vineetloomba.com
Difficulty: Easy Type: Integer Type

The value of definite integral $\int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)}{e^{x}+1} d x$ is equal to: mmmm

(a) $\frac{\pi}{2 \sqrt{3}}$

(b) $\frac{\pi}{\sqrt{3}}$

(c) $\frac{\pi}{4 \sqrt{3}}$

(d) $\frac{\pi}{3 \sqrt{3}}$

$\lim _{n \rightarrow \infty} \frac{n^{2}}{\left(\left(n^{2}+1^{2}\right)\left(n^{2}+2^{2}\right) \ldots \ldots \ldots\left(n^{2}+n^{2}\right)\right)^{\frac{1}{n}}}$

equals:

(a) $2 e^{2+\frac{\pi}{2}}$\

(b) $2 e^{2-\frac{\pi}{2}}$\

(c) $\frac{1}{2} e^{2-\frac{\pi}{2}}$\

(d) $\frac{1}{2} e^{2+\frac{\pi}{2}}$

\item The solution of differential equation $\frac{d y}{d x}=\frac{x^{2}+y^{2}+1}{2 x y}$ satisfying $y(1)=0$ is given by:\

(a) a circle\

(b) $y^{2}=x^{2}+x-10$\

(c) hyperbola\

(d) ellipse

Numerical Answer: 12

Solution

The solution of differential equation $\frac{d y}{d x}=\frac{x^{2}+y^{2}+1}{2 x y}$ satisfying $y(1)=0$ is given by:

(a) a circle

(b) $y^{2}=x^{2}+x-10$

(c) hyperbola

(d) ellipse