AP EAMCET · Maths · Continuity and Differentiability
If a function \(f\) defined by \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{array}\right.\) is continuous at \(\mathrm{x}=0\), then \(\mathrm{a}=\)
- A 8
- B 4
- C 3
- D 2
Answer & Solution
Correct Answer
(A) 8
Step-by-step Solution
Detailed explanation
\( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \frac{4^2}{2} = 8 \) \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} \cdot \frac{\sqrt{16 + \sqrt{x}} + 4}{\sqrt{16 + \sqrt{x}} + 4} \)…
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