Question

Assertion :For $K>0$, let $f\left(x\right)=\{\begin{array}{cc}\sqrt{x^2-1},& x\le -1\\ K\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{x+1},& -1<x<0\\ \sqrt{1-x^2},& 0\le x<1\\ \frac{1}{\sin (x-1)}-\frac{1}{\tan (x-1)},& x>1\end{array}$ then $f$ is a discontinuous function Reason: $f$ has a break point at $x=1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$\underset{x\to 1-}{lim}f\left(x\right)=K\underset{x\to 1-}{lim}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{\sqrt{x+1}}=K\underset{y\to \pi }{lim}\frac{\sqrt{\pi }-\sqrt{y}}{\sqrt{1+\cos y}}({\cos }^{-1}x=y)$

$=K\underset{y\to \pi }{lim}\frac{1}{\sqrt{2}}\frac{\pi -y}{\cos \frac{y}{2}}\frac{1}{\sqrt{p}+\sqrt{y}}$

$=K\frac{2}{\sqrt{2}}\underset{y\to \pi }{lim}\frac{\frac{\pi }{2}-\frac{y}{2}}{\sin (\frac{\pi }{2}-\frac{y}{2})}\times \frac{1}{\sqrt{\pi }+\sqrt{y}}$

$=\frac{K}{\sqrt{2\pi }}\ne 0=f(-1)$

So $f$ is discontinuous at $x=-1$

$\underset{x\to 1+}{lim}f\left(x\right)=\underset{x\to 1+}{lim}(\frac{1}{\sin (x-1)}-\frac{1}{\tan (x-1)})=\underset{u\to 0}{lim}(\frac{1}{\sin u}-\frac{1}{\tan u})$

$=\underset{u\to 0}{lim}\frac{1-\cos u}{\sin u}=0=f\left(1\right)=\underset{x\to 1-}{lim}f\left(x\right)$

Hence there is no break point at $x=1$ but there is a break point at $x=-1$