If a real valued function f(x) = begin{cases} | vedclass

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(C) For the function $f(x)$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$1$. Left-hand limit $(LHL

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$\sin 1$

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(C) For the function $f(x)$ to be continuous at $x = 1$,we must have $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$1$. Left-hand limit $(LHL)$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^-} e^{\frac{\sin a(x-[x])}{x-[x]}}$. Since $x $2$. Right-hand limit $(RHL)$: $\lim_{x 	o 1^+} f(x) = \lim_{x 	o 1^+} \frac{|x^2+x-2|}{x-1} = \lim_{x 	o 1^+} \frac{|(x-1)(x+2)|}{x-1}$. Since $x > 1$,$(x-1) > 0$,so $|x-1| = x-1$. Thus,$\lim_{x 	o 1^+} \frac{(x-1)(x+2)}{x-1} = \lim_{x 	o 1^+} (x+2) = 3$.$3$. Value at $x = 1$: $f(1) = b + 1$.Equating $LHL$,$RHL$,and $f(1)$:$e^{\sin a} = 3 \implies \sin a = \ln 3$.$b + 1 = 3 \implies b = 2$.Therefore,$b \sin a = 2 \ln 3 = \ln 3^2 = \ln 9$.

If a real valued function $f(x) = \begin{cases} e^{\frac{\sin a(x-[x])}{x-[x]}}, & \text{if } x < 1 \\ b+1, & \text{if } x = 1 \\ \frac{|x^2+x-2|}{x-1}, & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$,then $b \sin a =$ ([x] denotes the greatest integer function)

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