Question

If the function \( f \) defined by \[ f(x) = \begin{cases} \dfrac{1 - \cos 4x}{x^2}, & x<0 \\ a, & x = 0 \\ \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( a = \)

• A. \( 1 \)
• B. \( 2 \)
• C. \( 4 \)
• D. \( 8 \)

Hint:

Use trigonometric limits like \( \frac{1 - \cos x}{x^2} \to \frac{1}{2} \) and conjugate multiplication tricks for roots in limits.

Answer 1

The Correct Option is D

To ensure continuity at \( x = 0 \), we require: \[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) = a \] Step 1: Left-hand limit ( \( x \to 0^- \) ) \[ \lim_{x \to 0^-} \dfrac{1 - \cos 4x}{x^2} = \lim_{x \to 0^-} \dfrac{2 \sin^2 2x}{x^2} = \lim_{x \to 0^-} 4 . \dfrac{\sin^2 2x}{(2x)^2} . 4 = 4 . 4 = 16 \] \[ \Rightarrow \lim_{x \to 0^-} f(x) = 16 \] Step 2: Right-hand limit ( \( x \to 0^+ \) ) \[ \lim_{x \to 0^+} \dfrac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} \] Multiply numerator and denominator by the conjugate: \[ = \lim_{x \to 0^+} \dfrac{\sqrt{x} . (\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)} = \lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(16 + \sqrt{x}) - 16} \] \[ = \lim_{x \to 0^+} \dfrac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) = \sqrt{16} + 4 = 4 + 4 = 8 \] \[ \Rightarrow \lim_{x \to 0^+} f(x) = 8 \] Step 3: Continuity Condition Since: \[ \lim_{x \to 0^-} f(x) = 16,
\lim_{x \to 0^+} f(x) = 8 \] These are not equal, so **f is not continuous** at \( x = 0 \) unless we made a mistake. But the function is given as continuous, so something went wrong in LHL. Let’s re-check LHL: \[ \lim_{x \to 0^-} \dfrac{1 - \cos 4x}{x^2} = \lim_{x \to 0^-} \dfrac{2 \sin^2 2x}{x^2} = 2 . \lim_{x \to 0^-} \dfrac{\sin^2 2x}{x^2} = 2 . \lim_{x \to 0^-} \left( \dfrac{\sin 2x}{x} \right)^2 = 2 . (2)^2 = 8 \] Conclusion: \[ \lim_{x \to 0^-} f(x) = 8,
\lim_{x \to 0^+} f(x) = 8 \Rightarrow f(0) = a = 8 \] % Final Answer Value of \( a \): \( \boxed{8} \)