GUJCET 2023 Mathematics Question Paper with Answer and Solution

(B) Let the set be $A = \{\pi, \pi^2, \pi^3\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(a, a) \in R$ for all $a \in A$. Here,$(\pi, \pi) \in R$,$(\pi^2, \pi^2) \in R$,and $(\pi^3, \pi^3) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(a, b) \in R$,then $(b, a) \in R$. We have $(\pi, \pi^2) \in R$,but $(\pi^2, \pi) \notin R$. Thus,$R$ is not symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(a, b) \in R$ and $(b, c) \in R$,then $(a, c) \in R$. We have $(\pi, \pi^2) \in R$ and $(\pi^2, \pi^3) \in R$. For transitivity,$(\pi, \pi^3)$ must be in $R$. However,$(\pi, \pi^3) \notin R$. Thus,$R$ is not transitive.
Conclusion: $R$ is reflexive but neither symmetric nor transitive.

(B) Given the equation: $\cos \left(\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x\right)=1$
Taking $\cos ^{-1}$ on both sides:
$\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x = \cos ^{-1}(1)$
We know that $\cos ^{-1}(1) = 0$ and $\cos ^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}$.
Substituting these values:
$\frac{\pi}{6} + \sin ^{-1} x = 0$
$\sin ^{-1} x = -\frac{\pi}{6}$
$x = \sin \left(-\frac{\pi}{6}\right)$
Since $\sin(-\theta) = -\sin(\theta)$,we have:
$x = -\sin \left(\frac{\pi}{6}\right) = -\frac{1}{2}$
Thus,the value of $x$ is $-\frac{1}{2}$.

(C) We know that $\sec^{-1}(2) = \frac{\pi}{3}$ because $\sec(\frac{\pi}{3}) = 2$.
We know that $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$ because $\cot(\frac{\pi}{6}) = \sqrt{3}$.
We know that $\operatorname{cosec}^{-1}(\frac{2}{\sqrt{3}}) = \frac{\pi}{3}$ because $\operatorname{cosec}(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}$.
Substituting these values into the expression:
$\cos(\frac{\pi}{3}) + \tan(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = \frac{1}{2} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}$.
Combining the terms:
$\frac{1}{2} + \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}} = \frac{1+\sqrt{3}}{2} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}(1+\sqrt{3}) + 2}{2\sqrt{3}} = \frac{\sqrt{3} + 3 + 2}{2\sqrt{3}} = \frac{5+\sqrt{3}}{2\sqrt{3}}$.
Thus,the correct option is $C$.

(D) We need to evaluate the sum $S = \sum_{i=1}^3 \cot ^{-1} i = \cot ^{-1} 1 + \cot ^{-1} 2 + \cot ^{-1} 3$.
Using the identity $\cot ^{-1} x = \tan ^{-1} \frac{1}{x}$ for $x > 0$,we have $S = \tan ^{-1} 1 + \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{3}$.
Since $\tan ^{-1} 1 = \frac{\pi}{4}$,we calculate $\tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{3} = \tan ^{-1} \left( \frac{1/2 + 1/3}{1 - (1/2)(1/3)} \right) = \tan ^{-1} \left( \frac{5/6}{5/6} \right) = \tan ^{-1} 1 = \frac{\pi}{4}$.
Thus,$S = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.
Now,we substitute this into the expression: $\cos ^{-1} \{ \cot (S) \} = \cos ^{-1} \{ \cot (\frac{\pi}{2}) \}$.
Since $\cot (\frac{\pi}{2}) = 0$,the expression becomes $\cos ^{-1} (0)$.
We know that $\cos ^{-1} (0) = \frac{\pi}{2}$.
Therefore,the correct option is $D$.

(A) Given $A = \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix}$.
We need to calculate $A^2 = A \times A$.
$A^2 = \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -5 \\ 0 & -5 & 0 \\ -5 & 0 & 0 \end{bmatrix}$.
Performing matrix multiplication:
Row $1$: $(0)(0) + (0)(0) + (-5)(-5) = 25$,$(0)(0) + (0)(-5) + (-5)(0) = 0$,$(0)(-5) + (0)(0) + (-5)(0) = 0$.
Row $2$: $(0)(0) + (-5)(0) + (0)(-5) = 0$,$(0)(0) + (-5)(-5) + (0)(0) = 25$,$(0)(-5) + (-5)(0) + (0)(0) = 0$.
Row $3$: $(-5)(0) + (0)(0) + (0)(-5) = 0$,$(-5)(0) + (0)(-5) + (0)(0) = 0$,$(-5)(-5) + (0)(0) + (0)(0) = 25$.
Thus,$A^2 = \begin{bmatrix} 25 & 0 & 0 \\ 0 & 25 & 0 \\ 0 & 0 & 25 \end{bmatrix} = 25 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 25 I$.
Therefore,the correct option is $A$.

(D) Given $A = [2]$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$.
First,calculate the product $BA$:
$BA = \begin{bmatrix} 3 \\ 4 \end{bmatrix} [2] = \begin{bmatrix} 3 \times 2 \\ 4 \times 2 \end{bmatrix} = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$.
Now,find the transpose $(BA)'$:
$(BA)' = \begin{bmatrix} 6 & 8 \end{bmatrix}$.
Note: Based on the provided options,there seems to be a typo in the original question's matrix $A$. If $A = [2]$,the result is $[6 \quad 8]$. If the question intended $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and $A = [1 \quad 2]$,then $BA = \begin{bmatrix} 3 \\ 4 \end{bmatrix} [1 \quad 2] = \begin{bmatrix} 3 & 6 \\ 4 & 8 \end{bmatrix}$.
Then $(BA)' = \begin{bmatrix} 3 & 4 \\ 6 & 8 \end{bmatrix}$,which matches option $D$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula:
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Given vertices are $A(k, 1)$,$B(2, 4)$,and $C(1, 1)$,and the area is $6$ sq. units.
Substituting the values:
$6 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$.
$12 = |3k + 0 - 3|$.
$12 = |3k - 3|$.
This gives two cases:
Case $1$: $3k - 3 = 12 \implies 3k = 15 \implies k = 5$.
Case $2$: $3k - 3 = -12 \implies 3k = -9 \implies k = -3$.
Thus,the values of $k$ are $5$ and $-3$.

(C) The given determinant is $D = \sin \frac{11 \pi}{36} \cos \frac{2 \pi}{9} - \cos \frac{11 \pi}{36} \sin \frac{2 \pi}{9}$.
Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we set $A = \frac{11 \pi}{36}$ and $B = \frac{2 \pi}{9}$.
First,convert $B$ to a common denominator: $B = \frac{2 \pi}{9} = \frac{8 \pi}{36}$.
Now,$D = \sin \left( \frac{11 \pi}{36} - \frac{8 \pi}{36} \right) = \sin \left( \frac{3 \pi}{36} \right) = \sin \left( \frac{\pi}{12} \right)$.
Since $\sin \theta = \cos \left( \frac{\pi}{2} - \theta \right)$,we have $\sin \left( \frac{\pi}{12} \right) = \cos \left( \frac{\pi}{2} - \frac{\pi}{12} \right) = \cos \left( \frac{6 \pi - \pi}{12} \right) = \cos \frac{5 \pi}{12}$.
Thus,the correct option is $C$.

(B) Given the equation: $y = \sqrt{\sin^{-1} x + y}$
Squaring both sides,we get: $y^2 = \sin^{-1} x + y$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin^{-1} x + y)$
$2y \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{dy}{dx}$
Rearranging the terms to isolate $\frac{dy}{dx}$:
$2y \frac{dy}{dx} - \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$
$(2y - 1) \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$
Therefore,$\frac{dy}{dx} = \frac{1}{(2y - 1) \sqrt{1 - x^2}}$
Thus,the correct option is $B$.

(A) Given $x = a \cos \theta$ and $y = a \sin \theta$.
First,find $\frac{dx}{d\theta} = -a \sin \theta$ and $\frac{dy}{d\theta} = a \cos \theta$.
Then,$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta$.
Now,differentiate $\frac{dy}{dx}$ with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{d}{dx}(-\cot \theta) = \frac{d}{d\theta}(-\cot \theta) \cdot \frac{d\theta}{dx}$.
Since $\frac{d}{d\theta}(-\cot \theta) = \operatorname{cosec}^2 \theta$ and $\frac{d\theta}{dx} = \frac{1}{dx/d\theta} = \frac{1}{-a \sin \theta}$.
Therefore,$\frac{d^2 y}{dx^2} = \operatorname{cosec}^2 \theta \cdot \left( -\frac{1}{a \sin \theta} \right) = -\frac{1}{a} \operatorname{cosec}^3 \theta$.

(C) Let $f(x) = x^x + x^{x+1} + x^{x+2}$.
We know that $\frac{d}{dx}(x^x) = x^x(1 + \ln x)$.
For $x^{x+1}$,let $y = x^{x+1}$,then $\ln y = (x+1)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+1}{x} = \ln x + 1 + \frac{1}{x}$. So,$\frac{d}{dx}(x^{x+1}) = x^{x+1}(\ln x + 1 + \frac{1}{x})$.
For $x^{x+2}$,let $y = x^{x+2}$,then $\ln y = (x+2)\ln x$. Differentiating both sides: $\frac{1}{y} \frac{dy}{dx} = \ln x + \frac{x+2}{x} = \ln x + 1 + \frac{2}{x}$. So,$\frac{d}{dx}(x^{x+2}) = x^{x+2}(\ln x + 1 + \frac{2}{x})$.
Now,evaluate at $x=e$:
$\frac{d}{dx}(x^x)|_{x=e} = e^e(1 + \ln e) = e^e(1+1) = 2e^e$.
$\frac{d}{dx}(x^{x+1})|_{x=e} = e^{e+1}(\ln e + 1 + \frac{1}{e}) = e^{e+1}(2 + \frac{1}{e}) = 2e^{e+1} + e^e = e^e(2e+1)$.
$\frac{d}{dx}(x^{x+2})|_{x=e} = e^{e+2}(\ln e + 1 + \frac{2}{e}) = e^{e+2}(2 + \frac{2}{e}) = 2e^{e+2} + 2e^{e+1} = e^e(2e^2+2e)$.
Summing these: $e^e(2 + 2e + 1 + 2e^2 + 2e) = e^e(2e^2 + 4e + 3)$.

(C) To determine which function is decreasing in the interval $I = \left(0, \frac{\pi}{8}\right)$,we examine the derivative of each function:
$A) f(x) = \tan 4x \implies f'(x) = 4 \sec^2 4x$. Since $\sec^2 4x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.
$B) f(x) = \sin x \implies f'(x) = \cos x$. Since $\cos x > 0$ for all $x \in \left(0, \frac{\pi}{8}\right)$,$f'(x) > 0$,so the function is increasing.
$C) f(x) = \cos 4x \implies f'(x) = -4 \sin 4x$. For $x \in \left(0, \frac{\pi}{8}\right)$,$4x \in (0, \frac{\pi}{2})$. In this interval,$\sin 4x > 0$,so $f'(x) = -4 \sin 4x < 0$. Thus,the function is decreasing.
$D) f(x) = -\cos x \implies f'(x) = \sin x$. Since $\sin x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.
Therefore,the correct option is $C$.

(D) Let the radius of the sphere be $r$ and its diameter be $D = 2r$.
The volume $V$ of the sphere is given by $V = \frac{4}{3} \pi r^3$.
We need to find the rate of change of volume with respect to the diameter,which is $\frac{dV}{dD}$.
Since $D = 2r$,we have $r = \frac{D}{2}$.
Substituting $r$ in the volume formula: $V = \frac{4}{3} \pi (\frac{D}{2})^3 = \frac{4}{3} \pi (\frac{D^3}{8}) = \frac{1}{6} \pi D^3$.
Now,differentiate $V$ with respect to $D$: $\frac{dV}{dD} = \frac{d}{dD} (\frac{1}{6} \pi D^3) = \frac{1}{6} \pi (3D^2) = \frac{1}{2} \pi D^2$.
Substituting $D = 2r$ back into the expression: $\frac{dV}{dD} = \frac{1}{2} \pi (2r)^2 = \frac{1}{2} \pi (4r^2) = 2 \pi r^2$.
Thus,the correct option is $D$.

(C) We have the integral $I = \int_0^1 (0.001)^{\frac{x}{3}} e^x \, dx$.
First,simplify the term $(0.001)^{\frac{x}{3}}$.
Since $0.001 = 10^{-3}$,we have $(10^{-3})^{\frac{x}{3}} = 10^{-x} = \frac{1}{10^x}$.
Thus,the integral becomes $I = \int_0^1 \frac{e^x}{10^x} \, dx = \int_0^1 \left(\frac{e}{10}\right)^x \, dx$.
Using the formula $\int a^x \, dx = \frac{a^x}{\ln a} + C$,we get:
$I = \left[ \frac{(\frac{e}{10})^x}{\ln(\frac{e}{10})} \right]_0^1 = \frac{\frac{e}{10} - 1}{\ln e - \ln 10} = \frac{\frac{e-10}{10}}{1 - \ln 10}$.
Therefore,$I = \frac{e-10}{10(1-\ln 10)}$.
This matches option $C$.

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx$.
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we have:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\cot^4 x} dx$.
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\tan^4 x}{1+\tan^4 x} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1+\tan^4 x}{1+\tan^4 x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(A) To solve the integral $\int x^{2019} \cdot e^{x^{2020}} \, dx$,we use the method of substitution.
Let $u = x^{2020}$.
Then,differentiating both sides with respect to $x$,we get $\frac{du}{dx} = 2020 x^{2019}$.
This implies $du = 2020 x^{2019} \, dx$,or $x^{2019} \, dx = \frac{1}{2020} \, du$.
Substituting these into the original integral:
$\int e^u \cdot \frac{1}{2020} \, du = \frac{1}{2020} \int e^u \, du$.
The integral of $e^u$ is $e^u$,so we have $\frac{1}{2020} e^u + C$.
Finally,substituting $u = x^{2020}$ back,we get $\frac{1}{2020} e^{x^{2020}} + C$.
Therefore,the correct option is $A$.

(B) Let $I = \int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} dx$.
Since $\tan x = \frac{\sin x}{\cos x}$ and $\frac{1}{\cos x} = \sec x$,the integral becomes:
$I = \int \frac{\sec x \tan x}{(\sec x-1)(\sec x-2)} dx$.
Let $u = \sec x$,then $du = \sec x \tan x dx$.
The integral transforms to:
$I = \int \frac{1}{(u-1)(u-2)} du$.
Using partial fractions:
$\frac{1}{(u-1)(u-2)} = \frac{A}{u-1} + \frac{B}{u-2}$.
$1 = A(u-2) + B(u-1)$.
For $u=1$,$A = -1$. For $u=2$,$B = 1$.
So,$I = \int \left( \frac{1}{u-2} - \frac{1}{u-1} \right) du$.
$I = \log |u-2| - \log |u-1| + c = \log \left| \frac{u-2}{u-1} \right| + c$.
Substituting $u = \sec x$,we get:
$I = \log \left| \frac{\sec x - 2}{\sec x - 1} \right| + c$.

(B) Let $I = \int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx$.
Given that $\int \left\{ \cos^{-1} x - \frac{1}{\sqrt{1-x^2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$.
Let $f(x) = \cos^{-1} x$. Then $f'(x) = -\frac{1}{\sqrt{1-x^2}}$.
The integral becomes $\int k \left\{ f(x) + f'(x) \right\} dx = k \cdot \cos^{-1} x + c$.
We know that $\int e^x \{ f(x) + f'(x) \} dx = e^x f(x) + c$.
Comparing this with the given equation $\int k \{ f(x) + f'(x) \} dx = k \cdot f(x) + c$,we can see that $k$ must be a function of $x$ such that $k = e^x$.
Thus,$k = e^x$.

(C) The area $A$ is given by the integral of $|y|$ with respect to $x$ from $x = 2$ to $x = 5$.
$A = \int_{2}^{5} |3 - x| \, dx$.
Since the line $y = 3 - x$ crosses the $X$-axis at $x = 3$,we split the integral:
$A = \int_{2}^{3} (3 - x) \, dx + \int_{3}^{5} -(3 - x) \, dx$.
$A = \left[ 3x - \frac{x^2}{2} \right]_{2}^{3} + \left[ \frac{x^2}{2} - 3x \right]_{3}^{5}$.
$A = \left( (9 - 4.5) - (6 - 2) \right) + \left( (12.5 - 15) - (4.5 - 9) \right)$.
$A = (4.5 - 4) + (-2.5 + 4.5) = 0.5 + 2 = 2.5 = \frac{5}{2}$ square units.
Thus,the correct option is $C$.

(C) The given equation of the ellipse is $9x^2 + 4y^2 = 1$.
This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/2)^2} = 1$.
Here,$a = \frac{1}{3}$ and $b = \frac{1}{2}$.
The area of the entire ellipse is given by the formula $A = \pi ab$.
$A = \pi \times \frac{1}{3} \times \frac{1}{2} = \frac{\pi}{6}$.
Since the ellipse is symmetric about both axes,the area in the first quadrant is one-fourth of the total area.
Area in the first quadrant $= \frac{1}{4} \times \frac{\pi}{6} = \frac{\pi}{24}$.
Thus,the correct option is $C$.

(B) Given the differential equation $\frac{dy}{dx} = -4xy^2$.
Separating the variables,we get $\frac{dy}{y^2} = -4x \, dx$.
Integrating both sides,$\int y^{-2} \, dy = \int -4x \, dx$.
This gives $-\frac{1}{y} = -2x^2 + C$,or $\frac{1}{y} = 2x^2 - C$.
Using the initial condition $x = 0, y = 1$:
$\frac{1}{1} = 2(0)^2 - C \implies 1 = -C \implies C = -1$.
Substituting $C = -1$ into the equation $\frac{1}{y} = 2x^2 - C$,we get $\frac{1}{y} = 2x^2 + 1$.
Therefore,$y = \frac{1}{2x^2 + 1}$.

(A) The given differential equation is of the form $\frac{dy}{dx} + Py = Q$,where $P = \tan x$ and $Q = \sec x$.
The integrating factor $(IF)$ is given by the formula $IF = e^{\int P dx}$.
Substituting $P = \tan x$,we get $IF = e^{\int \tan x dx}$.
Since $\int \tan x dx = \ln|\sec x|$,we have $IF = e^{\ln|\sec x|}$.
Using the property $e^{\ln f(x)} = f(x)$,we get $IF = \sec x$.
Therefore,the correct option is $A$.

(D) Given the differential equation: $\left(\frac{d^3 y}{d x^3}\right)^{5/4} = \left(\frac{d^2 y}{d x^2}\right)^{4/3}$.
To eliminate the fractional exponents,raise both sides to the power of $12$ (the least common multiple of $4$ and $3$):
$\left(\left(\frac{d^3 y}{d x^3}\right)^{5/4}\right)^{12} = \left(\left(\frac{d^2 y}{d x^2}\right)^{4/3}\right)^{12}$.
This simplifies to: $\left(\frac{d^3 y}{d x^3}\right)^{15} = \left(\frac{d^2 y}{d x^2}\right)^{16}$.
The highest order derivative present is $\frac{d^3 y}{d x^3}$,so the order is $3$.
The degree is the power of the highest order derivative after the equation is made free from radicals and fractions,which is $15$.
Thus,the order is $3$ and the degree is $15$.

(A) We are given that $|\vec{a}|=3$,$|\vec{b}|=\frac{\sqrt{2}}{3}$,and $\vec{a} \times \vec{b}$ is a unit vector,which means $|\vec{a} \times \vec{b}|=1$.
We know the formula for the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$,where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Substituting the given values: $1 = 3 \times \frac{\sqrt{2}}{3} \times \sin \theta$.
This simplifies to $1 = \sqrt{2} \sin \theta$.
Therefore,$\sin \theta = \frac{1}{\sqrt{2}}$.
Since $\sin \theta = \frac{1}{\sqrt{2}}$,the angle $\theta = \frac{\pi}{4}$.

(C) Given vectors are $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.
We know that $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$.
First,calculate the magnitude squared of $\vec{a}$:
$|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 1 + 1 + 1 = 3$.
Next,calculate the magnitude squared of $\vec{b}$:
$|\vec{b}|^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$.
Now,substitute these values into the formula:
$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 3 - 14 = -11$.
Therefore,the correct option is $C$.

(B) Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$.
Then,$\vec{a} \times \hat{i} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \times \hat{i} = -a_2\hat{k} + a_3\hat{j}$.
So,$|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Since $|\vec{a}|^2 = a_1^2 + a_2^2 + a_3^2$,the expression equals $2|\vec{a}|^2$.

(B) The Cartesian equation of a line passing through a point $(x_1, y_1, z_1)$ and parallel to a vector $a \hat{i} + b \hat{j} + c \hat{k}$ is given by the formula:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
Here,the point is $(x_1, y_1, z_1) = (5, -2, 4)$ and the direction vector is $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$,so $(a, b, c) = (3, 2, -8)$.
Substituting these values into the formula,we get:
$\frac{x - 5}{3} = \frac{y - (-2)}{2} = \frac{z - 4}{-8}$
Simplifying this,we obtain:
$\frac{x - 5}{3} = \frac{y + 2}{2} = \frac{z - 4}{-8}$
Thus,option $B$ is the correct answer.

(D) The given constraints are:
$1) x + y \geq 8$
$2) x + y \leq 5$
$3) x \geq 0, y \geq 0$
Observe the first two inequalities: $x + y \geq 8$ and $x + y \leq 5$.
These two inequalities represent regions that do not overlap.
If $x + y$ is greater than or equal to $8$,it cannot simultaneously be less than or equal to $5$.
Therefore,there is no set of points $(x, y)$ that satisfies all the given constraints simultaneously.
Since there is no common region,the feasible region is empty (null set).
Thus,the objective function cannot be minimized as no feasible solution exists.

(A) To find the maximum value of the objective function $Z = 10x + 20y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At $(0, 10): Z = 10(0) + 20(10) = 0 + 200 = 200$
$2$. At $(5, 5): Z = 10(5) + 20(5) = 50 + 100 = 150$
$3$. At $(15, 15): Z = 10(15) + 20(15) = 150 + 300 = 450$
$4$. At $(0, 20): Z = 10(0) + 20(20) = 0 + 400 = 400$
Comparing these values,the maximum value is $450$ at the point $(15, 15)$.

(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 $(15, 20)$.

(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.
Since the coin is tossed $n = 5$ times,we use the binomial distribution formula: $P(X = k) = \binom{n}{k} p^k q^{n-k}$.
We want the probability of getting exactly $k = 3$ heads.
Substituting the values: $P(X = 3) = \binom{5}{3} (\frac{1}{2})^3 (\frac{1}{2})^{5-3}$.
Calculating the binomial coefficient: $\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10$.
Thus,$P(X = 3) = 10 \times (\frac{1}{2})^3 \times (\frac{1}{2})^2 = 10 \times (\frac{1}{2})^5$.
$P(X = 3) = 10 \times \frac{1}{32} = \frac{10}{32} = \frac{5}{16}$.
Therefore,the correct option is $B$.