GSEB 2021 Mathematics Question Paper with Answer and Solution

(B) To determine if $f(x) = x^3$ is one-one and onto for $f: N \rightarrow N$:
$1$. One-one check: Let $f(x_1) = f(x_2)$. Then $x_1^3 = x_2^3$. Since $x_1, x_2 \in N$,this implies $x_1 = x_2$. Thus,the function is one-one.
$2$. Onto check: For a function to be onto,the range must equal the codomain $(N)$. For $x = 2 \in N$,$f(2) = 2^3 = 8$. However,for an element like $y = 2 \in N$ (codomain),there is no $x \in N$ such that $x^3 = 2$ (since $\sqrt[3]{2} \notin N$). Thus,the function is not onto.
Therefore,the function is one-one but not onto.

(D) The given relation is $S = \{(a, b) : a < b^2\}$.
$1$. For reflexivity: Is $(a, a) \in S$? That is,is $a < a^2$? If we take $a = 0.5$,then $0.5 < (0.5)^2 = 0.25$,which is false. Thus,it is not reflexive.
$2$. For symmetry: Does $(a, b) \in S \implies (b, a) \in S$? Let $(1, 2) \in S$ because $1 < 2^2 = 4$ is true. However,$(2, 1) \notin S$ because $2 < 1^2 = 1$ is false. Thus,it is not symmetric.
$3$. For transitivity: Does $(a, b) \in S$ and $(b, c) \in S \implies (a, c) \in S$? Let $(3, 2) \in S$ $(3 < 4)$ and $(2, 1.5) \in S$ $(2 < 2.25)$. However,$(3, 1.5) \notin S$ because $3 < (1.5)^2 = 2.25$ is false. Thus,it is not transitive.
Since the relation is neither reflexive,nor symmetric,nor transitive,it is not an equivalence relation.

(B) We know that $\sin^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$ because $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.
Substituting this into the expression,we get $\sin^{-1} \left[\cos \left(\frac{\pi}{3}\right)\right]$.
Since $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$,the expression becomes $\sin^{-1} \left(\frac{1}{2}\right)$.
We know that $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$,therefore $\sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Thus,the correct option is $B$.

(A) Given the system of equations for matrices $X$ and $Y$:
$(1)$ $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$
$(2)$ $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
To find $2X$,we add equation $(1)$ and equation $(2)$:
$(X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
$2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix}$
$2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$
Thus,the correct option is $A$.

(A) Given $A^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,which is a $3 \times 1$ matrix. Therefore,$A$ is a $1 \times 3$ matrix.
Given $B^{\prime} = \begin{bmatrix} 4 & 3 & 2 \end{bmatrix}$,which is a $1 \times 3$ matrix. Therefore,$B$ is a $3 \times 1$ matrix.
We need to find $(BA)^{\prime}$.
Using the property of transpose,$(BA)^{\prime} = A^{\prime} B^{\prime}$.
However,$A^{\prime}$ is $3 \times 1$ and $B^{\prime}$ is $1 \times 3$.
Thus,$A^{\prime} B^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 4 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 2 \\ 8 & 6 & 4 \\ 12 & 9 & 6 \end{bmatrix}$.
This is a $3 \times 3$ matrix,which is a square matrix.
Therefore,the correct option is $A$.

(D) The inverse of a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,where $|A| = ad - bc$ and $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$.
First,calculate the determinant $|A| = (2)(3) - (-2)(4) = 6 - (-8) = 6 + 8 = 14$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{14} \begin{bmatrix} 3 & 2 \\ -4 & 2 \end{bmatrix}$.
This matches option $D$.

(A) 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 $(k, 0), (4, 0)$ and $(0, 2)$ and Area $= 4$.
Substituting the values:
$4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$.
$4 = \frac{1}{2} |-2k + 8|$.
$8 = |-2k + 8|$.
This implies two cases:
Case $1$: $-2k + 8 = 8 \implies -2k = 0 \implies k = 0$.
Case $2$: $-2k + 8 = -8 \implies -2k = -16 \implies k = 8$.
Thus,the values of $k$ are $0$ and $8$.

(D) Let $\Delta = \left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$.
Applying the column operation $C_1 \to C_1 + C_2$:
$\Delta = \left|\begin{array}{ccc}x+y+z & y+z & z+x \\ z+x & x & y \\ 2 & 1 & 1\end{array}\right|$.
This does not immediately simplify to zero. Let us apply $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc}2(x+y+z) & y+z & z+x \\ x+y+z & x & y \\ 3 & 1 & 1\end{array}\right|$.
Alternatively,apply $C_1 \to C_1 + C_2$ and then observe the rows. Let us perform $C_1 \to C_1 + C_2$:
$\Delta = \left|\begin{array}{ccc}x+y+z & y+z & z+x \\ x+z & x & y \\ 2 & 1 & 1\end{array}\right|$.
Actually,applying $C_1 \to C_1 + C_2$ is not helpful. Let us apply $C_1 \to C_1 + C_2 + C_3$ is also not helpful. Let us apply $R_1 \to R_1 + R_2$:
$\Delta = \left|\begin{array}{ccc}x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$.
Now,take $(x+y+z)$ as a common factor from $R_1$:
$\Delta = (x+y+z) \left|\begin{array}{ccc}1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1\end{array}\right|$.
Since $R_1$ and $R_3$ are identical,the value of the determinant is $0$.

(C) Given $x = a(1 - \cos \theta)$ and $y = a(\theta + \sin \theta)$.
First,differentiate $x$ with respect to $\theta$:
$\frac{dx}{d\theta} = a(0 - (-\sin \theta)) = a \sin \theta$.
Next,differentiate $y$ with respect to $\theta$:
$\frac{dy}{d\theta} = a(1 + \cos \theta)$.
Now,find $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a(1 + \cos \theta)}{a \sin \theta} = \frac{1 + \cos \theta}{\sin \theta}$.
Using trigonometric identities $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$:
$\frac{dy}{dx} = \frac{2 \cos^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = \cot \frac{\theta}{2}$.
Thus,the correct option is $C$.

(A) We know that the trigonometric identity for $\sin 2x$ is given by:
$\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$
Substituting this into the expression,we get:
$\frac{d}{dx} (\sin 2x)$
Using the chain rule for differentiation,the derivative of $\sin(ax)$ is $a \cos(ax)$:
$\frac{d}{dx} (\sin 2x) = 2 \cos 2x$
Therefore,the correct option is $A$.

(C) For a function $f(x)$ to be continuous at $x = a$,the limit $\lim_{x \to a} f(x)$ must equal $f(a)$.
Given $f(x)$ is continuous at $x = \frac{\pi}{2}$,we have $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2}) = \frac{1}{2}$.
Now,calculate the limit: $\lim_{x \to \frac{\pi}{2}} \frac{k \cos x}{\pi - 2x}$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
Substituting this into the limit: $\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-k \sin h}{-2h} = \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h}$.
Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$,the limit is $\frac{k}{2}$.
Equating this to $f(\frac{\pi}{2}) = \frac{1}{2}$,we get $\frac{k}{2} = \frac{1}{2}$,which implies $k = 1$.

(A) To find the interval where the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing,we first find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
$-6 - 4x > 0$
$-4x > 6$
Dividing by $-4$ reverses the inequality sign:
$x < -\frac{6}{4}$
$x < -\frac{3}{2}$.
Thus,the function is strictly increasing in the interval $(-\infty, -\frac{3}{2})$.
Therefore,the correct option is $A$.

(D) To solve the integral $I = \int \sqrt{3-2x-x^2} \, dx$,first complete the square inside the square root:
$3-2x-x^2 = 4 - (x^2+2x+1) = 2^2 - (x+1)^2$.
Now,the integral becomes $I = \int \sqrt{2^2 - (x+1)^2} \, dx$.
Using the standard formula $\int \sqrt{a^2 - u^2} \, du = \frac{u}{2} \sqrt{a^2 - u^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{u}{a}\right) + C$,where $u = x+1$ and $a = 2$:
$I = \frac{x+1}{2} \sqrt{2^2 - (x+1)^2} + \frac{2^2}{2} \sin^{-1}\left(\frac{x+1}{2}\right) + C$.
Simplifying this,we get:
$I = \frac{1}{2}(x+1) \sqrt{3-2x-x^2} + 2 \sin^{-1}\left(\frac{x+1}{2}\right) + C$.
Thus,the correct option is $D$.

(D) We need to evaluate the integral $I = \int \log x^2 \, dx$.
Using the property of logarithms,$\log x^2 = 2 \log x$.
So,$I = \int 2 \log x \, dx = 2 \int \log x \, dx$.
Using integration by parts,$\int \log x \, dx = x \log x - x + C_1$.
Therefore,$I = 2(x \log x - x) + C = 2x \log x - 2x + C$.
We can rewrite this as $2x(\log x - 1) = 2x(\log x - \log e) = 2x \log \left(\frac{x}{e}\right) + C$.
Thus,the correct option is $D$.

(C) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}}$.
We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.
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 get:
$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.
$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(B) To evaluate the integral $I = \int_0^1 \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = \sin^{-1} x$ and $dv = dx$.
Then $du = \frac{1}{\sqrt{1 - x^2}} \, dx$ and $v = x$.
Applying the formula:
$I = [x \sin^{-1} x]_0^1 - \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx$.
Evaluating the first part: $[1 \cdot \sin^{-1}(1) - 0 \cdot \sin^{-1}(0)] = 1 \cdot \frac{\pi}{2} = \frac{\pi}{2}$.
For the second part,let $t = 1 - x^2$,so $dt = -2x \, dx$ or $x \, dx = -\frac{1}{2} dt$.
When $x=0, t=1$; when $x=1, t=0$.
$-\int_1^0 \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int_0^1 t^{-1/2} \, dt = -\frac{1}{2} [2\sqrt{t}]_0^1 = -\frac{1}{2} [2] = -1$.
Thus,$I = \frac{\pi}{2} - 1$.

(B) The given curve is $y^2 = 4x$,which implies $x = \frac{y^2}{4}$.
Since the region is bounded by the $Y$-axis $(x = 0)$,the curve $x = \frac{y^2}{4}$,and the line $y = 3$,the area $A$ is given by the integral with respect to $y$ from $y = 0$ to $y = 3$.
$A = \int_{0}^{3} x \, dy = \int_{0}^{3} \frac{y^2}{4} \, dy$
$A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$
$A = \frac{1}{4} \left( \frac{3^3}{3} - 0 \right) = \frac{1}{4} \times \frac{27}{3} = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The area $A$ is given by the integral of the absolute value of the function $y = \cos x$ from $x = -\frac{\pi}{2}$ to $x = \pi$.
$A = \int_{-\frac{\pi}{2}}^{\pi} |\cos x| \, dx$.
Since $\cos x \ge 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ and $\cos x \le 0$ for $x \in [\frac{\pi}{2}, \pi]$,we split the integral:
$A = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} (-\cos x) \, dx$.
Evaluating the first part:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2}) = 1 - (-1) = 2$.
Evaluating the second part:
$\int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = [-\sin x]_{\frac{\pi}{2}}^{\pi} = -(\sin \pi - \sin \frac{\pi}{2}) = -(0 - 1) = 1$.
Total area $A = 2 + 1 = 3$ sq. units.

(D) particular solution of a differential equation is a solution obtained by assigning specific values to the arbitrary constants in the general solution.
By definition,a particular solution does not contain any arbitrary constants.
Therefore,for a differential equation of any order,including a fourth-order differential equation,the number of arbitrary constants in its particular solution is $0$.

(C) Given the differential equation: $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$.
Divide both sides by $\tan x \tan y$:
$\frac{\sec^2 x}{\tan x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0$.
Integrating both sides:
$\int \frac{\sec^2 x}{\tan x} \, dx + \int \frac{\sec^2 y}{\tan y} \, dy = C_1$.
Let $u = \tan x$,then $du = \sec^2 x \, dx$.
Let $v = \tan y$,then $dv = \sec^2 y \, dy$.
Substituting these into the integral:
$\int \frac{1}{u} \, du + \int \frac{1}{v} \, dv = C_1$.
$\ln|u| + \ln|v| = C_1$.
$\ln|\tan x| + \ln|\tan y| = C_1$.
Using the property $\ln a + \ln b = \ln(ab)$:
$\ln|\tan x \tan y| = C_1$.
Taking the exponential of both sides:
$|\tan x \tan y| = e^{C_1}$.
Let $e^{C_1} = c$,so $\tan x \tan y = c$.

(D) The degree of a differential equation is defined only when it is a polynomial equation in terms of its derivatives.
In the given equation $\left(\frac{d^2 y}{d x^2}\right)^5+\left(\frac{d y}{d x}\right)^2+\cos \left(\frac{d y}{d x}\right)+1=0$,the term $\cos \left(\frac{d y}{d x}\right)$ involves a transcendental function of the derivative $\frac{d y}{d x}$.
Because this term cannot be expressed as a polynomial in terms of the derivatives,the degree of the differential equation is not defined.
Therefore,the correct option is $D$.

(B) Given that $\vec{a}$,$\vec{b}$,and $\vec{a}-\vec{b}$ are unit vectors.
So,$|\vec{a}| = 1$,$|\vec{b}| = 1$,and $|\vec{a}-\vec{b}| = 1$.
We know that $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Substituting the given values,we get $1^2 = 1^2 + 1^2 - 2|\vec{a}||\vec{b}| \cos \theta$.
$1 = 1 + 1 - 2(1)(1) \cos \theta$.
$1 = 2 - 2 \cos \theta$.
$2 \cos \theta = 1$.
$\cos \theta = \frac{1}{2}$.
Since $\cos \theta = \frac{1}{2}$,we have $\theta = \frac{\pi}{3}$.

(C) The projection of a vector $\vec{a}$ on a vector $\vec{b}$ is given by the formula: $\text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
Let $\vec{a} = \hat{i}-\hat{j}$ and $\vec{b} = \hat{i}+\hat{j}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 - 1 = 0$.
Since the dot product is $0$,the projection of $\vec{a}$ on $\vec{b}$ is $\frac{0}{|\vec{b}|} = 0$.
Therefore,the correct option is $C$.

(B) We are given that the magnitude of vector $\vec{a}$ is $|\vec{a}| = 1$ and the magnitude of vector $\vec{b}$ is $|\vec{b}| = 2$.
The dot product of the two vectors is given as $\vec{a} \cdot \vec{b} = 1$.
The formula for the dot product of two vectors is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$,where $\theta$ is the angle between the vectors.
Substituting the given values into the formula:
$1 = (1)(2) \cos \theta$
$1 = 2 \cos \theta$
$\cos \theta = \frac{1}{2}$
Since $\cos \theta = \frac{1}{2}$,the angle $\theta$ is $\frac{\pi}{3}$ (or $60^{\circ}$).

(A) First,calculate the sum of the vectors: $\vec{a} + \vec{b} = (2 \hat{i} - \hat{j} + 2 \hat{k}) + (-\hat{i} + \hat{j} - \hat{k}) = (2-1) \hat{i} + (-1+1) \hat{j} + (2-1) \hat{k} = \hat{i} + \hat{k}$.
Next,find the magnitude of the resultant vector: $|\vec{a} + \vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1+1} = \sqrt{2}$.
The unit vector in the direction of $\vec{a} + \vec{b}$ is $\hat{u} = \frac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$.
The required vector with magnitude $\sqrt{2}$ is $\sqrt{2} \times \hat{u} = \sqrt{2} \times \frac{\hat{i} + \hat{k}}{\sqrt{2}} = \hat{i} + \hat{k}$.
Thus,the correct option is $A$.

(C) Given the cross product of two vectors is zero,the vectors must be collinear.
So,$(2 \hat{i} + 6 \hat{j} + 27 \hat{k}) = k (\hat{i} + \lambda \hat{j} + \mu \hat{k})$ for some scalar $k$.
Comparing the components:
$2 = k \implies k = 2$
$6 = k \lambda \implies 6 = 2 \lambda \implies \lambda = 3$
$27 = k \mu \implies 27 = 2 \mu \implies \mu = \frac{27}{2}$
Therefore,$\lambda + \mu = 3 + \frac{27}{2} = \frac{6 + 27}{2} = \frac{33}{2}$.

(A) First,rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line: $\frac{-(x-1)}{3} = \frac{7(y-2)}{2p} = \frac{z-3}{2} \implies \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$.
The direction ratios are $\vec{v_1} = (-3, \frac{2p}{7}, 2)$.
For the second line: $\frac{-7(x-1)}{3p} = \frac{y-5}{1} = \frac{-(z-6)}{5} \implies \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$.
The direction ratios are $\vec{v_2} = (-\frac{3p}{7}, 1, -5)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.
$\frac{11p}{7} = 10$.
$p = \frac{70}{11}$.

(B) For the minimum of the objective function $Z = px + qy$ to occur at two distinct points $(3,0)$ and $(1,1)$,the value of $Z$ must be equal at these two points.
At $(3,0)$: $Z_1 = p(3) + q(0) = 3p$.
At $(1,1)$: $Z_2 = p(1) + q(1) = p + q$.
Equating $Z_1$ and $Z_2$:
$3p = p + q$
$2p = q$
$p = \frac{q}{2}$.
Thus,the condition for the minimum to occur at both points is $p = \frac{q}{2}$.

(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 any two events $A$ and $B$,the probability of their union is given by the formula:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since $A$ and $B$ are independent events,$P(A \cap B) = P(A) P(B)$.
Substituting this into the formula,we get:
$P(A \cup B) = P(A) + P(B) - P(A) P(B)$.
Alternatively,we can express this using the complements $A'$ and $B'$:
$P(A \cup B) = 1 - P((A \cup B)') = 1 - P(A' \cap B')$.
Since $A$ and $B$ are independent,$A'$ and $B'$ are also independent.
Therefore,$P(A' \cap B') = P(A') P(B')$.
Thus,$P(A \cup B) = 1 - P(A') P(B')$.
Hence,the correct option is $A$.

(A) Given that $2 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26}$.
Given that $P(B) = \frac{5}{13}$.
Given that $P(A \mid B) = \frac{2}{5}$.
We know that $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Therefore,$P(A \cap B) = P(A \mid B) \times P(B) = \frac{2}{5} \times \frac{5}{13} = \frac{2}{13}$.
We use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the values: $P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.
$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{11}{26}$.

(C) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.
We know the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{5} = \frac{1}{2} + P(B) - P(A) \cdot P(B)$.
$\frac{3}{5} = \frac{1}{2} + P(B) - \frac{1}{2} P(B)$.
$\frac{3}{5} - \frac{1}{2} = P(B) (1 - \frac{1}{2})$.
$\frac{6-5}{10} = P(B) (\frac{1}{2})$.
$\frac{1}{10} = P(B) \cdot \frac{1}{2}$.
$P(B) = \frac{2}{10} = 0.2$.