If f(x) = begin{cases} mx+1, & x leq frac{pi} | vedclass

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(D) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function

Vedclass · Accepted Answer

$n=m\frac{\pi}{2} + 1 - 1 = m\frac{\pi}{2}$

Vedclass · Answer

(D) For the function $f(x)$ to be continuous at $x = \frac{\pi}{2}$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = \frac{\pi}{2}$ must be equal.$1$. Calculate the value of the function at $x = \frac{\pi}{2}$:$f(\frac{\pi}{2}) = m(\frac{\pi}{2}) + 1$$2$. Calculate the $LHL$ at $x = \frac{\pi}{2}$:$\lim_{x 	o \frac{\pi}{2}^-} f(x) = \lim_{x 	o \frac{\pi}{2}^-} (mx + 1) = m(\frac{\pi}{2}) + 1$$3$. Calculate the $RHL$ at $x = \frac{\pi}{2}$:$\lim_{x 	o \frac{\pi}{2}^+} f(x) = \lim_{x 	o \frac{\pi}{2}^+} (\sin x + n) = \sin(\frac{\pi}{2}) + n = 1 + n$$4$. Equate $LHL$ and $RHL$:$m(\frac{\pi}{2}) + 1 = 1 + n$$m(\frac{\pi}{2}) = n$Thus,the condition for continuity is $n = \frac{m \pi}{2}$.

If $f(x) = \begin{cases} mx+1, & x \leq \frac{\pi}{2} \\ \sin x+n, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,where $m, n \in \mathbb{Z}$,then:

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