TS EAMCET · Maths · Continuity and Differentiability
Define \(f: \mathbf{R} \rightarrow \mathbf{R}\) by \(f(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2}, & x \lt 0 \ a, & x=0 \ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4},}, & x\gt0\end{array}\right.\) Then the value of ' \(a\) ' so that \(f\) is continuous at \(x=0\) is
- A 8
- B 4
- C 2
- D 1
Answer & Solution
Correct Answer
(A) 8
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2}=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 2 x}{x^2}\)…
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