GUJCET 2020 Mathematics Question Paper with Answer and Solution

(D) The given parametric equations are $x = 4 \cos \theta$ and $y = 3 \sin \theta$.
This represents an ellipse of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a = 4$ and $b = 3$.
The area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values of $a$ and $b$, we get $A = \pi \times 4 \times 3 = 12 \pi$ sq. units.
Therefore, the correct option is $D$.

(C) The composite function $(g \circ f)(x)$ is defined as $g(f(x))$.
Given $f(x) = 2x^2 - 5$ and $g(x) = \frac{x}{x^2 + 1}$.
Substitute $f(x)$ into $g(x)$:
$(g \circ f)(x) = g(2x^2 - 5) = \frac{2x^2 - 5}{(2x^2 - 5)^2 + 1}$.
Now,expand the denominator:
$(2x^2 - 5)^2 + 1 = (4x^4 - 20x^2 + 25) + 1 = 4x^4 - 20x^2 + 26$.
Therefore,$(g \circ f)(x) = \frac{2x^2 - 5}{4x^4 - 20x^2 + 26}$.
Thus,the correct option is $C$.

(C) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $(2, \infty)$.
We can rewrite the function by completing the square:
$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.
Since the domain is $x \in (2, \infty)$,we have $x > 2$.
Subtracting $2$ from both sides,we get $x - 2 > 0$.
Squaring both sides,we get $(x - 2)^2 > 0$.
Adding $1$ to both sides,we get $(x - 2)^2 + 1 > 1$.
Therefore,$f(x) > 1$.
Thus,the range of the function $f$ is $(1, \infty)$.

(B) Let $\alpha = \sin^{-1}\left(\frac{3}{5}\right)$ and $\beta = \sin^{-1}\left(\frac{8}{17}\right)$.
Then $\sin \alpha = \frac{3}{5}$ and $\sin \beta = \frac{8}{17}$.
Using the identity $\cos \theta = \sqrt{1 - \sin^2 \theta}$,we get $\cos \alpha = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ and $\cos \beta = \sqrt{1 - (\frac{8}{17})^2} = \sqrt{\frac{225}{289}} = \frac{15}{17}$.
We use the formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$.
$\cos(\alpha - \beta) = \left(\frac{4}{5}\right)\left(\frac{15}{17}\right) + \left(\frac{3}{5}\right)\left(\frac{8}{17}\right) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$.
Therefore,$\alpha - \beta = \cos^{-1}\left(\frac{84}{85}\right)$.

(D) Let the expression be $E = \tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.
Step $1$: Simplify $\tan ^2(\sec ^{-1} 3)$.
Let $\sec ^{-1} 3 = \theta_1$,so $\sec \theta_1 = 3$. Then $\tan ^2 \theta_1 = \sec ^2 \theta_1 - 1 = 3^2 - 1 = 9 - 1 = 8$.
Step $2$: Simplify $\operatorname{cosec}^2(\cot ^{-1} 2)$.
Let $\cot ^{-1} 2 = \theta_2$,so $\cot \theta_2 = 2$. Then $\operatorname{cosec}^2 \theta_2 = 1 + \cot ^2 \theta_2 = 1 + 2^2 = 1 + 4 = 5$.
Step $3$: Simplify $\cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3})$.
We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Thus,$\cos ^2(\frac{\pi}{2}) = 0^2 = 0$.
Step $4$: Calculate the total sum.
$E = 8 + 5 + 0 = 13$.
Therefore,the correct option is $D$.

(C) In $\triangle ABC$,we know that $A + B + C = \pi$,so $B + C = \pi - A$ and $A + C = \pi - B$.
Substituting these into the determinant:
$\sin(B + C) = \sin(\pi - A) = \sin A$
$\tan(A + C) = \tan(\pi - B) = -\tan B$
The determinant becomes:
$D = \left|\begin{array}{ccc}0 & \sin A & \tan B \\ -\sin A & 0 & \cos C \\ -\tan B & -\cos C & 0\end{array}\right|$
This is a skew-symmetric determinant of order $3 \times 3$.
$A$ skew-symmetric matrix $M$ satisfies $M^T = -M$.
The determinant of a skew-symmetric matrix of odd order $n$ is given by $\det(M) = \det(M^T) = \det(-M) = (-1)^n \det(M)$.
Since $n = 3$ (odd),$\det(M) = -\det(M)$,which implies $2 \det(M) = 0$,so $\det(M) = 0$.
Thus,the value of the determinant is $0$.

(D) We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix of order $3 \times 3$.
Given $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{-2} \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -x & 3 & -1 \end{bmatrix}$.
Let $B = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -x & 3 & -1 \end{bmatrix}$. Then $A^{-1} = -\frac{1}{2} B$.
Thus,$A \cdot (-\frac{1}{2} B) = I$,which implies $A \cdot B = -2I = \begin{bmatrix} -2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix}$.
Multiplying the third row of $A$ with the first column of $B$ to find the element at position $(3, 1)$ of the product matrix:
$(3 \times -1) + (1 \times 8) + (1 \times -x) = -2$.
$-3 + 8 - x = -2$.
$5 - x = -2$.
$x = 5 + 2 = 7$.
Wait,let us re-calculate the determinant $|A|$:
$|A| = 0(2-3) - 1(1-9) + 2(1-6) = 0 - 1(-8) + 2(-5) = 8 - 10 = -2$.
Since $A^{-1} = \frac{1}{|A|} \text{adj}(A)$,the matrix $B$ must be $\text{adj}(A)$.
The element at $(3, 1)$ of $\text{adj}(A)$ is the cofactor $C_{13}$ of matrix $A$.
$C_{13} = (-1)^{1+3} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = 1(1 - 6) = -5$.
Comparing this with the matrix $B$,the element at $(3, 1)$ is $-x$.
So,$-x = -5$,which means $x = 5$.
Therefore,the correct option is $D$.

(A) Given $y = \sin^{-1}\left(\frac{2 \cdot 2^x}{1 + (2^x)^2}\right)$.
Let $2^x = \tan \theta$,then $\theta = \tan^{-1}(2^x)$.
$y = \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2 \tan^{-1}(2^x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x) = \frac{2}{1 + 4^x} \cdot 2^x \log 2 = \frac{2^{x+1} \log 2}{1 + 4^x}$.
Comparing this with $\frac{dy}{dx} = \frac{2^{x+1} \log 2}{f(x)}$,we get $f(x) = 1 + 4^x$.
Therefore,$f(0) = 1 + 4^0 = 1 + 1 = 2$.

(A) For a function $f(\alpha)$ to be continuous at $\alpha=0$,the limit of the function as $\alpha \to 0$ must equal the value of the function at $\alpha=0$.
Given $f(0) = k$.
We need to evaluate $\lim_{\alpha \to 0} f(\alpha) = \lim_{\alpha \to 0} \frac{1-\cos 6 \alpha}{36 \alpha^2}$.
Using the trigonometric identity $1-\cos \theta = 2 \sin^2(\theta/2)$,we have $1-\cos 6 \alpha = 2 \sin^2(3 \alpha)$.
Substituting this into the limit: $\lim_{\alpha \to 0} \frac{2 \sin^2(3 \alpha)}{36 \alpha^2} = \lim_{\alpha \to 0} \frac{2}{36} \left( \frac{\sin 3 \alpha}{\alpha} \right)^2$.
Using the standard limit $\lim_{x \to 0} \frac{\sin ax}{x} = a$,we get $\lim_{\alpha \to 0} \frac{1}{18} (3)^2 = \frac{9}{18} = \frac{1}{2}$.
Since the function is continuous,$k = 1/2$.

(A) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2019-x}{2019+x}\right) d x$.
Consider the function $f(x) = \log \left(\frac{2019-x}{2019+x}\right)$.
Check if the function is even or odd by evaluating $f(-x)$:
$f(-x) = \log \left(\frac{2019-(-x)}{2019+(-x)}\right) = \log \left(\frac{2019+x}{2019-x}\right)$.
Using the property $\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)$,we get:
$f(-x) = -\log \left(\frac{2019-x}{2019+x}\right) = -f(x)$.
Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.
For an odd function,the definite integral over a symmetric interval $[-a, a]$ is always $0$,i.e.,$\int_{-a}^{a} f(x) dx = 0$.
Therefore,$I = 0$.

(B) We use the standard integral formula: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$.
Given the integral: $I = \int \frac{x+100}{(x+101)^2} e^x \, dx$.
We can rewrite the numerator as: $x + 100 = (x + 101) - 1$.
So,the integral becomes: $I = \int \frac{(x+101) - 1}{(x+101)^2} e^x \, dx$.
$I = \int \left( \frac{x+101}{(x+101)^2} - \frac{1}{(x+101)^2} \right) e^x \, dx$.
$I = \int \left( \frac{1}{x+101} - \frac{1}{(x+101)^2} \right) e^x \, dx$.
Let $f(x) = \frac{1}{x+101}$.
Then $f'(x) = -\frac{1}{(x+101)^2}$.
Since the expression is in the form $\int e^x [f(x) + f'(x)] \, dx$,the result is $e^x f(x) + C$.
Therefore,$I = e^x \left( \frac{1}{x+101} \right) + C = \frac{e^x}{x+101} + C$.

(B) Given the differential equation $y \frac{dy}{dx} + x = k$.
Rearranging the terms,we get $y \frac{dy}{dx} = k - x$.
Integrating both sides with respect to $x$:
$\int y \, dy = \int (k - x) \, dx$.
This gives $\frac{y^2}{2} = kx - \frac{x^2}{2} + C$,where $C$ is the constant of integration.
Multiplying by $2$,we get $y^2 = 2kx - x^2 + 2C$.
Rearranging the terms: $x^2 - 2kx + y^2 = 2C$.
Completing the square for $x$: $(x^2 - 2kx + k^2) + y^2 = 2C + k^2$.
$(x - k)^2 + y^2 = 2C + k^2$.
This is the equation of a circle in the form $(x - h)^2 + (y - k_0)^2 = r^2$,where the center is $(k, 0)$ and the radius is $\sqrt{2C + k^2}$.
Therefore,the solution represents a circle.

(D) The given differential equation is $(\tan ^{-1} y - x) dy = (1+y^2) dx$.
Rearranging the terms,we get $\frac{dx}{dy} = \frac{\tan ^{-1} y - x}{1+y^2}$.
This can be written as $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan ^{-1} y}{1+y^2}$.
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{\tan ^{-1} y}{1+y^2}$.
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 $D$.

(D) Given that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.
Squaring both sides,we get $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = \vec{0} \cdot \vec{0}$.
This expands to $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Substituting the given magnitudes $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=2$:
$3^2 + 4^2 + 2^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$9 + 16 + 4 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
$29 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$.
Therefore,$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{29}{2}$.

(A) We are given the vectors: $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$.
First,we calculate the vector $\lambda \vec{b} + \vec{c}$:
$\lambda \vec{b} + \vec{c} = \lambda (\hat{i} + \hat{j} - 2 \hat{k}) + (\hat{i} + 3 \hat{j} - \hat{k})$
$= (\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}$.
Since $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,their dot product must be zero:
$\vec{a} \cdot (\lambda \vec{b} + \vec{c}) = 0$
$(2 \hat{i} - \hat{j} + \hat{k}) \cdot ((\lambda + 1) \hat{i} + (\lambda + 3) \hat{j} + (-2 \lambda - 1) \hat{k}) = 0$
$2(\lambda + 1) - 1(\lambda + 3) + 1(-2 \lambda - 1) = 0$
$2 \lambda + 2 - \lambda - 3 - 2 \lambda - 1 = 0$
$-\lambda - 2 = 0$
$\lambda = -2$.

(C) Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$.
Given that $|\vec{a}| = 3$,we have $a_1^2 + a_2^2 + a_3^2 = 3^2 = 9$.
Now,$\vec{a} \times \hat{i} = (a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) \times \hat{i} = -a_2 \hat{k} + a_3 \hat{j}$.
So,$|\vec{a} \times \hat{i}|^2 = a_2^2 + a_3^2$.
Similarly,$|\vec{a} \times \hat{j}|^2 = a_1^2 + a_3^2$ and $|\vec{a} \times \hat{k}|^2 = a_1^2 + a_2^2$.
Adding these,we get $|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = (a_2^2 + a_3^2) + (a_1^2 + a_3^2) + (a_1^2 + a_2^2) = 2(a_1^2 + a_2^2 + a_3^2)$.
Substituting the value,$2(9) = 18$.

(A) To find the maximum value of $Z = 3x + 4y$ subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0$,we identify the corner points of the feasible region.
The feasible region is a triangle with vertices at $(0, 0)$,$(4, 0)$,and $(0, 4)$.
Now,we evaluate $Z$ at each corner point:
$1$. At $(0, 0)$: $Z = 3(0) + 4(0) = 0$
$2$. At $(4, 0)$: $Z = 3(4) + 4(0) = 12$
$3$. At $(0, 4)$: $Z = 3(0) + 4(4) = 16$
Comparing these values,the maximum value of $Z$ is $16$ at the point $(0, 4)$.
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 $(15, 20)$.

(D) The expected value $E(X^2)$ is calculated using the formula $E(X^2) = \sum x_i^2 P(x_i)$.
Given the values:
$x_1 = 1, P(x_1) = \frac{1}{10} \implies x_1^2 P(x_1) = 1^2 \times \frac{1}{10} = \frac{1}{10}$
$x_2 = 2, P(x_2) = \frac{1}{5} \implies x_2^2 P(x_2) = 2^2 \times \frac{1}{5} = 4 \times \frac{1}{5} = \frac{4}{5} = \frac{8}{10}$
$x_3 = 3, P(x_3) = \frac{3}{10} \implies x_3^2 P(x_3) = 3^2 \times \frac{3}{10} = 9 \times \frac{3}{10} = \frac{27}{10}$
$x_4 = 4, P(x_4) = \frac{2}{5} \implies x_4^2 P(x_4) = 4^2 \times \frac{2}{5} = 16 \times \frac{2}{5} = \frac{32}{5} = \frac{64}{10}$
Summing these values:
$E(X^2) = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10} = \frac{1 + 8 + 27 + 64}{10} = \frac{100}{10} = 10$.
Therefore,the correct option is $D$.