JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a function given by \(f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2} & , x<0 \\ \alpha & , x=0, \text { where } \alpha, \beta \in R \text {. If } \\ \frac{\beta \sqrt{1-\cos x}}{x} & , x>0\end{cases}\) \(f\) is continuous at \(\mathrm{x}=0\), then \(\alpha^2+\beta^2\) is equal to :
- A \(48\)
- B \(12\)
- C \(3\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
\( f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 x}{x^2}=2=\alpha \) \( f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{2 \frac{x}{2}}=\frac{\beta}{\sqrt{2}}=2 \) \( \Rightarrow \beta=2 \sqrt{2} \)…
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