Q31 / 90 · JEE Main 2024 (04 Apr Shift 1) · Mathematics

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function given by $f(x)=\left\{\begin{array}{ll} \frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0 \end{array}\right.$ where $\alpha, \beta \in \mathbf{R}$. If $f$ is continuous at $x=0$, then $\alpha^2+\beta^2$ is equal to :

  • A
    3
  • Correct answer
    B
    12
  • C
    48
  • D
    6

Solution

$\begin{aligned} & f\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 x}{x^2}=2=\alpha \\ & f\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \beta \times \sqrt{2} \frac{\sin \frac{x}{2}}{2 \frac{x}{2}}=\frac{\beta}{\sqrt{2}}=2 \\ & \Rightarrow \beta=2 \sqrt{2} \\ & \alpha^2+\beta^2=4+8=12\end{aligned}$

Hence, the correct answer is option: 'B'.

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