Q29 / 60 · JEE Advanced 2011 (Paper 2) · Mathematics

If $f(x)=\left\{\begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array}\right.$, then

  • Correct answer
    A
    $f(x)$ is continuous at $x=-\frac{\pi}{2}$
  • Correct answer
    B
    $f(x)$ is not differentiable at $x=0$
  • Correct answer
    C
    $f(x)$ is differentiable at $x=1$
  • Correct answer
    D
    $f(x)$ is differentiable at $x=-\frac{3}{2}$

Solution

$f(x)=\left\{\begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array}\right.$ Continuity at $x=-\frac{\pi}{2}$, $$ \begin{aligned} & f\left(-\frac{\pi}{2}\right)=-\left(-\frac{\pi}{2}\right)-\frac{\pi}{2}=0 \\ & \mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}-\cos \left(-\frac{\pi}{2}+h\right)=0 \end{aligned} $$ $\therefore$ Continuous at $x=0$ Continuity at $x=0 \Rightarrow f(0)=-1$ $\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}(0+h)-1=-1$ $\therefore$ Continuous at $x=0$ Continuity at $x=1 ; f(1)=0$ $\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0} \log (1+h)=0$ $\therefore$ Continuous at $x=1$ Here, $f^{\prime}(x)=\left\{\begin{array}{cc}-1, & x \leq-\frac{\pi}{2} \\ \sin x, & -\frac{\pi}{2} < x \leq 0 \\ 1, & 0 < x \leq 1 \\ \frac{1}{x}, & x>1\end{array}\right.$ Differentiable at $x=0$, $\mathrm{LHD}=0, \mathrm{RHD}=1$ $\therefore \quad$ Not differentiable at $x=0$ Differentiable at $x=1$, $\mathrm{LHD}=1, \mathrm{RHD}=1$ $\therefore$ Differentiable at $x=1$ Also, for $x=-\frac{3}{2} \Rightarrow f(x)=-x-\frac{3}{2}$ $\therefore$ Differentiable at $x=-\frac{3}{2}$

Hence, the correct answer is options: 'A', 'B', 'C', 'D'.

Relevant Questions