MHT CET · Maths · Continuity and Differentiability
Given
\(\mathrm{f}(x)=\left\{\begin{array}{cc}\frac{1-\cos 4 x}{x^2} , \text { if } x < 0 \ \mathrm{a} , \text { if } x=0 \ \frac{\sqrt{x}}{\sqrt{16-\sqrt{x}-4}}, \text { if } x>0\end{array}\right.\)
If \(\mathrm{f}(x)\) is continuous at \(x=0\), then value of \(\mathrm{a}\) is
- A \(-8\)
- B \(2\)
- C \(-2\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
As \(\mathrm{f}(x)\) is continuous at \(x=0\), we get
\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a}=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \)
\( \therefore \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a} \)
\( \therefore \lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2}=\mathrm{a} \)
\( \therefore \lim _{x \rightarrow 0} 4 \frac{2 \sin ^2 2 x}{(2 x)^2}=\mathrm{a} \)
\( \therefore 8=\mathrm{a} \)
\(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a}=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x) \)
\( \therefore \lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\mathrm{a} \)
\( \therefore \lim _{x \rightarrow 0} \frac{1-\cos 4 x}{x^2}=\mathrm{a} \)
\( \therefore \lim _{x \rightarrow 0} 4 \frac{2 \sin ^2 2 x}{(2 x)^2}=\mathrm{a} \)
\( \therefore 8=\mathrm{a} \)
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