Question

lf $f\left(x\right)=\text{ sin}\left(\underset{t\to 0}{\lim }\frac{2x}{\pi }{\left(cot\right)}^{-1}\frac{x}{t^2}\right),$ then ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\left(x\right)dx$ is equal to (where, $x\ne 0$ )

• A. $-2$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

Let $y={\lim }_{t\to 0}\frac{2x}{\pi }{\cot }^{-1}\frac{x}{t^2}$
Case-I: when $x>0$ then $y=\frac{2x}{\pi }{\lim }_{t\to 0}{\cot }^{-1}\frac{x}{t^2}=\frac{2x}{\pi }\times 0=0$
Case-II : when $x<0$ then $y=\frac{2x}{\pi }{\lim }_{t\to 0}{\cot }^{-1}\frac{x}{t^2}=\frac{2x}{\pi }\times \pi =2x$
$f(x)=\{\begin{array}{ll}\sin 0& x>0\\ \sin 2x& x<0\end{array}$
Now, ${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}f(x)dx={\int }_{-\frac{\pi }{2}}^0\sin 2xdx+{\int }_0^{\frac{\pi }{2}}0dx=-{\left(\frac{\cos 2x}{2}\right)}_{-\frac{\pi }{2}}^0=-\frac{1}{2}(1-(-1))$