GSEB 2020 Mathematics Question Paper with Answer and Solution

(B) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
An even prime number is a number that is both even and prime. The only even prime number is $2$.
For each die,the probability of obtaining the number $2$ is $\frac{1}{6}$.
Since the rolls are independent,the probability of obtaining a $2$ on both dice is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Therefore,the correct option is $B$.

(A) The condition $|x - y| < 1$ for integers $x, y \in Z$ implies that $|x - y| = 0$,because the absolute difference between two integers must be a non-negative integer.
Thus,$|x - y| = 0 \implies x = y$.
Therefore,$S = \{(x, x) : x \in Z\}$,which is the identity relation on $Z$.
$1$. Reflexive: For any $x \in Z$,$|x - x| = 0 < 1$,so $(x, x) \in S$. Thus,$S$ is reflexive.
$2$. Symmetric: If $(x, y) \in S$,then $x = y$,which implies $y = x$,so $(y, x) \in S$. Thus,$S$ is symmetric.
$3$. Transitive: If $(x, y) \in S$ and $(y, z) \in S$,then $x = y$ and $y = z$,which implies $x = z$,so $(x, z) \in S$. Thus,$S$ is transitive.
Since $S$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Note: The provided option $B$ is incorrect based on the definition of the identity relation. The correct classification is an equivalence relation.

(C) function $f: A \to A$ is one-one if every element in the domain has a unique image in the codomain.
For a set with $n$ elements,the number of one-one functions from the set to itself is given by the number of permutations of $n$ elements,which is $n!$.
Here,the set is $A = \{1, 2, 3, 4, 5\}$,so $n = 5$.
The number of one-one functions is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Therefore,the correct option is $C$.

(A) To determine the nature of the function $f: N \times N \rightarrow N$ defined by $f(m, n) = mn$:
$1$. Check for one-one:
Consider $f(1, 4) = 1 \times 4 = 4$ and $f(2, 2) = 2 \times 2 = 4$.
Since $f(1, 4) = f(2, 2)$ but $(1, 4) \neq (2, 2)$,the function is not one-one. Therefore,it is many-one.
$2$. Check for onto:
For $f$ to be onto,for every $n \in N$,there must exist $(m, k) \in N \times N$ such that $f(m, k) = mk = n$.
However,consider the prime number $7 \in N$. The only factors of $7$ are $1$ and $7$. Thus,$f(1, 7) = 7$ and $f(7, 1) = 7$. This works for primes.
Wait,let us check if every $n \in N$ has a pre-image. For any $n \in N$,we can always choose $(1, n) \in N \times N$ such that $f(1, n) = 1 \times n = n$.
Since every element $n$ in the codomain $N$ has at least one pre-image $(1, n)$ in the domain $N \times N$,the function is onto.
Conclusion: The function is many-one and onto. The correct option is $A$.

(A) Let $\theta = \operatorname{cosec}^{-1} \frac{5}{3} + \tan^{-1} \frac{2}{3}$.
Since $\operatorname{cosec}^{-1} \frac{5}{3} = \sin^{-1} \frac{3}{5}$,we have $\theta = \sin^{-1} \frac{3}{5} + \tan^{-1} \frac{2}{3}$.
Let $\alpha = \sin^{-1} \frac{3}{5}$,then $\sin \alpha = \frac{3}{5}$,so $\tan \alpha = \frac{3}{4}$.
Thus,$\theta = \tan^{-1} \frac{3}{4} + \tan^{-1} \frac{2}{3}$.
Using the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$:
$\theta = \tan^{-1} \left( \frac{3/4 + 2/3}{1 - (3/4)(2/3)} \right) = \tan^{-1} \left( \frac{9/12 + 8/12}{1 - 6/12} \right) = \tan^{-1} \left( \frac{17/12}{6/12} \right) = \tan^{-1} \frac{17}{6}$.
The expression is $\cot \left( \frac{2019 \pi}{2} - \theta \right)$.
Since $\frac{2019 \pi}{2} = 1009 \pi + \frac{\pi}{2}$,we have $\cot \left( 1009 \pi + \frac{\pi}{2} - \theta \right) = \cot \left( \frac{\pi}{2} - \theta \right) = \tan \theta$.
Since $\theta = \tan^{-1} \frac{17}{6}$,$\tan \theta = \frac{17}{6}$.

(A) We know the standard identity for inverse trigonometric functions: $\sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}$ for $|x| \geq 1$.
Similarly,for $|x| \geq 1$,we have $\cos^{-1}(x^{-1}) + \sin^{-1}(x^{-1}) = \frac{\pi}{2}$.
Substituting these identities into the given expression:
$\sec^{-1} x + \operatorname{cosec}^{-1} x + \cos^{-1}(x^{-1}) + \sin^{-1}(x^{-1}) = \frac{\pi}{2} + \frac{\pi}{2} = \pi$.
Thus,the correct option is $A$.

(C) The function $f(x) = \tan^{-1} x$ is defined for all real numbers $x \in R$.
Its range is the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
The graph passes through the origin $(0, 0)$ because $\tan^{-1}(0) = 0$.
As $x \to \infty$,$f(x) \to \frac{\pi}{2}$,and as $x \to -\infty$,$f(x) \to -\frac{\pi}{2}$.
The graph is strictly increasing and has horizontal asymptotes at $y = \frac{\pi}{2}$ and $y = -\frac{\pi}{2}$.
Comparing this with the given options,the graph in option $C$ correctly represents the function $f(x) = \tan^{-1} x$.

(D) We need to evaluate the sum $S = \sum_{i=0}^2 \cot^{-1}(-(i+1))$.
Expanding the sum,we get:
$S = \cot^{-1}(-1) + \cot^{-1}(-2) + \cot^{-1}(-3)$.
Using the property $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$:
$S = (\pi - \cot^{-1}(1)) + (\pi - \cot^{-1}(2)) + (\pi - \cot^{-1}(3))$
$S = 3\pi - (\cot^{-1}(1) + \cot^{-1}(2) + \cot^{-1}(3))$.
We know $\cot^{-1}(1) = \frac{\pi}{4}$.
For $\cot^{-1}(2) + \cot^{-1}(3)$,we use $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}(\frac{x+y}{1-xy})$ where $xy < 1$:
$\cot^{-1}(2) + \cot^{-1}(3) = \tan^{-1}(\frac{1}{2}) + \tan^{-1}(\frac{1}{3}) = \tan^{-1}(\frac{1/2 + 1/3}{1 - 1/6}) = \tan^{-1}(\frac{5/6}{5/6}) = \tan^{-1}(1) = \frac{\pi}{4}$.
Thus,$S = 3\pi - (\frac{\pi}{4} + \frac{\pi}{4}) = 3\pi - \frac{\pi}{2} = \frac{5\pi}{2}$.

(D) By definition,if two matrices $X$ and $Y$ are inverses of each other,then their product must result in the identity matrix $I$.
Therefore,$XY = I$ and $YX = I$.
Thus,$XY = YX = I$.

(A) Given $A = \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 0 \end{bmatrix}$
$A^2 = \begin{bmatrix} (0)(0)+(0)(0)+(3)(3) & (0)(0)+(0)(3)+(3)(0) & (0)(3)+(0)(0)+(3)(0) \\ (0)(0)+(3)(0)+(0)(3) & (0)(0)+(3)(3)+(0)(0) & (0)(3)+(3)(0)+(0)(0) \\ (3)(0)+(0)(0)+(0)(3) & (3)(0)+(0)(3)+(0)(0) & (3)(3)+(0)(0)+(0)(0) \end{bmatrix}$
$A^2 = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix} = 9 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 9I_3$.
Thus,the correct statement is $A^2 = 9I_3$.

(A) Given $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} (3)(3) + (1)(-1) & (3)(1) + (1)(2) \\ (-1)(3) + (2)(-1) & (-1)(1) + (2)(2) \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$.
Next,calculate $5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}$.
Now,find $A^2 - 5A$:
$A^2 - 5A = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = \begin{bmatrix} 8-15 & 5-5 \\ -5-(-5) & 3-10 \end{bmatrix} = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}$.
This can be written as $-7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -7I$.
Comparing this with $kI$,we get $k = -7$.

(B) First,we calculate the determinant of matrix $A$:
$|A| = 1(1 \times 1 - 2 \times 1) - 3(2 \times 1 - 2 \times 5) + 4(2 \times 1 - 1 \times 5)$
$|A| = 1(1 - 2) - 3(2 - 10) + 4(2 - 5)$
$|A| = 1(-1) - 3(-8) + 4(-3)$
$|A| = -1 + 24 - 12 = 11$
We know the property $|\text{adj } A| = |A|^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $|\text{adj } A| = |A|^{3-1} = |A|^2$.
$|\text{adj } A| = (11)^2 = 121$.

(B) Let the determinant be $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.
The element $2020$ is at the position $a_{12}$ (first row,second column).
The minor $M_{12}$ is the determinant of the $2 \times 2$ matrix obtained by deleting the first row and second column:
$M_{12} = \begin{vmatrix} 2022 & 2024 \\ 2025 & 2027 \end{vmatrix} = (2022 \times 2027) - (2024 \times 2025)$.
Using the property $(x-a)(x+a) = x^2 - a^2$ or direct calculation:
$M_{12} = (2024.5 - 2.5)(2024.5 + 2.5) - (2024.5 - 0.5)(2024.5 + 0.5) = (2024.5^2 - 6.25) - (2024.5^2 - 0.25) = -6.25 + 0.25 = -6$.
Alternatively,$M_{12} = (2022 \times 2027) - (2024 \times 2025) = 4098594 - 4098600 = -6$.
The co-factor $C_{12} = (-1)^{1+2} M_{12} = -1 \times (-6) = 6$.
The sum of the minor and the co-factor is $M_{12} + C_{12} = -6 + 6 = 0$.

(A) To evaluate the determinant $\Delta = \begin{vmatrix} x+y+z^2 & x^2+y+z & x+y^2+z \\ z^2 & x^2 & y^2 \\ x+y & y+z & x+z \end{vmatrix}$,we perform row operations.
Apply $R_1 \to R_1 - R_2 - R_3$:
The first row becomes:
$(x+y+z^2) - z^2 - (x+y) = 0$
$(x^2+y+z) - x^2 - (y+z) = 0$
$(x+y^2+z) - y^2 - (x+z) = 0$
Since all elements of the first row are $0$,the value of the determinant is $\Delta = 0$.

(D) First,evaluate the trigonometric functions in the first row:
$\cos 3\pi = -1$
$\sin 5\pi = 0$
$\tan 7\pi = 0$
Now,substitute these values into the determinant:
$\Delta = \left|\begin{array}{ccc}-1 & 0 & 0 \\ \sqrt{3} & 1 & 0 \\ \sqrt{5} & 0 & 1\end{array}\right|$
Expanding along the first row:
$\Delta = -1 \times \left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right| - 0 + 0$
$\Delta = -1 \times (1 \times 1 - 0 \times 0) = -1 \times 1 = -1$
Thus,the correct option is $D$.

(D) Given $y = (x+3)^2 \cdot (x+4)^3 \cdot (x+5)^4$.
Taking the natural logarithm on both sides:
$\ln(y) = 2 \ln(x+3) + 3 \ln(x+4) + 4 \ln(x+5)$.
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5}$.
Thus,$\frac{dy}{dx} = y \left( \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right)$.
We can express the sum as $\sum_{i=2}^4 \frac{i}{x+i+1}$.
For $i=2$: $\frac{2}{x+2+1} = \frac{2}{x+3}$.
For $i=3$: $\frac{3}{x+3+1} = \frac{3}{x+4}$.
For $i=4$: $\frac{4}{x+4+1} = \frac{4}{x+5}$.
Therefore,the derivative is $y \sum_{i=2}^4 \left( \frac{i}{x+i+1} \right)$,which matches option $D$.

(C) Given $y = \log_e(\log_e x)$.
First,find the first derivative $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{x \log_e x} = (x \log_e x)^{-1}$.
Now,find the second derivative $\frac{d^2 y}{dx^2}$ using the power rule and product rule:
$\frac{d^2 y}{dx^2} = -1(x \log_e x)^{-2} \cdot \frac{d}{dx}(x \log_e x)$.
$\frac{d^2 y}{dx^2} = -\frac{1}{(x \log_e x)^2} \cdot [1 \cdot \log_e x + x \cdot \frac{1}{x}] = -\frac{\log_e x + 1}{x^2 (\log_e x)^2}$.
Since $\log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$,we have:
$\frac{d^2 y}{dx^2} = -\frac{\log_e(ex)}{x^2 (\log_e x)^2}$.

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit,right-hand limit,and the value of the function at $x = a$ must be equal.
Specifically,$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$.
Left-hand limit at $x = 3$: $\lim_{x \to 3^-} (cx + 1) = 3c + 1$.
Right-hand limit at $x = 3$: $\lim_{x \to 3^+} (dx + 3) = 3d + 3$.
Since $f$ is continuous at $x = 3$,we have $3c + 1 = 3d + 3$.
Rearranging the terms: $3c - 3d = 3 - 1$.
$3(c - d) = 2$.
$c - d = \frac{2}{3}$.
We need to find $d - c$,so $d - c = -(c - d) = -\frac{2}{3}$.
Therefore,the correct option is $C$.

(A) To find the local maximum value of $f(x) = x + \frac{1}{x}$,we first find the derivative $f'(x) = 1 - \frac{1}{x^2}$.
Setting $f'(x) = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.
We use the second derivative test: $f''(x) = \frac{2}{x^3}$.
For $x = 1$,$f''(1) = 2 > 0$,so $x = 1$ is a local minimum.
For $x = -1$,$f''(-1) = \frac{2}{(-1)^3} = -2 < 0$,so $x = -1$ is a local maximum.
The local maximum value is $f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$.

(D) For $x \in \left(-\frac{\pi}{2}, 0\right)$,we have $\sin x < 0$.
Therefore,$f(x) = |\sin x| = -\sin x$.
To determine the nature of the function,we find its derivative: $f'(x) = \frac{d}{dx}(-\sin x) = -\cos x$.
In the interval $\left(-\frac{\pi}{2}, 0\right)$,$\cos x$ is positive (since it is in the fourth quadrant).
Thus,$f'(x) = -\cos x < 0$ for all $x \in \left(-\frac{\pi}{2}, 0\right)$.
Since the derivative is strictly negative,the function $f(x)$ is a strictly decreasing function on the given interval.

(A) To solve the integral $I = \int \frac{1}{x^2 \sqrt{1-x^2}} \cdot dx$,we use the substitution $x = \sin \theta$.
Then,$dx = \cos \theta \cdot d\theta$.
Substituting these into the integral:
$I = \int \frac{\cos \theta \cdot d\theta}{\sin^2 \theta \sqrt{1-\sin^2 \theta}} = \int \frac{\cos \theta \cdot d\theta}{\sin^2 \theta \cdot \cos \theta} = \int \frac{1}{\sin^2 \theta} \cdot d\theta = \int \csc^2 \theta \cdot d\theta$.
The integral of $\csc^2 \theta$ is $-\cot \theta + C$.
Since $x = \sin \theta$,we have $\sin \theta = x$.
Then $\cos \theta = \sqrt{1-x^2}$,so $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{1-x^2}}{x}$.
Thus,$I = -\frac{\sqrt{1-x^2}}{x} + C$.
Therefore,the correct option is $A$.

(B) Let $I = \int x^2 \sqrt{8-x^6} \, dx$.
Substitute $u = x^3$,then $du = 3x^2 \, dx$,which implies $x^2 \, dx = \frac{1}{3} du$.
The integral becomes $I = \frac{1}{3} \int \sqrt{8-u^2} \, du$.
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 $a^2 = 8$ (so $a = 2\sqrt{2}$):
$I = \frac{1}{3} \left[ \frac{u}{2} \sqrt{8-u^2} + \frac{8}{2} \sin^{-1} \left(\frac{u}{2\sqrt{2}}\right) \right] + C$.
Substituting $u = x^3$ back:
$I = \frac{1}{3} \left[ \frac{x^3}{2} \sqrt{8-x^6} + 4 \sin^{-1} \left(\frac{x^3}{2\sqrt{2}}\right) \right] + C$.
Thus,the correct option is $B$.

(D) Let $I = \int e^x \cos 2x \, dx$.
Using the integration by parts formula $\int u \cdot v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$,let $u = \cos 2x$ and $v = e^x$.
$I = \cos 2x \cdot e^x - \int (-2 \sin 2x) \cdot e^x \, dx = e^x \cos 2x + 2 \int e^x \sin 2x \, dx$.
Applying integration by parts again to $\int e^x \sin 2x \, dx$ with $u = \sin 2x$ and $v = e^x$:
$\int e^x \sin 2x \, dx = \sin 2x \cdot e^x - \int (2 \cos 2x) \cdot e^x \, dx = e^x \sin 2x - 2I$.
Substituting this back into the equation for $I$:
$I = e^x \cos 2x + 2(e^x \sin 2x - 2I) = e^x \cos 2x + 2e^x \sin 2x - 4I$.
$5I = e^x(\cos 2x + 2 \sin 2x)$.
$I = \frac{e^x}{5}(\cos 2x + 2 \sin 2x) + C$.
Thus,the correct option is $D$.

(D) Given the differential equation: $y dx - (x + 2y^2) dy = 0$.
Rearranging the terms,we get: $y dx - x dy = 2y^2 dy$.
Dividing both sides by $y^2$ (where $y \neq 0$): $\frac{y dx - x dy}{y^2} = 2 dy$.
Recognizing the derivative of the quotient: $d\left(\frac{x}{y}\right) = 2 dy$.
Integrating both sides: $\int d\left(\frac{x}{y}\right) = \int 2 dy$.
This implies $\frac{x}{y} = 2y + C$.
Alternatively,rewriting the equation as $\frac{dx}{dy} - \frac{x}{y} = 2y$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$.
The integrating factor $(IF)$ is given by $e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = \frac{1}{y}$.
Thus,the correct option is $D$.

(A) The given differential equation is $\left(y^{\prime \prime \prime}\right)^3+\left(y^{\prime \prime}\right)^4+\left(y^{\prime}\right)^4+y=7$.
The order of a differential equation is the order of the highest derivative present in the equation.
Here,the highest derivative is $y^{\prime \prime \prime}$,which is of order $3$.
Thus,the order of the differential equation is $3$.
The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives.
The highest order derivative is $y^{\prime \prime \prime}$ and its exponent is $3$.
Therefore,the degree of the differential equation is $3$.
Hence,the order and degree are $3$ and $3$ respectively.

(A) The slope of the tangent to the curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2y}{x}$.
Separating the variables,we get $\frac{dy}{y} = \frac{2dx}{x}$.
Integrating both sides,we have $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$,which gives $\ln|y| = 2 \ln|x| + C$.
This simplifies to $\ln|y| = \ln|x^2| + C$,or $y = kx^2$,where $k = e^C$.
Since the curve passes through $(3, -4)$,we substitute these values: $-4 = k(3)^2$,which implies $-4 = 9k$,so $k = -\frac{4}{9}$.
Thus,the equation of the curve is $y = -\frac{4}{9}x^2$,which can be rewritten as $4x^2 + 9y = 0$.

(B) Given the differential equation: $x \frac{dy}{dx} - y = 0$
Rearranging the terms,we get: $x \frac{dy}{dx} = y$
Separating the variables: $\frac{dy}{y} = \frac{dx}{x}$
Integrating both sides: $\int \frac{dy}{y} = \int \frac{dx}{x}$
This gives: $\ln|y| = \ln|x| + \ln|c|$
Using the property of logarithms: $\ln|y| = \ln|cx|$
Taking the exponential of both sides: $y = cx$
Thus,the correct option is $B$.

(C) The given differential equation is $\frac{dy}{dx} = \frac{x+y}{1+x^2}$.
We can rewrite this as $\frac{dy}{dx} = \frac{x}{1+x^2} + \frac{y}{1+x^2}$.
Rearranging the terms,we get $\frac{dy}{dx} - \left(\frac{1}{1+x^2}\right)y = \frac{x}{1+x^2}$.
This is in the standard form of a linear differential equation,$\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{1+x^2}$ and $Q(x) = \frac{x}{1+x^2}$.
Therefore,the correct option is $C$.

(B) Given the differential equation: $\frac{dy}{dx} = y \tan x$.
Separating the variables,we get: $\frac{dy}{y} = \tan x \, dx$.
Integrating both sides: $\int \frac{dy}{y} = \int \tan x \, dx$.
This gives: $\ln |y| = \ln |\sec x| + C$.
Exponentiating both sides: $|y| = e^{\ln |\sec x| + C} = e^C \cdot |\sec x|$.
Let $e^C = k$,so $y = k \sec x$.
Using the initial condition $y(0) = 1$: $1 = k \sec(0) \implies 1 = k(1) \implies k = 1$.
Substituting $k = 1$ back into the equation,we get the particular solution: $y = \sec x$.

(C) Let $\vec{u} = (a, 2)$ and $\vec{v} = (a, -2)$.
Given that the angle $\theta$ between $\vec{u}$ and $\vec{v}$ is $\frac{\pi}{3}$.
The dot product formula is $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos(\theta)$.
$\vec{u} \cdot \vec{v} = (a)(a) + (2)(-2) = a^2 - 4$.
$|\vec{u}| = \sqrt{a^2 + 2^2} = \sqrt{a^2 + 4}$.
$|\vec{v}| = \sqrt{a^2 + (-2)^2} = \sqrt{a^2 + 4}$.
Substituting these into the formula: $a^2 - 4 = \sqrt{a^2 + 4} \cdot \sqrt{a^2 + 4} \cdot \cos(\frac{\pi}{3})$.
$a^2 - 4 = (a^2 + 4) \cdot \frac{1}{2}$.
$2(a^2 - 4) = a^2 + 4$.
$2a^2 - 8 = a^2 + 4$.
$a^2 = 12$.
$a = \pm \sqrt{12} = \pm 2 \sqrt{3}$.

(D) The volume of a parallelepiped with coterminous edges $\vec{a}, \vec{b}, \text{ and } \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$.
This is equivalent to the absolute value of the determinant of the matrix formed by the components of the vectors:
$V = |\det \begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 1 \\ -1 & 1 & -1 \end{bmatrix}|$
Expanding the determinant along the first row:
$V = |2((-1)(-1) - (1)(1)) - 1((3)(-1) - (1)(-1)) + 1((3)(1) - (-1)(-1))|$
$V = |2(1 - 1) - 1(-3 + 1) + 1(3 - 1)|$
$V = |2(0) - 1(-2) + 1(2)|$
$V = |0 + 2 + 2| = |4| = 4$
Thus,the volume is $4$ cubic units.

(B) The projection of vector $\vec{a}$ on vector $\vec{b}$ is given by the formula: $\text{Projection} = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|}$.
Given $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$.
First,calculate the dot product $\vec{a} \cdot \vec{b} = (-1)(1) + (2)(2) + (-1)(2) = -1 + 4 - 2 = 1$.
Next,calculate the magnitude of vector $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
The magnitude of the projection is $\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \frac{|1|}{3} = \frac{1}{3}$.
Note: Since the provided options do not contain $1/3$,and assuming the question intended to ask for the projection of $\vec{a}$ on $\vec{b}$ where $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ (magnitude $\sqrt{6}$),the result would be $1/\sqrt{6}$. Given the options,$B$ is the intended answer.

(C) We use the vector triple product formula: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.
Applying this to $(\vec{y} \times \vec{x}) \times \vec{x}$:
$(\vec{y} \times \vec{x}) \times \vec{x} = (\vec{y} \cdot \vec{x})\vec{x} - (\vec{x} \cdot \vec{x})\vec{y}$.
Given that $\vec{x} \cdot \vec{y} = 0$,we have $\vec{y} \cdot \vec{x} = 0$.
Also,$|\vec{x}| = 1$,so $\vec{x} \cdot \vec{x} = |\vec{x}|^2 = 1^2 = 1$.
Substituting these values:
$(\vec{y} \times \vec{x}) \times \vec{x} = (0)\vec{x} - (1)\vec{y} = -\vec{y}$.
Thus,the correct option is $C$.

(D) The area of a parallelogram with diagonals $\vec{d_1}$ and $\vec{d_2}$ is given by the formula:
Area $= \frac{1}{2} |\vec{d_1} \times \vec{d_2}|$.
Given $\vec{d_1} = 0\hat{i} + 1\hat{j} + 1\hat{k}$ and $\vec{d_2} = 1\hat{i} + 1\hat{j} + 0\hat{k}$.
First,calculate the cross product $\vec{d_1} \times \vec{d_2}$:
$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(0 - 1) + \hat{k}(0 - 1) = -\hat{i} + \hat{j} - \hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.
Finally,the area is $\frac{1}{2} \times \sqrt{3} = \frac{\sqrt{3}}{2}$ sq. units.
Thus,the correct option is $D$.

(C) The given equations of the lines are:
Line $1$: $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$.
Rewrite the first line in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$:
$\frac{2(x - 5/2)}{k} = \frac{y+2}{-5} = \frac{z}{1} \implies \frac{x - 5/2}{k/2} = \frac{y+2}{-5} = \frac{z}{1}$.
The direction ratios of the first line are $\vec{v_1} = (k/2, -5, 1)$.
Line $2$: $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$.
The direction ratios of the second line are $\vec{v_2} = (1, 2, 3)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero:
$(k/2)(1) + (-5)(2) + (1)(3) = 0$.
$k/2 - 10 + 3 = 0$.
$k/2 - 7 = 0$.
$k/2 = 7$.
$k = 14$.

(D) The equation of a line passing through a point $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Since the line passes through the origin,$(x_1, y_1, z_1) = (0, 0, 0)$.
Since the line is parallel to the $X$-axis,its direction ratios are proportional to $(1, 0, 0)$,so $(a, b, c) = (1, 0, 0)$.
Substituting these values into the equation,we get $\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$,which simplifies to $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$.
Therefore,the correct option is $D$.

(B) In a Linear Programming $(LP)$ problem,the objective function is a linear function of the form $Z = ax + by$,where $a$ and $b$ are constants. This function is the one that needs to be either maximized or minimized subject to certain constraints. Therefore,it is a function to be optimized.

(B) For the maximum value of $Z = px + qy$ to occur at two points $(x_1, y_1)$ and $(x_2, y_2)$,the value of $Z$ at these points must be equal.
Given points are $(3, 3)$ and $(1, 5)$.
$Z(3, 3) = p(3) + q(3) = 3p + 3q$.
$Z(1, 5) = p(1) + q(5) = p + 5q$.
Equating the two values: $3p + 3q = p + 5q$.
$3p - p = 5q - 3q$.
$2p = 2q$.
$p = q$.
Therefore,the condition is $p = q$.

(D) To find the minimum value of the objective function $Z = 10x - 20y + 30$,we evaluate $Z$ at each vertex of the feasible region:
$1$. At $O(0,0): Z = 10(0) - 20(0) + 30 = 30$
$2$. At $A(10,0): Z = 10(10) - 20(0) + 30 = 100 + 30 = 130$
$3$. At $B(0,20): Z = 10(0) - 20(20) + 30 = -400 + 30 = -370$
$4$. At $C(15,15): Z = 10(15) - 20(15) + 30 = 150 - 300 + 30 = -120$
Comparing these values $(30, 130, -370, -120)$,the minimum value is $-370$ at vertex $B(0,20)$.

(B) To find the maximum value of the objective function $Z = 2x + 3y$,we evaluate $Z$ at each corner point of the feasible region:
$1$. At point $A (20, 10)$: $Z = 2(20) + 3(10) = 40 + 30 = 70$.
$2$. At point $B (18, 12)$: $Z = 2(18) + 3(12) = 36 + 36 = 72$.
$3$. At point $C (12, 12)$: $Z = 2(12) + 3(12) = 24 + 36 = 60$.
Comparing the values $70$,$72$,and $60$,the maximum value is $72$. Therefore,the correct option is $B$.

(A) feasible solution is defined as any point $(x, y)$ in the feasible region that satisfies all the given constraints of the $LP$ problem simultaneously,including the non-negativity constraints.
Therefore,the correct option is $A$.

(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) Given that $P(E)=0.8$,$P(F)=0.5$,and $P(F \mid E)=0.4$.
Using the definition of conditional probability,$P(F \mid E) = \frac{P(E \cap F)}{P(E)}$.
Substituting the values,$0.4 = \frac{P(E \cap F)}{0.8}$.
Therefore,$P(E \cap F) = 0.4 \times 0.8 = 0.32$.
Now,we need to find $P(E \mid F)$.
Using the formula $P(E \mid F) = \frac{P(E \cap F)}{P(F)}$.
Substituting the values,$P(E \mid F) = \frac{0.32}{0.5} = 0.64$.
Thus,the correct option is $A$.

(B) This is a binomial distribution problem where $n = 5$ and $p = \frac{1}{5}$.
The probability of success (student is a singer) is $p = \frac{1}{5}$.
The probability of failure (student is not a singer) is $q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.
We want to find the probability that $x = 4$ students are singers.
The formula for binomial probability is $P(X = x) = \binom{n}{x} p^x q^{n-x}$.
Substituting the values: $P(X = 4) = \binom{5}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{5-4}$.
$P(X = 4) = 5 \times \left(\frac{1}{5}\right)^4 \times \frac{4}{5}$.
$P(X = 4) = 5 \times \frac{1}{5^4} \times \frac{4}{5} = \frac{4}{5^4} = 4 \left(\frac{1}{5}\right)^4$.
Thus,the correct option is $B$.

(D) For a probability distribution,the sum of all probabilities must be equal to $1$.
Thus,$\sum_{x=1}^{4} P(x) = 1$.
Given $P(x) = \frac{c}{3} \binom{4}{x}$,we have:
$\frac{c}{3} [\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}] = 1$.
We know that $\sum_{x=0}^{n} \binom{n}{x} = 2^n$.
Therefore,$\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 2^4 = 16$.
Since $\binom{4}{0} = 1$,we have $\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 16 - 1 = 15$.
Substituting this into the equation:
$\frac{c}{3} \times 15 = 1$.
$5c = 1$.
$c = \frac{1}{5}$.

(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 independent event property: $P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B)$.
Given $P(A \cup B) = 0.5$ and $P(A) = 0.2$,we substitute these values:
$0.5 = 0.2 + P(B) - 0.2 \cdot P(B)$.
$0.5 - 0.2 = P(B)(1 - 0.2)$.
$0.3 = 0.8 \cdot P(B)$.
$P(B) = \frac{0.3}{0.8} = \frac{3}{8}$.
Therefore,the correct option is $C$.