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(A) For the function $f(x)$ to be continuous at $x = \pi$,the left-hand limit,right-hand limit,and the value of the function at $x = \pi$ must be equa

Vedclass · Accepted Answer

$-\frac{2}{\pi}$

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(A) For the function $f(x)$ to be continuous at $x = \pi$,the left-hand limit,right-hand limit,and the value of the function at $x = \pi$ must be equal.$1$. Value of the function at $x = \pi$:$f(\pi) = k(\pi) + 1 = k\pi + 1$.$2$. Left-hand limit $(LHL)$ at $x = \pi$:$\lim_{x 	o \pi^-} f(x) = \lim_{x 	o \pi^-} (kx + 1) = k\pi + 1$.$3$. Right-hand limit $(RHL)$ at $x = \pi$:$\lim_{x 	o \pi^+} f(x) = \lim_{x 	o \pi^+} \cos x = \cos(\pi) = -1$.For continuity,$LHL$ = $RHL$ = $f(\pi)$:$k\pi + 1 = -1$.Solving for $k$:$k\pi = -1 - 1$$k\pi = -2$$k = -\frac{2}{\pi}$.Thus,the required value of $k$ is $-\frac{2}{\pi}$.

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx + 1, & \text{if } x \le \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$.

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