boxed{a=1}$

Let the three numbers in arithmetic series be
$a-d,\ a,\ a+d$

Given: sum = 15
$(a-d)+a+(a+d)=15$ $3a=15 \Rightarrow a=5$

So the numbers are $5-d,\ 5,\ 5+d$.

Given: geometric mean between the first and third number is 4.
$\sqrt{(5-d)(5+d)}=4$ Squaring both sides, $(5-d)(5+d)=16$ $25-d^2=16$ $d^2=9$ $d=\pm 3$

Case 1: $d=3$
Numbers: $5-3,\ 5,\ 5+3 = 2,\ 5,\ 8$

Case 2: $d=-3$
Numbers: $5+3,\ 5,\ 5-3 = 8,\ 5,\ 2$

Hence, the required numbers are
$\boxed{2,\ 5,\ 8}$
(or $8,\ 5,\ 2$, which are the same numbers in reverse order).

Since $(x-2)$ is a factor of $f(x)$, by the factor theorem:

$f(2)=0$

$(2)^3 - 5(2)^2 + (k+1)(2) + 8 = 0$

$8 - 20 + 2k + 2 + 8 = 0$

$2k - 2 = 0$

$2k = 2$

$\therefore k = 1$

Let the three numbers in arithmetic series be
$a-d,\ a,\ a+d$

Given: sum = 18
$(a-d)+a+(a+d)=18$ $3a=18 \Rightarrow a=6$

So the numbers are $6-d,\ 6,\ 6+d$.

Geometric mean between the first and third numbers is $4\sqrt{2}$: $\sqrt{(6-d)(6+d)}=4\sqrt{2}$ $\sqrt{36-d^2}=4\sqrt{2}$

Squaring both sides: $36-d^2=32$ $d^2=4 \Rightarrow d=\pm 2$

Taking $d=2$, the numbers are: $6-2,\ 6,\ 6+2 = 4,\ 6,\ 8$

Taking $d=-2$, we get $8,\ 6,\ 4$, which is the same set.

$\boxed{\text{The required numbers are }4,\ 6,\ 8.}$

Let $p(x)=x^3-3x^2+kx+k$. Dividing by $x+2$, the remainder is $p(-2)$ by the remainder theorem.

$p(-2)=(-2)^3-3(-2)^2+k(-2)+k=-8-12-2k+k=-20-k$

Given the remainder is $k$,

$-20-k=k\Rightarrow 2k=-20$

$\boxed{k=-10}$

Let $f(x)=x^3+4x^2+kx-30$. Since $x-3$ is a factor, $f(3)=0$.

$27+36+3k-30=0$ $3k+33=0$ $k=-11$

Let the three numbers in arithmetic series be $a-d,\ a,\ a+d$.

Sum: $(a-d)+a+(a+d)=3a=30 \Rightarrow a=10$.

The GM between the first and third numbers is 8:

$\sqrt{(a-d)(a+d)}=8 \Rightarrow a^2-d^2=64 \Rightarrow 100-d^2=64 \Rightarrow d^2=36 \Rightarrow d=\pm6.$

For $d=6$ the numbers are $4,\ 10,\ 16$ (for $d=-6$ the same numbers in reverse order).

$\boxed{4,\ 10,\ 16}$

Check: $4+10+16=30$ and $\sqrt{4\times16}=8$.

Write the equations in standard form:

$4x - 3y = 11$ $3x + 7y = -1$

In matrix form,

Let

Determinant of $A$:

So inverse exists.

Now,

Hence,

1

The determinant of an identity matrix is always 1. For example, for the $2 \times 2$ identity matrix:

$\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$

Step 1: Boundary line. Take $2x+y=2$.

Plot $(0,2)$ and $(1,0)$ and join them with a solid line, because the inequality is $\le$ (points on the line are included).

Step 2: Region. Test the origin: $2(0)+0=0\le 2$ is true, so the origin lies in the solution set.

Shade the half-plane containing the origin, i.e. the region below and to the left of the line $2x+y=2$. This shaded half-plane together with the line represents $2x+y\le 2$.

For the system $ax-5y=3,\qquad 4x+3y=4$ the determinant is $D=\begin{vmatrix} a & -5 \\ 4 & 3 \end{vmatrix}=a(3)-(-5)(4)=3a+20$

Given $D=26$, $3a+20=26$ $3a=6$ $a=2$

So, $\boxed{a=2}$.

The line of symmetry of a parabola $y=ax^2+bx+c$ is

$x=-\frac{b}{2a}$

Here, $a=2,\ b=-4$, so

$x=-\frac{-4}{2(2)}=\frac{4}{4}=1$

Hence, the equation of the line of symmetry is

$\boxed{x=1}$

The feasible region is bounded by
$x-2y=1,\quad x+y=4,\quad y=0,\quad x=0.$

The vertices of the feasible region are found by intersecting the boundary lines:

• Intersection of $x-2y=1$ and $y=0$:
  $x=1 \Rightarrow (1,0)$
• Intersection of $x+y=4$ and $y=0$:
  $x=4 \Rightarrow (4,0)$
• Intersection of $x-2y=1$ and $x+y=4$:
  $x=1+2y,\quad 1+2y+y=4 \Rightarrow y=1,\ x=3 \Rightarrow (3,1)$

Now evaluate $P=3x+y$ at each vertex:

The maximum value is therefore
$\boxed{12}$
attained at $(x,y)=(4,0)$.

By the remainder theorem, when $f(x)$ is divided by $x-3$, the remainder is $f(3)$.

Given remainder = 4, so

$f(3)=4$

$(3)^3 + m(3)^2 - 3 + 7 = 4$

$27 + 9m - 3 + 7 = 4$

$31 + 9m = 4$

$9m = 4 - 31 = -27$

$m = -3$

$\boxed{m = -3}$

For $y=ax^2+bx+c$ the line of symmetry is $x=-\dfrac{b}{2a}$.

Here $a=1,\ b=-4$:

$x=-\frac{-4}{2(1)}=2$

$\boxed{x=2}$

Constraints: $x+y\le4,\ x+2y\le6,\ x\ge0,\ y\ge0$.

Boundary lines: $x+y=4$ (through $(4,0),(0,4)$) and $x+2y=6$ (through $(6,0),(0,3)$). Solving $x+y=4$ and $x+2y=6$ together gives $y=2,\ x=2$.

The vertices of the feasible region are $(0,0),\ (4,0),\ (2,2),\ (0,3)$.

The maximum value of $P$ is $\boxed{15}$, at $(0,3)$.

Draw the boundary line $3x-2y=6$.

• Put $y=0$: $x=2$, so the point is $(2,0)$.
• Put $x=0$: $y=-3$, so the point is $(0,-3)$.

Join these points with a solid line, because the sign is $\le$ (points on the line are included).

Test the origin: $3(0)-2(0)=0\le 6$ is true. So shade the side of the line that contains the origin. This shaded half-plane, together with the line, is the required graph of $3x-2y\le 6$.

For $3x+y=7$ and $x+2y=4$:

$D=\begin{vmatrix}3&1\\1&2\end{vmatrix}=6-1=5 \quad\text{(as given)}$

$D_x=\begin{vmatrix}7&1\\4&2\end{vmatrix}=14-4=10$

$x=\frac{D_x}{D}=\frac{10}{5}=2$

Constraints: $x-y\le0$ (that is $y\ge x$), $x+y\ge0$ (that is $y\ge -x$) and $y\le2$.

The boundary lines meet at:

• $y=x$ and $y=-x$: $(0,0)$
• $y=x$ and $y=2$: $(2,2)$
• $y=-x$ and $y=2$: $(-2,2)$

The feasible region is the triangle with vertices $(0,0)$, $(2,2)$ and $(-2,2)$.

Values of $Z=3x+2y+k$ at the corners:

The maximum is $10+k$ at $(2,2)$. Given that it is $15$:

$10+k=15\Rightarrow k=5$

This satisfies $k>0$. Hence $k=5$.

Let $p(x)=x^3-7x^2+11x-k$. Since the remainder on division by $(x-2)$ is $0$, by the remainder theorem $p(2)=0$:

$2^3-7(2)^2+11(2)-k=0$ $8-28+22-k=0 \Rightarrow 2-k=0$

$\boxed{k=2}$

Constraints: $2x+y\le20$, $2x+3y\le24$, $x\ge0$, $y\ge0$.

Corner points of the feasible region:

• $O(0,0)$
• On the $x$-axis with $2x+y=20$: $A(10,0)$
• On the $y$-axis with $2x+3y=24$: $B(0,8)$
• Intersection of $2x+y=20$ and $2x+3y=24$: subtracting gives $2y=4$, so $y=2$, $x=9$. $C(9,2)$

Values of $P=5x+3y$:

The maximum value of $P$ is $\boxed{51}$, at $x=9,\ y=2$.

A function $f: A \to A$ defined by $f(x) = x$ for all $x \in A$ is called an identity function.

For two positive numbers $a$ and $b$,

$\text{AM} \ge \text{GM}, \quad \text{i.e.} \quad \frac{a+b}{2} \ge \sqrt{ab}$

Equality holds only when $a = b$.

$\begin{pmatrix}1&0\\0&1\end{pmatrix}$

The given matrix is the identity matrix $I$. Since $I\cdot I=I$, the identity matrix is its own inverse.

Tables for plotting

For $y=x^2$:

For $y=2x+3$:

Plot these points on the same graph paper and join them. The parabola $y=x^2$ and the straight line $y=2x+3$ intersect at the points

So the $x$-coordinates of the intersection points give the required solution:

Check:

• For $x=-1$, $y=(-1)^2=1$ and $y=2(-1)+3=1$.
• For $x=3$, $y=3^2=9$ and $y=2(3)+3=9$.

Hence the graphical solution is $\boxed{x=-1,\ 3}$.

The given equations are:

$2x - 3y = 7$ $3x - 4y = 10$

In matrix form,

Let
$A=\begin{bmatrix}2&-3\\3&-4\end{bmatrix},\quad X=\begin{bmatrix}x\\y\end{bmatrix},\quad B=\begin{bmatrix}7\\10\end{bmatrix}$

Now,

$|A| = 2(-4) - (-3)(3) = -8 + 9 = 1 \neq 0$

So, $A^{-1}$ exists.

Now,

Hence,

$\boxed{x=2,\quad y=-1}$

Let the three consecutive terms of the GP be
$a,\ ar,\ ar^2$

Given, $a+ar+ar^2=56$ $a(1+r+r^2)=56 \quad (1)$

After subtracting $1,7,21$ respectively, the terms become $a-1,\quad ar-7,\quad ar^2-21$ These form an AP, so $2(ar-7)=(a-1)+(ar^2-21)$

Simplifying, $2ar-14=a+ar^2-22$ $a(r^2-2r+1)=8$ $a(r-1)^2=8 \quad (2)$

Dividing (1) by (2), $\frac{1+r+r^2}{(r-1)^2}=\frac{56}{8}=7$ $1+r+r^2=7(r^2-2r+1)$ $1+r+r^2=7r^2-14r+7$ $6r^2-15r+6=0$ $2r^2-5r+2=0$ $(2r-1)(r-2)=0$

So, $r=2 \quad \text{or} \quad r=\frac12$

Case 1: $r=2$
From (1), $a(1+2+4)=56$ $7a=56\Rightarrow a=8$ The numbers are $8,\ 16,\ 32$

Check: subtracting $1,7,21$ gives $7,\ 9,\ 11$ which is an AP.

Case 2: $r=\frac12$
From (1), $a\left(1+\frac12+\frac14\right)=56$ $a\cdot \frac74=56\Rightarrow a=32$ The numbers are $32,\ 16,\ 8$

Check: subtracting $1,7,21$ gives $31,\ 9,\ -13$ which is also an AP.

Therefore, the required numbers are
$\boxed{8,\ 16,\ 32 \quad \text{or} \quad 32,\ 16,\ 8}$

A composite function is a function formed by combining two functions so that the output of one becomes the input of the other. If $f$ and $g$ are two functions, their composite is written

$(f\circ g)(x)=f(g(x)),$

defined for those $x$ in the domain of $g$ for which $g(x)$ lies in the domain of $f$.

The last term of the arithmetic series.

In $S_n=\frac{n}{2}(a+l)$, $a$ is the first term and $l$ is the last term (the $n$th term) of the arithmetic series.

$B$ is the inverse of $A$ (and $A$ is the inverse of $B$).

Since $AB=I$, the product of the two matrices gives the identity matrix, so $B=A^{-1}$ and $A=B^{-1}$.

Let $f(x)=2x+3$.

Let $y=f(x)$, so
$y=2x+3$

Interchanging $x$ and $y$ for the inverse: $x=2y+3$ $2y=x-3$ $y=\frac{x-3}{2}$

So, $f^{-1}(x)=\frac{x-3}{2}$

Now, $f^{-1}(3)=\frac{3-3}{2}=0$

Answer: $f^{-1}(3)=0$