If f(x) = begin{cases} -x-frac{pi}{2}, & x le | vedclass

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Question

(A) Given $f(x) = \begin{cases} -x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \ -\cos x, & -\frac{\pi}{2} < x \leq 0 \ x-1, & 0 < x \leq 1 \ \ln x, & x

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$(A, B, C, D)$

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(A) Given $f(x) = \begin{cases} -x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \ -\cos x, & -\frac{\pi}{2}  1 \end{cases}$.$1$. Continuity at $x=-\frac{\pi}{2}$:$f(-\frac{\pi}{2}) = -(-\frac{\pi}{2}) - \frac{\pi}{2} = 0$.$RHL = \lim_{h 	o 0} -\cos(-\frac{\pi}{2}+h) = -\cos(-\frac{\pi}{2}) = 0$.Since $LHL = RHL = f(-\frac{\pi}{2})$,$f(x)$ is continuous at $x=-\frac{\pi}{2}$. (Statement $A$ is true).$2$. Differentiability at $x=0$:$LHD = \frac{d}{dx}(-\cos x)|_{x=0} = \sin(0) = 0$.$RHD = \frac{d}{dx}(x-1)|_{x=0} = 1$.Since $LHD 
eq RHD$,$f(x)$ is not differentiable at $x=0$. (Statement $B$ is true).$3$. Differentiability at $x=1$:$LHD = \frac{d}{dx}(x-1)|_{x=1} = 1$.$RHD = \frac{d}{dx}(\ln x)|_{x=1} = \frac{1}{1} = 1$.Since $LHD = RHD$,$f(x)$ is differentiable at $x=1$. (Statement $C$ is true).$4$. Differentiability at $x=-\frac{3}{2}$:Since $-\frac{3}{2} Thus,all statements $A, B, C, D$ are true.

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