If f(x) = begin{cases} k cos x - x cos k, & x | vedclass

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(D) For $f(x)$ to be differentiable at $x = \frac{\pi}{2}$,it must be continuous at $x = \frac{\pi}{2}$.Continuity at $x = \frac{\pi}{2}$ implies $\li

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$k \in \phi$ (Null set)

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(D) For $f(x)$ to be differentiable at $x = \frac{\pi}{2}$,it must be continuous at $x = \frac{\pi}{2}$.Continuity at $x = \frac{\pi}{2}$ implies $\lim_{x 	o \frac{\pi}{2}^-} f(x) = \lim_{x 	o \frac{\pi}{2}^+} f(x) = f(\frac{\pi}{2})$.$k \cos(\frac{\pi}{2}) - \frac{\pi}{2} \cos k = k \sin(\frac{\pi}{2}) + \frac{\pi}{2} \sin k$.Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$,we get $-\frac{\pi}{2} \cos k = k + \frac{\pi}{2} \sin k$,which simplifies to $k = -\frac{\pi}{2}(\sin k + \cos k)$.For differentiability,the left-hand derivative $(LHD)$ must equal the right-hand derivative $(RHD)$ at $x = \frac{\pi}{2}$.$f'(x) = -k \sin x - \cos k$ for $x  \frac{\pi}{2}$.$LHD$ at $x = \frac{\pi}{2}$ is $-k \sin(\frac{\pi}{2}) - \cos k = -k - \cos k$.$RHD$ at $x = \frac{\pi}{2}$ is $k \cos(\frac{\pi}{2}) + \sin k = \sin k$.Equating $LHD$ and $RHD$: $-k - \cos k = \sin k$,or $k = -(\sin k + \cos k)$.Comparing the two conditions: $-\frac{\pi}{2}(\sin k + \cos k) = -(\sin k + \cos k)$.This implies $(\sin k + \cos k)(\frac{\pi}{2} - 1) = 0$. Since $\frac{\pi}{2} 
eq 1$,we must have $\sin k + \cos k = 0$,which means $	an k = -1$,so $k = n\pi - \frac{\pi}{4}$.Substituting this back into $k = -(\sin k + \cos k)$,if $k = -\frac{\pi}{4}$,then $-(\sin(-\frac{\pi}{4}) + \cos(-\frac{\pi}{4})) = -(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = 0 
eq -\frac{\pi}{4}$.Thus,there is no real value of $k$ that satisfies both conditions simultaneously. Therefore,$k \in \phi$.

If $f(x) = \begin{cases} k \cos x - x \cos k, & x \in [0, \frac{\pi}{2}] \\ k \sin x + x \sin k, & x \in (\frac{\pi}{2}, \pi] \end{cases}$ is differentiable in $(0, \pi)$,then:

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