Question

Assertion :If $f$ and $g$ are defined on $[0,\infty ]$ by $f\left(x\right)=\underset{n\to \infty }{lim}\frac{x^n-1}{x^n+1}$ and $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ then $g$ is continuous but not differentiable at $x=1$ Reason: $f\left(x\right)=sgn(x-1)$

• 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 B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

$f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\underset{n\to \infty }{lim}\frac{x^n-1}{x^n+1}& 0<x<1\end{array}\\ \begin{array}{cc}0& x=1\end{array}\\ \begin{array}{cc}\underset{n\to \infty }{lim}\frac{1-\frac{1}{x^n}}{1+\frac{1}{x^n}}& x>1\end{array}\end{array}=\{\begin{array}{c}\begin{array}{cc}-1& 0<x<1\end{array}\\ \begin{array}{cc}0& x=1\end{array}\\ \begin{array}{cc}1& x>1\end{array}\end{array}=sgn(x-1)$
$g\left(x\right)={\int }_0^x-1dt=-x$ for $x\le 1$

$g\left(x\right)={\int }_0^1(-1)dt+{\int }_0^{x}f\left(t\right)dt=-x+x-1=-1$ for $x>1$
$\underset{x\to 1}{lim}g\left(x\right)=-1$ so $g$ is continous at $x=1$
$g^{\prime }(1-)=-1g^{\prime }(1+)=0$