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need solution for RD Sharma maths class 12 chapter 8 Continuity exercise Fill in the blanks question 7

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Answer: 0
Hint: Use L-Hospital’s rule
Given:f(x)=\left\{\begin{array}{cc} \frac{1-\sin x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ k & , x=\frac{\pi}{2} \end{array}\right.    is continuous at x=\frac{\pi }{2}


            If f(x) is continuous at x=\frac{\pi }{2} , then

            \begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right) \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin x}{\pi-2 x}=k \end{aligned}

            Applying L-Hospital’s rule on LHS that is differentiating both numerator and denominator,

            \begin{aligned} &\lim _{x \rightarrow \frac{\pi}{2}} \frac{0-(\cos x)}{0-2(1)}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos x}{-2}=k \\ &\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x}{2}=k \\ &\frac{\cos \frac{\pi}{2}}{2}=k \end{aligned}

            k=0 \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{2}=0\right]

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