Q107 / 160 · TS EAMCET 2022 (20 Jul Shift 1) · Mathematics

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 $\lim _{x \rightarrow 0} A(x)=$

  • A
    0
  • B
    4
  • Correct answer
    C
    6
  • D
    8

Solution

(c) Given $\mathrm{f}(\mathrm{x})$ is differentiable function. Take, $\lim _{x \rightarrow 0} A(x)=\lim _{x \rightarrow 0}\left[\begin{array}{l}2 f(x) \operatorname{cosec} 4 x+ \\ 4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x\end{array}\right]$ $$ \lim _{x \rightarrow 0} A(x)=2 \lim _{x \rightarrow 0} f(x) \lim _{4 x \rightarrow 0} \frac{1}{\left(\frac{\sin 4 x}{4 x}\right) 4 x}+4 \lim _{x \rightarrow 0} f(x) $$ $\lim _{x \rightarrow 0}\left(\cos ^2 x+1\right)-4 \lim _{x \rightarrow 0} \cos ^2 x$ $\lim _{x \rightarrow 0} A(x)=\frac{2}{4} \lim _{x \rightarrow 0} \frac{f(x)}{x} \times \frac{1}{1}+4 \lim _{x \rightarrow 0} f(x)\left(\cos ^2 0+1\right)$ $-4 \cos 0$ $$ \lim _{x \rightarrow 0} \mathrm{aA}(\mathrm{x})=\frac{1}{2} \lim _{x \rightarrow 0} \frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}}+4 \lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(\mathrm{x})(2)-4 $$ Apply L'Hospital Rule $\lim A(x)=\frac{1}{2} \frac{f^{\prime}(x)}{1}+8 \lim _{x \rightarrow 0} f(x)-4$. Apply the limit $$ \begin{aligned} & \lim _{x \rightarrow 0} A(x)=\frac{1}{2} f^{\prime}(0)+8 f(0)-4 \\ & \frac{1}{2} \times 20+8 \times 0-4 \\ & =10-4=6 \end{aligned} $$ Therefore, option (c) is correct

Hence, the correct answer is option: 'C'.

Relevant Questions