GUJCET 2024 Mathematics Question Paper with Answer and Solution

(D) 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 two dice are rolled independently,the probability of obtaining an even prime number on each die is the product of their individual probabilities:
$P = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Thus,the correct option is $D$.

(B) To determine if $f(x) = x^3 + 2$ is one-one and onto on the set of integers $\mathbb{Z}$:
$1$. One-one check: Let $f(x_1) = f(x_2)$. Then $x_1^3 + 2 = x_2^3 + 2$,which implies $x_1^3 = x_2^3$. Since the cube function is strictly increasing,$x_1 = x_2$. Thus,$f$ is one-one.
$2$. Onto check: For $f$ to be onto,for every $y \in \mathbb{Z}$,there must exist $x \in \mathbb{Z}$ such that $y = x^3 + 2$. This means $x^3 = y - 2$,or $x = \sqrt[3]{y - 2}$. For $x$ to be an integer,$y - 2$ must be a perfect cube. For example,if $y = 3$,then $x^3 = 3 - 2 = 1$,so $x = 1 \in \mathbb{Z}$. However,if $y = 0$,then $x^3 = 0 - 2 = -2$. Since $-2$ is not a perfect cube of any integer,there is no $x \in \mathbb{Z}$ such that $f(x) = 0$. Therefore,$f$ is not onto.
Conclusion: $f$ is one-one but not onto.

(B) To determine the properties of the relation $R$ on set $A = \{a, b, c\}$:
$1$. Reflexivity: For a relation to be reflexive,$(x, x) \in R$ for all $x \in A$. Here,$(a, a), (b, b), (c, c) \in R$,so $R$ is reflexive.
$2$. Symmetry: For a relation to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Here,$(a, c) \in R$ but $(c, a) \notin R$. Therefore,$R$ is not symmetric.
$3$. Transitivity: For a relation to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. The pairs are $(a, a), (b, b), (c, c), (a, c)$. Checking $(a, c) \in R$ and $(c, c) \in R$,we get $(a, c) \in R$. All conditions for transitivity are satisfied. Thus,$R$ is transitive.
Conclusion: The relation is reflexive and transitive but not symmetric. The correct option is $B$.

(B) We know that the range of the principal value branch of $\sin ^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Given expression is $\sin ^{-1}(\sin \frac{23 \pi}{6})$.
First,simplify the angle $\frac{23 \pi}{6}$:
$\frac{23 \pi}{6} = \frac{24 \pi - \pi}{6} = 4 \pi - \frac{\pi}{6}$.
Since $\sin(4 \pi - \theta) = -\sin \theta$,we have:
$\sin(\frac{23 \pi}{6}) = \sin(4 \pi - \frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = \sin(-\frac{\pi}{6})$.
Now,$\sin ^{-1}(\sin(-\frac{\pi}{6})) = -\frac{\pi}{6}$,which lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Thus,the correct option is $B$.

(D) We know the principal value branches of the inverse trigonometric functions:
$1$. $\tan ^{-1}(-x) = -\tan ^{-1}(x)$,so $\tan ^{-1}(-1) = -\frac{\pi}{4}$.
$2$. $\sec ^{-1}(-x) = \pi - \sec ^{-1}(x)$,so $\sec ^{-1}(-2) = \pi - \sec ^{-1}(2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
$3$. $\sin ^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}$.
Adding these values together:
$-\frac{\pi}{4} + \frac{2\pi}{3} + \frac{\pi}{4} = \frac{2\pi}{3}$.

(C) The function $y = \tan^{-1} x$ represents the inverse tangent function.
By definition,the principal value branch of the inverse tangent function is defined as the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.
Therefore,for any real number $x$,the range of $y = \tan^{-1} x$ is $-\frac{\pi}{2} < y < \frac{\pi}{2}$.

(C) Given that $A^2 = A$ (idempotent matrix).
We need to evaluate $(I + A)^2 - 3A$.
Expanding the expression: $(I + A)^2 = I^2 + IA + AI + A^2 = I + A + A + A^2 = I + 2A + A^2$.
Since $A^2 = A$,we substitute it into the expression: $I + 2A + A = I + 3A$.
Now,subtract $3A$: $(I + 3A) - 3A = I$.
Therefore,the correct option is $C$.

(D) Given the matrix $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$.
First,we calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$
Performing matrix multiplication:
Row $1$: $(0)(0) + (0)(0) + (-1)(-1) = 1$,$(0)(0) + (0)(-1) + (-1)(0) = 0$,$(0)(-1) + (0)(0) + (-1)(0) = 0$
Row $2$: $(0)(0) + (-1)(0) + (0)(-1) = 0$,$(0)(0) + (-1)(-1) + (0)(0) = 1$,$(0)(-1) + (-1)(0) + (0)(0) = 0$
Row $3$: $(-1)(0) + (0)(0) + (0)(-1) = 0$,$(-1)(0) + (0)(-1) + (0)(0) = 0$,$(-1)(-1) + (0)(0) + (0)(0) = 1$
Thus,$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I$.
Now,we calculate $I + A^2 = I + I = 2I$.

(A) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Then,the transpose $A^{\prime} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Given the condition $A+A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Adding $A$ and $A^{\prime}$:
$A+A^{\prime} = \begin{bmatrix} \sin \alpha + \sin \alpha & -\cos \alpha + \cos \alpha \\ \cos \alpha - \cos \alpha & \sin \alpha + \sin \alpha \end{bmatrix} = \begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix}$.
Equating this to the identity matrix $I$:
$\begin{bmatrix} 2\sin \alpha & 0 \\ 0 & 2\sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $2\sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.
Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we have $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,$\cos \alpha = \pm \frac{\sqrt{3}}{2}$.
Looking at the options,$\frac{\sqrt{3}}{2}$ is provided.

(D) To find the inverse of a matrix $A$,we first calculate its determinant,$|A|$.
Given $A = \begin{bmatrix} 2 & -4 \\ -3 & 6 \end{bmatrix}$.
$|A| = (2 \times 6) - (-4 \times -3) = 12 - 12 = 0$.
Since the determinant of matrix $A$ is $0$,the matrix is singular.
$A$ singular matrix does not have an inverse.
Therefore,$A^{-1}$ does not exist.

(B) Let the given determinants be $D_1$ and $D_2$.
$D_1 = \left|\begin{array}{ll}2017 & 2018 \\ 2019 & 2020\end{array}\right| = (2017 \times 2020) - (2018 \times 2019)$.
Using the property $a(a+3) - (a+1)(a+2) = a^2 + 3a - (a^2 + 3a + 2) = -2$,we get $D_1 = -2$.
Similarly,$D_2 = \left|\begin{array}{ll}2021 & 2022 \\ 2023 & 2024\end{array}\right| = (2021 \times 2024) - (2022 \times 2023)$.
Using the same property,$D_2 = -2$.
Given $D_1 + D_2 = 2k$,so $-2 + (-2) = 2k$.
$-4 = 2k \implies k = -2$.
Therefore,$k^3 = (-2)^3 = -8$.

(C) 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 $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$,and the area is $3$ sq. units.
Substituting the values:
$3 = \frac{1}{2} |k(4 - 1) + 2(1 - 1) + 1(1 - 4)|$
$6 = |3k + 0 - 3|$
$6 = |3k - 3|$
This implies two cases:
Case $1$: $3k - 3 = 6 \implies 3k = 9 \implies k = 3$.
Case $2$: $3k - 3 = -6 \implies 3k = -3 \implies k = -1$.
Thus,$k = -1$ or $k = 3$.

(C) Given the equation $y = 7 \sin x + 5 \cos x$.
First,find the first derivative with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(7 \sin x + 5 \cos x) = 7 \cos x - 5 \sin x$.
Next,find the second derivative with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(7 \cos x - 5 \sin x) = -7 \sin x - 5 \cos x$.
Factor out $-1$ from the expression:
$\frac{d^2y}{dx^2} = -(7 \sin x + 5 \cos x)$.
Since $y = 7 \sin x + 5 \cos x$,we can substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -y$.
Rearranging gives $\frac{d^2y}{dx^2} + y = 0$.
Comparing this with the given equation $\frac{d^2y}{dx^2} - my = 0$,we have $-m = 1$,which implies $m = -1$.
Therefore,the correct option is $C$.

(B) 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 $B$.

(A) Let $y = e^{x \log x} + e^3$.
Since $e^{x \log x} = e^{\log(x^x)} = x^x$,the expression becomes $y = x^x + e^3$.
Now,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(x^x) + \frac{d}{dx}(e^3)$.
Since $e^3$ is a constant,its derivative is $0$.
To differentiate $x^x$,let $u = x^x$. Then $\log u = x \log x$.
Differentiating both sides with respect to $x$: $\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$.
Thus,$\frac{du}{dx} = u(1 + \log x) = x^x(1 + \log x)$.
Therefore,$\frac{dy}{dx} = x^x(1 + \log x) + 0 = x^x(1 + \log x)$.

(C) Since $f(x)$ is continuous at $x=\frac{\pi}{2}$,we have $\lim_{x \to \frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$.
Given $f(\frac{\pi}{2}) = 2024$.
Now,$\lim_{x \to \frac{\pi}{2}} \frac{2k \cos x}{\pi-2x} = 2024$.
Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$,$h \to 0$.
$\lim_{h \to 0} \frac{2k \cos(\frac{\pi}{2} + h)}{\pi - 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{2k(-\sin h)}{\pi - \pi - 2h} = \lim_{h \to 0} \frac{-2k \sin h}{-2h} = k \lim_{h \to 0} \frac{\sin h}{h} = k(1) = k$.
Equating the limit to the function value,we get $k = 2024$.

(D) To find the absolute maximum value of $f(x) = \sin x + \cos x$ on the interval $[0, \pi]$,we first find the critical points by setting the derivative to zero.
$f'(x) = \cos x - \sin x$.
Setting $f'(x) = 0$,we get $\cos x = \sin x$,which implies $\tan x = 1$.
For $x \in [0, \pi]$,the solution is $x = \pi/4$.
Now,we evaluate $f(x)$ at the critical point and the endpoints of the interval:
$1$. At $x = 0$: $f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$.
$2$. At $x = \pi/4$: $f(\pi/4) = \sin(\pi/4) + \cos(\pi/4) = 1/\sqrt{2} + 1/\sqrt{2} = 2/\sqrt{2} = \sqrt{2}$.
$3$. At $x = \pi$: $f(\pi) = \sin(\pi) + \cos(\pi) = 0 - 1 = -1$.
Comparing these values,the absolute maximum value is $\sqrt{2}$.

(D) To determine the intervals of increase and decrease,we find the derivative of $f(x) = \sin 3x$.
$f'(x) = \frac{d}{dx}(\sin 3x) = 3 \cos 3x$.
For the function to be increasing,$f'(x) > 0$,so $3 \cos 3x > 0$,which implies $\cos 3x > 0$.
Since $x \in [0, \frac{\pi}{2}]$,$3x \in [0, \frac{3\pi}{2}]$.
$\cos 3x > 0$ when $3x \in [0, \frac{\pi}{2})$,which means $x \in [0, \frac{\pi}{6})$.
For the function to be decreasing,$f'(x) < 0$,so $3 \cos 3x < 0$,which implies $\cos 3x < 0$.
$\cos 3x < 0$ when $3x \in (\frac{\pi}{2}, \frac{3\pi}{2}]$,which means $x \in (\frac{\pi}{6}, \frac{\pi}{2}]$.
Thus,the function is increasing on $[0, \frac{\pi}{6})$ and decreasing on $(\frac{\pi}{6}, \frac{\pi}{2}]$.
Therefore,the correct option is $D$.

(C) The surface area $A$ of a sphere with radius $r$ is given by the formula: $A = 4 \pi r^2$.
To find the rate of change of the area with respect to the radius, we differentiate $A$ with respect to $r$:
$\frac{dA}{dr} = \frac{d}{dr}(4 \pi r^2) = 4 \pi (2r) = 8 \pi r$.
Given that the radius $r = 6 \text{ cm}$, we substitute this value into the derivative:
$\frac{dA}{dr} = 8 \pi (6) = 48 \pi$.
Therefore, the rate of change of the area is $48 \pi \text{ cm}^2/\text{cm}$.

(B) To evaluate the integral $\int_0^1 x e^x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.
Let $u = x$ and $dv = e^x \, dx$.
Then $du = dx$ and $v = e^x$.
Applying the formula:
$\int_0^1 x e^x \, dx = [x e^x]_0^1 - \int_0^1 e^x \, dx$
$= (1 \cdot e^1 - 0 \cdot e^0) - [e^x]_0^1$
$= (e - 0) - (e^1 - e^0)$
$= e - (e - 1)$
$= e - e + 1$
$= 1$.
Thus,the correct option is $B$.

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$.
We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x - 3$.
This does not immediately show $f(x)$ is even or odd,so we split the integral:
$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^5 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$.
Since $x^5$,$x^3 \cos x$,and $\sin^3 x$ are odd functions,their integrals over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $0$.
Thus,$I = 0 - 0 + 0 - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$.
$I = -3 \times [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -3 \times (\frac{\pi}{2} - (-\frac{\pi}{2})) = -3 \times \pi = -3\pi$.

(B) We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$.
Given integral is $I = \int e^x \left( \frac{1 + \sin x}{1 + \cos x} \right) dx$.
Using trigonometric identities: $1 + \sin x = 1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $1 + \cos x = 2 \cos^2 \frac{x}{2}$.
Substituting these into the expression: $\frac{1 + \sin x}{1 + \cos x} = \frac{1 + 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2 \cos^2 \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \frac{1}{2} \sec^2 \frac{x}{2} + \tan \frac{x}{2}$.
Let $f(x) = \tan \frac{x}{2}$,then $f'(x) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} = \frac{1}{2} \sec^2 \frac{x}{2}$.
Thus,the integral becomes $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c = e^x \tan \frac{x}{2} + c$.

(B) To solve the integral $I = \int \frac{1}{\sqrt{4x-x^2}} dx$,we first complete the square for the quadratic expression inside the square root.
$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = -( (x-2)^2 - 4 ) = 4 - (x-2)^2$.
Substituting this back into the integral,we get:
$I = \int \frac{1}{\sqrt{4 - (x-2)^2}} dx$.
Using the standard integral formula $\int \frac{1}{\sqrt{a^2 - u^2}} du = \sin^{-1}(\frac{u}{a}) + c$,where $a = 2$ and $u = x-2$:
$I = \sin^{-1}\left(\frac{x-2}{2}\right) + c$.
Thus,the correct option is $B$.

(A) To solve the integral $I = \int \frac{e^{2x}-1}{e^{2x}+1} dx$,we can rewrite the integrand by dividing the numerator and denominator by $e^x$:
$I = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$
Let $u = e^x + e^{-x}$. Then,the derivative is $du = (e^x - e^{-x}) dx$.
Substituting these into the integral,we get:
$I = \int \frac{1}{u} du = \log|u| + C$
Substituting $u$ back,we have $I = \log|e^x + e^{-x}| + C$.
Alternatively,we can write $e^x + e^{-x} = \frac{e^{2x}+1}{e^x}$.
So,$I = \log|\frac{e^{2x}+1}{e^x}| + C = \log|e^{2x}+1| - \log|e^x| + C = \log(e^{2x}+1) - x + C$.
Comparing this with the given options,the correct choice is $A$.

(D) The area $A$ is given by the integral of the absolute value of the function over the given interval:
$A = \int_{-\pi/2}^{\pi/2} |\cos x| \, dx$.
Since $\cos x \geq 0$ for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,we have:
$A = \int_{-\pi/2}^{\pi/2} \cos x \, dx$.
Evaluating the integral:
$A = [\sin x]_{-\pi/2}^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})$.
$A = 1 - (-1) = 1 + 1 = 2$.
Thus,the area is $2$ square units.

(B) The given equation of the parabola is $x^2 = 4y$,which implies $y = \frac{x^2}{4}$.
To find the area bounded by the curve $y = \frac{x^2}{4}$,the $X$-axis,and the line $x = 3$,we integrate the function with respect to $x$ from $x = 0$ to $x = 3$.
Area $= \int_{0}^{3} y \, dx = \int_{0}^{3} \frac{x^2}{4} \, dx$.
Area $= \frac{1}{4} \left[ \frac{x^3}{3} \right]_{0}^{3}$.
Area $= \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right) = \frac{1}{4} \left( \frac{27}{3} \right) = \frac{1}{4} \times 9 = \frac{9}{4}$ sq. units.
Therefore,the correct option is $B$.

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$.
This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1$.
Here,$a = \frac{1}{3}$ and $b = \frac{1}{4}$.
The total area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\pi ab$.
The area in the first quadrant is $\frac{1}{4}$ of the total area.
Area $= \frac{1}{4} \pi ab = \frac{1}{4} \pi \left(\frac{1}{3}\right) \left(\frac{1}{4}\right) = \frac{\pi}{48}$ sq. units.
Thus,the correct option is $A$.

(A) Given the differential equation: $(\tan ^{-1} y - x) dy = (1 + y^2) dx$
Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$:
$\frac{dx}{dy} = \frac{\tan ^{-1} y - x}{1 + y^2}$
$\frac{dx}{dy} = \frac{\tan ^{-1} y}{1 + y^2} - \frac{x}{1 + y^2}$
$\frac{dx}{dy} + \frac{1}{1 + y^2} x = \frac{\tan ^{-1} y}{1 + y^2}$
Here,$P(y) = \frac{1}{1 + y^2}$.
The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy}$.
$IF = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan ^{-1} y}$.
Thus,the correct option is $A$.

(C) Given the differential equation: $\frac{x dy - y dx}{y} = 0$.
Multiplying by $y$ (assuming $y \neq 0$),we get: $x dy - y dx = 0$.
Rearranging the terms: $x dy = y dx$.
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 logarithmic properties: $\ln|y| = \ln|cx|$.
Taking the exponential of both sides: $y = cx$.

(A) Given the differential equation: $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left(\frac{d y}{d x}\right)^4+2}$.
To find the order and degree,we first remove the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$):
$\left(\left(\frac{d^2 y}{d x^2}\right)^{1/2}\right)^6 = \left(\left(\left(\frac{d y}{d x}\right)^4+2\right)^{1/3}\right)^6$.
This simplifies to: $\left(\frac{d^2 y}{d x^2}\right)^3 = \left(\left(\frac{d y}{d x}\right)^4+2\right)^2$.
The highest order derivative present is $\frac{d^2 y}{d x^2}$,so the order is $2$.
The power of the highest order derivative after rationalizing is $3$,so the degree is $3$.
Therefore,the order is $2$ and the degree is $3$.

(C) We know the properties of unit vectors in cross product:
$\hat{i} \times \hat{j} = \hat{k}$,$\hat{j} \times \hat{k} = \hat{i}$,$\hat{k} \times \hat{i} = \hat{j}$.
Also,$\hat{a} \times \hat{b} = -(\hat{b} \times \hat{a})$ and $\hat{a} \times \hat{a} = 0$.
$1$. $\hat{j} \cdot(\hat{i} \times \hat{k}) = \hat{j} \cdot(-\hat{j}) = -(\hat{j} \cdot \hat{j}) = -1$.
$2$. $\hat{i} \cdot(\hat{j} \times \hat{j}) = \hat{i} \cdot(0) = 0$.
$3$. $\hat{k} \cdot(\hat{j} \times \hat{i}) = \hat{k} \cdot(-\hat{k}) = -(\hat{k} \cdot \hat{k}) = -1$.
$4$. $\hat{i} \cdot(\hat{k} \times \hat{j}) = \hat{i} \cdot(-\hat{i}) = -(\hat{i} \cdot \hat{i}) = -1$.
Summing these values: $(-1) + 0 + (-1) + (-1) = -3$.
Thus,the correct option is $C$.

(A) The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product,$|\vec{a} \times \vec{b}|$.
First,we calculate the cross product $\vec{a} \times \vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{vmatrix}$
$= \hat{i}(3(-2) - 4(-1)) - \hat{j}(2(-2) - 4(0)) + \hat{k}(2(-1) - 3(0))$
$= \hat{i}(-6 + 4) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0)$
$= -2\hat{i} + 4\hat{j} - 2\hat{k}$.
Now,we find the magnitude of this vector:
$|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2}$
$= \sqrt{4 + 16 + 4} = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$.
Thus,the area is $2\sqrt{6}$ sq. units.

(B) The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Given $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - \hat{k}$.
First,calculate the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1$.
Next,calculate the magnitudes: $|\vec{a}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\vec{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.
Now,substitute these values into the formula: $\cos \theta = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = \frac{-1}{3}$.
Therefore,$\theta = \cos^{-1} \left(-\frac{1}{3}\right)$.

(D) The given lines are $L_1: \frac{x-3}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $L_2: \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$.
The direction ratios of $L_1$ are $\vec{a_1} = (-3, 2k, 2)$.
The direction ratios of $L_2$ are $\vec{a_2} = (3k, 1, -5)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero:
$\vec{a_1} \cdot \vec{a_2} = 0$
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k = 10$
$k = -\frac{10}{7}$.

(B) The direction ratios of the first line are $\vec{b_1} = (1, 2, 2)$.
The direction ratios of the second line are $\vec{b_2} = (3, 2, 6)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Substituting these values into the formula: $\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}\left(\frac{19}{21}\right)$.

(B) 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 required line is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$,it will have the same direction ratios,which are $(3, 5, 6)$.
Given that the line passes through the point $(1, -3, 5)$,we substitute these values into the formula:
$\frac{x-1}{3} = \frac{y-(-3)}{5} = \frac{z-5}{6}$
This simplifies to $\frac{x-1}{3} = \frac{y+3}{5} = \frac{z-5}{6}$.
Thus,the correct option is $B$.

(D) The given constraints are $x + y < 5$,$x + y < 10$,$x > 0$,and $y > 0$.
Since $x + y < 5$ is a subset of $x + y < 10$,the effective constraint is $x + y < 5$ in the first quadrant $(x > 0, y > 0)$.
The feasible region is an open triangular region with vertices approaching $(0,0)$,$(5,0)$,and $(0,5)$.
Since the region is open and does not include the boundary points (due to the strict inequality $<$),the minimum value of the objective function $t = 7x + 3y$ cannot be attained at any specific point within the region.
As $x$ and $y$ approach $0$,the value of $t$ approaches $0$,but since $x > 0$ and $y > 0$,$t$ is always greater than $0$.
Thus,the minimum value does not exist.

(A) To find the maximum value of the objective function $z = 6x + 12y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At $(0, 6)$: $z = 6(0) + 12(6) = 0 + 72 = 72$
$2$. At $(3, 3)$: $z = 6(3) + 12(3) = 18 + 36 = 54$
$3$. At $(9, 9)$: $z = 6(9) + 12(9) = 54 + 108 = 162$
$4$. At $(0, 12)$: $z = 6(0) + 12(12) = 0 + 144 = 144$
Comparing these values,the maximum value is $162$ at the point $(9, 9)$.

(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) Given that $P(A \mid B) = 1$.
By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.
Since $P(A \mid B) = 1$,we have $\frac{P(A \cap B)}{P(B)} = 1$,which implies $P(A \cap B) = P(B)$.
This equality holds if and only if $B \subset A$.
Therefore,the correct option is $B$.

(A) Given that $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \times P(B)$.
We know the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{5} = P + \frac{1}{2} - (P \times \frac{1}{2})$.
$\frac{3}{5} = P + \frac{1}{2} - \frac{P}{2}$.
$\frac{3}{5} = \frac{P}{2} + \frac{1}{2}$.
Subtract $\frac{1}{2}$ from both sides: $\frac{3}{5} - \frac{1}{2} = \frac{P}{2}$.
$\frac{6-5}{10} = \frac{P}{2}$.
$\frac{1}{10} = \frac{P}{2}$.
$P = \frac{2}{10} = \frac{1}{5}$.
Thus,the value of $P$ is $\frac{1}{5}$.