mathematics-class-12-unit-08-chapter-06-integral-calculus-l-6-6-teckf9x5nvi

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \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 . $

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<footer style = "font-size: 18px">Integral calculus lecture-6 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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)  $0<x<1$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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.

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 $y(1)=0 \Rightarrow c=0$ $\therefore \quad y=-\sqrt{1-x^2}+\ln \left(\frac{1+\sqrt{\left.1-x^2\right)}}{x}\right)$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 $y=u(x+2) \Rightarrow \frac{d y}{d x}=u+(x+2) \frac{d u}{d x}$

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

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\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 $
- $y(1)=3 \Rightarrow \ln 3+\frac{3}{1+2}$
- $=c \rightarrow c=1+\ln 3$

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

- 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 $
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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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 $

- $\therefore\left(\right)$ does not hold for any $x>0$.
- $\therefore$ The solution curve does not intersets $y=(x+3)^2$.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- 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}}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- (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 $(-1,0)$.

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- 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$ $\Rightarrow \quad f^{\prime}(x)  =\frac{1}{2} f(x)$    $\Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}$  
- $=\frac{1}{2} \Rightarrow \ln f(x)=\frac{x}{2}+c$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Solution $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $f(0)=1 \Rightarrow c=0$

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

- 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.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore \quad y d y=-u d u$

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

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

- $ \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

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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>