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JEE Advanced · Mathematics · 23. C&D

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

  1. A
    \(f(x)\) is continuous at \(x=-\frac{\pi}{2}\)
  2. B
    \(f(x)\) is not differentiable at \(x=0\)
  3. C
    \(f(x)\) is differentiable at \(x=1\)
  4. D
    \(f(x)\) is differentiable at \(x=-\frac{3}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A)
\(f(x)\) is continuous at \(x=-\frac{\pi}{2}\)

Step-by-step Solution

Detailed explanation

\(f(x)=\left\{\begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array}\right.\)
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 \(x=-\frac{3}{2}\)
From JEE Advanced
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