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(B) Given, $f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x 0 \end{cases}$ Since $f(x)$

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$8$

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(B) Given, $f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x 0 \end{cases}$ Since $f(x)$ is continuous at $x=0$, we must have $\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^+} f(x) = f(0)$. First, calculate the Left Hand Limit $(LHL)$: $\lim_{x ightarrow 0^-} f(x) = \lim_{x ightarrow 0^-} \frac{1-\cos 4x}{x^2} = \lim_{x ightarrow 0^-} \frac{2\sin^2(2x)}{x^2} = 2 \lim_{x ightarrow 0^-} \left(\frac{\sin 2x}{2x}ight)^2 	imes 4 = 2 	imes 1^2 	imes 4 = 8$. Next, calculate the Right Hand Limit $(RHL)$: $\lim_{x ightarrow 0^+} f(x) = \lim_{x ightarrow 0^+} \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}$. Rationalizing the denominator: $= \lim_{x ightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)} = \lim_{x ightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{16+\sqrt{x}-16} = \lim_{x ightarrow 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{\sqrt{x}} = \lim_{x ightarrow 0^+} (\sqrt{16+\sqrt{x}}+4) = \sqrt{16}+4 = 8$. Since $LHL = RHL = 8$, for continuity at $x=0$, we must have $f(0) = a = 8$.

The value of $a$ for which the function $f(x) = \begin{cases} \frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$, is

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