TS EAMCET · Maths · Continuity and Differentiability
The value of ' \(a\) ' for which the function \(f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2}, & x < 0 \ a, & x=0 \text { is continuous at } \ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0\end{array}\right.\) \(x=0\), is
- A 2
- B 8
- C 4
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) 8
Step-by-step Solution
Detailed explanation
Given, \(f(x)=\left\{\begin{array}{ccc}\frac{1-\cos 4 x}{x^2}, & \text { when } x 0\end{array}\right.\) Since, \(f(x)\) is continuous at \(x=0\) \(\therefore \quad(\mathrm{LHL})_{x=0}=(\mathrm{RHL})_{x=0}=f(0)\) Now,…
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