AP EAMCET · Maths · Continuity and Differentiability
If the 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 \(1\)
- B \(2\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1-\cos 4 x}{x^2} = \frac{4^2}{2} = 8\) \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}}+4)}{(\sqrt{16+\sqrt{x}}-4)(\sqrt{16+\sqrt{x}}+4)}\)…
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