VITEEE 2022 Question Paper with Solutions

Let $P$ be the image of the origin with respect to the mirror line $x+y=1$.

Since, $O A=O B$, therefore $Q$ is the mid-point of $A B$.

So, coordinates of $Q$ are $\left(\frac{1}{2}, \frac{1}{2}\right)$.

Let the coordinates of $P$ be $\left(x_1, y_1\right)$.

Since, $Q$ is the mid-point of $O P$.

$\therefore \frac{0+x_1}{2}=\frac{1}{2}$ and $\frac{0+y_1}{2}=\frac{1}{2} \Rightarrow x_1=1, y_1=1$

Hence, the coordinates of $P$ are $(1,1)$.

Let, the circle be $x^2+y^2+2 g x+2 f y=0$, we have, $g^2+f^2=r^2$

On putting $x=0$ and $y=0$ in succession in the equation of the given circle, we get

$$\begin{aligned} & y^2+2 f y=0 \text { and } x^2+2 g x=0 \\ & \Rightarrow A \equiv(-2 g, 0) \text { and } B \equiv(0,-2 f) \end{aligned}$$

Let $P(h, k)$ be the centroid of $\triangle O A B$

$$\begin{aligned} & \therefore 3 h=-2 g \text { and } 3 k=-2 f \\ & \Rightarrow g^2+f^2=\frac{9}{4}\left(h^2+k^2\right) \end{aligned}$$

$\Rightarrow x^2+y^2=\frac{9}{4} r^2$, which is the required locus

Given curve is

$y=\frac{1}{2} x^4-5 x^3+18 x^2-19 x \quad \text{.... (i)}$

Slope of the curve $=\frac{d y}{d x}$

Differentiating Eq. (i) w.r.t. $x$,

$$\begin{aligned} & \therefore \frac{d y}{d x}=\frac{1}{2}\left(4 x^3\right)-5\left(3 x^2\right)+18(2 x)-19 \\ & \frac{d y}{d x}=2 x^3-15 x^2+36 x-19=f(x) \text { (say) } \quad \text{.... (ii)} \end{aligned}$$

Again differentiating w.r.t. $x$, we get

$$\begin{aligned} f^{\prime}(x) & =2\left(3 x^2\right)-15(2 x)+36 \\ & =6 x^2-30 x+36 \quad \text{.... (iii)} \end{aligned}$$

Equating $f^{\prime}(x)=0$

$$\begin{array}{ll} \Rightarrow & 6 x^2-30 x+36=0 \\ \Rightarrow & x^2-5 x+6=0 \\ \Rightarrow & x^2-3 x-2 x+6=0 \\ \Rightarrow & x(x-3)-2(x-3)=0 \\ \Rightarrow & (x-3)(x-2)=0 \\ & x=3 \text { and } x=2 \end{array}$$

Differentiating Eq. (iii) w.r.t. $x$, we get $f^{\prime \prime}(x)=12 x-30$

Now, $f^{\prime \prime}(2)=12(2)-30=24-30=-6<0$

and $f^{\prime \prime}(3)=12(3)-30=36-30=6>0$

Since $f^{\prime \prime}(2)<0 \Rightarrow 2$ is point of maximum

$\therefore$ At $x=2$, slope will be maximum.

$\therefore$ From Eq. (i)

$$\begin{aligned} y= & \frac{1}{2}(2)^4-5(2)^3+18(2)^2-19(2) \\ \quad & \quad \text { substituting } x=2 \text { in Eq. (i)] } \\ y= & 8-40+72-38 \\ = & 80-78=2 \end{aligned}$$

$\therefore$ Maximum slope occurs at point $(2,2)$

Given, $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}=\int \frac{d x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}}$

$$\begin{aligned} & \text { Put } 1+\frac{1}{x^4}=t^4 \\ & \Rightarrow \quad \frac{-4}{x^5} d x=4 t^3 d t \\ & \Rightarrow \quad \frac{d x}{x^5}=-t^3 d t \\ & \therefore \text { Required integral }=\int \frac{-t^3}{t^3} d t \\ & =-\int d t=-t+C \\ & =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+C \\ \end{aligned}$$

Six faces have numbers $(1,2,3,5,7,11)$ Sum of numbers on top faces of both the sides is less than or equal to 8 is

$$\begin{aligned} & =(1,1)(1,2)(1,3)(1,5)(1,7) \rightarrow 5 \\ & (2,1)(2,2)(2,3)(2,5) \rightarrow 4 \\ & (3,1)(3,2)(3,3)(3,5) \rightarrow 4 \\ & (5,1)(5,2)(5,3) \rightarrow 3 \\ & (7,1) \rightarrow 1 \\ & n(S)=5+4+4+3+1=17 \\ & n(T)=6 \times 6=36 \\ & \therefore \text { Required probability }=\frac{17}{36} \end{aligned}$$

Given, \lim _\limits{x \rightarrow \infty}\left[\frac{p^{1 / x}+q^{1 / x}+r^{1 / x}+s^{1 / x}}{4}\right]^{3 x}, p, q,r, s>0

Let us put $\frac{1}{x}=z$ so that $x \rightarrow \infty \Rightarrow z \rightarrow 0$

\therefore \lim _\limits{x \rightarrow \infty}\left[\frac{p^{1 / x}+q^{1 / x}+r^{1 / x}+s^{1 / x}}{4}\right]^{3 x}

=\lim _\limits{z \rightarrow 0}\left[\frac{p^z+q^z+r^z+s^z}{4}\right]^{\frac{3}{z}}, which is the form of $1^{\infty}$.

\begin{aligned} & =e^{\lim _\limits{z \rightarrow 0}\left[\frac{p^z+q^z+r^z+s^z}{4}-1\right] 3 / z} \\ & =e^{\lim _\limits{z \rightarrow 0}\left[\frac{p^z-1}{z}+\frac{q^z-1}{z}+\frac{r^z-1}{z}+\frac{s^z-1}{z}\right] 3 / 4} \\ & =e^{\frac{3}{4}\left(\log e^p+\log e^q+\log e^r+\log e^s\right)} \\ & =e^{\frac{3}{4} \log _e(p q r s)}=(p q r s)^{\frac{3}{4}} \end{aligned}

Let a variable plane cuts the coordinate axes is $A(a, 0,0), B(0, b, 0), C(0,0, c)$ and $(\alpha, \beta, \gamma)$ be the coordinates of centroid of the tetrahedron $O A B C$

$\therefore \alpha=\frac{a}{4}, \beta=\frac{b}{4}, \gamma=\frac{c}{4} \Rightarrow a=4 \alpha, b=4 \beta, c=4 \gamma$

Since, the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ is at constant distance $p$ from the origin.

$\therefore p=\frac{0+0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$

$$\begin{aligned} & \Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \frac{1}{p^2}=\frac{1}{(4 \alpha)^2}+\frac{1}{(4 \beta)^2}+\frac{1}{(4 \gamma)^2} \end{aligned}$$

Hence, required locus is $16 p^{-2}=x^{-2}+y^{-2}+z^{-2}$

Given, $\frac{d y}{d x}=\frac{1}{\cos (x-y)}=\sec (x-y)$

Put $t=x-y$

$$\begin{array}{ll} & \frac{d t}{d x}=1-\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d t}{d x} \\ \Rightarrow & 1-\frac{d t}{d x}=\sec t \Rightarrow \frac{d t}{d x}=1-\sec t \\ \Rightarrow \quad & \frac{d t}{1-\sec t}=d x \\ \Rightarrow \quad & \frac{\cos t}{\cos t-1} d t=d x \\ \Rightarrow \quad & \frac{\cos t-1+1}{\cos t-1} d t=d x \end{array}$$

$$\begin{array}{ll} \Rightarrow & \left(1+\frac{1}{\cos t-1}\right) d t=d x \\ \Rightarrow & \left(1-\frac{1}{1-\cos t}\right) d t=d x \\ \Rightarrow & \left(1-\frac{1}{2 \sin ^2 t / 2}\right) d t=d x \\ \Rightarrow & \left(1-\frac{1}{2} \operatorname{cosec}^2 \frac{t}{2}\right) d t=d x \end{array}$$

Taking integration on both sides,

$\int\left(1-\frac{1}{2} \operatorname{cosec}^2 \frac{t}{2}\right) d t=\int d x+C$

$$\begin{array}{ll} \Rightarrow & t-\frac{1}{2} \frac{\left(-\cot \frac{t}{2}\right)}{1 / 2}=x+C \\ \Rightarrow & t+\cot \frac{t}{2}=x+C \\ \Rightarrow & x-y+\cot \frac{x-y}{2}=x+C \\ \Rightarrow \quad & y=\cot \left(\frac{x-y}{2}\right)+C \end{array}$$

Let $P\left(Z_1\right), Q\left(Z_2\right)$ and $R\left(Z_3\right)$ be the vertices an isosceles triangle $P Q R$ right angled at $Q$. Let us rotate $Q\left(Z_2\right)$ about $P$ in anti-clockwise direction, we get

$$\begin{aligned} & Z_3-Z_1=\frac{\left|Z_3-Z_1\right|}{\left|Z_2-Z_1\right|}\left(Z_2-Z_1\right) e^{\frac{i \pi}{4}} \\ & Z_3-Z_1=\frac{P R}{P Q}\left(Z_2-Z_1\right) e^{\frac{i \pi}{4}}=\sqrt{2}\left(Z_2-Z_1\right) e^{\frac{i \pi}{4}} \ldots \text { (i) } \end{aligned}$$

Also by rotating $Q\left(Z_2\right)$ about $R\left(Z_3\right)$ in clockwise direction.

$$\begin{aligned} Z_1-Z_3 & =\frac{\left|Z_1-Z_3\right|}{\left|Z_2-Z_3\right|}\left(Z_2-Z_3\right) e^{-i \frac{\pi}{4}} \\ \quad \quad \quad \quad Z_1-Z_3 & =\frac{P R}{Q R}\left(Z_2-Z_3\right) e^{-i \frac{\pi}{4}} \\ & =\sqrt{2}\left(Z_2-Z_3\right) e^{-i \frac{\pi}{4}}\quad \text{.... (ii)} \end{aligned}$$

Multiply Eqs. (i) and (ii), we get

$$\begin{aligned} & \left(Z_3-Z_1\right)\left(Z_1-Z_3\right)=2\left(Z_2-Z_1\right)\left(Z_2-Z_3\right) \\ & \Rightarrow\left(Z_3-Z_1\right)^2=2\left(Z_1-Z_2\right)\left(Z_2-Z_3\right) \end{aligned}$$

Which is the required condition.

$$\begin{array}{ll} \text { Given, } & X \phi(x)=\int_5^x 3 t^2-2 \phi^{\prime}(t) d t \\ \Rightarrow & X \phi(x)=\left[t^3-2 \phi(t)\right]_5^x \\ \Rightarrow & X \phi(x)=\left[x^3-125\right]-2[\phi(x)-\phi(5)] \end{array}$$

Now, $\phi(0)=4$

$$\begin{array}{ll} \Rightarrow & 0=-125-2[4-\phi(5)] \\ \Rightarrow & \phi(5)=\frac{133}{2} \end{array}$$

$$\begin{aligned} & \begin{array}{l} \text { For } \phi(2), \\ \Rightarrow 2 \phi(2)=(8-125)-2\left[\phi(2)-\frac{133}{2}\right] \end{array} \\ & \Rightarrow \quad 4 \phi(2)=16 \\ & \Rightarrow \quad \phi(2)=4 \\ \end{aligned}$$

Let $Q\left(x_1, y_1, z_1\right)$ be the coordinates of the image of the point $P(2,-3,4)$ with respect to the given plane $4 x+2 y-4 z+3=0$.

Then, the mid-point $\left(\frac{x_1+2}{2}, \frac{y_1-3}{2}, \frac{z_1+4}{2}\right)$

$\therefore 4\left(\frac{x_1+2}{2}\right)+2\left(\frac{y_1-3}{2}\right)-4\left(\frac{z_i+4}{2}\right)+3=0$

$\Rightarrow 2 x_1+y_1-2 z_1=4$ and the line $P Q$ is parallel to the normal to the given plane $4 x+2 y-4 z+3=0$

$$\begin{array}{ll} \therefore & \frac{x_1-2}{4}=\frac{y_1+3}{2}=\frac{z_1-4}{-4}=k \text { (say) } \\ \Rightarrow & x_1=2+2 k, y_1=k-3, z_1=4-2 k \end{array}$$

Now,

$$\begin{aligned} & 2 x_1+y_1-2 z_1=4 \text { gives } \\ & 4+4 k-3+k-8+4 k=4 \\ & \Rightarrow \quad k=\frac{11}{9} \\ & \therefore \quad x_1=2+\frac{22}{9}=\frac{40}{9} \text {, } \\ & y_1=\frac{11}{9}-3=-\frac{16}{9} \\ & \text { and } \quad z_1=4-\frac{22}{9}=\frac{14}{9} \\ \end{aligned}$$

Hence, the image of the point $(2,-3,4)$ with respect to the given plane is $\left(\frac{40}{9}, \frac{-16}{9}, \frac{14}{9}\right)$

$\because$ Displacement $A B$

$$\begin{aligned} &=(3-2) \hat{\mathbf{i}}+(1+1) \hat{\mathbf{j}}+(2-1) \hat{\mathbf{k}} \\ &=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & \text { and force, } \mathbf{F}=\frac{\sqrt{6}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{6}}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \therefore \text { Work done }=\mathrm{F} \cdot \mathrm{AB} \\ &=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=6 \end{aligned}$$

Given, equation is

$$\begin{aligned} & \cos ^{-1} x-\cos ^{-1} \frac{y}{2}=\alpha \text { where, }-1 \leq x \leq 1, -2 \leq y \leq 2 \text { and } x \leq \frac{y}{2} \\ & \therefore \cos ^{-1}\left(x \cdot \frac{y}{2}+\sqrt{1-x^2} \sqrt{1-\left(\frac{y}{2}\right)^2}\right)=\alpha \\ & (\because \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}(x y+\sqrt{1-x^2} \sqrt{1-y^2})\\ & \Rightarrow \frac{x y^{\prime}}{2}+\sqrt{1-x^2} \sqrt{1-\left(\frac{y}{2}\right)^2}=\cos \alpha \\ & \sqrt{1-x^2} \sqrt{1-(y / 2)^2}=\cos \alpha-\frac{x y}{2} \end{aligned}$$

On squaring both sides, we get

$$\begin{aligned} & \left(1-x^2\right)\left(1-\frac{y^2}{4}\right)=\cos ^2 \alpha+\frac{x^2 y^2}{4}-\frac{2 x y}{2} \cos \alpha \\ & \Rightarrow\left(1-x^2\right)-\frac{y^2}{4}+\frac{x^2 y^2}{4}=\cos ^2 \alpha+\frac{x^2 y^2}{4} -xy \cos \alpha\\ & \Rightarrow x^2+\frac{y^2}{4}-x y \cos \alpha=1-\cos ^2 \alpha \\ & \Rightarrow 4 x^2-4 x y \cos \alpha+y^2=4 \sin ^2 \alpha \end{aligned}$$

Order The order of a differential equation is the highest derivative appearing in it.

Degree The degree of a differential equation is the power of the highest derivatives

Here, $\left(\frac{d^3 y}{d x^3}\right)^3\left(\frac{d^2 y}{d x^2}\right)^4+4\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d^3 y}{d x^3}\right)^4$

$=\left(x^2-1\right)\left(\frac{d^3 y}{d x^3}\right)^3$

So, here, $p=3, r=4$

$\frac{d y}{d x}+x(x+y)=x^3(x+y)^3-1$

$\Rightarrow\left(\frac{d y}{d x}+1\right)+x(x+y)=x^3(x+y)^3$

or $\frac{d y+d x}{d x}+x(x+y)=x^3(x+y)^3$

$\begin{gathered} \text{Put}\quad x+y=v \\ \qquad d x+d y=d v \end{gathered}$

Then, $\frac{d v}{d x}+v x=v^3 x^3$

or $\quad v^{-3} \frac{d v}{d x}+v^{-2} x=x^3\quad \text{.... (i)}$

Now, put $v^{-2}=t$

$\therefore \quad-2 v^{-3} \frac{d v}{d x}=\frac{d t}{d x}$

Then, from Eq. (i)

$$\begin{aligned} \frac{d t}{d x}-2 t x & =-2 x^3 \\ I F & =e^{\int-2 x d x}=e^{-x^2} \end{aligned}$$

$\therefore$ Solution is

$$\begin{aligned} & t\left(e^{-x^2}\right)=\int\left(-2 x^3\right)\left(e^{-x^2}\right) d x \\ & \Rightarrow \quad t\left(e^{-x^2}\right)=x^2 e^{-x^2}+e^{-x^2}+C \\ & \text { or } \quad t=x^2+1+c e^{x^2} \\ & \text { or } \quad \frac{1}{(x+y)^2}=x^2+1+c e^{x^2} \\ \end{aligned}$$

Given, $\mathop {\lim }\limits_{t \to \infty } {{\ln \left( {{3 \over 2}t} \right)} \over {{t^2}}}$

by use L' Hospital rule,

$\mathop {\lim }\limits_{t \to \infty } {{{d \over {dt}}\left( {\ln \left( {{3 \over 2}t} \right)} \right)} \over {{d \over {dt}}({t^2})}} \Rightarrow \mathop {\lim }\limits_{t \to \infty } {{{{{1 \over {3t}}} \over 2}\left( {{3 \over 2}} \right)} \over {2t}}$

$\Rightarrow \mathop {\lim }\limits_{t \to \infty } {1 \over {2{t^2}}} = 0\,\left( {{1 \over \infty } = 0} \right)$

Given, $\frac{d y}{d x}=\frac{1}{\cos (x+y)}=\sec (x+y)$

Put $t=x+y$

$$\begin{array}{ll} \Rightarrow & \frac{d t}{d x}=1+\frac{d y}{d x} \\ & \frac{d y}{d x}=\frac{d t}{d x}-1 \\ \Rightarrow & \frac{d t}{d x}-1=\sec t \\ \Rightarrow & \frac{d t}{d x}=1+\sec t \\ \Rightarrow & \frac{d t}{1+\sec t}=d x \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{\cos t}{\cos t+1} d t=d x \\ \Rightarrow & \frac{\cos t+1-1}{\cos t+1} d t=d x \\ \Rightarrow & \left(1-\frac{1}{1+\cos t}\right) d t=d x \end{array}$$

Taking integration both sides

$$\begin{array}{ll} & \int\left(1-\frac{1}{1+\cos t}\right) d t=\int d x+C \\ \Rightarrow & \int 1 \cdot d t-\int \frac{1}{2 \cos ^2 \frac{t}{2}} d t=x+C \\ \Rightarrow & t-\frac{1}{2} \int \sec ^2 \frac{t}{2} d t=x+C \\ \Rightarrow & x+y-\tan \frac{t}{2}=x+C \\ \Rightarrow & y=\tan \left(\frac{x+y}{2}\right)+C \end{array}$$

According to the question,

$$\begin{aligned} & t_2 \text { in }(a+b)^n={ }^n C_1 a^{n-1} b=240 \quad \text{.... (i)}\\ & t_3 \text { in }(a+b)^n={ }^n C_2 a^{n-2} b^2=720 \quad \text{.... (ii)}\\ & t_4 \text { in }(a+b)^n={ }^n C_3 a^{n-3} b^3=1080 \quad \text{... (iii)} \end{aligned}$$

Now, dividing Eq. (i) by Eq. (ii), we get

$\frac{n a^{n-1} b}{\frac{n(n-1)}{2} a^{n-2} b^2}=\frac{240}{720}$

$\Rightarrow \quad \frac{2 a}{(n-1) b}=\frac{1}{3} \quad \text{.... (iv)}$

Also dividing Eq. (ii) by Eq. (iii), we get

$$\begin{aligned} & \frac{\frac{n(n-1)}{2} a^{n-2} b^2}{\frac{n(n-1)(n-2)}{6} a^{n-3} b^3}=\frac{720}{1080} \\ \Rightarrow \quad & \frac{3 a}{(n-2) b}=\frac{2}{3} \quad \text{.... (v)} \end{aligned}$$

On dividing Eq. (iv) by Eq. (v), we get

$$\begin{aligned} & \Rightarrow \quad \frac{\frac{2 a}{(n-1) b}}{\frac{3 a}{(n-2) b}}=\frac{\frac{1}{3}}{\frac{2}{3}} \\ & \Rightarrow \quad \frac{2(n-2)}{3(n-1)}=\frac{1}{2} \\ & \Rightarrow \quad 4 n-8=3 n-3 \Rightarrow n=5 \\ \end{aligned}$$

From Eq. (iv)

$$\begin{aligned} & 2 a=\frac{4 b}{3} \\ & \Rightarrow \quad a=\frac{2 b}{3} \Rightarrow \quad b=\frac{3 a}{2} \end{aligned}$$

From Eq. (i),

$$\begin{array}{ll} & { }^5 C_1 a^4 b=240 \\ \Rightarrow & 5 \times a^4 \times \frac{3 a}{2}=240 \\ \Rightarrow & a^5=32=2^5 \\ \Rightarrow & a=2 \Rightarrow b=\frac{3 a}{2}=3 \end{array}$$

Hence, $(n, b, a)=(5,3,2)$