Question

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

• A. $-2$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

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