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(D) For the function $f(x)$ to be continuous at $x = -1$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(-1)$ mu

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$\pi / 2$

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(D) For the function $f(x)$ to be continuous at $x = -1$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function $f(-1)$ must be equal.$1$. Calculate the value of the function at $x = -1$:$f(-1) = \sin^{-1}[-1] = \sin^{-1}(-1) = -\pi / 2$.$2$. Calculate the right-hand limit $(RHL)$ at $x = -1$:$\lim_{x 	o -1^+} f(x) = \lim_{x 	o -1^+} \sin^{-1}[x]$.Since $x$ approaches $-1$ from the right,$-1 Thus,$\lim_{x 	o -1^+} \sin^{-1}(-1) = -\pi / 2$.$3$. Calculate the left-hand limit $(LHL)$ at $x = -1$:$\lim_{x 	o -1^-} f(x) = \lim_{x 	o -1^-} k([x] + |x|)$.Since $x$ approaches $-1$ from the left,$x Thus,$\lim_{x 	o -1^-} k(-2 - x) = k(-2 - (-1)) = k(-2 + 1) = -k$.$4$. Equate $LHL$ and $f(-1)$:$-k = -\pi / 2 \implies k = \pi / 2$.

If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \geq 0 \\ \sin^{-1}[x], & -1 \leq x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$,then $k =$

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