WBJEE 2023 Paper 1 (Mathematics) Question Paper with Solutions

We have, $x=\sin \theta$ and $y=\sin k \theta$

Differentiate $x$ and $y$ w.r.t. $\theta$,

$$\begin{aligned} & \frac{d x}{d \theta}=\cos \theta \text { and } \frac{d y}{d \theta}=k \cos k \theta \\ & y_1=\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=\frac{k \cos k \theta}{\cos \theta} \end{aligned}$$

Again, differentiate $\frac{d y}{d x}$ w.r.t. $x$,

$$\begin{aligned} y_2 & =\frac{d^2 y}{d x^2} \\ & =\frac{\cos \theta\left[-k^2 \sin k \theta\right]-k \cos k \theta(-\sin \theta) }{\cos ^2 \theta}\cdot \frac{d \theta}{d x} \\ & =\frac{k \cos k \theta \sin \theta-k^2 \cos \theta \sin k \theta}{\cos ^3 \theta} \end{aligned}$$

$\text { Now, }\left(1-x^2\right) y_2-x y_1-\alpha y=0$

$\left(1-\sin ^2 \theta\right)\left[\frac{k \cos k \theta \sin \theta-k^2 \cos \theta \sin k \theta}{\cos ^3 \theta}\right] -\sin \theta \times \frac{k \cos k \theta}{\cos \theta}-\alpha \sin k \theta=0$

$\Rightarrow \quad k \cos k \theta \sin \theta-k^2 \cos \theta \sin k \theta -k \cos k \theta \sin \theta-\alpha \sin k \theta \cos \theta=0$

$\Rightarrow \quad \alpha=-k^2$

$$\begin{aligned} & \text { } y=e^{k x} \Rightarrow \frac{d y}{d x}=k e^{k x}=k y \\ & \Rightarrow \quad \frac{d^2 y}{d x^2}=k^2 e^{k x}=k^2 y \\ & \Rightarrow \quad\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=y \frac{d y}{d x} \\ & \Rightarrow \quad\left(k^2 y+k y\right)(k y-y)=y \times k y \\ & \Rightarrow \quad\left(k^2+k\right)(k-1) y^2=k y^2 \\ & \Rightarrow \quad k(k+1)(k-1)-k=0 \\ & \Rightarrow \quad k\left[k^2-1-1\right]=0 \\ & \Rightarrow \quad k\left(k^2-2\right)=0 \\ & \Rightarrow \quad k=0, \pm \sqrt{2} \end{aligned}$$

Given lines are $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_2: \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

$\Rightarrow$ Direction ratios of $L_1:< 2,3,4>$

Since, plane contains these two lines, so normal to plane $\| \overrightarrow{L_1} \times \overrightarrow{L_2}$

$$\begin{aligned} \mathbf{L}_1 \times \mathbf{L}_2 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right| \\ & =\hat{\mathbf{i}}(15-16)-\hat{\mathbf{j}}(10-12)+\hat{\mathbf{k}}(8-9) \\ & =-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

$\therefore$ Direction ratios of normal to plane are $<-1,2,-1>$.

$L$, passes through $(1,2,3)$.

So, plane also passes through $(1,2,3)$

$\therefore$ Its equation is

$(-1)(x-1)+2(y-2)+(-1)(z-3)=0$

i.e., $x-2 y+z=0$

Other parallel plane is $\alpha x-2 y+z-k=0$

Since, they are parallel

$\therefore \frac{\alpha}{1}=\frac{-2}{-2}=\frac{1}{1} \Rightarrow \alpha=1$

Distance between these planes $=\sqrt{6}$ units

$\Rightarrow \frac{|0-(-k)|}{\sqrt{\left(1^2\right)+(-2)^2+(1)^2}}=\sqrt{6} \Rightarrow|k|=6$

$$\begin{aligned} & \text { } y=\frac{x}{\log _e|c x|} \\ & \Rightarrow \quad y \log _e|c x|=x \\ & \Rightarrow \quad y \times \frac{1}{c x} \times c+\log _e|c x| \frac{d y}{d x}=1 \\ & \Rightarrow \quad \frac{y}{x}+\frac{x}{y}\left(\frac{y}{x}+\phi\left(\frac{x}{y}\right)\right)=1 \\ & \Rightarrow \quad \frac{y}{x}+1+\frac{x}{y} \phi\left(\frac{x}{y}\right)=1 \Rightarrow \phi\left(\frac{x}{y}\right)=\frac{-y^2}{x^2} \end{aligned}$$

We have, hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$

$A(2 \sec \theta, 3 \tan \theta), B(2 \sec \phi, 3 \tan \phi)$

Equation of normal in parametric form gives equation of normal at $A$ and $B$ as:

$$\begin{aligned} & 2 x \cos \theta+3 y \cot \theta=13 \quad \text{.... (i)}\\ & 2 x \cos \phi+3 y \cot \phi=13 \quad \text{.... (ii)}\\ & \text { Also, } \theta+\phi=\frac{\pi}{2} \Rightarrow \phi=\frac{\pi}{2}-\theta \end{aligned}$$

$\therefore$ Eq. (ii) becomes,

$$\begin{aligned} & 2 x \cos \left(\frac{\pi}{2}-\theta\right)+3 y \cot \left(\frac{\pi}{2}-\theta\right)=13 \\ & \Rightarrow \quad 2 x \sin \theta+3 y \tan \theta=13 \quad \text{.... (iii)}\\ & \sin \theta \times(1)-\cos \theta \times(3), \\ & 3 y(\cos \theta-\sin \theta)=13(\sin \theta-\cos \theta) \\ & \Rightarrow \quad y=\frac{-13}{3} \Rightarrow \beta=\frac{-13}{3} \end{aligned}$$

Let $p(x)=x^2+p x-q^2$

Since, one of root $p(x)$ be less than 2 and other be greater than 2

$$\begin{array}{lrl} & \therefore & p(2)<0 \\ & \Rightarrow & (2)^2+p(2)-q^2<0 \\ & \Rightarrow & 4+2 p-q^2<0 \end{array}$$

$$\begin{aligned} & \text { } \begin{aligned} \alpha & =\hat{\mathbf{i}}+a \hat{\mathbf{j}}+\hat{\mathbf{k}}, \boldsymbol{\beta}=0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+a \hat{\mathbf{k}}, \\ \gamma=a \hat{\mathbf{i}} & +0 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ {[\alpha, \beta \gamma] } & =\left|\begin{array}{lll} 1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1 \end{array}\right| \\ & =1(1-0)-0+a\left(a^2-1\right) \\ & =a^3-a+1=f(a) \quad \text{(Let)} \end{aligned} \end{aligned}$$

For maximum scalar triple product,

$$\begin{aligned} f^{\prime}(a) & =0 \Rightarrow 3 a^2-1=0 \\ \Rightarrow a & = \pm \frac{1}{\sqrt{3}} \end{aligned}$$

Now, $f^{\prime \prime}(a)=6 a$

$\Rightarrow f^{\prime \prime}\left(\frac{1}{\sqrt{3}}\right)>0, f^{\prime \prime}\left(\frac{-1}{\sqrt{3}}\right)<0$

$\therefore a=\frac{-1}{\sqrt{3}}$ is the point of maxima.

Given, ${P}(n)=3^{2 n+1}+2^{n+2}, n \in N$

At $n=1, P(1)=3^3+2^3=35$

Which is divisible by $(3+2)=5$

i.e. a prime number.

$$\begin{aligned} & \text { Given, } r^2 \cos ^2\left(\theta-\frac{\pi}{3}\right)=2 . \\ & \Rightarrow \quad r^2\left(\cos \left(\theta-\frac{\pi}{3}\right)\right)^2=2 \\ & \Rightarrow \quad r^2\left(\cos \theta \cos \frac{\pi}{3}+\sin \theta \sin \frac{\pi}{3}\right)^2=2 \\ & \Rightarrow \quad \quad r^2\left(\frac{\cos \theta}{2}+\frac{\sqrt{3}}{2} \sin \theta\right)^2=2 \\ & \Rightarrow \quad \frac{1}{4}(r \cos \theta+\sqrt{3} r \sin \theta)^2=2 \end{aligned}$$

$$\begin{array}{lrl} \Rightarrow & (x+\sqrt{3} y)^2=8 \\ \Rightarrow & (x+\sqrt{3} y)^2-(2 \sqrt{2})^2 & =0 \\ \Rightarrow & (x+\sqrt{3} y+2 \sqrt{2})(x+\sqrt{3} y-2 \sqrt{2}) & =0 \end{array}$$

Which represents a pair of lines.

$$\begin{aligned} & \text { Given, } I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2} \\ & =\int \frac{x}{(x \sin x+\cos x)^2} \times \frac{x \cos x}{\cos x} d x \\ & \quad \frac{d}{d x}(x \sin x+\cos x) \\ & =x \cos x+\sin x-\sin x \\ & =x \cos x \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad I= & \frac{x}{\cos x} \int \frac{x \cos x}{(x \sin x+\cos x)^2} d x \\ & \quad-\int\left(\frac{d}{d x}\left(\frac{x}{\cos x}\right) \int \frac{x \cos d x}{(x \sin x+\cos x)^2}\right) d x \quad \text{[Integrating by part]} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad I=\frac{x}{\cos x}\left(\frac{-1}{x \sin x+\cos x}\right) \\ & -\int \frac{(\cos x+x \sin x)}{\cos ^2 x}\left(\frac{-1}{x \sin x+\cos x}\right) d x \\ & \Rightarrow \quad I=\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^2 x \\ & \Rightarrow \quad I=\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+C \\ & \therefore f(x)=\frac{-x}{\cos x(x \sin x+\cos x)} \\ & \end{aligned}$$

$\text { Let } I_1=\int_0^n[x] d x$

$$\begin{aligned} & =\int_0^1 0 \cdot d x+\int_1^2 1 \cdot d x+\int_2^3 2 d x+\ldots \\ & \quad+\int_{n-2}^{n-1}(n-2) d x+\int_{n-1}^n(n-1) d x \end{aligned}$$

$$\begin{aligned} &=1(x)_0^1+2(x)_2^3+3(x)_3^4+\ldots+(n-2)(x)_{n-2}^{n-1} +(n-1)(x)_{n-1}^n \end{aligned}$$

$=1+2+3+\ldots+(n-1)=\frac{n(n-1)}{2}$

$$\begin{aligned} \text { Let } & I_2=\int_0^n\{x\} d x=\int_0^n(x-[x]) d x \\ &=\int_0^n x d x-\int_0^n[x] d x \\ &=\left(\frac{x^2}{2}\right)_0^n-l_1 \\ &=\frac{n^2}{2}-\frac{n(n-1)}{2}=\frac{n^2-n^2+n}{2}=\frac{n}{2} \\ & \therefore \quad \frac{\int_0^n[x] d x}{\int_0^n\{x\} d x}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}}=n-1 \end{aligned}$$

$$\begin{aligned} \text { Let } I & =\int_0^{2 \pi} \theta \sin ^6 \theta \cos \theta d \theta \quad \text{.... (i)}\\ & =\int_0^{2 \pi}(2 \pi-\theta) \sin ^6(2 \pi-\theta) \cos (2 \pi-\theta) d \theta \\ & =\int_0^{2 \pi}(2 \pi-\theta) \sin ^6 \theta \cos \theta d \theta \quad \text{.... (ii)} \end{aligned}$$

On adding Eqs. (i) and (ii), we get

$$\begin{aligned} 2 I & =\int_0^{2 \pi} 2 \pi \sin ^6 \theta \cos \theta d \theta \\ \Rightarrow \quad 2 I & =2 \pi \int_0^{2 \pi} \sin ^6 \theta \cos \theta d \theta \\ & =2 \pi \times 2 \int_0^\pi \sin ^6 \theta \cos \theta d \theta \quad [\because f(2 a-x)=f(x)] \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & I=2 \pi \int_0^\pi \sin ^6 \theta \cos \theta d \theta \\ \Rightarrow & I=2 \pi \times 0 \text { [since, } f(2 a-x)=-f(x) \end{array}$$

$$\begin{array}{ll} \therefore & \int_0^{2 a} f(x) d x=0] \\ \Rightarrow & I=0 \end{array}$$

Direction ratios of normal to plane $2 x-y+2 z-1=0$ are $<2,-1,2>$

Direction ratios of $X$-axis are $<1,0,0>$

Let $\theta$ be angle between the normal and $X$-axis,

$\therefore \cos \theta=\frac{|(2)(1)+(-1)(0)+(2)(0)|}{\sqrt{(2)^2+(-1)^2+\left(2^2\right)} \cdot \sqrt{1^2+0^2+0^2}}=\frac{2}{3}$

$\therefore \quad \theta=\cos ^{-1}(2 / 3)$ is the required angle.

$$\left[x^2\right]=\left\{\begin{array}{ll} 0, & 0 \leq x^2<1 \\ 1, & 1 \leq x^2<2 \\ 2, & 2 \leq x^2<3 \\ 3, & 3 \leq x^2<4 \\ 4, & 4 \leq x^2<5 \\ \vdots & \end{array} \quad\left(\because x^2 \geq 0\right)\right.$$

$$\Rightarrow\left[x^2\right]=\left\{\begin{array}{lc} 0, & 0 \leq x<1 \\ 1, & 1 \leq x<\sqrt{2} \\ 2, & \sqrt{2} \leq x<\sqrt{3} \\ 3, & \sqrt{3} \leq x<2 \\ 4, & 2 \leq x<\sqrt{5} \end{array} \quad(\because x>0)\right.$$

$$f(x)=\left[x^2\right] \sin \pi x=\left\{\begin{array}{cc} 0, & 0 \leq x<1 \\ \sin \pi x, & 1 \leq x<\sqrt{2} \\ 2 \sin \pi x, & \sqrt{2} \leq x<\sqrt{3} \\ 3 \sin \pi x, & \sqrt{3} \leq x<2 \\ 4 \sin \pi x, & 2 \leq x<\sqrt{5} \end{array}\right.$$

$$\begin{aligned} & \text { at } x=1, \mathrm{LHL}=0, \\ & \mathrm{RHL}=\sin \pi=0=f(\mathrm{l}) \end{aligned}$$

$\therefore f(x)$ is continuous at $x=1$

at $x=2, \mathrm{LHL}=3 \sin (2 \pi)=0$

$\mathrm{RHL}=4 \sin (2 \pi)=0=f(2)$

$\therefore f(x)$ is continuous at $x=2$ and so on.

$\therefore f(x)$ is continuous at only those points which are perfect square.

We have, $E_k=\{(a, b) \in S: a b=k\}$ and $p_k=P\left(E_k\right)$

$$\begin{aligned} \text { and } p_k & =P\left(E_k\right) \\ \text { Here, } E_1 & =\{(1,1)\} \\ E_2 & =\{(1,2),(2,1)\} \\ E_4 & =\{(2,2),(1,4),(4,1)\} \\ E_5 & =\{(1,5),(5,1)\} \\ E_8 & =\{(2,4),(4,2)\} \\ E_{10} & =\{(2,5),(5,2)\} \end{aligned}$$

$E_{14}, E_{16}, E_{17}$ are not possible

$$\begin{aligned} \therefore \quad p_1 & =\frac{1}{36}, p_2=\frac{2}{36}, \\ p_4 & =\frac{3}{36}, p_5=\frac{2}{36}, p_8=\frac{2}{36} \\ \text { and } p_{10} & =\frac{2}{36} \end{aligned}$$

and $p_{10}=\frac{2}{36}$

$\therefore p_1< p_{10}< p_4$ which is correct.

We have, volume of parallelopiped with $a \times b,\mathrm{b} \times \mathrm{c}$ and $\mathrm{c} \times \mathrm{a}$ is 9 cu. units:

$$\begin{array}{cc} \Rightarrow & {[\mathbf{a} \times \mathbf{b} \mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a}]=9} \\ \Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot\{(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})\}=9 \\ \Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot\{[\mathbf{b} \mathbf{c} \mathbf{a}] \mathbf{c}-[\mathbf{b} \mathbf{c} \mathbf{c}] \mathbf{a}\}=9 \\ & \{\because(\mathbf{a} \times \mathbf{b}) \times(\mathbf{c} \times \mathbf{d})=[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d}\} \\ \Rightarrow & (\mathbf{a} \times \mathbf{b})\{[\mathbf{b} \mathbf{c} \mathbf{a}] \mathbf{c}\}=9 \quad \{\because[\mathbf{b} c \mathbf{a}]=0\} \end{array}$$

$\Rightarrow \quad (\mathrm{a\times b) [b c a]=9}$

$\Rightarrow \quad[a b c][a b c]=9\quad \{\because[\mathbf{b} \mathbf{c} \mathbf{a}]=[\mathbf{a} \mathbf{b} c] \}$

$\Rightarrow \quad[\mathbf{a} \mathbf{b} \mathbf{c}]^2=\mathbf{9}$

$$\begin{aligned} & \because \quad(a \times b) \times(c \times d)=\left[\begin{array}{lll} a & b & d \end{array}\right] c-\left[\begin{array}{lll} a & b & c \end{array}\right] d \\ & \because \quad(a \times b) \times(b \times c)=\left[\begin{array}{lll} a & b & c \end{array}\right] b-\left[\begin{array}{lll} a & b & b \end{array}\right] c=\left[\begin{array}{lll} a & b & c \end{array}\right] b \quad \{\because[\mathbf{a} \mathbf{b} \mathbf{c}]=0\} \end{aligned}$$

Similarly,

$$\begin{aligned} (\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a}) & =\left[\begin{array}{lll} \mathbf{b} & \mathbf{a} \end{array}\right] \mathbf{c} \\ \text { and }(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b}) & =\left[\begin{array}{lll} \mathbf{c} & \mathbf{a} & \mathbf{b} \end{array}\right] \mathbf{a} \end{aligned}$$

Volume of required parallel $\theta$ piped

$$\begin{aligned} & =[(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})] \\ & =[[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b}[\mathbf{b} \mathbf{c} \mathbf{a}] \mathbf{c}[\mathbf{c} \mathbf{a} \mathbf{b}] \mathbf{a}] \\ & =[\mathbf{a} \mathbf{b} \mathbf{c}]^3[\mathbf{b} \mathbf{c} \mathbf{a}] \end{aligned}$$

$$=\left[\begin{array}{lll} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{array}\right]^4\left[\because\left[\begin{array}{lll} \mathbf{b} & \mathbf{c} & \mathbf{a} \end{array}\right]=\left[\begin{array}{lll} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{array}\right]\right]$$

$$\begin{aligned} & =9^2 \\ & =81 \text { cu units } \end{aligned}$$

Given, $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in GP.

$$\begin{array}{ll} \therefore & \cos ^2 \theta=\frac{1}{6} \sin \theta \tan \theta \\ \Rightarrow & 6 \cos ^2 \theta=\frac{\sin ^2 \theta}{\cos \theta} \\ \Rightarrow & 6 \cos ^3 \theta-\sin ^2 \theta-1=0 \\ \Rightarrow & (2 \cos \theta-1)\left(3 \cos ^2 \theta+2 \cos \theta+2\right)=0 \\ \Rightarrow & 2 \cos \theta-1=0 \quad \left[\because 3 \cos ^2 \theta+2 \cos \theta+2 \neq 0\right] \end{array}$$

$\Rightarrow \quad \cos \theta=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta=2 n \pi \pm \frac{\pi}{3}$

$$f(x)=\left\{\begin{array}{cc} 1, & x=1 \\ e^{\left(x^{10}-1\right)}+(x-1)^2 \sin \left(\frac{1}{x-1}\right), & x \neq 1 \end{array}\right.$$

\lim _\limits{x \rightarrow 1} f(x)=e^{1^{10}-1}+(1-1)^2 \times \quad oscillating number between ($-1$ and $1$)$=1+0=1=f(1)$

$\therefore f(x) \text { is continuous at } f(l)$