Let $f$ be a differentiable function on $\mathbb{R}$ such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then find the value of $f(-3)$.

Solution: Slope of tangent at $P(x, y)$ is $\frac{d y}{d x}$

$\therefore$ Eq of the tangent at $P(x, y)$ is given by

$y-y=\frac{d y}{d x}(X-x)$

Putting $X=0$, we get $Y=y-x \frac{d y}{d x}$

This gives the $y$-intercept abscissa of $P(x, y)$ is $x$,

given $y-x \frac{d y}{d x}=x^3$

$\Rightarrow \quad \frac{d y}{d x}-\frac{1}{x} y=-x^2 \leftarrow \text { linear 1st order ODE }$

$\text { I.F. }=e^{\int \frac{-1}{x} d x}=e^{-\ln x}=\frac{1}{x}$

So, $y \cdot \frac{1}{x}=\int \frac{1}{x}\left(-x^2\right) d x=\frac{-x^2}{2}+C$

$\Rightarrow \quad y=c x-\frac{x^3}{2}$

when $x=1, y=1 \Rightarrow 1=c-\frac{1}{2} \Rightarrow c=\frac{3}{2}$

$\therefore \quad y =f(x)=\frac{3}{2} x-\frac{x^3}{2}$

$\therefore \quad f(-3) =\frac{3}{2}(-3)-\frac{(-3)^3}{2}$

$=\frac{-9}{2}+\frac{27}{2}=9 .$

Q2. Let P denote a curve $y=y(x)$ which is in the first quadrant and let the point $(1,0)$ lie as it.Let the tangent to P at a point $P$ intersect the $y$-axis at $Y_p$. If $P Y_p$ has length 1 for each point $P$ on P, then

$(A)\quad y=\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$

$(B) \quad x y^{\prime}+{\sqrt{1-x^2}}^2=0$

$(C) \quad y=-\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$

$(D) \quad x y^{\prime}-\sqrt{1-x^2}=0$

Solution: Let $P \equiv(x, y)$ Then $Y_p \equiv\left(0, y-x \frac{d y}{d x}\right)$ (By the previous problem) given $P Y_P=1$

$\Rightarrow 1=p y_p^2=x^2+\left(x \frac{d y}{d x}\right)^2$

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

$\text { Since }\left(\frac{d y}{d x}\right)^2 \geqslant 0 \Rightarrow x^2 \leq 1 \Rightarrow-1 \leq x \leq 1$

$\therefore \quad \frac{d y}{d x}= \pm \frac{\sqrt{1-x^2}}{x},$

(because P lies in the $1^{st}$ quadrant, x>0) $0<x<1$

If $\frac{d y}{d x}>0$ for some $x_1 \in(0,1)$ and $\frac{d y}{d x}<0$ for some $x_2 \in(0,1)$ then by the continuity $\frac{d y}{d x}$ must be zero at some $x \in(0,1)$ However, $\left(\frac{d y}{d x}\right)^2=\frac{1-x^2}{x^2} \neq 0 \quad \forall x \in(0,1)$

$\therefore$ either $\frac{d y}{d x}=\frac{\sqrt{1-x^2}}{x}$ or $\frac{d y}{d x}=-\frac{\sqrt{1-x^2}}{x}$ for all $x \in(0,1)$.

Now if $\frac{d y}{d x}>0$ in $(0,1)$, then $y(x)$ is increasing in $(0,1)$

Also, it is given that $y(1)=0 \Rightarrow y(x)<0$ in $(0,1)$

$\Rightarrow P$ lies in the forth quadrant.

This is not true. $\frac{d y}{d x}=-\frac{\sqrt{1-x^2}}{x} \text { for } x \in(0,1)$ $\Rightarrow x y^{\prime}+\sqrt{1-x^2}=0$

Integrating, we get $y=-\int \frac{\sqrt{1-x^2}}{x} d x$

$[\text { Put } x=\sin \theta, \\ \text { Then } d x=\cos \theta d \theta, \\ \sqrt{1-x^2}=\cos \theta ]$

$=-\int \frac{\cos \theta}{\sin \theta} \cos \theta d \theta$ $=\int \frac{\sin ^2 \theta-1}{\sin \theta} d \theta$ $=\int \sin \theta d \theta-\int \operatorname{cosec} \theta d \theta$ $=-\cos \theta+\ln |\operatorname{cosec} \theta+\cot \theta|+c$

$\Rightarrow y=-\sqrt{1-x^2}+\ln \left(\frac{1+\sqrt{1-x^2}}{x}\right)+c$

since $y(1)=0 \Rightarrow c=0$ $\therefore \quad y=-\sqrt{1-x^2}+\ln \left(\frac{1+\sqrt{\left.1-x^2\right)}}{x}\right)$

A solution curve of the differential equation

$\left(x^2+x y+4 x+2 y+4\right) \frac{d y}{d x}=y^2, \quad x>0$

passes though the point $(1,3)$. Then the solution curve

(A) intersets $y=x+2$ at exactly one point.

(B) intersets $y=x+2$ at exactly tor points.

(C) intersets $y=(x+2)^2$

(D) does not intersect $y=(x+3)^2.$

Solution.

$\frac{d y}{d x} =\frac{y^2}{x^2+x y+4 x+2 y+4}$

$=\frac{y^2}{(x+2)^2+y(x+2)}=\frac{y^2}{(x+2)(x+2+y)}$

$=\frac{(y / x+2)^2}{\left(1+\frac{y}{x+2}\right)}$

put $y=u(x+2) \Rightarrow \frac{d y}{d x}=u+(x+2) \frac{d u}{d x}$

$\therefore \quad u+(x+2) \frac{d u}{d x}=\frac{u^2}{1+u}$

$\Rightarrow \quad(x+2) \frac{d u}{d x}=\frac{u}{1+u}-u=\frac{-u}{1+u}$

$\Rightarrow \quad \int \frac{1+u}{u} d u$

$=\int \frac{-1}{x+2} d x$

$\Rightarrow \quad \ln |u|+u=-\ln |x+2|+c$

$\Rightarrow \quad \ln |u(x+2)|+u=c$

$\Rightarrow \quad \ln |y|+\frac{y}{x+2}=c$

$y(1)=3 \Rightarrow \ln 3+\frac{3}{1+2}$

$=c \rightarrow c=1+\ln 3$

$\therefore \quad \ln | y \mid+\frac{y}{x+2}=1+\ln 3$

Putting $y=x+2$, we get

$\ln |x+2|+1=1+\ln 3$

$\Rightarrow \quad \ln |x+2|=\ln 3$

$\Rightarrow \quad \ln (x+2)=\ln 3 \quad(since \quad x>0)$

$\Rightarrow \quad x+2=3 \Rightarrow x=1$

Putting $y=(x+2)^2$ in $(*)$, we get

$\ln (x+2)^2+\frac{(x+2)^2}{x+2}=1+\ln 3$

$\Rightarrow 2 \ln (x+2)+(x+2)=1+\ln 3$$\text { for any } x>0, \text { L.H.S. }>2+2 \ln 2$

$=2+\ln 4>1+\ln 3$

$\therefore$ L.H.S $\neq$ R.H.S for every $x>0$

Now, putting $y=(x+3)^2$ in $(*)$, we get

$\ln (x+3)^2+\frac{(x+3)^2}{x+2}=1+\ln 3$

$\Rightarrow 2 \ln (x+3)+\frac{(x+3)^2}{x+2}=1+\ln 3-(**)$

$x>0 \Rightarrow 2 \ln (x+3)+\frac{(x+3)^2}{x+2}>2 \ln 3+1$

Question-4 Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy $y(2)=\frac{2}{\sqrt{3}}$. Consider two statements:

(1) $y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$ (2) $y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$.

Then (A) Both (1) & (2) are true (B) (1) is true but (2) is false (C) (1) is false but (2) is true

(2) Both (I) & (2) are false.

Solution: We have $\frac{d y}{y \sqrt{y^2-1}}=\frac{d x}{x \sqrt{x^2-1}}$ Note that $\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{|x| \sqrt{x^2-1}} \ldots$

By integrating, we get $\sec ^{-1} y =\sec ^{-1}(x)+C$

$y(2)=\frac{2}{\sqrt{3}} \Rightarrow \sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\sec ^{-1}(2)+C$

$\Rightarrow C=\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{6}$

$\therefore \sec ^{-1} y =\sec ^{-1} x-\frac{\pi}{6}$ $\Rightarrow \quad y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$ $\Rightarrow \frac{1}{y} =\cos \left(\sec ^{-1} x-\frac{\pi}{6}\right)$

$=\cos \left(\operatorname{sec}^{-1} x\right)$

$\cos \frac{\pi}{6}+\sin \left(\sec ^{-1}(x)\right) \sin \frac{\pi}{6}$

$=\frac{1}{x} \cdot \frac{\sqrt{3}}{2}+\sqrt{1-\frac{1}{x^2}} \cdot \frac{1}{2}$ $=\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}}$

Let $y(x)$ be $a$ solution of the differential equation.

$\left(1+e^x\right) y^{\prime}+y e^x=1 \text {. If } y(0)=2 \text {, then }$

(A) $\quad y(-4)=0$

(B) $\quad y(-2)=0$

(C) $y(x)$ has a critical point in $(-1,0)$

(D) $y(m)$ has no critical point in $(-1,0)$

Solution'$\left(1+e^x\right) y^{\prime}+y e^x=1$

$\Rightarrow \quad \frac{d}{d x}\left[\left(1+e^x\right) y\right]=1$

$\Rightarrow \quad\left(1+e^x\right) y=x+c$

$\Rightarrow \quad y=\frac{x+c}{1+e^x}$

$y(0)=2 \Rightarrow 2=\frac{c}{2} \Rightarrow c=4$

$\therefore \quad y=\frac{x+4}{1+e^x} \Rightarrow y(-4)=0$

$y(-2)=\frac{2}{1+e^{-2}}>0$

$\therefore(A)$ is correct

(B) is wrong$y(x)=\frac{x+4}{1+e^x}$ $\Rightarrow y^{\prime}(x)=\frac{\left(1+e^x\right) \cdot 1-(x+4) e^x}{\left(1+e^x\right)^2}$

$\therefore y^{\prime}(0)=\frac{2-4}{2^2}=\frac{-2}{4}=\frac{-1}{2}<0$

$y^{\prime}(-1)=\frac{\left(1+e^{-1}\right)-3 e^{-1}}{\left(1+e^{-1}\right)^2}$

$=\frac{1-\frac{2}{e}}{\left(1+e^{-1}\right)^2}=\frac{e-2}{e\left(1+\frac{1}{e}\right)^2}>0$

By the intermediate value theorem , $y^{\prime}(x)=0$ for some $x \in(-1,0)$.

ie. $y(x)$ has a critical point in $(-1,0)$.

Suppose $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y) \quad \forall x, y \in \mathbb{R}$ and $f(0)=1$. The find the value of $\ln f(u)$.

Solution: $f(x+y)=f(x) f^{\prime}(y)+f^{\prime}(x) f(y) \quad \forall x, y \in \mathbb{R}$

Putting $x=0, y=0$, we get

$f(0) =f(0) f^{\prime}(0)+f^{\prime}(0) f(0)=2 f(0) f^{\prime}(0)$

$\Rightarrow \quad 1 =2 f^{\prime}(0) \Rightarrow f^{\prime}(0)=\frac{1}{2}$

putting $y=0$ is given equation, we get

$\quad f(x) =f(x) f^{\prime}(0)+f^{\prime}(x) f(0)$ $=f(x) \cdot \frac{1}{2}+f^{\prime}(x) \cdot 1$ $\Rightarrow \quad f^{\prime}(x) =\frac{1}{2} f(x)$ $\Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}$

$=\frac{1}{2} \Rightarrow \ln f(x)=\frac{x}{2}+c$

$f(0)=1 \Rightarrow c=0$

$\therefore \quad \ln f(x) =\frac{x}{2} \Rightarrow \ln f(u)=\frac{4}{2} = 2$

Let's check that this $f(x)$ satisfies the given functional equation.

$\ln f(x)=\frac{x}{2} \Rightarrow f(x)=e^{x / 2}$ $f(x+y)=e^{\frac{x+y}{2}}=e^{x / 2} \cdot e^{y / 2}$

$f(x) f^{\prime}(y)+f^{\prime}(x) f(y)$ $=e^{x / 2} \cdot \frac{1}{2} e^{y / 2}+\frac{1}{2} e^{x / 2} \cdot e^{y / 2}$

$=e^{x / 2} \cdot e^{y / 2}$ $\therefore$ It is satisfied.

Q7. The differential equation. $\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of circles with

(A) variable radii and a fixed centre at $(0,1)$

(B) variable radii and a fixed centre at $(0,-1)$

(C) fixed radius 1 and variable centres along the $x$-axis.

(D) fixed radius 1 and variable centres along the $y$-axis.

Solution $\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y} \quad \text { Variables separable DE }$

$\therefore \int \frac{y^2}{\sqrt{1-y^2}} d y=\int d x$

Put $1-y^2=u^2 \Rightarrow-2 y d y=2 u d u$

$\therefore \quad y d y=-u d u$

$\int \frac{y}{\sqrt{1-y^2}} d y=\int \frac{-u d u}{u}$ $=-u+c=-\sqrt{1-y^2}+c$

$\therefore \quad x=-\sqrt{1-y^2}+c$

$\Rightarrow \sqrt{1-y^2}=c-x$ $\Rightarrow 1-y^2=(c-x)^2=(x-c)^2$ $(x-c)^2+y^2=1$

where a c is arbitrary constant.A circle with centre at $(c, 0)$ and radius 1 . lies on the x-axis