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(C) Given that $f(x)$ is a differentiable function with $f(0)=0$ and $f^{\prime}(0)=20$. We need to find $\lim _{x \rightarrow 0} A(x) = \lim _{x \rig

Vedclass · Accepted Answer

$6$

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(C) Given that $f(x)$ is a differentiable function with $f(0)=0$ and $f^{\prime}(0)=20$. We need to find $\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} [2 f(x) \operatorname{cosec} 4 x + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x]$. Rewrite the expression as: $\lim _{x \rightarrow 0} A(x) = \lim _{x \rightarrow 0} \left[ \frac{2 f(x)}{\sin 4x} + 4 f(x)(\cos ^2 x + 1) - 4 \cos ^2 x \right]$. Since $f(0)=0$,the first term is an indeterminate form of type $\frac{0}{0}$. Using the limit $\lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} = \lim _{x \rightarrow 0} \frac{f(x)}{x} \cdot \frac{4x}{\sin 4x} \cdot \frac{1}{4} = f^{\prime}(0) \cdot 1 \cdot \frac{1}{4} = \frac{20}{4} = 5$. Now,evaluate the limit: $\lim _{x \rightarrow 0} A(x) = 2 \left( \lim _{x \rightarrow 0} \frac{f(x)}{\sin 4x} \right) + 4 \lim _{x \rightarrow 0} f(x)(\cos ^2 x + 1) - 4 \lim _{x \rightarrow 0} \cos ^2 x$. $= 2(5) + 4(0)(1+1) - 4(1)^2$. $= 10 + 0 - 4 = 6$. Therefore,the correct option is $C$.

Let $f(x)$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(0)=20$. For $x \in \left(0, \frac{\pi}{2}\right]$,if $A(x)=2 f(x) \operatorname{cosec} 4 x+4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x$,then $\lim _{x \rightarrow 0} A(x)=$

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