Let $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , x \lt 0 \\ \mathrm{a} & , x=0 \\…

Let $\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} & , x \lt 0 \\ \mathrm{a} & , x=0 \\ \frac{\sqrt{2}}{\sqrt{16+\sqrt{x-4}}} & , x\gt0 .\end{array}\right.$ If $\mathrm{f}(x)$ is continuous at $x=0$, then the value of a is
  1. 8
  2. 4
  3. $\frac{1}{2}$
  4. 2

Solution

$\begin{array}{ll} & \mathrm{f}(x) \text { is continuous at } x=0 . \\ \therefore \quad & \mathrm{f}(0)=\lim _{x \rightarrow 0^{-}} \mathrm{f}(x) \\ \therefore \quad & \mathrm{a}=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2} \\ & =2 \lim _{x \rightarrow 0^{-}} \frac{\sin ^2 2 x}{(2 x)^2} \times 4=2(1)^2 \times 4=8\end{array}$

Asked in: MHT CET 2024 (11 May Shift 2)

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