Differential Calculus for JEE Main and Advanced

Problem 13

$x^{2}+y^{2}-\frac{10}{3} y+1=0$
$x^{2}+\left(y-\frac{5}{3}\right)^{2}=\left(\frac{4}{3}\right)^{2}$
$y^{2}=x^{3}$
$2 y y_{1}=3 x^{2}$
$y_{1}=\frac{3 x^{2}}{2 y}$
Normal eqn $\quad y-y_{1}=-\frac{2 y_{1}}{3 x_{1}^{2}}\left(x-x_{1}\right)$
as it passes through $\left(0, \frac{5}{3}\right)$
$\frac{5}{3}-y_{1}=\frac{2 y_{1}}{3 x_{1}}$
$5-3 y_{1}=2 y^{1 / 3}$
$5-9 y^{3}-225 y+135 y^{2}=8 y$
$9 y^{3}+233 y-135 y^{2}-25=0$

Akshaya Rs

Numerade Educator

03:19

Problem 14

$x^{3}+y^{3}=a^{3}$
$x^{2}+y^{2} \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{-x^{2}}{y^{2}}$
Eqn of tangent $y-y_{1}=\frac{-x^{2}}{y_{1}^{2}}\left(y-x_{1}\right)$
$y_{1}^{2} y-y_{1}^{3}=-x_{1}^{2} x+x_{1}^{3}$
$y_{1}^{2} y+x_{1}^{2} x=a^{3}$
Now, $y_{1}^{2} k+x_{1}^{2} h=a^{3}$
$\frac{k}{y_{1} x_{1}^{3}}+\frac{h}{x_{1} y_{1}^{3}}=\frac{a^{3}}{x_{1}^{3} y_{1}^{3}}$
$\frac{B}{x_{1}^{3}}+\frac{A}{y_{1}^{3}}=\frac{a^{3}}{x_{1}^{3}\left(a^{3}-x_{1}^{3}\right)}$
$\frac{B}{x_{1}^{3}}+\frac{A}{a^{3}-x_{1}^{3}}=\frac{a^{3}}{x_{1}^{3}\left(a^{3}-x_{1}^{3}\right)}$
$a^{3} B-x_{1}^{3} B+x_{1}^{3} A=a^{3}$
$x_{1}^{3}=\frac{a^{3}(1-B)}{A-B}$

Varsha Aggarwal

Numerade Educator

Problem 15

$\mathrm{h}=\mathrm{ar}+\mathrm{b}$
$\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{a} \frac{\mathrm{dr}}{\mathrm{dt}} \Rightarrow \mathrm{a}=3, \mathrm{~b}=3$
$\mathrm{v}=3 \pi \mathrm{r}^{2}(\mathrm{r}+1)$
$\frac{\mathrm{dv}}{\mathrm{dt}}=3 \pi\left(3 \mathrm{r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}}+2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}\right)$
$1=3 \pi\left(3-6^{2}+12\right) \frac{\mathrm{dr}}{\mathrm{dt}}$
$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{360 \pi}$
Now, when $\mathrm{r}=36$
$\frac{d v}{d t}=3 \pi\left(3 \cdot(36)^{2}+72\right) \times \frac{1}{360 \pi}$
$=33$

Varsha Aggarwal

Numerade Educator

01:38

Problem 16

At $x=\frac{\pi}{3}, y=\frac{3 \sqrt{3}}{2}$
$\frac{d y}{d x}=2 \cos x+2 \cos 2 x$
$\frac{\mathrm{dy}}{\mathrm{dx}}=0$
Tangent $\Rightarrow y=\frac{3 \sqrt{3}}{2}$
Normal $\Rightarrow x=\frac{\pi}{3}$
Area $=\frac{\pi}{3} \times \frac{3 \sqrt{3}}{2}=\frac{\sqrt{3} \pi}{2}$

Akshaya Rs

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02:55

Problem 17

$\mathrm{r}=\mathrm{OA}=\mathrm{OQ}=10$
Let $<\mathrm{BOQ}=\theta$
$\angle \mathrm{OQP}=\theta / 2, \quad \angle \mathrm{OPQ}=\frac{\pi}{2}+\frac{\theta}{2}$
In $\Delta \mathrm{OPQ}$,
$\frac{\mathrm{OP}}{\sin \theta / 2}=\frac{10}{\cos \theta / 2}$
$\mathrm{OP}=10 \tan \theta / 2$
$\frac{\mathrm{d}(0 \mathrm{p})}{\mathrm{dt}}=\frac{10}{2} \sec ^{2} \theta_{2} \frac{\mathrm{d} \theta}{\mathrm{dt}}$
$\frac{\mathrm{d} \theta}{\mathrm{dt}}=\cos ^{2} \theta / 2$
Now, $\operatorname{arcB} \theta=\frac{\theta}{360^{\circ}} \times 2 \pi \mathrm{r}$
$\mathrm{l}=\theta \mathrm{r}$
$\frac{\mathrm{dl}}{\mathrm{dt}}=10 \frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \cos ^{2} \theta / 2$
As $\angle \mathrm{BA} \theta=45^{\circ}, \theta=\pi / 2$
$\frac{\mathrm{dl}}{\mathrm{dt}}=\frac{10}{2}=5 \mathrm{~cm} / \mathrm{sec}$

Akshaya Rs

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02:25

Problem 18

$y^{2}-x+2=0$
$2 y \frac{d y}{d x}-1=0$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}_{1}}$, Slope of normal $\rightarrow-2 \mathrm{y}_{1}$
Now, $x^{2}-y+2=0$
$2 x-\frac{d y}{d x}=0$
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}$, slope of normal $=\frac{-1}{2 \mathrm{x}_{2}}$
where $\mathrm{P} \rightarrow\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ \& $\theta \rightarrow\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$
As both the curveare inverse of each other. Hence, they are mirror image about $\mathrm{y}=\mathrm{x}$
$\Rightarrow$ Slope of normal $=-1$ $\Rightarrow \mathrm{y}_{1}=\frac{1}{2}, \mathrm{x}_{1}=\frac{9}{4}$
$\mathrm{x}_{2}=\frac{1}{2}, \mathrm{y}_{1}=\frac{9}{4}$
Shortest distance $(\mathrm{PQ})=\frac{7}{2 \sqrt{2}}$

Varsha Aggarwal

Numerade Educator

02:20

Problem 19

Let $\mathrm{P}$ be $\left(\mathrm{t}^{2}, 2 \mathrm{t}\right)$
$y^{2}=4 x$
$2 y \frac{d y}{d x}=4$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{t}}$
If $Q$ is a pt where normal meets the parabola again wit $\mathrm{t}_{1}$ as parameter $\mathrm{t}_{1}=-\mathrm{t}-\frac{2}{\mathrm{t}}$
Coordinate of $Q \rightarrow\left(\left(t+\frac{2}{t}\right)^{2},-2 t-\frac{4}{t}\right)$
Distance $=\sqrt{\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{4}+4\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{2}}$
Distance $=\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right) \sqrt{\left(\mathrm{t}+\frac{2}{\mathrm{t}}\right)^{2}+4}$
Min Distance $=2 \sqrt{2} \sqrt{12}=4 \sqrt{6}$

Varsha Aggarwal

Numerade Educator

02:00

Problem 20

$\mathrm{y}=\mathrm{x}^{2}+\mathrm{bx}-\mathrm{b}, \quad(1,1)$ lies on curve
$\frac{d y}{d x}=2 x+b=b+2$
eqn of tangent $\Rightarrow y-1=(b+2)(x-1)$
$x$ int $\Rightarrow 1-\frac{1}{b+2}$
$y$ int $\Rightarrow 1-(b+2)$
Area of $\Delta=\frac{1}{2} \frac{(b+1)^{2}}{(b+2)}$
$\Rightarrow(b+1)^{2}=-4 b-8$
$\Rightarrow b^{2}+6 b+9=0$
$(b+3)^{2}=0$
$\Rightarrow b=-3$

Varsha Aggarwal

Numerade Educator

01:23

Problem 21

$y=\ln x$
$\frac{d y}{d x}=\frac{1}{x}$
Slope of normal $\Rightarrow-x_{1}$
Slope of 1 chord $=\frac{1}{e-1}$
$\Rightarrow \frac{x_{1}}{e-1}=1$
$\Rightarrow \quad x_{1}=e-1$
$y_{1}=\ln (e-1)$

Varsha Aggarwal

Numerade Educator

01:23

Problem 22

If $\mathrm{y}=-2$ is a tangent
$\Rightarrow \mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}+2=0 \quad$ will have only one real root
$\Rightarrow \mathrm{q}^{2}-4 \mathrm{p}(\mathrm{r}+2)=0$
$=\mathrm{q}^{2}-4 \mathrm{pr}-8 \mathrm{p}=0$
$\Rightarrow \mathrm{p}<0$ as $\left(\mathrm{q}^{2}-4 \mathrm{pr}<0\right)$
$\Rightarrow \mathrm{r}<0$

Varsha Aggarwal

Numerade Educator

01:41

Problem 23

$y=\sin ^{-1} 2 x \sqrt{1-x^{2}}= \begin{cases}2 \sin ^{-1} x & |x| \leq \frac{1}{\sqrt{2}} \\ \pi-2 \sin ^{-1} x & x>\frac{1}{\sqrt{2}} \\ -\left(\pi+2 \sin ^{-1} x\right) & x<\frac{-1}{\sqrt{2}}\end{cases}$
$\mathrm{x}=0, \mathrm{y}=0$ is only integral point where the function has a unique tangent

Akshaya Rs

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01:42

Problem 24

$y=\tan ^{-1} \frac{2 x}{1-x^{2}}=2 \tan ^{-1} x$
$\frac{d y}{d x}=\frac{2}{1+x^{2}}=2$
$\mathrm{y}=2 \mathrm{x}$ is the eqn of tangent
Now, if tangent intersects the curve again, $\Rightarrow \tan ^{-1} x=x$
$y=\tan ^{-1} \frac{2 x}{1-x^{2}}=2 \tan ^{-1} x$
$\frac{d y}{d x}=\frac{2}{1+x^{2}}=2$
$\mathrm{y}=2 \mathrm{x}$ is the eqn of tangent
Now, if tangent intersects the curve again, $\Rightarrow \tan ^{-1} x=x$

Varsha Aggarwal

Numerade Educator

02:00

Problem 25

slope of normal $\Rightarrow 3 x-y+3=0$
$$
x=0 \& y=3
$$
Pt of normal $=(0,3)$
$\frac{d y}{d x}=\frac{-1}{3}=f^{\prime}(0)$
$\lim _{x \rightarrow 0} \frac{x^{2}}{f\left(x^{2}\right)+4 f\left(7 x^{2}\right)-5 f\left(4 x^{2}\right)}$
$\lim _{x \rightarrow 0} \frac{2 x}{x\left[2 f^{\prime}\left(x^{2}\right)+56 f^{\prime}\left(7 x^{2}\right)-40 f^{\prime}\left(4 x^{2}\right)\right]}$
$=\frac{2}{-6}=\frac{-1}{3}$

Akshaya Rs

Numerade Educator

01:38

Problem 26

$y=6-x-x^{2}$
$\frac{d y}{d x}=-1-2 x$
eqn of tangent $y-6+x_{1}+x_{1}^{2}=-\left(1+2 x_{1}\right)\left(x-x_{1}\right)$
$x y=x+3$
$x \frac{d y}{d x}+y=1$
$\frac{d y}{d x}=\frac{1-y_{2}}{x}$
eqn of tangent $\rightarrow$
$y-\frac{x_{2}+3}{x_{2}}=\frac{1-\frac{x_{2}+3}{x_{2}}}{x_{2}}\left(x-x_{2}\right)$
$x_{2} y-\left(x_{2}+3\right)=\frac{-3}{x_{2}}\left(x-x_{2}\right)$
Comparing the two eqns
$\frac{1}{x_{2}}=\frac{-\left(1+2 x_{1}\right)}{-\frac{3}{x_{2}}}=\frac{x_{1}\left(1+2 x_{1}\right)+6-x_{1}-x_{1}^{2}}{3+x_{2}+3}$
$\frac{1}{x_{2}}=\frac{x_{2}\left(1+2 x_{1}\right)}{3} \& \frac{1}{x_{2}}=\frac{x_{1}^{2}+6}{x_{2}+6}$
$3=x_{2}^{2}+2 x_{1} x_{2}^{2} \quad$ \& $x_{2}+6=x_{1}^{2} x_{2}+6 x_{2}$
$3=\left(1+2 \mathrm{x}_{1}\right) \mathrm{x}_{2}^{2} \quad \& \quad \mathrm{x}_{2}=\frac{6}{5-\mathrm{x}_{1}^{2}}$
$\Rightarrow 3=\frac{\left(1+2 x_{1}\right) 36}{\left(5-x_{1}^{2}\right)^{2}}$
$\Rightarrow\left(5-\mathrm{x}_{1}^{2}\right)^{2}=12\left(1+2 \mathrm{x}_{1}\right)$
Let $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ be the common tangent
$m x+c=6-x-x^{2}$
$\Rightarrow \mathrm{x}^{2}+(\mathrm{m}+1) \mathrm{x}+(\mathrm{c}-6)=0$
$\mathrm{D}=0$
$(m+1)^{2}-4(c-6)=0$
For $x y=x+3$ $x(m x+c)=x+3$
$m x^{2}+(c-1) x-3=0$
$\mathrm{D}=0$
$(c-1)^{2}+12 m=0$
Using (I) \& (II)
$(m+1)^{2}+24=4 c$
$\mathrm{c}-1=\frac{(\mathrm{m}+1)^{2}+20}{4}$
$\Rightarrow\left(\frac{(\mathrm{m}+\mathrm{l})^{2}+20}{4}\right)^{2}+12 \mathrm{~m}=0$
$\left((m+1)^{2}+20\right)^{2}+192 m=0$
Upon solving, $\mathrm{m}=-3$
$\Rightarrow \mathrm{c}=7$
eqn of tangent $=y=-3 x+7$

Akshaya Rs

Numerade Educator

01:04

Problem 27

$y=2 x^{2}-x+1$
$\frac{d y}{d x}=4 x_{1}-1$
$4 x_{1}-1=3$
$x_{1}=1$
$y_{1}=2$

Varsha Aggarwal

Numerade Educator

01:45

Problem 28

$y^{2}-2 x^{2}-4 y+8=0$
$2 y \frac{d y}{d x}-4 x-4 \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{2 x}{y-2}$
Now, $y^{2}-2 x^{2}-4 y+8=0$
$(y-2)^{2}-2\left(x^{2}-2\right)=0$
eqn of tangent $y-y_{1}=\frac{2 x_{1}}{y_{1}-2}\left(x-x_{1}\right)$
$-\left(2-y_{1}\right)^{2}=2 x_{1}\left(1-x_{1}\right)$
Using (II) $-2\left(x_{1}^{2}-2\right)=2 x_{1}-2 x_{1}^{2}$
$2 \mathrm{x}_{1}=4$
$\mathrm{x}_{1}=2$ \& $\mathrm{x}_{1}=0$ (Horizontal tangent)

Akshaya Rs

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01:24

Problem 29

$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$
$\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}+\frac{n}{b}\left(\frac{y}{b}\right)^{n-t} \frac{d y}{d x}=0$
$\frac{d y}{d x}=-\frac{b^{n} x^{n-1}}{a^{n} y^{n-1}}=-\frac{b}{a}$

Akshaya Rs

Numerade Educator

01:33

Problem 30

$y=1-a x^{2}$
$\frac{d y}{d x}=-2 a x_{1}$
\& $y=x^{2}$
$\frac{d y}{d x}=2 x_{1}$
If curves are orthogonal, $\mathrm{m}_{1} \mathrm{~m}_{2}=-1$
$+4 a x^{2}=1$
Now $1-a x^{2}=x^{2} \Rightarrow \frac{1}{1+a}=x^{2}$
Using eq (1), we get $\frac{4 a}{1+a}=1$
$\mathrm{a}=\frac{1}{3}$

Varsha Aggarwal

Numerade Educator

00:58

Problem 31

$y=x^{3}+3 x^{2}+3 x-1$
$\frac{d y}{d x}=3 x^{2}+6 x+3=3\left(x^{2}+2 x+1\right)^{2}$
$=3(x+1)^{2}$
Minslope of curveis 0

Varsha Aggarwal

Numerade Educator

02:32

Problem 32

$f_{1}^{\prime}(x)=2 x-1 \quad \& \quad f_{2}^{\prime}(x)=3 x^{2}-2 x-2$
$\Rightarrow 2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2$
$\Rightarrow 3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0$
For $\mathrm{x}_{2}$ be real, $\mathrm{D} \geq 0$
$4-4\left(2 x_{1}+1\right)(3)$
$4-24 x_{1}-12$
$\Rightarrow-\left(24 x_{1}+8\right)$
There can be infinite such values of $\mathrm{x}_{\mathrm{l}}$

Varsha Aggarwal

Numerade Educator

01:35

Problem 33

$x^{2}-4 y^{2}+c=0$
$2 x-8 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{x}{4 y}$
\& $y^{2}=4 x$
$\frac{d y}{d x}=\frac{2}{y}$
For orthogonal, $\mathrm{m}_{1} \mathrm{~m}_{2}=-1$
$\frac{\mathrm{x}}{2 \mathrm{y}^{2}}=-1$
$\Rightarrow y^{2}=4\left(-2 y^{2}\right) \Rightarrow y=0$

Varsha Aggarwal

Numerade Educator

00:58

Problem 34

Slope of secant $=\frac{9 a^{2}-a^{2}}{3 a-a}=4 a$
$\frac{d y}{d x}=2 x=4 a$
$x=2 a$
$y=4 a^{2}$

Varsha Aggarwal

Numerade Educator

01:25

Problem 35

$y=e^{2 x}+x^{2}$
$\frac{d y}{d x}=2 e^{2 x}+2 x$
at $x=0, \quad \frac{d y}{d x}=2$
$-\frac{d x}{d y}=-\frac{1}{2}$
eqn of normal $\rightarrow y-1=-\frac{1}{2} x$
$\Rightarrow 2 y+x-2=0$
Distance from $(0,0)=\frac{2}{\sqrt{5}}$

Varsha Aggarwal

Numerade Educator

02:17

Problem 36

$y^{3}-x^{2} y+5 y-2 x=0$
$3 y^{2} \frac{d y}{d x}-x^{2} \frac{d y}{d x}-2 x y+\frac{5 d y}{d x}-2=0$
$\frac{d y}{d x}=\frac{2(x y+1)}{3 y^{2}-x^{2}+5}=\frac{2}{5}$
For $x^{4}-x^{3} y^{2}+5 x+2 y=0$
$4 x^{3}-3 x^{2} y^{2}-2 x^{3} y \frac{d y}{d x}+5+\frac{2 d y}{d x}=0$
$\frac{d y}{d x}=\frac{4 x^{3}-3 x^{2} y^{2}+5}{2 x^{3} y-2}=\frac{5}{-2}$ at $(0,0)$

Varsha Aggarwal

Numerade Educator

02:05

Problem 37

$1^{2}=x^{2}+y^{2}$
$1 \frac{d l}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t}$
$y=x^{3 / 2}$
$\frac{d y}{d t}=\frac{3}{2} x^{1 / 2} \frac{d x}{d t}$
Using (1) \& (2)
$11 \sqrt{x^{2}+x^{3}}=x \frac{d x}{d t}+\frac{3}{2} x^{3 / 2} x^{1 / 2} \frac{d x}{d t}$
$\frac{d x}{d t}=\frac{66}{3+\frac{27}{2}}=\frac{66 \times 2}{33}=4$

Varsha Aggarwal

Numerade Educator

Problem 38

$y^{3}=27 x$
$3 y^{2} \frac{d y}{d x}=27 \frac{d x}{d t}$
$\frac{3 y^{2}}{27}<1$
$y^{2}<9 \Rightarrow y \in(-3,3)$

Varsha Aggarwal

Numerade Educator

01:37

Problem 39

$y=x \ln x+\sin (\pi \ln x)$
$\frac{d y}{d x}=1+\ln x+\cos (\pi \ln x) \frac{\pi}{x}$
at $x=e$
$\frac{d y}{d x}=2-\frac{\pi}{e}=\frac{2 e-\pi}{e}$
at $x=e^{2}$
$\frac{d y}{d x}=3+\frac{\pi}{e^{2}}=\frac{3 e^{2}+\pi}{e^{2}}$

Akshaya Rs

Numerade Educator

01:28

Problem 40

$\mathrm{l}_{1}=\mathrm{l}_{2}+\mathrm{l}_{2}^{3}+6$
$\frac{\mathrm{d} \mathrm{l}_{1}}{\mathrm{dt}}=\frac{\mathrm{dl}_{2}}{\mathrm{dt}}+3 \mathrm{l}_{2}^{2} \frac{\mathrm{dl}_{2}}{\mathrm{dt}}$
$\frac{\mathrm{d} \mathrm{S}_{2}}{\mathrm{dS}_{1}}=\frac{\mathrm{dS}_{2}}{\mathrm{dt}} \times \frac{\mathrm{dt}}{\mathrm{dS}_{1}}$
$=\frac{2 \mathrm{l}_{2} \mathrm{~d}_{2} / \mathrm{dt}}{2 \mathrm{l}_{1} \mathrm{dl}_{1} / \mathrm{dt}}$
$=\frac{1}{4} \times \frac{1}{8}$

Akshaya Rs

Numerade Educator

02:50

Problem 41

$y^{2}=x\left(2-x^{2}\right)$
$2 y \frac{d y}{d x}=2-3 x^{2}$
$\frac{d y}{d x}=\frac{-1}{2} \quad$ at $(1,1)$
eqn of tangent
$y-1=\frac{-1}{2}(x-1)$
$2 y+x-3=0$
Solving with curve $\left(\frac{3-x}{2}\right)^{2}=2 x-x^{3}$
$4 x^{3}+x^{2}-14 x+9=0$

Varsha Aggarwal

Numerade Educator

02:42

Problem 42

$x^{3}-y^{2}=0$
$3 x^{2}-2 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{3 x^{2}}{2 y}=\frac{3 \times 16 m^{4}}{2 \times 8 m^{3}}=3 m$
Slope of normal $=\frac{-1}{3 \mathrm{~m}_{1}}$ at point $\mathrm{m}_{1}$
$y-8 m^{3}=3 m\left(x-4 m^{2}\right) \rightarrow$ eqn of tangent
$y-8 m_{1}^{3}=\frac{-1}{3 m_{1}}\left(x-4 m_{1}^{2}\right) \rightarrow$ eqn of normal.
$\Rightarrow \frac{3 m}{-1 / 3 m_{1}}=1$
$\Rightarrow 9 \mathrm{~mm}_{1}=-1$
Now, $\frac{8 m^{3}-12 m^{3}}{-4 m^{3}}=8 m_{1}^{3}+\frac{4}{3} m_{1}$
$=\frac{-8}{(9 \mathrm{~m})^{3}}-\frac{4}{27 \mathrm{~m}}$
$\Rightarrow+4 m^{6}=\frac{8}{9^{3}}+\frac{4}{27} m^{2}$
$=\frac{8+108 \mathrm{~m}^{2}}{9^{3}}$

Akshaya Rs

Numerade Educator

01:26

Problem 43

$y=1-a x^{2}$
$\frac{d y}{d x}=-2 a x$
\& $\mathrm{y}=\mathrm{x}^{2}$
$\frac{d y}{d x}=2 x$
For orthogonal $\Rightarrow 4 a x^{2}=1$
Now, $1-a x^{2}=x^{2}$
$x^{2}=\frac{1}{1+a}$
Using eq (1)
$\frac{4 a}{1+a}=1$
$a=\frac{1}{3}$

Varsha Aggarwal

Numerade Educator

03:04

Problem 44

Let $y=m x+\frac{1}{m}$ be tangent to $y^{2}=4 x$
eqn of normal at $\left(x_{1}, y_{1}\right)$ to $x^{2}=4$ by is $y-y_{1}=-\frac{2 b}{x_{1}}\left(x-x_{1}\right)$
$\Rightarrow y=-\frac{2 b}{x_{1}} x+\frac{x_{1}^{2}}{4 b}+2 b$
Comparing two eqn $\mathrm{m}=-\frac{2 \mathrm{~b}}{\mathrm{x}_{1}}$
$\frac{\mathrm{x}_{1}^{2}}{4 \mathrm{~b}}+2 \mathrm{~b}=\frac{1}{\mathrm{~m}}$
Using (I) \& (II)

Varsha Aggarwal

Numerade Educator

Problem 45

eqn of normal $y=m x-2 m-m^{3}$
It passes through $(6,0)$
$\mathrm{m}^{3}-4 \mathrm{~m}=0$
$\mathrm{m}\left(\mathrm{m}^{2}-4\right)=0$
$\mathrm{m}=0, \pm 2$
Pts are $\left(a m^{2}-2 a m\right)$
$(0,0),(4,-4),(4,4)$
Minimum distance $=\sqrt{21}-\sqrt{5}$

Akshaya Rs

Numerade Educator

01:07

Problem 46

$y=a x^{2}+b x+c$
$\frac{d y}{d x}=2 a x+b=10 a+b=0 \quad(a t 5,4)$
$\mathrm{Now}_{3} \cdot 4=25 a+5 \mathrm{~b}+\mathrm{c}$
using (I) $4=25 a+5(-10 a)+c$
$c=4+25 a$
$\Rightarrow$ Maximum value of $c=104$

Varsha Aggarwal

Numerade Educator

01:05

Problem 47

$|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}$
$|\mathbf{x}|-\mathrm{a}>1 \quad$ for no solution $|x|>1+a$

Akshaya Rs

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01:19

Problem 48

Ellipse $\rightarrow \frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
$\Rightarrow \frac{x}{8} \frac{d x}{d t}+\frac{2 y}{9} \frac{d y}{d t}=1$
$\Rightarrow \frac{4 \sqrt{1-y^{2} / 9}}{8} \frac{d x}{d t}+\frac{2 y}{9} \times \frac{d y}{d x} \times \frac{d x}{d t}=1$
Now put $\frac{d y}{d x}=1 \quad \& \quad y=1$
$\Rightarrow \frac{4}{6 \sqrt{2}} \frac{d x}{d t}+\frac{2}{9} \frac{d x}{d t}=1$
$\Rightarrow \frac{d x}{d t}\left(\frac{\sqrt{2}}{3}+\frac{2}{9}\right)=1$
$\Rightarrow \frac{d x}{d t}=\frac{9}{3 \sqrt{2}+2}$

Akshaya Rs

Numerade Educator

01:20

Problem 49

$y=x^{3}$
$\frac{d v}{d t}=3 x^{2} \frac{d x}{d t}$
$\quad=900$

Varsha Aggarwal

Numerade Educator

01:43

Problem 50

Similar to Q.34

Akshaya Rs

Numerade Educator

01:38

Problem 51

$x^{2}-4 y^{2}+c=0$
$2 x-8 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{x}{4 y}$
$y^{2}=4 x$
$y \frac{d y}{d x}=2$
$\frac{d y}{d x}=\frac{2}{y}$
For orthogonal, $\frac{\mathrm{x}}{2 \mathrm{y}^{2}}=-1$
There is no $x, y$ which will satisfy eq. (I)

Akshaya Rs

Numerade Educator

01:07

Problem 52

$x=t^{2}+t+1$
$\frac{d x}{d t}=2 t+1$
$y=t^{2}-t+1$
$\frac{d y}{d t}=2 t-1$
$\frac{d y}{d x}=\frac{2 t-1}{2 t+1}$
$\Rightarrow y-\left(t^{2}-t+1\right)=\frac{2 t-1}{2 t+1}\left(x-\left(t^{2}+t+1\right)\right)$
$\Rightarrow\left(t-t^{2}\right)=\left(\frac{2 t-1}{2 t+1}\right)\left(-\left(t^{2}+t\right)\right)$
$\Rightarrow 2 t^{2}+t-2 t^{3}-t^{2}=t^{2}+t-2 t^{3}-2 t^{2}$
$\Rightarrow 2 t^{2}=0$
$\Rightarrow t=0$

Akshaya Rs

Numerade Educator

Problem 53

$\mathrm{x}=\mathrm{t}^{2}+\mathrm{t}+1$
$\frac{\mathrm{d} \mathrm{x}}{\mathrm{dt}}=2 \mathrm{t}+1$
$\mathrm{y}=\mathrm{t}^{2}-\mathrm{t}+1$
$\frac{\mathrm{dy}}{\mathrm{dt}}=2 \mathrm{t}-1$
$\frac{d y}{\mathrm{~d} \mathrm{x}}=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}$
$\Rightarrow \mathrm{y}-\left(\mathrm{t}^{2}-\mathrm{t}+1\right)=\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\left(\mathrm{x}-\left(\mathrm{t}^{2}+\mathrm{t}+1\right)\right)$
$\Rightarrow\left(\mathrm{t}-\mathrm{t}^{2}\right)=\left(\frac{2 \mathrm{t}-1}{2 \mathrm{t}+1}\right)\left(-\left(\mathrm{t}^{2}+\mathrm{t}\right)\right)$
$\Rightarrow 2 \mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-\mathrm{t}^{2}=\mathrm{t}^{2}+\mathrm{t}-2 \mathrm{t}^{3}-2 \mathrm{t}^{2}$
$\Rightarrow 2 \mathrm{t}^{2}=0$
$\Rightarrow \mathrm{t}=0$

Akshaya Rs

Numerade Educator

01:29

Problem 54

$y=\left(x_{1}-x_{2}\right)^{2}+\left(x_{1}-4-\frac{x_{2}^{2}}{4}\right)^{2}$
Min value of $y$ is shortest distance between
$y=x-4$
\& $y=\frac{x^{2}}{4}$
slope of normal $=-1$
$\frac{d y}{d x}=\frac{x}{2}$
$-\frac{d x}{d y}=\frac{-2}{x}=-1$
$\mathrm{x}=2$

Akshaya Rs

Numerade Educator

01:41

Problem 55

$x=\cos ^{3} t$
$\frac{d x}{d t}=-3 \cos ^{2} t \sin t$
$y=\sin ^{2} t$
$\frac{d y}{d t}=\sin 2 t$
$\frac{d y}{d x}=\frac{2 \sin t \cos t}{-3 \cos ^{2} t \sin t}=\frac{-2}{3} \sec t=\frac{-4}{3}$

Varsha Aggarwal

Numerade Educator

Problem 56

$A=\frac{1}{2} b \sqrt{x^{2}-\frac{b^{2}}{4}}$
$\frac{d A}{d t}=\frac{b}{2} \frac{x 2 x}{x 2 \sqrt{x^{2}-b^{2} / 4}} \frac{d x}{d t}$
$=\sqrt{3 b}$

Akshaya Rs

Numerade Educator

Problem 57

$y=x^{3}+a x$
$\frac{d y}{d x}=3 x^{2}+a$
$\frac{d y}{d x}=3+a$
$y=b x^{2}+c$
$\frac{d y}{d x}=2 b x=-2 b$
$\Rightarrow 3+a=-2 b$
$\Rightarrow 3+a+2 b=0$
As $-1-a=0$
$\Rightarrow a=-1$
$b+c=0$
$b=-1$
$c=1$
Then, $\left(a+b+c^{2}\right)=-1-1+1=-1$

Varsha Aggarwal

Numerade Educator

01:04

Problem 58

$y^{2}-2 y-8 x+17=0$
$2 y \frac{d y}{d x}-2 \frac{d y}{d x}-8=0$
$\frac{d y}{d x}=\frac{y}{y-1}=1$
$\Rightarrow y=5, \quad x=4$

Akshaya Rs

Numerade Educator

01:13

Problem 59

$y=\frac{x^{2}}{4}-2$
$\frac{d y}{d x}=\frac{x}{2}$
$\left(y-\frac{x_{\perp}^{2}}{4}+2\right)=\frac{-2}{x}\left(x-x_{1}\right)$
$-1-\frac{x_{1}^{2}}{4}+2=\frac{-2}{x_{1}}\left(1-x_{1}\right)$
$\Rightarrow 2 x_{1}-\frac{x_{1} 3}{4}=-2+2 x_{1}$
$x_{1}^{3}=8$
$x_{1}=2, \quad y=-1$

Akshaya Rs

Numerade Educator

01:23

Problem 60

$g(x)=x f(x)$
$g^{\prime}(x)=f(x)+x f^{\prime}(x)$
$g(2)=13$
\& $g(2)=6$
$y-6=13(x-2) \rightarrow$ eqn of tangent
For y int, $\mathrm{x}=0$
$y=-20$

Varsha Aggarwal

Numerade Educator

01:40

Problem 61

$x \sin y+y \sin x=\pi$
$\sin y+x \cos y \frac{d y}{d x}+\frac{d y}{d x} \sin x+y \cos x=0$
$\frac{d y}{d x}=\frac{-(\sin y+y \cos x)}{x \cos y+\sin x}=-1$
eqn of tangent $\rightarrow y=-x+\pi$

Varsha Aggarwal

Numerade Educator

02:09

Problem 62

$y=-t+e^{a t}=0$
$x=t+e^{a}$
$\frac{d y}{d t}=-1+a e^{u r}$
$\frac{d x}{d t}=1+a e^{u}$
$\frac{d y}{d x}=\frac{a e^{a t}-1}{a e^{a t}+1}=0$
$\Rightarrow \mathrm{e}^{a t}=\frac{1}{a}$
$\Rightarrow a t=-\ln a$
$\mathrm{t}=-\frac{\ln \mathrm{a}}{\mathrm{a}}$
$\mathrm{x}=-\frac{\ln \mathrm{a}}{\mathrm{a}}+\frac{1}{\mathrm{a}}=\frac{\ln \mathrm{e} / \mathrm{a}}{\mathrm{a}}=\frac{2}{\mathrm{a}}$

Varsha Aggarwal

Numerade Educator

02:18

Problem 63

$x y^{2}=1$
$y^{2}+2 x y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{-y}{2 x}$
$-\frac{d x}{d y}=\frac{2 x}{y}=\frac{2}{y^{3}}$
$y-y_{1}=\frac{2}{y_{1}^{3}}\left(x-x_{1}\right)$
$+y_{1}^{4}=2 x_{1}$
$y_{1}^{6}=2$
$y_{1}=\pm 2^{1 / 6}$
$x_{1}=\pm 2^{-\sqrt{3}}$

Varsha Aggarwal

Numerade Educator

Problem 64

$\mathrm{xy}=1$
$\mathrm{y}+\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=0$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}}{\mathrm{x}}$
$-\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}=\mathrm{x}_{1}^{2}$
$\Rightarrow \mathrm{x}_{1}^{2}=\frac{\mathrm{a}-3}{\mathrm{a}} \geq 0$
$\mathrm{a} \in(-\infty, 0) \mathrm{U}(3, \infty)$

Akshaya Rs

Numerade Educator

01:40

Problem 65

$y=\frac{a x}{1+x}=a-\frac{a}{1+x}$
$\frac{d y}{d x}=\frac{a}{(1+x)^{2}}=-1$
$a=-\left(1+x_{1}\right)^{2}$
eqn of tangent
$y-\frac{a x_{1}}{1+x_{1}}=-1\left(x-x_{1}\right)$
$\mathrm{y}+\mathrm{x}=\mathrm{x}_{1}+\frac{\mathrm{ax}_{1}}{1+\mathrm{x}_{1}}$
\& $y-3=-x \Rightarrow y+x=3$
$\Rightarrow \mathrm{x}_{1}+\frac{\mathrm{ax}_{1}}{1+\mathrm{x}_{1}}=3$
$\Rightarrow \mathrm{x}_{1}-\mathrm{x}_{1}\left(1+\mathrm{x}_{1}\right)=3$

Akshaya Rs

Numerade Educator

Problem 66

$y=x \sqrt{1-x^{2}}$
$\frac{d y}{d x}=\sqrt{1-x^{2}}-\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}$
$\Rightarrow x=\pm 1$

Varsha Aggarwal

Numerade Educator

02:31

Problem 67

$y=2 x^{2}-x, \quad y=2-x^{3}$
Pt of intersection are $2 \mathrm{x}^{2}-\mathrm{x}=2-\mathrm{x}^{3}$
$\Rightarrow 2\left(x^{2}-1\right)+x\left(x^{2}-1\right)=0$
$\Rightarrow\left(x^{2}-1\right)(x+2)=0$
$\Rightarrow x=\pm 1,-2$
$\Rightarrow(1,1),(-1,3),(-2,10)$
For $\mathrm{y}=2 \mathrm{x}^{2}-\mathrm{x}$
$\frac{d y}{d x}=4 x-1$
For $y=2-x^{3}$
$\frac{d y}{d x}=-3 x^{2}$
At $(1,1)$
$\theta=\tan ^{-1}\left(\frac{3+3}{1-9}\right)=\tan ^{-1}\left(\frac{3}{-4}\right)$
At $(-1,3)$
$\theta=\tan ^{-1}\left(\frac{-5+3}{1+15}\right)=\tan ^{-1}\left(\frac{1}{8}\right)$
At $(-2 ; 10)$
$\theta=\tan ^{-1}\left(\frac{-9+12}{1+108}\right)=\tan ^{-1} \frac{3}{109}$

Varsha Aggarwal

Numerade Educator

01:03

Problem 68

$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-1}{2-\mathrm{x}}=-1+\frac{1}{2-\mathrm{x}}$
$\mathrm{f}^{\prime}(\mathrm{x})=+\frac{1}{(2-\mathrm{x})^{2}}=\frac{\mathrm{t}}{4}$
$\Rightarrow \quad \mathrm{x}-2=\pm 2$
$\mathrm{x}=0$ or 4
$\mathrm{Pt} \rightarrow\left(0, \frac{-1}{2}\right), \quad \&\left(4, \frac{-3}{2}\right)$

Varsha Aggarwal

Numerade Educator

03:12

Problem 69

$\frac{d h_{4}}{d t}=0.5 t+2$
Integrating $=\frac{t^{2}}{4}+2 t+5$
$\frac{d h_{b}}{d t}=t+1$
Integrating, $h_{b}=\frac{t^{2}}{2}+t+5$

Varsha Aggarwal

Numerade Educator

01:44

Problem 70

$x^{2}+2 x y+2 y^{2}=45$
$x+y+x \frac{d y}{d x}+2 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{-(x+y)}{x+2 y}=-2$
$\Rightarrow x+y=2 x+4 y$
$\Rightarrow x+3 y=0$
Put eq (II) in eq (I)
$9 y^{2}-6 y^{2}+2 y^{2}=45$
$y=\pm 3$
$x=\pm 9$
Pt $\rightarrow(9,-3), \quad(-9,3)$

Varsha Aggarwal

Numerade Educator

01:36

Problem 71

$f(x)=\frac{x}{1-x^{2}}$
$f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$
$1+x^{2}=1+x^{4}-2 x^{2}$
$\Rightarrow x^{2}\left(x^{2}-3\right)=0$
$\Rightarrow x=0, \pm \sqrt{3}$
$P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right),\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

Akshaya Rs

Numerade Educator

01:45

Problem 71

Similar to $Q-67$

Akshaya Rs

Numerade Educator

01:22

Problem 72

$f(x)=\frac{x}{1-x^{2}}$
$f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$
$1+x^{2}=1+x^{4}-2 x^{2}$
$\Rightarrow \quad x^{2}\left(x^{2}-3\right)=0$
$\Rightarrow x=0, \pm \sqrt{3}$
$P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right) \cdot\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

Akshaya Rs

Numerade Educator

02:18

Problem 72

$f(x)=\frac{x}{1-x^{2}}$
$f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$
$1+x^{2}=1+x^{4}-2 x^{2}$
$\Rightarrow \quad x^{2}\left(x^{2}-3\right)=0$
$\Rightarrow x=0, \pm \sqrt{3}$
$P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right)\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

Varsha Aggarwal

Numerade Educator

01:51

Problem 73

For $y=x(c-x)$
$\frac{d y}{d x}=c-2 x$
For $y=x^{2}+a x+b$
$\frac{d y}{d x}=2 x+a$
At $(1,0)$
$0=(C-1) \Rightarrow C=1$
$0=1+a+b, 1-2=2+a$
$\Rightarrow \quad a=-3$
$\Rightarrow b=2$

Varsha Aggarwal

Numerade Educator

01:49

Problem 74

$\ln \left(x^{2}+y^{2}\right)=\operatorname{ctan}^{-1} y / x$
$\frac{2 x+2 y d y / d x}{x^{2}+y^{2}}=c \times \frac{x^{2}}{x^{2}+y^{2}} \times \frac{(x d y / d x-y)}{x^{2}}$
$2 x+2 y \frac{d y}{d x}=c\left(x \frac{d y}{d x}-y\right)$
$\tan \theta=\frac{\frac{2 x+c y}{c x-2 y}-\frac{y}{x}}{1+\left(\frac{2 x+c y}{c x-2 y}\right) \frac{y}{x}}$
$\frac{d y}{d x}=\frac{2 x+c y}{c x-2 y}$
$=\frac{2 x^{2}+2 y^{2}}{c\left(x^{2}+y^{2}\right)}=\frac{2}{c}$

Varsha Aggarwal

Numerade Educator

01:37

Problem 75

$\alpha y^{2}=(x+\beta)^{3}$
$2 \alpha y \frac{d y}{d x}=3(x+\beta)^{2}$
$y \frac{d y}{d x}=\frac{3(x+\beta)^{2}}{2 \alpha}=$ Sub Normal
$y \frac{d x}{d y}=\frac{2 \alpha y^{2}}{3(x+\beta)^{2}}=$ SubTangent
Square of sub tangent $=\frac{4 \alpha^{2}(x+\beta)^{4}}{9}$

Akshaya Rs

Numerade Educator

Problem 76

$y=x^{3}-x^{2}-x+2$
$\frac{d y}{d x}=3 x^{2}-2 x-1=0$ at $x=1$
eqn of tangent
$y=1$
solving with curve, $x^{3}-x^{2}-x+1=0$
$(x-1)\left(x^{2}-1\right)=0$
$\mathrm{x}=\pm \mathrm{l}$

Akshaya Rs

Numerade Educator

Problem 77

$y=a x^{2}+b x+c$
$\frac{d y}{d x}=2 a x+b$
$A t x=1, \quad y=1$
$\Rightarrow a+b+c=1$
$2 a+b=1$
$f^{\prime \prime}(1)+f^{\prime}(0)=2 a+b=1$

Akshaya Rs

Numerade Educator

01:36

Problem 78

$y \frac{d y}{d x}=y \frac{d x}{d y}$
$\Rightarrow \frac{d y}{d x}=\pm 1$
Length of normal $=y_{1} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$
$=\sqrt{2 y}$

Akshaya Rs

Numerade Educator

02:42

Problem 79

$(x-4)^{2}+y^{2}=c^{2}$
$2(x-4)+2 y \frac{d y}{d x}=0$
$\frac{d y}{d x}=\frac{4-x}{y}$
For $y^{2}=x^{3}+1$
$2 y \frac{d y}{d x}=3 x^{2}$
$\frac{d y}{d x}=\frac{3 x^{2}}{2 y}$
$\Rightarrow \frac{4-x}{y}=\frac{3 x^{2}}{2 y}$
$\Rightarrow 3 x^{2}+2 x-8=0$
$\Rightarrow(3 x-4)(x+2)=0$
$x=\frac{4}{3}$ or $-2$

Akshaya Rs

Numerade Educator

01:05

Problem 80

$x=2-3 \sin \theta, \quad y=3+2 \cos \theta$
$\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-3}{2}\right)^{2}=1 \Rightarrow$ eqn of ellipse
end pts of major axis $\rightarrow(-1,3) \&(5,3)$

Akshaya Rs

Numerade Educator

01:33

Problem 81

$\mathrm{s}=30\left(1-\mathrm{e}^{\mathrm{ln}}\right)$
$5=30\left(\mathrm{l}-\mathrm{e}^{k}\right)$
$\mathrm{e}^{k}=\frac{5}{6}$
$\mathrm{k}=\ln (5 / 6)$

Varsha Aggarwal

Numerade Educator

01:11

Problem 82

$\mathrm{S}=30\left(1-\mathrm{e}^{\mathrm{k} \mathrm{a}}\right)$
$\frac{\mathrm{dS}}{\mathrm{dt}}=-30 \mathrm{ke}^{\mathrm{k} 1}>0 \quad($ as $\mathrm{k}<0)$
Max 5 when $t \rightarrow \infty$ Max $\mathrm{S}=30$

Varsha Aggarwal

Numerade Educator

01:48

Problem 83

$\mathrm{S}=30\left(1-\mathrm{e}^{(\ln 5 / 6) 5}\right)=30\left(1-\left(\frac{5}{6}\right)^{5}\right)$

Varsha Aggarwal

Numerade Educator

01:19

Problem 84

$f(0)<0$
$\Rightarrow \mathrm{a}^{2}+\mathrm{a}-2<0$
$\Rightarrow a \in(-2,1)$
Integral values of a are $-1,0$.

Varsha Aggarwal

Numerade Educator

Problem 85

$y=\left(\frac{x}{2}-a\right)^{2}+a-2$
$4(y-(a-2))=(x-2 a)^{2}$
Vertex $\Rightarrow \mathrm{h}=2 \mathrm{a}, \mathrm{k}=\mathrm{a}-2$
Locus of vertex $\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}-2$
$\Rightarrow 2 y=x-4$

Akshaya Rs

Numerade Educator

Problem 86

$y=\frac{x^{2}}{4}-3 x+10$
For $y=2-\frac{x^{2}}{4}$
eqn of tangent
$y-2=m x+m^{2}$
Solving (II) \& (I) $m x+2+m^{2}=\frac{x^{2}}{4}-3 x+10$
$\Rightarrow \frac{x^{2}}{4}-(m+3) x+\left(8-m^{2}\right)=0$
$\mathrm{D}=0$
$(m+3)^{2}-\left(8-m^{2}\right)=0$
$2 m^{2}+6 m+1=0$
$\mathrm{m}_{1}+\mathrm{m}_{2}=-3$

Akshaya Rs

Numerade Educator

01:13

Problem 87

eqn of $\mathrm{P} \theta$ $y-a t_{1}^{3}=\frac{a\left(t_{2}^{3}-t_{1}^{3}\right)}{a\left(t_{2}^{2}-t_{1}^{2}\right)}\left(x-a t_{1}^{2}\right)$
$y-a t_{1}^{3}=\frac{t_{2}^{2}+t_{1}^{2}+t_{1} t_{2}}{t_{1}+t_{2}}\left(x-a t_{1}^{2}\right)$
It passes through $\mathrm{R}\left(\mathrm{at}^{2}, \mathrm{at}^{3}\right)$
$\Rightarrow \mathrm{t}^{3}-\mathrm{t}_{1}^{3}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\left(\mathrm{t}^{2}-\mathrm{t}_{1}^{2}\right)$
$\Rightarrow\left(\mathrm{t}^{2}+\mathrm{t}_{1}^{2}+\mathrm{tt}_{1}\right)\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\left(\mathrm{t}+\mathrm{t}_{1}\right)\left(\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}\right)$
$\Rightarrow \mathrm{t}^{2} \mathrm{t}_{1}+\mathrm{t}_{1}^{3}+\mathrm{tt}_{1}^{2}+\mathrm{t}^{2} \mathrm{t}_{2}+\mathrm{t}_{1}^{2} \mathrm{t}_{2}+\mathrm{tt}_{1} \mathrm{t}_{2}=\mathrm{tt}_{1}^{2}+\mathrm{tt}_{2}^{2}$
$+\mathrm{tt}_{1} \mathrm{t}_{2}+\mathrm{t}_{1}^{3}+\mathrm{t}_{1} \mathrm{t}_{2}^{2}+\mathrm{t}_{1}^{2} \mathrm{t}_{2}$
$\Rightarrow \mathrm{t}^{2} \mathrm{t}_{1}+\mathrm{t}^{2} \mathrm{t}_{2}=\mathrm{tt}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}^{2}$
$\Rightarrow\left(t_{1}+t_{2}\right) t^{2}-t_{2}^{2}(t)-t_{1} t_{2}^{2}=0$
$\Rightarrow \mathrm{t}=\frac{\mathrm{t}_{2}^{2} \pm \sqrt{t_{2}^{4}+4\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)} \mathrm{t}_{1} \mathrm{t}_{2}^{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}$
$\mathrm{t}=\frac{\mathrm{t}_{2}^{2}+\mathrm{t}_{2} \sqrt{t_{2}^{2}+4 \mathrm{t}_{1}^{2}+4 \mathrm{t}_{1} \mathrm{t}_{2}}}{2\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}$
$\mathrm{t}_{2}=\frac{\mathrm{t}_{2}^{2} \pm\left(\mathrm{t}_{2}+2 \mathrm{t}_{1}\right) \mathrm{t}_{2}}{2\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)}$
$t=\frac{-t_{1} t_{2}}{t_{1}+t_{2}}$

Akshaya Rs

Numerade Educator

01:22

Problem 88

$a y^{2}=x^{3}$
2ay $\frac{d y}{d x}=3 x^{2}$
$\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}=\frac{3 a^{2} t^{4}}{2 a^{2} t^{3}}=\frac{3}{2} t$
eqn of tangent $y=a t^{3}=\frac{3}{2} t\left(x-a t^{2}\right)$
$\Rightarrow 2 y-2 a t^{3}=3 t x-3 a t^{3}$
$2 y=3 t x-a t^{3}$
$\Rightarrow a t^{3}-(3 x) t+2 y=0$
$\mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}=0$
$\mathrm{t}_{1} \mathrm{t}_{2}+\mathrm{t}_{1} \mathrm{t}_{3}+\mathrm{t}_{2} \mathrm{t}_{3}=-\frac{3 \mathrm{x}}{\mathrm{a}}$
Using (II) \& (I) $\mathrm{t}_{1} \mathrm{t}_{2}+\mathrm{t}_{3}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=\frac{-3 \mathrm{x}}{\mathrm{a}}$
$\mathrm{t}_{1} \mathrm{t}_{2}-\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}=\frac{-3 \mathrm{x}}{\mathrm{a}}$
$\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}=\frac{3 \mathrm{x}}{\mathrm{a}}$

Akshaya Rs

Numerade Educator

Problem 89

eqn of tangent $\mathrm{P} \& \mathrm{Q}$ are
$2 y=3 t_{1} x-a t_{1}^{3}$
$2 y=3 t 2 x-a t_{2}^{3}$
Solving $3\left(t_{1}-t_{2}\right) x+a\left(t_{2}^{3}-t_{1}^{3}\right)=0$
$3 x=a\left(t_{1}^{2}+t_{2}^{2}+t_{1} t_{2}\right)$
$2\left(t_{2}-t_{1}\right) y=a\left(t_{1} t_{2}^{3}-t_{1}^{3} t_{2}\right)$
$2 y=a t_{1} t_{2}\left(t_{2}+t_{1}\right)$
from (I) \& (II) $3 \mathrm{tx}+2 \mathrm{y}=\operatorname{at}\left(\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}\right)+a \mathrm{t}_{1} \mathrm{t}_{2}\left(\mathrm{t}_{2}+\right.$
$=\frac{-\mathrm{at}_{1} \mathrm{t}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\left[\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}-\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}\right]$
$=\frac{+\mathrm{a}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)^{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}$
$\begin{gathered}\Rightarrow(3 \mathrm{tx}+2 \mathrm{y})^{2} & =\mathrm{a}^{2} \mathrm{t}^{2}\left(\mathrm{t}_{1} \mathrm{t}_{2}\right)^{2} \\ & =-\frac{2 \mathrm{yt}}{\mathrm{a}} \mathrm{a}^{2} \mathrm{t}^{2} \\ = & -2 \mathrm{at}^{3} \mathrm{y}\end{gathered}$

Akshaya Rs

Numerade Educator

01:40

Problem 90

$\begin{aligned} \text { Volume inhaled } &=\int_{0}^{6} 0.85 \sin \frac{\pi t}{3} \mathrm{dt} \\ &=-0.85 \cos \frac{\pi t}{3} \times \frac{3}{\pi} \int_{0}^{6} \\ &=\frac{2.3}{\pi}(0.85)=\frac{5.1}{\pi} \end{aligned}$

Varsha Aggarwal

Numerade Educator

01:39

Problem 91

At exercise, respiratory cycle has a period of 4 . Hence difference in frequency is $\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$

Akshaya Rs

Numerade Educator

01:50

Problem 92

Volume inhaled while exercising $=\int_{0}^{4} 1.75 \sin \frac{\pi t}{2}$ dt
$=-1.75\left(\cos \frac{\pi t}{2}\right) \times \frac{2}{\pi}$
$=\frac{7}{\pi}$
Difference $=\frac{7}{\pi}-\frac{5.1}{\pi}$
$=\frac{1.9}{\pi}$
Hence, B is correct Comprehension $5:$ $\begin{aligned} x=& a(2 \cos t+\cos 2 t), \quad y=a(2 \sin t-\sin 2 t\\ \frac{d x}{d t} &=a(-2 \sin t-2 \sin 2 t) \\ &=-2 a[\sin t+2 \sin 2 t] \\ &=-2 a \sin t[1+2 \cos t] \\ \frac{d y}{d t} &=a[2 \cos t-2 \cos 2 t] \\ &=-2 a\left[2 \cos ^{2} t-\cos t-1\right] \\ &=-2 a(\cos t-1)(2 \cos t+1) \end{aligned}$
$\frac{d y}{d x}=\frac{(\cos t-1)(2 \cos t+1)}{(\sin t)(1+2 \cos t)}=-\tan t / 2$
slope of normal $=\cot t / 2 .$

Varsha Aggarwal

Numerade Educator

Problem 93

eqn of normal $\sin t / 2 y-a\left[2 \sin \frac{t}{2} \sin t-\sin 2 t \sin \frac{t}{2}\right]$
$$
=x \cos \frac{t}{2}-a\left[2 \cos t \cos \frac{t}{2}+\cos 2 t \operatorname{tos} \frac{t}{2}\right]
$$
$\Rightarrow x \cos \frac{t}{2}-y \sin \frac{t}{2}=a\left[\begin{array}{c}2 \cos t \cos \frac{t}{2}-2 \sin t \sin \frac{t}{2} \\ +\cos 2 \operatorname{t} \cos \frac{t}{2}+\sin 2 t \sin \frac{t}{2}\end{array}\right]$
$=a\left[2 \cos \frac{3 t}{2}+\cos \frac{3 t}{2}\right]$
$=3 a \cos \frac{3 t}{2}$

Akshaya Rs

Numerade Educator

02:55

Problem 94

Subnormal $=y \frac{d y}{d x}=\left|y \tan \frac{t}{2}\right|$

Akshaya Rs

Numerade Educator

02:00

Problem 95

Eqn of tangent $\cos \frac{\mathrm{t}}{2} \mathrm{y}-\mathrm{a}\left[2 \cos \frac{\mathrm{t}}{2} \sin \mathrm{t}-\sin 2 \operatorname{t} \cos \frac{\mathrm{t}}{2}\right]=$
$-\sin \frac{\mathrm{t}}{2} \mathrm{~s}+\mathrm{a}\left[2 \sin \frac{\mathrm{t}}{2} \cos \mathrm{t}+\cos 2 \mathrm{t} \sin \frac{\mathrm{t}}{2}\right]$
$\left(\cos \frac{t}{2}\right) y+\left(\sin \frac{t}{2}\right) x=a\left[\begin{array}{l}2\left[\sin \frac{t}{2} \cos t+\cos \frac{t}{2} \sin t\right] \\ +\cos 2 \operatorname{t} \sin \frac{t}{2}-\sin 2 t \operatorname{t} 0 s \frac{t}{2}\end{array}\right]$
$$
=a\left[2 \sin \frac{3 t}{2}-\sin \frac{3 t}{2}\right]
$$
$=a \sin \frac{3 t}{2} .$
so, $p=a \sin \frac{3 t}{2}, p_{1}=3 a \cos \frac{3 t}{2}$
$9 p^{2}+p_{1}^{2}=9 a^{2}$

Akshaya Rs

Numerade Educator

01:24

Problem 96

A) As $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$
$\Rightarrow a=2 R \sin A$
$\Rightarrow \frac{d a}{d A}=2 R \cos A$
$\Rightarrow \frac{d a}{\cos A}=2 R d A$
Similarly, $\frac{d b}{\cos B}=2 R d B, \frac{d c}{\cos c}=2 R d C$
$\quad \frac{d a}{\cos A}+\frac{d b}{\cos B}+\frac{d c}{\cos C}+1=$
$2 R(d A+d B+d C)+1$
$\quad A s A+B+C=\pi$
$\Rightarrow d A+d B+d C=0$
using eq (I) $|\mathbf{m}|=1$
B) $\begin{aligned} & \Rightarrow \mathrm{m}=\pm 1 \\ \mathrm{x}^{2} \mathrm{y}^{2}=& 16 \end{aligned}$
$2 x^{2} y \frac{d y}{d x}+2 x y^{2}=0$
$\frac{d y}{d x}=-\frac{y}{x}=1$
Subtangent $=y / d y / d x=2=|\mathrm{k}|$
$\Rightarrow \mathrm{k}=\pm 2$
C) $y=2 e^{2 x}$
$\frac{d y}{d x}=4 e^{2 x}=4$ for yaxis
$\tan ^{-1} 4=\frac{\pi}{2}-\cot ^{-1}\left(\frac{8 n-4}{3}\right)=\tan ^{-1}\left(\frac{8 n+4}{3}\right)$
$12=|8 n-4|$
$\mathrm{n}=2$ or $-1$
D) $\mathrm{x}=\underline{\mathrm{e}^{\sin y}}$
$1=e^{\sin y} \cos y \frac{d y}{d x}$
At $(1,0) \Rightarrow \frac{d y}{d x}=1$
eqn of Normal $\rightarrow y=-(x-1)$
$y+x-1=0$
Area of $\Delta=\frac{1}{2} \times 1 \times 1=\frac{1}{2}$
$\Rightarrow|2 t+1|=3$
$2 t+1=\pm 3 \Rightarrow t=1$ or $-2$
$\mathrm{A} \rightarrow(\mathrm{PQ}), \mathrm{B} \rightarrow(\mathrm{RS}), \mathrm{C} \rightarrow(\mathrm{RQ}), \mathrm{D} \rightarrow(\mathrm{PS})$

Akshaya Rs

Numerade Educator

01:28

Problem 97

A) $\mathrm{A}=\pi \mathrm{\pi}^{2}$
$\mathrm{dA}=2 \pi \mathrm{rdr}$
Approximate increase in Area $=0.72 \pi$
B) $\mathrm{v}=\mathrm{x}^{3}$
$d v=3 x^{2} d x=3(0.02) x^{3}$
$\%$ ine $=\frac{d v}{v} \times 100=6$
C) $y=\frac{x^{2}}{2}-2 x+5$
$\frac{d y}{d t}=x \frac{d x}{d t}-2 \frac{d x}{d t}$
$\Rightarrow+3 \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}-2 \frac{\mathrm{dx}}{\mathrm{dt}}$
$\Rightarrow \mathrm{x}=5$
D) $\mathrm{A}=\frac{\sqrt{3}}{4} \mathrm{x}^{2}$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\sqrt{3}}{2} \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\sqrt{3}}{2} \times 30 \times 0.1=\frac{3 \sqrt{3}}{2}$
$\mathrm{~A} \rightarrow(\mathrm{Q}), \mathrm{B} \rightarrow(\mathrm{R}), \mathrm{C} \rightarrow(\mathrm{P}), \mathrm{D} \rightarrow(\mathrm{S})$

Akshaya Rs

Numerade Educator