5 Marks Questions - MATHS STD 12 Science Questions - Vidyadip

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15 questions · timed · auto-graded

Question 15 Marks

If $y=x^x+x^a+a^x+a^a$ then find $\frac{d y}{d x}$.

Answer

Suppose that
$y=x^x+x^a+a^x+a^a$
differentiating w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a+0$
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x^x\right)+a x^{a-1}+a^x \log _e a$
Suppose that $u=x^x$
$ \therefore \log u =\log _e x^x=x \log _e x$
$\therefore \frac{1}{u} \frac{d u}{d x} =1 \cdot \log _e x+x \cdot \frac{1}{x}=1+\log _e x$
$ \frac{d}{d x}\left(x^x\right) =\frac{d u}{d x}=u\left(1+\log _e x\right)=x^x\left(1+\log _e x\right)$
Put the value of $\frac{d}{d x}\left(x^x\right)$ in equation (1)$
\frac{d y}{d x}=x^x\left(1+\log _e x\right)+a x^{a-1}+a^x \log _e a$

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Question 25 Marks

If $x^y+y^x=b^a+a^b$ then find $\frac{d y}{d x}$.

Answer

$\text{ATQ}$
$x^y+y^x=b^a+a^b$
or $ u+v=b^a+a^b$ when suppose $u=x^y, v=y^x$ taking logarithmic
$\log u=\log x^y=y \log x \quad\left[\log m^n=n \log m\right]$
now differentiating $\text{w.r.t. x}$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \frac{d}{d x}(\log x)$
$\frac{1}{u} \frac{d u}{d x} =\left(\frac{d y}{d x}\right) \log x+y \cdot \frac{1}{x}$
$\therefore \frac{d u}{d x} =u\left[\left(\frac{d y}{d x}\right) \log x+\frac{y}{x}\right]$

again
$v=y^x$
taking $\operatorname{logarithmic}$
$\log v=\log y^x=x \log y\ \left[\because \log m^n=n \log m\right]$
$\log v=x \log y$
now differentiating $\text{w.r.t. x}$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=1 \cdot \log y+x \cdot \frac{d}{d x}(\log y)$
$\Rightarrow \frac{1}{v} \frac{d v}{d x}=\log y+x \cdot \frac{1}{y} \frac{d y}{d x}$
$\Rightarrow \frac{d v}{d x}=\log y+\frac{x}{y} \frac{d y}{d x}$
$\therefore \frac{d v}{d x}\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]$

now $ u+v=b^a+a^b$
$\therefore \frac{d u}{d x}+\frac{d v}{d x}=\frac{d}{d x}\left(b^a+a^b\right)=0$
from equation $(1)$ and $(2),$
Put the value of $\frac{d u}{d x}, \frac{d v}{d x}$
$x^y\left[(\log x) \frac{d y}{d x}+\frac{y}{x}\right]+y^x\left[\log y+\frac{x}{y} \frac{d y}{d x}\right]=0$
$\Rightarrow \quad\left(x^y \log x+\frac{x \cdot y^x}{y}\right) \frac{d y}{d x}+x^y \cdot \frac{y}{x}+y^x \log y=0$
$\text { or } \quad\left(x^y \log x+x \cdot y^{x-1}\right) \frac{d y}{d x}+y \cdot x^{y-1}+y^x \log y=0$
$\therefore \frac{d y}{d x}=-\frac{y x^{y-1}+y^x \log y}{x^y \log x+x y^{x-1}}$

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Question 35 Marks

Find $\frac{d y}{d x}$
$(a)\ \sqrt{x^2+y^2}=\log _e\left(x^2-y^2\right)$
$(b)\ y=x^{x^{x^x} \ldots......\infty}$
$(c)\ x y \log (x+y)=1$

Answer

$(a) \ \sqrt{x^2+y^2}=\log _c\left(x^2-y^2\right)$
differentiating $\text{w.r.t. x}$
$\frac{1}{2 \sqrt{x^2+y^2}}\left(2 x+2 y \frac{d y}{d x}\right)=\frac{1}{\left(x^2-y^2\right)}\left(2 x-2 y \frac{d y}{d x}\right)$
$\text { or } \frac{x+y \frac{d y}{d x}}{\sqrt{x^2+y^2}}=\frac{2 x}{x^2-y^2}-\frac{2 y}{x^2-y^2} \frac{d y}{d x}$
$\text { or } \frac{x}{\sqrt{x^2+y^2}}+\frac{y}{\sqrt{x^2+y^2}} \frac{d y}{d x}=\frac{2 x}{x^2-y^2}-\frac{2 y}{x^2-y^2} \frac{d y}{d x}$
$\left(\frac{y}{\sqrt{x^2+y^2}}+\frac{2 y}{x^2-y^2}\right) \frac{d y}{d x}=\frac{2 x}{x^2-y^2}-\frac{x}{\sqrt{x^2+y^2}}$
$\text { or } y\left(x^2-y^2+2 \sqrt{x^2+y^2}\right) \frac{d y}{d x}$
$=x\left(2 \sqrt{x^2+y^2}-\left(x^2-y^2\right)\right)$
or $ \frac{d y}{d x}=\frac{x\left[2 \sqrt{x^2+y^2}-\left(x^2-y^2\right)\right]}{y\left[x^2-y^2+2 \sqrt{x^2+y^2}\right]}$
$(b)\ y=x^{x^{x^x \ldots \infty}}$
$\Rightarrow y =x^y$
$\Rightarrow \log _e y =\log _e e^y=y \log _e x$
now differentiate $\text{w.r.t. x}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}-\log _c x \cdot \frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \frac{\left(1-y \log _c x\right)}{y} \frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}=\frac{y^2}{x\left(1-y \log _c x\right)}$
$(c)\ x y \log _e(x+y)=1$
differetiate $\text{w.r.t. x}$
$\Rightarrow(x y) \frac{d}{d x}\{\log (x+y)\}+\log (x+y) \frac{d}{d x}(x y)=\frac{d}{d x}(1)$
$\Rightarrow x y \cdot \frac{1}{x+y}\left[1+\frac{d y}{d x}\right]+\log (x+y)\left[x \frac{d y}{d x}+y\right]=0$
$\Rightarrow \frac{x y}{x+y}+\left(\frac{x y}{x+y}\right) \frac{d y}{d x}+x \log (x+y) \frac{d y}{d x}+y \log (x+y)$
$=0$
$\Rightarrow \frac{d y}{d x}\left[\frac{x y}{x+y}+x \log (x+y)\right]$
$=-\left[y \log (x+y)+\frac{x y}{x+y}\right]$
$\Rightarrow \frac{d y}{d x}=-\left[\frac{y \log (x+y)+\frac{x y}{x+y}}{\frac{x y}{x+y}+x \log (x+y)}\right]$
$\frac{d y}{d x}=\frac{-y}{x}\left[\frac{(x+y) \log (x+y)+x}{y+(x+y) \log (x+y)}\right]$

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Question 45 Marks

Examine the continuity of function at $x=2$ and $x =3 , f(x)=|x-2|+|x-3| $

Answer

Given function $f(x)=|x-2|+|x-3|$, It is written as :
$\begin{array}{l} f(x)=\left\{\begin{array}{cc} -(x-2)-(x-3), & 0 \leq x<2 \\ x-2-(x-3), & 2 \leq x<3 \\ x-2+x-3, & x \geq 3 \end{array}\right. \end{array} $
$ \Rightarrow f(x)=\left\{\begin{array}{cl} 5-2 x, & \text { if } 0 \leq x>2 \\ 1, & \text { if } 2 \leq x<3 \\ 2 x-5, & \text { if } x \geq 3 \end{array}\right. $
$(i)$ Examining the continuity at $x=2$.
$\begin{array}{l} \text { R.H.L. }=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} 1=1\end{array} $
$ \begin{aligned} \text { L.H.L. } & =\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0}[5-2(2-h)] \end{aligned} $
$ =\lim _{h \rightarrow 0}(5-4+2 h)=1$
at $x=2$
$ f(x)=5-2 x$
$\therefore f(2)=5-2 \times 2$
$=5-4$
$=1$
hence at $x=2$
$f(2)=\lim _{h \rightarrow 0} f(2+h)=\lim _{h \rightarrow 0} f(2-h)=1$
hence function is continuous at $x=2$.
$(ii)$ Examining the continuity at $x=3$
$\text{R.H.L.}$
$\lim _{h \rightarrow 0} f(3+h) =\lim _{h \rightarrow 0}[2(3+h)-5]$
$ =\lim _{h \rightarrow 0}(6+2 h-5)=1$
$\text{L.H.L.}$
$\lim _{h \rightarrow 0} f(3-h)=\lim _{h \rightarrow 0} 1=1$
at $x=3$
$f(x)=2 x-5,$
$\therefore f(3)=2 \times 3-5=6-5=1$
$\therefore f(3)=\lim _{h \rightarrow 0} f(3+h)=\lim _{h \rightarrow 0} f(3-h)=1$
Hence function is continuous at $x=3$.

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Question 55 Marks

If $x=a \cos \theta+b \sin \theta$ and $y=a \sin \theta-b \cos \theta$ then prove that $y^2 y_2-x y_1+y=0$.

Answer

Given that : $x=a \cos \theta+b \sin \theta$
and $y=a \sin \theta-b \cos \theta$
$\Rightarrow x^2+y^2=(a \cos \theta+b \sin \theta)^2+(a \sin \theta-b \cos \theta)^2$
$=a^2 \cos ^2 \theta+2 a b \sin \theta \cos \theta+b^2 \sin ^2 \theta+a^2 \sin ^2$
$\Rightarrow x^2+y^2=a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+b^2\left(\cos ^2 \theta+\sin ^2 \theta\right)$
$\Rightarrow x^2+y^2=a^2+b^2$
now differentiating $\text{w.r.t. x}$
$\Rightarrow 2 x+2 y y_1$
$= 0$
$\Rightarrow y_1=\frac{-x}{y}$
again differentiating $\text{w.r.t. x}$
$y_2 =-\left[\frac{y \cdot 1-x \cdot y_1}{y^2}\right]$
$ =-\left(\frac{y+x \cdot \frac{x}{y}}{y^2}\right)\ [$from equation $(1)]$
$\Rightarrow y_2 =-\left(\frac{y^2+x^2}{y^3}\right)$
now $y^2 y_2-x y_1 +y$
$= -y^2\left(\frac{y^2+x^2}{y^3}\right)+x \cdot \frac{x}{y}+y$
$ =-\frac{\left(y^2+x^2\right)}{y}+\frac{x^2}{y}+y$
$ =-y-\frac{x^2}{y}+\frac{x^2}{y}+y$
$\Rightarrow y^2 y_2-x y_1+y =0$
Hence proved.

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Question 65 Marks

If given function is continuous at $x=1$, then find $a$ and $b. f(x)=\left\{\begin{array}{cl} 3 a x+b, & \text { if } x>1 \\ 11, & \text { if } x=1 \\ 5 a x-2 b, & \text { if } x<1 \end{array}\right. $

Answer

Given function
$ f(x)=\left\{\begin{array}{cl} 3 a x+b, & \text { if } x>1 \\ 11, & \text { if } x=1 \\ 5 a x-2 b, & \text { if } x<1 \end{array}\right. $
value of $\text{L.H.L.}$
$\lim _{x \rightarrow 1^{-}} f(x) =\lim _{x \rightarrow 1^{-}}[5 a x-2 b]$
$ =\lim _{h \rightarrow 0}[5 a(1-h)-2 b]$
$ =\lim _{h \rightarrow 0}[5 a-2 b-5 a h]$
$ =5 a-2 b$
$\therefore \lim _{x \rightarrow 1^{-}} f(x) =5 a-2 b .$
value of $\text{R.H.L.}$
$=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}(3 a x+b)$
$ =\lim _{h \rightarrow 0}[3 a(1+h)+b]$
$ =\lim _{h \rightarrow 0}(3 a+b+3 a h)=3 a+b$
$\therefore \lim _{x \rightarrow l^{+}} =3 a+b$
and $f(1) =11$
because function is continuous at $x=1$
$ f(1) =\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0} f(1-h)$
$\Rightarrow 11=3 a+b=5 a-2 b$
$\Rightarrow 5 a-2 b=11....(1)$
and $3 a+b=11....(2)$
Solving equation $(1)$ and $(2)$
hence $ a=3, b=2$

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Question 75 Marks

If $(a x+b) e^{y / x}=x$ then show that : $x^3 \frac{d^2 y}{d x^2}=\left(x \frac{d y}{d x}-y\right)^2$

Answer

Given :
$(a x+b) e^{y / x}=x$
taking logarithmic both sides
$\log _e\{(a x+b)\} e^{y / x}=\log _e x$
$\Rightarrow \log _e(a x+b)+\log _e e^{y / x}=\log _e x$
$\Rightarrow \log _e(a x+b)+\frac{y}{x} \log _e e=\log _e x$
$\Rightarrow \log _e(a x+b)+\frac{y}{x}=\log _e x \ \left[\because \log _e e=1\right]$
$\Rightarrow \frac{y}{x}=\log _e x-\log _e(a x+b)$ differentiating $\text{w.r.t. x}$
$\frac{x \frac{d y}{d x}-y \cdot 1}{x^2}=\frac{1}{x}-\frac{a}{a x+b}$
$\Rightarrow x \frac{d y}{d x}-y=x^2\left[\frac{1}{x}-\frac{a}{a x+b}\right]$
$\Rightarrow x \frac{d y}{d x}-y=x^2\left[\frac{a x+b-a x}{x(a x+b)}\right]=\frac{b x}{(a x+b)}$
$\Rightarrow x \frac{d y}{d x}-y=\frac{b x}{(a x+b)}$
again differentiating w.r.t. $x$$x \frac{d^2 y}{d x^2}+\frac{d y}{d x}(1)-\frac{d y}{d x}=\frac{(a x+b) \frac{d}{d x}(b x)-b x \frac{d}{d x}(a x+b)}{(a x+b)^2}$
$ x \frac{d^2 y}{d x^2}=\frac{(a x+b) \times b-b x \times a}{(a x+b)^2}$
$\Rightarrow x \frac{d^2 y}{d x^2}=\frac{a b x+b^2-a b x}{(a x+b)^2}=\frac{b^2}{(a x+b)^2}$
multiplying by $x^2$
$x^3 \frac{d^2 y}{d x^3}=\frac{x^2 b^2}{(a x+b)^2}=\left(\frac{b x}{(a x+b)}\right)^2$
Putting the value from equation $(1)$
$x^3 \frac{d^2 y}{d x^3}=\left(x \frac{d y}{d x}-y\right)^2$
Hence proved.

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Question 85 Marks

If $y^3+3 a x^2+x^3=0$ then prove that:$
\frac{d^2 y}{d x^2}+\frac{2 a^2 x^2}{y^5}=0
$

Answer

Given that :
$
y^3+3 a x^2+x^3=0
....(1)$
differentiating both sides w.r.t. $x$
$
3 y^2 \frac{d y}{d x}+6 a x+3 x^2=0
$
or $\quad$$
y^2 \frac{d y}{d x}+x^2+2 a x=0
......(2)$
or$\quad$$
\frac{d y}{d x}=-\left(\frac{x^2+2 a x}{y^2}\right) .
$
again differentiating both sides of (2) w.r.t. $x$$
\begin{aligned}
y^2 \frac{d^2 y}{d x^2}+2 y\left(\frac{d y}{d x}\right)^2 & =-2 x-2 a \\
y^2 \frac{d^2 y}{d x^2}+2 y\left(\frac{d y}{d x}\right)^2+2 x+2 a & =0 \\
y^2 \frac{d^2 y}{d x^2}+2\left[y\left(\frac{d y}{d x}\right)^2+x+a\right] & =0
\end{aligned}
$
from equation (3) put the value of $\frac{d y}{d x}$ :
or $y^2 \frac{d^2 y}{d x^2}+2\left[y\left(-\left(\frac{x^2+2 a x}{y^2}\right)\right)^2+x+a\right]=0$
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3+(a+x) y^3\right]=0$
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3+(a+x)\right.$
$\left.\left(-3 a x^2-x^3\right)\right]=0[$ from equation (1)]
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3-3 a^2 x^2-3 a x^3\right.$
$\left.-a x^3-x^4\right]=0$
$y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left(a^2 x^2\right)=0$
or $\quad \frac{d^2 y}{d x^2}+\frac{2 a^2 x^2}{y^5}=0 \quad$
Hence proved.

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Question 95 Marks

If $y=x \cos (\log x)$ then prove that :$
x^2 y_2-x y_1+2 y=0
$

Answer

Given :
$
y=x \cos (\log x)
.....(1)$
differentiating both sides w.r.t. $x$$
y_1=1 \cdot \cos (\log x)+x\left[-\sin (\log x) \cdot \frac{1}{x}\right]
$
or$\quad$$
y_1=\cos (\log x)-\sin (\log x)
$
again differentiating both sides w.r.t. $x$
or $\quad y_2=\frac{-\sin (\log x)}{x}-\frac{\cos (\log x)}{x}$
or $\quad x y_2=-\sin (\log x)-\cos (\log x)$
multiplying by $x$ both sides
or $\quad x^2 y_2=-x \sin (\log x)-x \cos (\log x)$
from equation (1) and (2)
or $\quad x^2 y_2=-y+x\left[y_1-\cos (\log x)\right]$
or $\quad x^2 y_2=-y+x y_1-y \quad$ [from equation (1)]
or $\quad x^2 y_2-x y_1+2 y=0$ Hence proved.

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Question 105 Marks

Prove that $\frac{d}{d x}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]=\sqrt{a^2-x^2}$

Answer

Suppose $y=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$
Put $x=a \sin \theta$
$y =\frac{a \sin \theta}{2} \sqrt{a^2-a^2 \sin ^2 \theta}+\frac{a^2}{2} \sin ^{-1}\left(\frac{a \sin \theta}{a}\right)$
$y =\frac{a \sin \theta}{2} \times a \sqrt{1-\sin ^2 \theta}+\frac{a^2}{2} \sin ^{-1}(\sin \theta)$
$ =\frac{a^2}{2} \sin \theta \cos \theta+\frac{a^2}{2} \cdot \theta\left[\because \sin ^{-1}(\sin \theta)=\theta\right]$
$y=\frac{a^2}{2}[\sin \theta \cos \theta+\theta]$
differentiting $\text{w.r.t.}$
$\theta \frac{d y}{d \theta}=\frac{a^2}{2}[\cos \theta \cdot \cos \theta+\sin \theta(-\sin \theta)+1]$
$\frac{d y}{d \theta}=\frac{a^2}{2}\left[\cos ^2 \theta-\sin ^2 \theta+1\right]$
now $x=a \sin \theta$
$\frac{d x}{d \theta}=a \cos \theta$
hence $ \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$
$=\frac{a^2}{2}\left[\cos ^2 \theta-\sin ^2 \theta+1\right] \times \frac{1}{a \cos \theta}$
$=\frac{a}{2}\left[\cos ^2 \theta+\cos ^2 \theta\right] \times \frac{1}{\cos \theta}$
$=\frac{2 a \cos ^2 \theta}{2 \cos \theta}=a \cos \theta$
$=a \sqrt{1-\sin ^2 \theta}$
$=a \sqrt{1-\left(\frac{x}{a}\right)^2}=a \sqrt{\frac{a^2-x^2}{a^2}}$
$\frac{d y}{d x}=\sqrt{a^2-x^2}$
hence
$\frac{d}{d x}\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]=\sqrt{a^2-x^2}$
Hence proved.

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Question 115 Marks

If $x=a \cos ^3 \theta, y=a \sin ^3 \theta$ then find $\left(\frac{d^2 y}{d x^2}\right)_{\theta=\frac{\pi}{4}}$

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Question 125 Marks

If $x^3+y^3=t-\frac{1}{t}$ and $x^6+y^6=t^2+\frac{1}{t^2}$ then prove that $x^4 y^2 \frac{d y}{d x}=1$

Answer

Given:
$x^3+y^3=t-\frac{1}{t}......(1)$
$x^6+y^6=t^2+\frac{1}{t^2}........(2)$
Squaring both sides of equation $(1)
\left(x^3+y^3\right)^2 =\left(t-\frac{1}{t}\right)^2$
$\Rightarrow x^6+2 x^3 y^3+y^6 =t^2-2+\frac{1}{t^2}$
from equation $(2)\ x^6+y^6=t^2+\frac{1}{t^2}$
So, $ 2 x^3 y^3=-2$
or$x^3 y^3=-1.....(3)$
now differentiating $w.r.t. x$
$x^3 \cdot 3 y^2 \frac{d y}{d x}+y^3 \cdot 3 x^2=0$
$\Rightarrow 3 x^3 y^2 \frac{d y}{d x}+3 x^2 y^3=0$
or $\quad$$ x^3 y^2 \frac{d y}{d x}+x^2 y^3=0$
multiplying by $x$
$\therefore x^4 y^2 \frac{d y}{d x}+x^3 y^3 =0$
$x^4 y^2 \frac{d y}{d x}-1 =0$
Put the value from equation $(3)$
hence $ x^4 y^2 \frac{d y}{d x}=1 \quad$ Hence proved.

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Question 135 Marks

If $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$ Prove that $x \frac{d y}{d x}+y=0$.

Answer

Given : $x^2+y^2=t-\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$
So, squaring $\quad\left(x^2+y^2\right)^2=\left(t-\frac{1}{t}\right)^2$
$\Rightarrow \quad x^4+y^4+2 x^2 y^2=t^2+\frac{1}{t^2}-2$
but given that
$x^4+y^4=t^2+\frac{1}{t^2}$
So, $\quad 2 x^2 y^2=-2$
or $\quad x^2 y^2=-1$
now differentiating w.r.t. $x$
$\begin{aligned}
x^2 \cdot 2 y \frac{d y}{d x}+2 x \cdot y^2 & =0 \\
\Rightarrow \quad 2 x^2 y \frac{d y}{d x}+2 x y^2 & =0
\end{aligned}$
Dividing $2 x y$ both sides$
x \frac{d y}{d x}+y=0
$
Hence proved.

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Question 145 Marks

Find $\frac{d y}{d x}$
$(a) \ y=\sin x^{\sin x^{\sin x ...... \infty}}$
$(b) \ y=\sqrt{\log _e x+\sqrt{\log _e x+\sqrt{\log _e x+\ldots \ldots \ldots \infty}}}$
$(c)\ y=e^{x+e^{x+e^{x+\ldots \infty}}}$

Answer

$(a)\ y=\sin x^{\sin x^{\sin x}}$
or $y=(\sin x)^y$
taking logarithmic both sides
$\log _e y =\log _e(\sin x)^y$
$\log _e y =y \log _e(\sin x)$
differentiate both sides $\text{w.r.t. x}$
$\frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{\sin x} \cdot \cos x+\log _e \sin x \cdot \frac{d y}{d x}$
$\Rightarrow\left(\frac{1}{y}-\log _e \sin x\right) \frac{d y}{d x}=y \cot x$
$\Rightarrow \frac{d y}{d x}=\frac{y^2 \cot x}{1-y \log _e \sin x}$
$(b)\ y=\sqrt{\log _c x+\sqrt{\log _c x+\sqrt{\log _c x+\ldots \ldots \infty}}}$
$ \text { or } y =\sqrt{\log _c x+y}$
$\text { or } y^2 =\log _e x+y$
now differentiate $\text{w.r.t. x}$
$2 y \frac{d y}{d x} =\frac{1}{x}+\frac{d y}{d x}$
$(2 y-1) \frac{d y}{d x} =\frac{1}{x}$
$\Rightarrow \frac{d y}{d x} =\frac{1}{x(2 y-1)}$
$(c) \ y=e^{x+e^{x+e^x.....\infty}}$
or $y=e^{x+y}$
taking logarithmic both sides
$\therefore \log _e y =\log _e e^{x+y}$
$\log _c y =(x+y) \log _e e$
or $ \log _c y=x+y \ \left(\because \log _c e=1\right)$
differentiate $\text{w.r.t. x}$
$\frac{1}{y} \frac{d y}{d x}=1+\frac{d y}{d x}$
or $\quad\left(\frac{1}{y}-1\right) \frac{d y}{d x}=1$
$\frac{(1-y)}{y} \frac{d y}{d x}=1$
or $ \frac{d y}{d x}=\frac{y}{1-y}$

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Question 155 Marks

If $x=\sin ^{-1}\left(\frac{2 t}{1+t^2}\right), y=\cos ^{-1}\left(\frac{1-t^2}{1+t^2}\right)$ then find $\frac{d y}{d x}$.

Answer

Given :
$
x=\sin ^{-1}\left(\frac{2 t}{1+t^2}\right), y=\cos ^{-1}\left(\frac{1-t^2}{1+t^2}\right)
$
Put $x=\tan \theta$
So,$
x=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right), y=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)
$
$
\begin{aligned}
& \\
& x=\sin ^{-1}(\sin 2 \theta), y=\cos ^{-1}(\cos 2 \theta) \\
& x=2 \theta, y=2 \theta \\
\therefore \quad & x=2 \tan ^{-1} t, y=2 \tan ^{-1} t
\end{aligned}
$
now differentiate w.r.t. $t$.
$
\begin{aligned}
\frac{d x}{d t} & =\frac{2}{1+t^2} \\
\text { and } \quad \frac{d y}{d t} & =\frac{2}{1+t^2}
\end{aligned}
$
So, $\quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\frac{2}{1+t^2}}{\frac{2}{1+t^2}}=1$

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