GSEB 2024 Mathematics Question Paper with Answer and Solution

(B) To find the maximum value of $f(x) = \sin x + \cos x$,we can rewrite the function by multiplying and dividing by $\sqrt{2}$:
$f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right)$
Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we can write:
$f(x) = \sqrt{2} \left( \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} \right)$
$f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right)$
Since the maximum value of the sine function $\sin \theta$ is $1$,the maximum value of $f(x)$ is $\sqrt{2} \times 1 = \sqrt{2}$.

(D) Let the point on the curve $x^2=2y$ be $P(x, y)$. Since $x^2=2y$,we have $y = \frac{x^2}{2}$.
Thus,the point $P$ is $(x, \frac{x^2}{2})$.
The distance $D$ between $P(x, \frac{x^2}{2})$ and $(0, 5)$ is given by $D^2 = (x-0)^2 + (\frac{x^2}{2}-5)^2$.
Let $f(x) = D^2 = x^2 + (\frac{x^2}{2}-5)^2 = x^2 + \frac{x^4}{4} - 5x^2 + 25 = \frac{x^4}{4} - 4x^2 + 25$.
To find the minimum distance,we differentiate $f(x)$ with respect to $x$ and set it to $0$:
$f'(x) = x^3 - 8x = x(x^2 - 8) = 0$.
This gives $x = 0$ or $x^2 = 8$,so $x = \pm 2\sqrt{2}$.
For $x=0$,$y=0$,$D^2 = 25$,$D=5$.
For $x^2=8$,$y = \frac{8}{2} = 4$. Then $D^2 = 8 + (4-5)^2 = 8 + 1 = 9$,so $D=3$.
Since $3 < 5$,the nearest point is $(2\sqrt{2}, 4)$ or $(-2\sqrt{2}, 4)$.
Comparing with the given options,the correct point is $(2\sqrt{2}, 4)$.

(D) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 4$ and $b^2 = 9$.
Thus,$a = 2$ and $b = 3$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,we get $A = \pi \times 2 \times 3 = 6 \pi$.
Therefore,the correct option is $D$.

(B) Given: $P(A)=0.5$,$P(A \cup B)=0.6$,and $P(B)=K$.
Since $A$ and $B$ are independent events,the probability of their intersection is given by $P(A \cap B) = P(A) \times P(B) = 0.5K$.
Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $0.6 = 0.5 + K - 0.5K$.
Simplifying the equation: $0.6 = 0.5 + 0.5K$.
Subtracting $0.5$ from both sides: $0.1 = 0.5K$.
Solving for $K$: $K = \frac{0.1}{0.5} = 0.2$.

(C) When a pair of dice is rolled,the total number of possible outcomes is $6 \times 6 = 36$.
An even number on a die can be $2, 4,$ or $6$.
So,there are $3$ favorable outcomes for each die.
The number of favorable outcomes for both dice to show an even number is $3 \times 3 = 9$.
The probability is given by the ratio of favorable outcomes to total outcomes:
$P = \frac{9}{36} = \frac{1}{4}$.

(A) To find the inverse function $f^{-1}(x)$,let $y = f(x)$.
Given $y = 4x + 3$.
Now,solve for $x$ in terms of $y$:
$y - 3 = 4x$
$x = \frac{y - 3}{4}$.
Since $f^{-1}(y) = x$,we have $f^{-1}(y) = \frac{y - 3}{4}$.
Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{x - 3}{4}$.
Therefore,the correct option is $A$.

(C) To check if $f$ is one-one:
Let $f(x_1) = f(x_2)$.
If $x_1$ is odd and $x_2$ is even,then $x_1+1 = x_2-1$,so $x_2 - x_1 = 2$. Since $x_1$ is odd,$x_1+1$ is even. Since $x_2$ is even,$x_2-1$ is odd. This contradicts $f(x_1) = f(x_2)$.
If both are odd,$x_1+1 = x_2+1 \implies x_1 = x_2$.
If both are even,$x_1-1 = x_2-1 \implies x_1 = x_2$.
Thus,$f$ is one-one.
To check if $f$ is onto:
For any $y \in N$,if $y$ is odd,we can choose $x = y+1$ (which is even),then $f(y+1) = (y+1)-1 = y$.
If $y$ is even,we can choose $x = y-1$ (which is odd),then $f(y-1) = (y-1)+1 = y$.
Since for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is one-one and onto.

(B) $1$. Reflexive: $A$ relation $R$ is reflexive if $(a, a) \in R$ for all $a \in \mathbb{R}$. Here,$a < a$ is false for any real number $a$. Thus,$R$ is not reflexive.
$2$. Symmetric: $A$ relation $R$ is symmetric if $(a, b) \in R \implies (b, a) \in R$. If $a < b$,it does not imply $b < a$. For example,$1 < 2$ is true,but $2 < 1$ is false. Thus,$R$ is not symmetric.
$3$. Transitive: $A$ relation $R$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$. If $a < b$ and $b < c$,then by the transitive property of inequality,$a < c$. Thus,$R$ is transitive.
Therefore,$R$ is transitive but not reflexive and symmetric.

(C) For a relation $R$ on $A = \{1, 2, 3\}$ to be an equivalence relation containing $(1, 2)$,it must satisfy reflexivity,symmetry,and transitivity.
Since it is reflexive,$(1, 1), (2, 2), (3, 3) \in R$.
Since it contains $(1, 2)$ and is symmetric,$(2, 1) \in R$.
Now we have $R_0 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. This is an equivalence relation.
To add more elements while maintaining equivalence,we must consider the partitions of $A$. The partition corresponding to $R_0$ is $\{\{1, 2\}, \{3\}\}$.
The only other partition of $A$ that contains the subset $\{1, 2\}$ is the trivial partition $\{\{1, 2, 3\}\}$.
The equivalence relation corresponding to the partition $\{\{1, 2, 3\}\}$ is the universal relation $A \times A$,which contains $(1, 2)$.
Thus,there are $2$ such equivalence relations: $\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$ and $A \times A$.

(C) Given $f: N \rightarrow N$ defined by $f(x) = x^6$.
$1$. Check for one-one:
Let $f(x_1) = f(x_2)$ for $x_1, x_2 \in N$.
$x_1^6 = x_2^6$.
Since $x_1, x_2 \in N$ (natural numbers are positive),we have $x_1 = x_2$.
Thus,$f$ is one-one.
$2$. Check for onto:
$A$ function is onto if the range equals the codomain $(N)$.
For $f(x) = x^6$,the range is ${1^6, 2^6, 3^6, \dots} = {1, 64, 729, \dots}$.
Since the range is not equal to the codomain $N$ (e.g.,$2 \in N$ but there is no $x \in N$ such that $x^6 = 2$),$f$ is not onto.
Therefore,$f$ is one-one but not onto.

(D) Given the set $A = \{a, b, c\}$ and the relation $R = \{(a, a), (b, b), (c, c), (a, b), (b, a)\}$.
$1$. Reflexivity: For $R$ to be reflexive,$(a, a), (b, b), (c, c)$ must be in $R$. Since all these are present,$R$ is reflexive.
$2$. Symmetry: For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Here,$(a, b) \in R$ and $(b, a) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Here,$(a, b) \in R$ and $(b, a) \in R$,but $(a, a) \in R$. Also $(b, a) \in R$ and $(a, b) \in R$,and $(b, b) \in R$. All conditions are satisfied,so $R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation. The original question provided an incomplete relation; assuming the standard definition for this problem type,the correct classification is an equivalence relation.

(B) We know that the principal value branch of $\cos ^{-1} x$ is $[0, \pi]$ and $\tan ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
First,consider $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$.
Since $\frac{13 \pi}{6} = 2 \pi + \frac{\pi}{6}$,we have $\cos \frac{13 \pi}{6} = \cos \frac{\pi}{6}$.
Thus,$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) = \cos ^{-1}\left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6}$.
Next,consider $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$.
Since $\frac{7 \pi}{6} = \pi + \frac{\pi}{6}$,we have $\tan \frac{7 \pi}{6} = \tan \frac{\pi}{6}$.
Thus,$\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) = \tan ^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6}$.
Adding these results,we get $\frac{\pi}{6} + \frac{\pi}{6} = \frac{2 \pi}{6} = \frac{\pi}{3}$.

(D) Let $x = \sec \theta$. Since $x > 1$,we have $0 < \theta < \frac{\pi}{2}$.
Then,$\sqrt{x^2-1} = \sqrt{\sec^2 \theta - 1} = \sqrt{\tan^2 \theta} = \tan \theta$.
Substituting this into the expression,we get $\cot ^{-1}\left(\frac{1}{\tan \theta}\right) = \cot ^{-1}(\cot \theta)$.
Since $0 < \theta < \frac{\pi}{2}$,$\cot ^{-1}(\cot \theta) = \theta$.
Substituting back $\theta = \sec ^{-1} x$,we get the simplest form as $\sec ^{-1} x$.

(A) Let $\tan^{-1} x = \theta$.
Then,$\tan \theta = x = \frac{x}{1}$.
In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$.
The hypotenuse is $\sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Now,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$.
Therefore,$\sin (\tan^{-1} x) = \frac{x}{\sqrt{1+x^2}}$.
The correct option is $A$.

(C) We know that the principal value branch of $\tan^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$ and for $\sec^{-1} x$ is $[0, \pi] - \{\frac{\pi}{2}\}$.
First,evaluate $\tan^{-1}(-\sqrt{3})$:
Since $\tan(\frac{\pi}{3}) = \sqrt{3}$,we have $\tan(-\frac{\pi}{3}) = -\sqrt{3}$.
Thus,$\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
Next,evaluate $\sec^{-1}(-2)$:
Since $\sec(\frac{\pi}{3}) = 2$,we have $\sec(\pi - \frac{\pi}{3}) = \sec(\frac{2\pi}{3}) = -2$.
Thus,$\sec^{-1}(-2) = \frac{2\pi}{3}$.
Finally,calculate the expression:
$\tan^{-1}(-\sqrt{3}) - \sec^{-1}(-2) = -\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{3\pi}{3} = -\pi$.
Therefore,the correct option is $C$.

(C) Let $\theta = \tan^{-1} x$.
Then,$\tan \theta = x = \frac{x}{1}$.
In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$.
The hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Therefore,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{1+x^2}}$.
Thus,$\cos(\tan^{-1} x) = \frac{1}{\sqrt{1+x^2}}$.
The correct option is $C$.

(D) The principal value branch of the inverse cosine function,denoted by $\cos^{-1} x$,is defined as the interval $[0, \pi]$.
Therefore,if $\cos^{-1} x = y$,then the range of $y$ must be $0 \leq y \leq \pi$.
Thus,the correct option is $D$.

(A) We know that the range of the principal value branch of $\tan ^{-1} x$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Given expression is $\tan ^{-1}(\tan \frac{31 \pi}{6})$.
First,simplify the angle $\frac{31 \pi}{6}$:
$\frac{31 \pi}{6} = \frac{30 \pi + \pi}{6} = 5 \pi + \frac{\pi}{6}$.
Since $\tan(n \pi + \theta) = \tan \theta$,we have $\tan(5 \pi + \frac{\pi}{6}) = \tan \frac{\pi}{6}$.
Therefore,$\tan ^{-1}(\tan \frac{31 \pi}{6}) = \tan ^{-1}(\tan \frac{\pi}{6})$.
Since $\frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2})$,the value is $\frac{\pi}{6}$.
Thus,the correct option is $A$.

(D) We know that $\sin ^{-1}(-x) = -\sin ^{-1}(x)$ for $x \in [-1, 1]$.
Therefore,$\sin ^{-1}\left(-\frac{1}{2}\right) = -\sin ^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}$.
We also know that $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$ because $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.
Adding these values,we get: $-\frac{\pi}{6} + \frac{\pi}{6} = 0$.
Thus,the correct option is $D$.

(D) Given that $A$ and $B$ are skew-symmetric matrices of the same order,we have $A^{\prime} = -A$ and $B^{\prime} = -B$.
Using the property of the transpose of a product,$(AB)^{\prime} = B^{\prime}A^{\prime}$.
Substituting the values of $A^{\prime}$ and $B^{\prime}$,we get $(AB)^{\prime} = (-B)(-A)$.
Therefore,$(AB)^{\prime} = BA$.

(C) Given the matrix $A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$.
We are given that $A^2 = I$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}$
$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & \alpha \beta - \beta \alpha \\ \gamma \alpha - \alpha \gamma & \beta \gamma + \alpha^2 \end{bmatrix}$
$A^2 = \begin{bmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \beta \gamma \end{bmatrix}$
Since $A^2 = I$,we have:
$\begin{bmatrix} \alpha^2 + \beta \gamma & 0 \\ 0 & \alpha^2 + \beta \gamma \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Comparing the elements,we get $\alpha^2 + \beta \gamma = 1$.
Rearranging the terms,we get $1 - \alpha^2 - \beta \gamma = 0$.
Therefore,the correct option is $C$.

(B) In general,matrix multiplication is not commutative. For two square matrices $A$ and $B$ of the same order,the product $AB$ is not necessarily equal to $BA$. Therefore,$AB \neq BA$ is the correct statement representing the general property of matrix multiplication.

(A) Let the order of matrix $B$ be $p \times q$.
Then the order of $B^{\prime}$ is $q \times p$.
For $AB^{\prime}$ to be defined,the number of columns in $A$ must equal the number of rows in $B^{\prime}$.
Thus,$n = q$.
For $B^{\prime}A$ to be defined,the number of columns in $B^{\prime}$ must equal the number of rows in $A$.
Thus,$p = m$.
Therefore,the order of matrix $B$ is $m \times n$.

(B) Given $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = 2A$.
Now,calculate $A^3 = A^2 \times A = (2A) \times A = 2(A^2) = 2(2A) = 2^2 A$.
Similarly,$A^4 = A^3 \times A = (2^2 A) \times A = 2^2 (A^2) = 2^2 (2A) = 2^3 A$.
By induction,we can generalize that $A^n = 2^{n-1} A$.
For $n = 10$,$A^{10} = 2^{10-1} A = 2^9 A$.

(C) Given that $A^2 = A$ and $I$ is the identity matrix.
We need to evaluate the expression $(I + A)^2 - 3A$.
Expanding the expression using the property $(I + A)^2 = I^2 + IA + AI + A^2$:
Since $IA = AI = A$ and $I^2 = I$,we have:
$(I + A)^2 = I + A + A + A^2$
$(I + A)^2 = I + 2A + A^2$
Substitute $A^2 = A$ into the expression:
$(I + A)^2 = I + 2A + A = I + 3A$
Now,substitute this back into the original expression:
$(I + A)^2 - 3A = (I + 3A) - 3A$
$(I + A)^2 - 3A = I$
Therefore,the correct option is $C$.

(D) The matrix $A$ is a $2 \times 2$ identity matrix,denoted as $I_2$.
The matrix $B$ is a $3 \times 2$ matrix.
For the product $AB$ to be defined,the number of columns in $A$ must be equal to the number of rows in $B$.
Here,the number of columns in $A$ is $2$,and the number of rows in $B$ is $3$.
Since $2 \neq 3$,the matrix multiplication $AB$ is not defined.
Therefore,$AB$ does not exist.

(B) matrix of order $3 \times 2$ has $3 \times 2 = 6$ entries.
Each entry can be filled in $2$ ways (either $1$ or $2$).
Since there are $6$ independent positions to fill,the total number of such matrices is $2^6$.
$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$.
Therefore,the correct option is $B$.

(A) Given that $A$ and $B$ are matrices of order $3 \times 3$,so $n=3$.
We know the property of determinants: $|kA| = k^n |A|$,where $n$ is the order of the matrix.
Also,$|AB| = |A| |B|$.
Therefore,$|3AB| = 3^3 |AB| = 27 |A| |B|$.
Substituting the given values: $|3AB| = 27 \times 5 \times 3$.
$|3AB| = 27 \times 15 = 405$.
Thus,the correct option is $A$.

(C) Given the diagonal matrix $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$.
Since $A$ is a diagonal matrix,its inverse $A^{-1}$ is given by the reciprocal of its diagonal elements:
$A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix}$.
Now,we need to find $(A^{-1})^2 = A^{-1} \times A^{-1}$:
$(A^{-1})^2 = \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix} \times \begin{bmatrix} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & 0 & 0 \\ 0 & \frac{1}{9} & 0 \\ 0 & 0 & \frac{1}{16} \end{bmatrix}$.
Thus,the correct option is $C$.

(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:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(2, -6)$,$(5, 4)$,and $(k, 4)$ with $\text{Area} = 35$.
Substituting the values:
$35 = \frac{1}{2} |2(4 - 4) + 5(4 - (-6)) + k(-6 - 4)|$
$70 = |2(0) + 5(10) + k(-10)|$
$70 = |50 - 10k|$
This implies two cases:
$1) 50 - 10k = 70 \implies -10k = 20 \implies k = -2$
$2) 50 - 10k = -70 \implies -10k = -120 \implies k = 12$
Thus,the values of $k$ are $12$ and $-2$.

(B) We know that for any non-singular matrix $A$,the product of $A$ and its inverse $A^{-1}$ is the identity matrix $I$.
$A \cdot A^{-1} = I$
Taking the determinant on both sides:
$\det(A \cdot A^{-1}) = \det(I)$
Using the property $\det(AB) = \det(A) \cdot \det(B)$,we get:
$\det(A) \cdot \det(A^{-1}) = \det(I)$
Since $\det(I) = 1$ for an identity matrix,we have:
$\det(A) \cdot \det(A^{-1}) = 1$
Therefore,$\det(A^{-1}) = \frac{1}{\det(A)}$.

(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:
$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
Given vertices are $(3, 5), (2, 2)$ and $(k, 2)$ and $\text{Area} = 3$.
Substituting the values:
$3 = \frac{1}{2} |3(2 - 2) + 2(2 - 5) + k(5 - 2)|$
$3 = \frac{1}{2} |3(0) + 2(-3) + k(3)|$
$3 = \frac{1}{2} |0 - 6 + 3k|$
$6 = |-6 + 3k|$
This implies two cases:
$1) -6 + 3k = 6 \implies 3k = 12 \implies k = 4$
$2) -6 + 3k = -6 \implies 3k = 0 \implies k = 0$
Thus,the values of $k$ are $0$ and $4$.

(C) We know that for any square matrix $A$ of order $n$,the property $A(\operatorname{adj} A) = |A| I$ holds,where $I$ is the identity matrix of order $n$.
Given $A = \begin{bmatrix} 5 & -2 \\ 4 & 3 \end{bmatrix}$.
The determinant of $A$ is $|A| = (5 \times 3) - (-2 \times 4) = 15 - (-8) = 15 + 8 = 23$.
Since the order of matrix $A$ is $2 \times 2$,we have $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Therefore,$A(\operatorname{adj} A) = |A| I = 23 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 23 I$.

(B) Given the matrix $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 3 \\ 1 & 0 & 1 \end{bmatrix}$.
First,we calculate the determinant of $A$,denoted as $|A|$.
Expanding along the first row: $|A| = 1(2 \times 1 - 3 \times 0) - 0 + 0 = 1(2) = 2$.
The order of the matrix $A$ is $n = 3$.
We use the property of the adjoint of a matrix: $|\operatorname{adj} A| = |A|^{n-1}$.
Substituting the values: $|\operatorname{adj} A| = 2^{3-1} = 2^2 = 4$.
Therefore,the correct option is $B$.

(A) Given the function $f(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}$.
To find $f\left(\frac{\pi}{6}\right)$,we substitute $\theta = \frac{\pi}{6}$ into the matrix.
$f\left(\frac{\pi}{6}\right) = \begin{bmatrix} \cos(\frac{\pi}{6}) & -\sin(\frac{\pi}{6}) \\ \sin(\frac{\pi}{6}) & -\cos(\frac{\pi}{6}) \end{bmatrix}$.
We know that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
Substituting these values,we get:
$f\left(\frac{\pi}{6}\right) = \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & -\frac{\sqrt{3}}{2} \end{bmatrix}$.
Thus,the correct option is $A$.

(A) Given the equation $xy = e^{x-y}$.
Taking the natural logarithm on both sides,we get:
$\ln(xy) = \ln(e^{x-y})$
$\ln x + \ln y = x - y$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{d}{dx}(y)$
$\frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx}$
Rearranging the terms to collect $\frac{dy}{dx}$:
$\frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x}$
$\frac{dy}{dx} (\frac{1}{y} + 1) = \frac{x-1}{x}$
$\frac{dy}{dx} (\frac{1+y}{y}) = \frac{x-1}{x}$
$\frac{dy}{dx} = \frac{y(x-1)}{x(y+1)}$
Thus,the correct option is $A$.

(C) Given $x = \sin y$,we have $y = \arcsin x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} = (1 - x^2)^{-1/2}$.
Now,differentiating again with respect to $x$ using the chain rule:
$\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot \frac{d}{dx}(1 - x^2)$.
$\frac{d^2 y}{dx^2} = -\frac{1}{2}(1 - x^2)^{-3/2} \cdot (-2x)$.
$\frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{3/2}}$.
Thus,the correct option is $C$.

(C) Let $u = \sin ^2 x$ and $v = \cos ^2 x$.
We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$.
First,differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(\sin ^2 x) = 2 \sin x \cos x = \sin(2x)$.
Next,differentiate $v$ with respect to $x$:
$\frac{dv}{dx} = \frac{d}{dx}(\cos ^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x = -\sin(2x)$.
Now,calculate the derivative:
$\frac{du}{dv} = \frac{\sin(2x)}{-\sin(2x)} = -1$.

(A) The given series is the expansion of the exponential function $y = e^x$.
We know that $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$.
Therefore,$y = e^x$.
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$.
Since $y = e^x$,it follows that $\frac{dy}{dx} = y$.
Thus,the correct option is $A$.

(D) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = \frac{\pi}{2}$ must be equal.
$1$. Value of the function at $x = \frac{\pi}{2}$:
$f(\frac{\pi}{2}) = k(\frac{\pi}{2}) + 1$
$2$. Left-hand limit $(LHL)$:
$\lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} (kx + 1) = k(\frac{\pi}{2}) + 1$
$3$. Right-hand limit $(RHL)$:
$\lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \sin x = \sin(\frac{\pi}{2}) = 1$
For continuity,$LHL = RHL$:
$k(\frac{\pi}{2}) + 1 = 1$
$k(\frac{\pi}{2}) = 0$
$k = 0$
Thus,the correct option is $D$.

(A) To determine the interval where the function $y = f(x) = 6 - 9x - x^2$ is strictly increasing,we find its derivative:
$f'(x) = \frac{d}{dx}(6 - 9x - x^2) = -9 - 2x$.
For the function to be strictly increasing,we must have $f'(x) > 0$.
So,$-9 - 2x > 0$.
$-2x > 9$.
Dividing by $-2$ reverses the inequality: $x < -4.5$.
Thus,the function is strictly increasing on the interval $(-\infty, -4.5)$.

(D) Let $r$ be the radius and $C$ be the circumference of the circle.
Given that the rate of change of the radius is $\frac{dr}{dt} = 0.7 \text{ cm/s}$.
The formula for the circumference of a circle is $C = 2 \pi r$.
Differentiating both sides with respect to time $t$,we get:
$\frac{dC}{dt} = 2 \pi \frac{dr}{dt}$.
Substituting the given value $\frac{dr}{dt} = 0.7 \text{ cm/s}$:
$\frac{dC}{dt} = 2 \pi (0.7) = 1.4 \pi \text{ cm/s}$.
Thus,the circumference of the circle is increasing at a rate of $1.4 \pi \text{ cm/s}$.

(C) Given that $x = at^2$ and $y = 2at$.
We need to find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Differentiating $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a$.
Differentiating $x$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
Now,$\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.
Since $y = 2at$,we have $t = \frac{y}{2a}$.
Substituting $t$ in the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{y/(2a)} = \frac{2a}{y}$.
Alternatively,from $x = at^2$,we have $t^2 = \frac{x}{a}$,so $t = \sqrt{\frac{x}{a}}$.
Thus,$\frac{dy}{dx} = \frac{1}{\sqrt{x/a}} = \sqrt{\frac{a}{x}}$.
Looking at the options,if we express $\frac{1}{t}$ in terms of $x$ and $y$,we note that $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
Given $y = 2at$,then $t = \frac{y}{2a}$.
Substituting $t$ into $\frac{dy}{dx} = \frac{1}{t}$ gives $\frac{2a}{y}$.
However,checking the ratio $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
If we evaluate $\frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$.
Comparing with options,$\frac{1}{t}$ is the derivative. If the question implies expressing in terms of $x$ and $y$,$\frac{dy}{dx} = \frac{2a}{y}$.

(A) To determine the interval where the function $f(x) = x^2 - 6x + 10$ is increasing,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x^2 - 6x + 10) = 2x - 6$.
$A$ function is increasing when $f'(x) > 0$.
$2x - 6 > 0 \implies 2x > 6 \implies x > 3$.
Thus,the function is increasing in the interval $(3, \infty)$.
Therefore,the correct option is $A$.

(C) The marginal revenue is defined as the rate of change of total revenue with respect to the number of units sold,which is given by the derivative $R'(x)$.
Given $R(x) = x^2 + 6x + 5$.
Differentiating with respect to $x$,we get:
$R'(x) = \frac{d}{dx}(x^2 + 6x + 5) = 2x + 6$.
To find the marginal revenue when $x = 20$,we substitute $x = 20$ into the derivative:
$R'(20) = 2(20) + 6 = 40 + 6 = 46$.
Thus,the marginal revenue is $46$ Rupees.

(A) The area of a circle $A$ is given by the formula $A = \pi r^2$,where $r$ is the radius.
To find the rate of change of the area with respect to the radius,we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2 \pi r$.
Now,we evaluate this derivative at $r = 3 \text{ cm}$:
$\left. \frac{dA}{dr} \right|_{r=3} = 2 \pi (3) = 6 \pi$.
Thus,the rate of change of the area with respect to its radius at $r = 3 \text{ cm}$ is $6 \pi \text{ cm}^2/\text{cm}$.

(D) Let $I = \int \tan ^8 x \cdot \sec ^4 x \, dx$.
We can write $\sec ^4 x$ as $\sec ^2 x \cdot \sec ^2 x$.
So,$I = \int \tan ^8 x \cdot \sec ^2 x \cdot \sec ^2 x \, dx$.
Since $\sec ^2 x = 1 + \tan ^2 x$,we have:
$I = \int \tan ^8 x (1 + \tan ^2 x) \sec ^2 x \, dx$.
Let $u = \tan x$,then $du = \sec ^2 x \, dx$.
Substituting these into the integral:
$I = \int u^8 (1 + u^2) \, du = \int (u^8 + u^{10}) \, du$.
Integrating with respect to $u$:
$I = \frac{u^9}{9} + \frac{u^{11}}{11} + C$.
Substituting back $u = \tan x$:
$I = \frac{\tan ^9 x}{9} + \frac{\tan ^{11} x}{11} + C$.
Comparing this with the given options,the correct option is $D$.

(A) We know that $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the integral:
$\int \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} d x = \int \tan^2 \frac{x}{2} d x$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$:
$\int (\sec^2 \frac{x}{2} - 1) d x$.
Integrating term by term:
$= \int \sec^2 \frac{x}{2} d x - \int 1 d x$.
$= 2 \tan \frac{x}{2} - x + C$.

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

(B) We know that the integral of the form $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.
Given the integral is $\int e^x \cdot \sec x(1 + \tan x) \, dx$.
This can be rewritten as $\int e^x (\sec x + \sec x \tan x) \, dx$.
Here,let $f(x) = \sec x$.
Then,the derivative $f'(x) = \sec x \tan x$.
Substituting these into the formula,we get $\int e^x (f(x) + f'(x)) \, dx = e^x \sec x + C$.
Therefore,the correct option is $B$.

(A) Let $I = \int \frac{e^x(1+x)}{\sin^2(x \cdot e^x)} dx$.
Substitute $u = x \cdot e^x$.
Then,$du = (1 \cdot e^x + x \cdot e^x) dx = e^x(1+x) dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{\sin^2(u)} du = \int \csc^2(u) du$.
Since $\int \csc^2(u) du = -\cot(u) + C$,
we have $I = -\cot(x \cdot e^x) + C$.
Thus,the correct option is $A$.

(B) To evaluate the integral $I = \int \frac{x}{\sqrt{x+4}} \, dx$,let $u = x+4$. Then $du = dx$ and $x = u-4$.
Substituting these into the integral,we get:
$I = \int \frac{u-4}{\sqrt{u}} \, du$
$I = \int (\frac{u}{\sqrt{u}} - \frac{4}{\sqrt{u}}) \, du$
$I = \int (u^{1/2} - 4u^{-1/2}) \, du$
Integrating term by term:
$I = \frac{u^{3/2}}{3/2} - 4 \frac{u^{1/2}}{1/2} + C$
$I = \frac{2}{3} u^{3/2} - 8 u^{1/2} + C$
$I = \frac{2}{3} u^{1/2} (u - 12) + C$
Substituting $u = x+4$ back:
$I = \frac{2}{3} \sqrt{x+4} (x+4 - 12) + C$
$I = \frac{2}{3} \sqrt{x+4} (x-8) + C$
Thus,the correct option is $B$.

(A) To evaluate the integral $I = \int \frac{\sin (\tan ^{-1} x)}{1+x^2} d x$,we use the method of substitution.
Let $u = \tan ^{-1} x$.
Then,differentiating both sides with respect to $x$,we get $du = \frac{1}{1+x^2} dx$.
Substituting these into the integral,we have $I = \int \sin(u) du$.
The integral of $\sin(u)$ is $-\cos(u) + C$.
Substituting back $u = \tan ^{-1} x$,we get $I = -\cos (\tan ^{-1} x) + C$.

(A) Given that $\frac{d}{d x}(f(x))=4 x^3-3 x^{-4}$.
To find $f(x)$,we integrate both sides with respect to $x$:
$f(x) = \int (4 x^3-3 x^{-4}) d x$.
$f(x) = 4 \frac{x^4}{4} - 3 \frac{x^{-3}}{-3} + C$.
$f(x) = x^4 + x^{-3} + C = x^4 + \frac{1}{x^3} + C$.
Given $f(2) = 0$,substitute $x=2$:
$f(2) = 2^4 + \frac{1}{2^3} + C = 0$.
$16 + \frac{1}{8} + C = 0$.
$C = - (16 + \frac{1}{8}) = - \frac{128+1}{8} = - \frac{129}{8}$.
Therefore,$f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$.

(D) Let $I = \int_0^{2 \pi} \sin ^3 x \cos ^2 x \, dx$.
Using the property $\int_0^{2a} f(x) \, dx = 0$ if $f(2a - x) = -f(x)$.
Here,$f(x) = \sin ^3 x \cos ^2 x$.
$f(2 \pi - x) = \sin ^3(2 \pi - x) \cos ^2(2 \pi - x) = (-\sin x)^3 (\cos x)^2 = -\sin ^3 x \cos ^2 x = -f(x)$.
Since $f(2 \pi - x) = -f(x)$,the integral evaluates to $0$.