Question

Let $f\left(x\right)=\{\begin{array}{cc}\frac{(1-\cos 4x)\tan x}{x^3}& \text{if}x<0\\ 2a^2& \text{if}x=0\\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}}& \text{if}x>0\end{array}$
The possible value of $a$ so that $f$ is continuous is

• A. $4$
• B. $2$
• C. $-2$
• D. not possible

Answer 1

Answer: (B), (C)

The correct options are
B $2$
C $-2$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{(1-\cos 4x)}{x^2}\frac{\tan x}{x}=\underset{x\to 0^{-}}{lim}\frac{2{\sin }^{2}2x}{x^2}\frac{\tan x}{x}$
$=\underset{x\to 0^{-}}{lim}\frac{{\sin }^{2}2x}{(2x{)}^2}.4\frac{\tan x}{x}$
$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$
$=\underset{x\to 0^{+}}{lim}\frac{\sqrt{x}}{16+\sqrt{x}-16}.\sqrt{16+\sqrt{x}}+4$
$=\underset{x\to 0^{+}}{lim}(\sqrt{16+\sqrt{x}}+4)=8$,
Thus $2a^2=8\Rightarrow a=\pm 2$