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The area (in sq units) of the region ${(x,y):y^{2}\geq2x, and x^{2}+y^{2}\leq 4x, x\geq0, and  y\geq 0}$ is

Answer:

Explanation:

Given the equation  of curve  are

           $y^{2}=2x$   ........(i)

 which is parabola with vertex (0,0) and axis parallel to X-axis

and    x²+y² =4x

 which is circle with centre (2,0) and radius=2                .......(ii)

  on substuting y²=2x in Eq (ii) we get

    x²+2x= 4x

$\Rightarrow x^{2}=2x$

$\Rightarrow x=0$

  or x=2

$\Rightarrow y=0 $ or $y=\pm2$   [ using  Eq.. (i)]

Now, the required area is the area of shaded region . ie,

 Required area=  $\frac{Area of a circle}{4}$- $\int_{0}^{2} \sqrt{2x}dx$

$\frac{\pi(2)^{2}}{4}-\sqrt{2}\int_{0}^{2} x^{\frac{1}{2}}dx$

$\pi -\sqrt{2}\left[\frac{x^{\frac{3}{2}}}{{\frac{3}{2}}}\right]_0^2$

$=\pi -\frac{2\sqrt{2}}{3}[2\sqrt{2}-0]$

$=(\pi -\frac{8}{3})$ sq units

$\lim_{n \rightarrow \infty}[\frac{(n+1)(n+2)...3n}{n^{2n}}]^{1/n}$ is equal to 

Answer:

Explanation:

Let l=$\lim_{n \rightarrow \infty}[\frac{(n+1)(n+2)...3n}{n^{2n}}]^{1/n}$

$\lim_{n \rightarrow \infty}[\frac{(n+1)(n+2)...(n+2n)}{n^{2n}}]^{1/n}$

$\lim_{n \rightarrow \infty}[(\frac{n+1}{n})(\frac{n+2}{n})....(\frac{n+2n}{n})]^{\frac{1}{n}}$

On taking long both sides, we get

   $\log I=\lim_{n \rightarrow \infty}\frac{1}{n}[\log {(1+\frac{1}{n})(1+\frac{2}{n}).....(1+\frac{2n}{n})}]$

$\Rightarrow \log l= \lim_{n\rightarrow \infty}\frac{1}{n}$   $[\log (1+\frac{1}{n})+log(1+\frac{2}{n})+.....+log(1+\frac{2n}{n})]$

$\Rightarrow \log l=\lim_{n \rightarrow \infty}\frac{1}{n}\sum_{r=1}^{2n}\log(1+\frac{r}{n})$

$\Rightarrow \log l= \int_{0}^{2} \log(1+x)dx$

$\Rightarrow \log l=[\log(1+x).x-\int_{}^{} \frac{1}{1+x}.xdx]_0^2$

$\Rightarrow \log l=[\log(1+x).x]_2^0-\int_{0}^{2} \frac{x+1-x}{1+x}.xdx$

$\Rightarrow \log l=2.\log 3-\int_{0}^{2} (1-\frac{1}{1+X})dx$

$\Rightarrow \log l=2.\log 3-[x-\log\mid1+x\mid]_0^2 $

$Rightarrow \log l=2.\log 3-[ 2-\log3]$

$\Rightarrow \log l=3.\log3-2$

$\Rightarrow \log l=.\log27-2$

$\therefore l=e^{\log 27-2}=27.e^{-2}$

 =$\frac{27}{e^{2}}$

The intergral $\int_{}^{}\frac{2x^{12}+5x^{9}}{(x^{5}+x^{3}+1)^{3}} dx$ is equal to 

where C is an arbitrary constant

Answer:

Explanation:

Let $I=\int_{}^{} \frac{2x^{2}+5x^{9}}{(x^{5}+x^{3}+1)^{3}}dx$

$=\int_{}^{} \frac{2x^{12}+5x^{9}}{x^{15}(1+x^{-2}+x^{-5})^{3}}dx$

$=\int_{}^{} \frac{2x^{-3}+5x^{-6}}{(1+x^{-2}+x^{-5})^{3}}dx$

Now put 1+x^-2+x^-5=t

$\Rightarrow  (-2x^{-3}-5x^{-6})dx= dt$

 $\Rightarrow  (2x^{-3}+5x^{-6})dx=- dt$

$\therefore   I= -\int_{}^{} \frac{dt}{t^{3}}=-\int_{}^{} t^{-3}dt$ 

$=-\frac{t^{-3+1}}{-3+1}+C= \frac{I}{2t^{2}}+C$

$\frac{x^{10}}{2(x^{5}+x^{3}+1)^{2}}+C$

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side =x units and a circle of radius =r units. If the sum of the areas of the square and the circle so formed is minimum, then

Answer:

Explanation:

According to the given information, we have

 the perimeter of square + perimeter of circle=2 units

  $\Rightarrow 4x+2\pi r=2\Rightarrow r=\frac{1-2x}{\pi}..(i)$

Now, let A be the sum of the areas of the square and the circle, Then

$A=x^{2}+\pi r^{2}$

   $A(x)=x^{2}+\pi\frac{(1-2x)^{2}}{\pi^{2}}$

 $\Rightarrow x^{2}+\frac{(1-2x)^{2}}{\pi^{}}$

Now, for minimum value of A(x). $\frac{\text{d}A}{\text{d}x}=0$

$\Rightarrow 2x+\frac{2(1-2x)}{\pi}.(-2)=0$

$\Rightarrow x=\frac{2-4x}{\pi}$

$\Rightarrow \pi x+4x=2$

$\Rightarrow x=\frac{2}{\pi+4}$   .........(ii)

Now, from Eq.(i), we get

$r=\frac{1-2.\frac{2}{\pi+4}}{\pi}$

$r=\frac{\pi+4-4}{\pi(\pi+4)}=\frac{1}{\pi+4}$     ..........(iii)

From Eq(ii) and (iii)

 we get x=2r

 Consider $f(x)= tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},x\epsilon(0,\frac{\pi}{2})$. A normal to y= f(x) at x= $\frac{\pi}{6}$ also passes through the point

Answer:

Explanation:

We have

$f(x)= tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},x\epsilon(0,\frac{\pi}{2})$

       $f(x)= tan^{-1}\sqrt{\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})^{2}}{(\cos\frac{x}{2}-\sin \frac{x}{2})^{2}}}$

   $f(x)= tan^{-1}({\frac{\cos\frac{x}{2}+\sin\frac{x}{2}^{}}{\cos\frac{x}{2}-\sin \frac{x}{2}^{}}})$

(  $\cos \frac{x}{2}>\sin \frac{x}{2} for$  $0< \frac{x}{2}. \frac{\pi}{4} $ 

$=\tan^{-1}(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})$

$=\tan^{-1}[\tan(\frac{\pi}{4}+\frac{x}{2})]=\frac{\pi}{4}+\frac{x}{2}$

$\Rightarrow f'(x)=\frac{1}{2}\Rightarrow f'(\frac{\pi}{6})=\frac{1}{2}$

  Now, equation of normal at x= $\frac{\pi}{6}$ is given by

  $(y-f(\frac{\pi}{6}))=-2(x-\frac{\pi}{6})$

$\Rightarrow (y-\frac{\pi}{3})=-2(x-\frac{\pi}{6})$

$[\because f(\frac{\pi}{6})=\frac{\pi}{4}+\frac{\pi}{12}=\frac{4\pi}{12}=\frac{\pi}{3}$

which passes  through (0, $\frac{2\pi}{3}$)

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