GUJCET 2018 Mathematics Question Paper with Answer and Solution

(A) For a relation $S$ on set $A = \{3, 4, 5\}$ to be reflexive,$(a, a)$ must be in $S$ for all $a \in A$. Here,$(5, 5) \notin S$,so it is not reflexive.
For $S$ to be symmetric,if $(a, b) \in S$,then $(b, a) \in S$. Since $(3, 3)$ and $(4, 4)$ are in $S$,their reverses are also in $S$. Thus,it is symmetric.
For $S$ to be transitive,if $(a, b) \in S$ and $(b, c) \in S$,then $(a, c) \in S$. For $(3, 3)$ and $(3, 3)$,$(3, 3) \in S$. Similarly for $(4, 4)$. Thus,it is transitive.
Therefore,the relation is symmetric and transitive but not reflexive.

(C) To check if the function $f: N \rightarrow Z$ is one-one and onto,we analyze the mapping:
$1$. One-one check:
If $n$ is even,$f(n) = \frac{n}{2}$. For $n \in \{2, 4, 6, \dots\}$,the values are $f(2)=1, f(4)=2, f(6)=3, \dots$,which maps to the set of positive integers $\{1, 2, 3, \dots\}$.
If $n$ is odd,$f(n) = -\left(\frac{n-1}{2}\right)$. For $n \in \{1, 3, 5, \dots\}$,the values are $f(1)=0, f(3)=-1, f(5)=-2, \dots$,which maps to the set of non-positive integers $\{0, -1, -2, \dots\}$.
Since every distinct input $n \in N$ produces a distinct output in $Z$,the function is one-one.
$2$. Onto check:
For any integer $y \in Z$,if $y > 0$,we can choose $n = 2y$ (which is even),so $f(2y) = \frac{2y}{2} = y$. If $y \le 0$,we can choose $n = -2y + 1$ (which is odd),so $f(-2y+1) = -\left(\frac{-2y+1-1}{2}\right) = -(-y) = y$. Since every $y \in Z$ has a pre-image in $N$,the function is onto.
Therefore,the function is one-one and onto.

(B) Given the equation $2 \cos \left(2 \tan ^{-1} x\right)=1$.
Dividing by $2$,we get $\cos \left(2 \tan ^{-1} x\right) = \frac{1}{2}$.
We know that $\cos \theta = \frac{1}{2}$ when $\theta = \frac{\pi}{3}$.
Therefore,$2 \tan ^{-1} x = \frac{\pi}{3}$.
Dividing by $2$,we get $\tan ^{-1} x = \frac{\pi}{6}$.
Taking the tangent of both sides,$x = \tan \left(\frac{\pi}{6}\right)$.
Since $\tan \left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$,we have $x = \frac{1}{\sqrt{3}}$.
Thus,the correct option is $B$.

(C) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$.
Let $\sin^{-1} x = \theta$,then $x = \sin \theta$. Thus,$\cos(\sin^{-1} x) = \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$.
So,$\sin^{-1}(\cos(\sin^{-1} x)) = \sin^{-1}(\sqrt{1 - x^2})$.
Similarly,let $\cos^{-1} x = \phi$,then $x = \cos \phi$. Thus,$\sin(\cos^{-1} x) = \sin \phi = \sqrt{1 - \cos^2 \phi} = \sqrt{1 - x^2}$.
So,$\cos^{-1}(\sin(\cos^{-1} x)) = \cos^{-1}(\sqrt{1 - x^2})$.
Now,the expression becomes $\sin^{-1}(\sqrt{1 - x^2}) + \cos^{-1}(\sqrt{1 - x^2})$.
Using the identity $\sin^{-1} A + \cos^{-1} A = \frac{\pi}{2}$ where $A = \sqrt{1 - x^2}$,we get the result $\frac{\pi}{2}$.

(C) Let $x = \tan \theta$,where $\theta = \tan^{-1} x$.
Then,$\sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta$.
The expression becomes $\cot^{-1}\left(\frac{\sec \theta - 1}{\tan \theta}\right)$.
$= \cot^{-1}\left(\frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}}\right) = \cot^{-1}\left(\frac{1 - \cos \theta}{\sin \theta}\right)$.
Using half-angle formulas,$1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$.
$= \cot^{-1}\left(\frac{2 \sin^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) = \cot^{-1}\left(\tan \frac{\theta}{2}\right)$.
Since $\cot^{-1}(\tan \alpha) = \frac{\pi}{2} - \tan^{-1}(\tan \alpha) = \frac{\pi}{2} - \alpha$,we have:
$= \frac{\pi}{2} - \frac{\theta}{2} = \frac{\pi}{2} - \frac{1}{2} \tan^{-1} x$.

(A) Let $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$.
First,calculate $AB$:
$AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ 0 & x \end{bmatrix} = \begin{bmatrix} 1(x) + 1(0) & 1(y) + 1(x) \\ 0(x) + 1(0) & 0(y) + 1(x) \end{bmatrix} = \begin{bmatrix} x & x+y \\ 0 & x \end{bmatrix}$.
Next,calculate $BA$:
$BA = \begin{bmatrix} x & y \\ 0 & x \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} x(1) + y(0) & x(1) + y(1) \\ 0(1) + x(0) & 0(1) + x(1) \end{bmatrix} = \begin{bmatrix} x & x+y \\ 0 & x \end{bmatrix}$.
Since $AB = BA$ for any $x, y$,the matrix $B$ must be of the form $\begin{bmatrix} x & y \\ 0 & x \end{bmatrix}$.
Thus,option $A$ is the correct choice.

(B) Given $A = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix}$.
First,calculate $A^2 = A \times A$:
$A^2 = \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 3 & 3 & 3 \\ 3 & 3 & 3 \\ 3 & 3 & 3 \end{bmatrix} = \begin{bmatrix} 9+9+9 & 9+9+9 & 9+9+9 \\ 9+9+9 & 9+9+9 & 9+9+9 \\ 9+9+9 & 9+9+9 & 9+9+9 \end{bmatrix} = \begin{bmatrix} 27 & 27 & 27 \\ 27 & 27 & 27 \\ 27 & 27 & 27 \end{bmatrix} = 9A$.
Now,calculate $A^3 = A^2 \times A$:
$A^3 = (9A) \times A = 9(A^2) = 9(9A) = 81A$.
Therefore,the correct option is $B$.

(A) Given that $x, y, z \in \mathbb{R}$ and $x^4+y^4+z^4=0$.
Since the sum of even powers of real numbers is zero,each term must be zero individually: $x^4=0, y^4=0, z^4=0$.
This implies $x=0, y=0, z=0$.
Now,substitute $x=0, y=0, z=0$ into the determinant:
$\Delta = \left|\begin{array}{ccc}1 & (0)(0) & (0)(0) \\ (0)(0) & 1 & (0)(0) \\ (0)(0) & (0)(0) & 1\end{array}\right| = \left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|$.
The determinant of the identity matrix is $1(1-0) - 0 + 0 = 1$.
Thus,the value is $1$.

(B) Let $\Delta = \left|\begin{array}{ccc} S+c & a & b \\ c & S+a & b \\ c & a & S+b \end{array}\right|$.
Applying the operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left|\begin{array}{ccc} S+c+a+b & a & b \\ c+S+a+b & S+a & b \\ c+a+S+b & a & S+b \end{array}\right|$.
Since $a+b+c = S$,we have $S+c+a+b = S+S = 2S$.
$\Delta = \left|\begin{array}{ccc} 2S & a & b \\ 2S & S+a & b \\ 2S & a & S+b \end{array}\right| = 2S \left|\begin{array}{ccc} 1 & a & b \\ 1 & S+a & b \\ 1 & a & S+b \end{array}\right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = 2S \left|\begin{array}{ccc} 1 & a & b \\ 0 & S & 0 \\ 0 & 0 & S \end{array}\right|$.
Expanding along the first column:
$\Delta = 2S \times [1(S \times S - 0 \times 0)] = 2S \times S^2 = 2S^3$.

(A) We know that $\log _{|x|} e = \frac{1}{\log _e |x|} = \frac{1}{\ln |x|}$.
Let $y = \frac{1}{\ln |x|}$.
Using the chain rule,$\frac{d y}{d x} = \frac{d}{d x}(\ln |x|)^{-1} = -1 \cdot (\ln |x|)^{-2} \cdot \frac{d}{d x}(\ln |x|)$.
Since $\frac{d}{d x}(\ln |x|) = \frac{1}{x}$,we have $\frac{d y}{d x} = -(\ln |x|)^{-2} \cdot \frac{1}{x} = \frac{-1}{x(\ln |x|)^2}$.
Thus,the correct option is $A$.

(D) Given $x = at^2$ and $y = 2at$.
First,find $\frac{dx}{dy}$ using the chain rule:
$\frac{dx}{dy} = \frac{dx/dt}{dy/dt}$.
Since $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$,
$\frac{dx}{dy} = \frac{2at}{2a} = t$.
Now,differentiate $\frac{dx}{dy}$ with respect to $y$:
$\frac{d^2 x}{dy^2} = \frac{d}{dy}(t) = \frac{d}{dt}(t) \times \frac{dt}{dy}$.
Since $\frac{dt}{dy} = \frac{1}{dy/dt} = \frac{1}{2a}$,
$\frac{d^2 x}{dy^2} = 1 \times \frac{1}{2a} = \frac{1}{2a}$.
Thus,the correct option is $D$.

(A) Let $y = \tan^{-1} \left( \frac{1-x}{1+x} \right)$.
We know that $\tan^{-1} \left( \frac{a-b}{1+ab} \right) = \tan^{-1} a - \tan^{-1} b$.
Here,$a = 1$ and $b = x$,so $y = \tan^{-1}(1) - \tan^{-1}(x)$.
Since $\tan^{-1}(1) = \frac{\pi}{4}$,we have $y = \frac{\pi}{4} - \tan^{-1}(x)$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} \right) - \frac{d}{dx} \left( \tan^{-1} x \right)$.
$\frac{dy}{dx} = 0 - \frac{1}{1+x^2} = \frac{-1}{1+x^2}$.
Thus,the correct option is $A$.

(A) To find where the function $f(x) = x + \sqrt{1 - x}$ decreases,we find its derivative $f'(x)$.
$f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}((1 - x)^{1/2}) = 1 + \frac{1}{2}(1 - x)^{-1/2}(-1) = 1 - \frac{1}{2\sqrt{1 - x}}$.
For the function to decrease,we set $f'(x) < 0$.
$1 - \frac{1}{2\sqrt{1 - x}} < 0 \implies 1 < \frac{1}{2\sqrt{1 - x}}$.
Since $0 < x < 1$,$\sqrt{1 - x}$ is positive,so we can multiply by $2\sqrt{1 - x}$ without changing the inequality sign:
$2\sqrt{1 - x} < 1 \implies \sqrt{1 - x} < \frac{1}{2}$.
Squaring both sides:
$1 - x < \frac{1}{4} \implies 1 - \frac{1}{4} < x \implies x > \frac{3}{4}$.
Given the domain $0 < x < 1$,the function decreases on the interval $\left(\frac{3}{4}, 1\right)$.
Thus,the correct option is $A$.

(C) The given equation is $|x| + y = 1$,which can be written as $y = 1 - |x|$.
This equation represents two lines:
$1$) For $x \ge 0$,$y = 1 - x$.
$2$) For $x < 0$,$y = 1 - (-x) = 1 + x$.
The region is bounded by these lines and the $x$-axis $(y = 0)$.
The vertices of the region are $(0, 1)$,$(1, 0)$,and $(-1, 0)$.
This forms a triangle with base length $b = 1 - (-1) = 2$ and height $h = 1$.
The area of the triangle is given by $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1$ sq. units.

(B) The given differential equation is $2x \frac{dy}{dx} - y = 0$.
Rearranging the terms,we get $2x \frac{dy}{dx} = y$.
Separating the variables,we have $\frac{dy}{y} = \frac{dx}{2x}$.
Integrating both sides,$\int \frac{dy}{y} = \frac{1}{2} \int \frac{dx}{x}$,which gives $\ln|y| = \frac{1}{2} \ln|x| + C$.
This can be written as $\ln|y| = \ln|x^{1/2}| + C$,so $y = k \sqrt{x}$,where $k = e^C$.
Using the condition $y(1) = 2$,we get $2 = k \sqrt{1}$,so $k = 2$.
The solution is $y = 2 \sqrt{x}$,which implies $y^2 = 4x$.
This equation represents a parabola.

(A) The rate of change of population $p$ with respect to time $t$ is proportional to the population itself,given by the differential equation: $\frac{dp}{dt} = \frac{3}{100} p$.
Separating the variables,we get $\frac{dp}{p} = \frac{3}{100} dt$.
Integrating both sides,we have $\int \frac{dp}{p} = \int \frac{3}{100} dt$.
This results in $\ln(p) = \frac{3t}{100} + K$,where $K$ is the constant of integration.
Exponentiating both sides,we get $p = e^{\frac{3t}{100} + K} = e^K \cdot e^{\frac{3t}{100}}$.
Letting $C = e^K$,we obtain the equation $p = C e^{\frac{3t}{100}}$.

(B) Given the differential equation $e^{\frac{y}{x}} = x$.
Taking the natural logarithm on both sides,we get $\frac{y}{x} = \log x$.
Thus,$y = x \log x$.
However,the problem specifies $y(1) = 3$.
Let us re-evaluate the differential equation form. If the equation is $\frac{dy}{dx} = \frac{y}{x} + 1$,then the solution is $y = x \log x + Cx$.
Using the condition $y(1) = 3$:
$3 = 1 \cdot \log(1) + C(1) \implies 3 = 0 + C \implies C = 3$.
Therefore,the particular solution is $y = x \log x + 3x$.

(B) The first line is given by $\vec{r} = (3\hat{i} + \hat{j} - 2\hat{k}) + t(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_1} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_1(3, 1, -2)$.
The second line is given by $x = 4+k, y = -k, z = -4-2k$,which can be written as $\vec{r} = (4\hat{i} - 4\hat{k}) + k(\hat{i} - \hat{j} - 2\hat{k})$. The direction vector is $\vec{b_2} = \hat{i} - \hat{j} - 2\hat{k}$ and it passes through $P_2(4, 0, -4)$.
Since $\vec{b_1} = \vec{b_2}$,the lines are parallel.
To check if they are coincident,we check if $P_1$ lies on the second line. Substituting $P_1(3, 1, -2)$ into the second line equations:
$3 = 4+k \implies k = -1$
$1 = -k \implies k = -1$
$-2 = -4-2k \implies -2 = -4-2(-1) = -4+2 = -2$
Since $k = -1$ satisfies all equations,the point $P_1$ lies on the second line. Therefore,the lines are coincident.