GUJCET 2022 Mathematics Question Paper with Answer and Solution

(D) function $f: A \to A$ is one-one if every element in the domain has a unique image in the codomain.
Since the set $A$ has $n = 4$ elements,the number of one-one functions from $A$ to $A$ is given by the number of permutations of $n$ elements,which is $n!$.
Here,$n = 4$,so the number of one-one functions is $4! = 4 \times 3 \times 2 \times 1 = 24$.
Thus,the correct option is $D$.

(A) Let $h(x) = (f+g)(x) = \sin x + \cos x = \sqrt{2} \sin(x + \frac{\pi}{4})$.
For $x \in [0, \frac{\pi}{2}]$,$x + \frac{\pi}{4} \in [\frac{\pi}{4}, \frac{3\pi}{4}]$.
In this interval,the sine function is not monotonic (it increases then decreases),so $f+g$ is not one-one.
Let $k(x) = (fg)(x) = \sin x \cos x = \frac{1}{2} \sin(2x)$.
For $x \in [0, \frac{\pi}{2}]$,$2x \in [0, \pi]$.
In this interval,the sine function is not monotonic (it increases then decreases),so $fg$ is not one-one.
Thus,both $f+g$ and $fg$ are not one-one.

(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$.
We know that $\cos \theta = 0$ implies $\theta = \frac{\pi}{2}$ (considering the principal value branch).
Therefore,$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$.
Rearranging the terms,we get $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$.
Using the identity $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$,we have $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$.
Thus,$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$.
Comparing both sides,we get $x = \frac{1}{5}$.

(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.
Let $x = \sin \theta$,where $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Then $2 \sin^{-1}(\sin \theta) = \sin^{-1}(2 \sin \theta \sqrt{1-\sin^2 \theta}) = \sin^{-1}(2 \sin \theta \cos \theta) = \sin^{-1}(\sin 2\theta)$.
For $2\theta = 2 \sin^{-1} x$ to be valid,$2\theta$ must lie in the principal value branch of $\sin^{-1}$,which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
So,$-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2} \implies -\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.
Taking sine on all sides,$\sin(-\frac{\pi}{4}) \le \sin \theta \le \sin(\frac{\pi}{4})$,which gives $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.
Thus,$x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

(C) Given that $\sin ^{-1} a = \alpha + \beta$ and $\sin ^{-1} b = \alpha - \beta$.
This implies $a = \sin(\alpha + \beta)$ and $b = \sin(\alpha - \beta)$.
Using the product-to-sum formula for sine:
$a b = \sin(\alpha + \beta) \sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$.
We need to find $\sin^2 \alpha + \cos^2 \beta$.
Since $\cos^2 \beta = 1 - \sin^2 \beta$,we substitute this into the expression:
$\sin^2 \alpha + \cos^2 \beta = \sin^2 \alpha + (1 - \sin^2 \beta) = 1 + (\sin^2 \alpha - \sin^2 \beta)$.
Substituting the value of $ab$ from the product formula:
$1 + (\sin^2 \alpha - \sin^2 \beta) = 1 + a b$.
Therefore,the correct option is $C$.

(C) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix}$.
Since $A$ is a diagonal matrix,its inverse $A^{-1}$ is also a diagonal matrix where the diagonal elements are the reciprocals of the diagonal elements of $A$.
Thus,$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{1} & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix}$.
The sum of all elements of $A^{-1}$ is $\frac{1}{2} + 1 + \frac{1}{3}$.
To add these,find the common denominator,which is $6$.
Sum $= \frac{3}{6} + \frac{6}{6} + \frac{2}{6} = \frac{11}{6}$.

(A) The transpose of a matrix $A$,denoted by $A'$,is obtained by interchanging its rows and columns.
Given $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
To find $A'$,we swap the elements of the first row with the first column and the second row with the second column.
Thus,$A' = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Comparing this with the given options,option $A$ is the correct answer.

(B) The given determinant is $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$.
Expanding the determinant: $a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0$.
This simplifies to $abc - a^3 - b^3 + abc + abc - c^3 = 0$,which is $3abc - (a^3 + b^3 + c^3) = 0$.
Rearranging gives $a^3 + b^3 + c^3 - 3abc = 0$.
Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$.
Since $a, b, c$ are sides of a triangle,$a+b+c \neq 0$.
Thus,$a^2 + b^2 + c^2 - ab - bc - ca = 0$,which implies $\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$.
This holds only if $a = b = c$.
Since $a = b = c$,the triangle is equilateral,so $A = B = C = 60^\circ$.
Therefore,$\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 60^\circ + \sin^2 60^\circ + \sin^2 60^\circ = (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} + \frac{3}{4} = \frac{9}{4}$.

(B) Given the determinant equation:
$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$
We can split the third column:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + \left|\begin{array}{ccc}x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3\end{array}\right|=0$
In the second determinant,take $x, y, z$ common from rows $1, 2, 3$ respectively:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| + xyz \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|=0$
For the first determinant,swap column $3$ with column $2$,then column $2$ with column $1$:
$\left|\begin{array}{ccc}x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1\end{array}\right| = -\left|\begin{array}{ccc}x & 1 & x^2 \\ y & 1 & y^2 \\ z & 1 & z^2\end{array}\right| = \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right|$
Substituting this back:
$(1 + xyz) \left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| = 0$
Since the second determinant is given as non-zero,we must have $1 + xyz = 0$,which implies $xyz = -1$.

(C) Given $B = A^{-1}$,we have $AB = I$,where $I$ is the identity matrix.
Given $10B = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$,so $B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}$.
Since $AB = I$,we have $A(10B) = 10I$.
$\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}$.
We need to find $\alpha$. Let us multiply the second row of $A$ with the third column of $10B$:
$(2)(2) + (1)(\alpha) + (-3)(3) = 0$.
$4 + \alpha - 9 = 0$.
$\alpha - 5 = 0$.
$\alpha = 5$.

(C) Given $x = \sqrt{10^{\sin^{-1} t}}$ and $y = \sqrt{10^{\cos^{-1} t}}$.
Multiply $x$ and $y$:
$xy = \sqrt{10^{\sin^{-1} t} \cdot 10^{\cos^{-1} t}} = \sqrt{10^{\sin^{-1} t + \cos^{-1} t}}$.
Since $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have:
$xy = \sqrt{10^{\pi/2}}$.
Differentiating both sides with respect to $x$:
$x \frac{dy}{dx} + y(1) = 0$.
$x \frac{dy}{dx} = -y$.
Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

(A) For $x \in [1.2, 1.9]$,the greatest integer function $[x]$ is constant because $1 \le x < 2$.
Specifically,for any $x$ in the interval $[1.2, 1.9]$,$[x] = 1$.
Since $f(x) = 1$ is a constant function on the interval $[1.2, 1.9]$,it is continuous and differentiable at every point in the interval.
The derivative of a constant function is zero,so $f'(x) = 0$ for all $x \in [1.2, 1.9]$.
Therefore,option $A$ is correct.

(D) Given the equation $y = 100 e^{2x} + 200 e^{-2x}$.
First,find the first derivative $\frac{dy}{dx}$:
$\frac{dy}{dx} = 100 \cdot 2 e^{2x} + 200 \cdot (-2) e^{-2x} = 200 e^{2x} - 400 e^{-2x}$.
Next,find the second derivative $\frac{d^2 y}{dx^2}$:
$\frac{d^2 y}{dx^2} = 200 \cdot 2 e^{2x} - 400 \cdot (-2) e^{-2x} = 400 e^{2x} + 800 e^{-2x}$.
Factor out $4$ from the expression:
$\frac{d^2 y}{dx^2} = 4(100 e^{2x} + 200 e^{-2x})$.
Since $y = 100 e^{2x} + 200 e^{-2x}$,we can substitute $y$ into the equation:
$\frac{d^2 y}{dx^2} = 4y$.
Comparing this with $\frac{d^2 y}{dx^2} = ay$,we get $a = 4$.

(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.
Using the product rule,$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$.
$f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x) e^{-x}$.
For the function to be increasing,we require $f'(x) > 0$.
Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.
$x(2 - x) > 0$ implies $x(x - 2) < 0$.
This inequality holds when $x$ is between the roots $0$ and $2$.
Thus,the function is increasing on the interval $(0, 2)$.

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x + 1) \, dx$.
Using the property $\int_{-a}^{a} f(x) \, dx = 0$ if $f(x)$ is an odd function,and $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ if $f(x)$ is an even function.
Let $f(x) = x^{13} + x \cos x + \tan^{15} x + 1$.
We can split the integral as $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^{13} + x \cos x + \tan^{15} x) \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx$.
Since $x^{13}$,$x \cos x$,and $\tan^{15} x$ are all odd functions,their integral over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is $0$.
Thus,$I = 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

(B) Since $f(x) = \sin^2 x$ is an even function,we can write:
$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^2 x \, dx = 2 \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx$
Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$:
$2 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx = \int_{0}^{\frac{\pi}{4}} (1 - \cos(2x)) \, dx$
Integrating term by term:
$[x - \frac{\sin(2x)}{2}]_{0}^{\frac{\pi}{4}}$
Substituting the limits:
$(\frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2}) - (0 - \frac{\sin(0)}{2})$
$= \frac{\pi}{4} - \frac{1}{2} - 0 = \frac{\pi}{4} - \frac{1}{2}$
Thus,the correct option is $B$.

(B) To evaluate the integral $\int \frac{x}{(x-1)(x-2)} dx$,we use partial fraction decomposition.
Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$.
Multiplying both sides by $(x-1)(x-2)$,we get $x = A(x-2) + B(x-1)$.
Setting $x = 1$,we get $1 = A(1-2) \implies A = -1$.
Setting $x = 2$,we get $2 = B(2-1) \implies B = 2$.
Thus,$\int \frac{x}{(x-1)(x-2)} dx = \int \left( \frac{-1}{x-1} + \frac{2}{x-2} \right) dx$.
$= -\log |x-1| + 2 \log |x-2| + C$.
$= \log |x-2|^2 - \log |x-1| + C$.
$= \log \left| \frac{(x-2)^2}{x-1} \right| + C$.
Therefore,the correct option is $B$.

(A) Let $I = \int (x+1)(x+3)(x+2)^7 \, dx$.
Substitute $u = x+2$,then $du = dx$.
Also,$x+1 = u-1$ and $x+3 = u+1$.
Substituting these into the integral:
$I = \int (u-1)(u+1)u^7 \, du$
$I = \int (u^2-1)u^7 \, du$
$I = \int (u^9 - u^7) \, du$
Integrating with respect to $u$:
$I = \frac{u^{10}}{10} - \frac{u^8}{8} + C$
Substituting back $u = x+2$:
$I = \frac{(x+2)^{10}}{10} - \frac{(x+2)^8}{8} + C$.
Thus,the correct option is $A$.

(A) Let $I = \int \sqrt{\frac{\cos x(1 - \cos^2 x)}{1 - \cos^3 x}} \, dx = \int \sqrt{\frac{\cos x \sin^2 x}{1 - \cos^3 x}} \, dx$.
Since $\sin x = \sqrt{1 - \cos^2 x}$,we have $I = \int \sin x \sqrt{\frac{\cos x}{1 - \cos^3 x}} \, dx$.
Let $u = \cos^{3/2} x$. Then $du = \frac{3}{2} \cos^{1/2} x (-\sin x) \, dx$,which implies $\sin x \sqrt{\cos x} \, dx = -\frac{2}{3} du$.
Also,$u^2 = \cos^3 x$.
Substituting these into the integral: $I = \int \sqrt{\frac{1}{1 - u^2}} \left(-\frac{2}{3} du\right) = -\frac{2}{3} \int \frac{1}{\sqrt{1 - u^2}} \, du$.
This evaluates to $-\frac{2}{3} \sin^{-1}(u) + C$.
Since $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$,we have $-\frac{2}{3} (\frac{\pi}{2} - \cos^{-1}(u)) = \frac{2}{3} \cos^{-1}(u) - \frac{\pi}{3}$.
Absorbing the constant $-\frac{\pi}{3}$ into $C$,we get $\frac{2}{3} \cos^{-1}(\cos^{3/2} x) + C$.

(A) Let $I = \int e^{\sin x} \sin 2x \, dx$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we get:
$I = \int e^{\sin x} (2 \sin x \cos x) \, dx$.
Let $t = \sin x$,then $dt = \cos x \, dx$.
Substituting these into the integral:
$I = \int e^t (2t) \, dt = 2 \int t e^t \, dt$.
Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$:
$I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + c$.
$I = 2 e^t (t - 1) + c$.
Substituting $t = \sin x$ back:
$I = 2 e^{\sin x} (\sin x - 1) + c$.
Thus,the correct option is $A$.

(C) We know that $x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$.
Therefore,the integral becomes $\int \frac{(x+1)(x^4 - x^3 + x^2 - x + 1)}{x+1} \, dx = \int (x^4 - x^3 + x^2 - x + 1) \, dx$.
Integrating term by term,we get $\frac{x^5}{5} - \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + x + c$.
This can be written in summation notation as $\sum_{n=1}^5 (-1)^{n+1} \cdot \frac{x^n}{n} + c$.
Comparing this with the given options,the correct option is $C$.

(C) The given curve is $y^2 = 4x$ and the line is $x = 3$.
Since the curve is symmetric about the $x$-axis,the total area $A$ is given by $2 \times \int_{0}^{3} y \, dx$.
From $y^2 = 4x$,we have $y = \sqrt{4x} = 2\sqrt{x}$.
Thus,$A = 2 \int_{0}^{3} 2\sqrt{x} \, dx = 4 \int_{0}^{3} x^{1/2} \, dx$.
Evaluating the integral: $A = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3} = 4 \times \frac{2}{3} \times [x^{3/2}]_{0}^{3}$.
$A = \frac{8}{3} \times (3)^{3/2} = \frac{8}{3} \times 3\sqrt{3} = 8\sqrt{3}$ sq. units.
Therefore,the correct option is $C$.

(B) The given differential equation is $\frac{dy}{dx} = e^{x-y}$.
We can rewrite this as $\frac{dy}{dx} = e^x \cdot e^{-y}$.
Separating the variables,we get $e^y \, dy = e^x \, dx$.
Integrating both sides,we have $\int e^y \, dy = \int e^x \, dx$.
This results in $e^y = e^x + C$,where $C$ is the constant of integration.

(B) To find the integrating factor,we first write the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.
Dividing the given equation $x \frac{dy}{dx} - y = x^2$ by $x$,we get:
$\frac{dy}{dx} - \frac{1}{x}y = x$.
Here,$P(x) = -\frac{1}{x}$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P(x) dx}$.
$IF = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = e^{\ln|x^{-1}|} = x^{-1} = \frac{1}{x}$.
Thus,the correct option is $B$.

(D) The length of the subnormal at any point $(x, y)$ on a curve is given by $|y \frac{dy}{dx}| = k$,where $k$ is a constant.
This can be written as $y \frac{dy}{dx} = \pm k$.
Integrating both sides with respect to $x$,we get $\int y \, dy = \pm \int k \, dx$.
This yields $\frac{y^2}{2} = \pm kx + C$.
If the curve passes through the origin,$C = 0$,so $y^2 = \pm 2kx$.
This equation represents a parabola.
Therefore,the correct option is $D$.

(B) The area of a triangle with vertices $A$,$B$,and $C$ is given by the formula $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Given vertices are $A(1, 1, 1)$,$B(1, 2, 3)$,and $C(2, 3, 1)$.
First,find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k} = \hat{j} + 2\hat{k}$.
$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k} = \hat{i} + 2\hat{j}$.
Now,calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - 1\hat{k}$.
The magnitude of the cross product is $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
Therefore,the area of the triangle is $\frac{1}{2} \sqrt{21}$ sq. units.

(C) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors,we have $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.
We are given the equation $\vec{a}+\vec{b}+\vec{c} = \vec{0}$.
Squaring both sides of the equation,we get $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = \vec{0} \cdot \vec{0}$.
Expanding the dot product,we get $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Substituting the magnitudes,we get $1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
This simplifies to $3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Therefore,$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$.
Thus,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -3/2$.

(C) To find the minimum value of the objective function $Z = 3x + 9y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0, 10)$: $Z = 3(0) + 9(10) = 0 + 90 = 90$
$2$. At $(5, 5)$: $Z = 3(5) + 9(5) = 15 + 45 = 60$
$3$. At $(15, 15)$: $Z = 3(15) + 9(15) = 45 + 135 = 180$
$4$. At $(0, 20)$: $Z = 3(0) + 9(20) = 0 + 180 = 180$
Comparing these values,the minimum value of $Z$ is $60$ at the point $(5, 5)$.

(A) To find the minimum value of the objective function $Z = 6x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $(2, 72)$: $Z = 6(2) + 3(72) = 12 + 216 = 228$
$2$. At point $(15, 20)$: $Z = 6(15) + 3(20) = 90 + 60 = 150$
$3$. At point $(40, 15)$: $Z = 6(40) + 3(15) = 240 + 45 = 285$
Comparing the values $228$,$150$,and $285$,the minimum value is $150$,which occurs at the point $(15, 20)$.

(A) For a Binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.
Here,$n = 10$ and $p = \frac{1}{2}$,so $1-p = \frac{1}{2}$.
Thus,$P(X = k) = \binom{10}{k} \left( \frac{1}{2} \right)^k \left( \frac{1}{2} \right)^{10-k} = \binom{10}{k} \left( \frac{1}{2} \right)^{10}$.
We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.
$P(X=0) = \binom{10}{0} \left( \frac{1}{2} \right)^{10} = 1 \times \left( \frac{1}{2} \right)^{10}$.
$P(X=1) = \binom{10}{1} \left( \frac{1}{2} \right)^{10} = 10 \times \left( \frac{1}{2} \right)^{10}$.
$P(X=2) = \binom{10}{2} \left( \frac{1}{2} \right)^{10} = \frac{10 \times 9}{2} \times \left( \frac{1}{2} \right)^{10} = 45 \times \left( \frac{1}{2} \right)^{10}$.
Adding these probabilities: $P(X \leq 2) = (1 + 10 + 45) \left( \frac{1}{2} \right)^{10} = 56 \left( \frac{1}{2} \right)^{10}$.
Comparing this with $m \left( \frac{1}{2} \right)^{10}$,we get $m = 56$.