Let [f: R R] be a function given by [f(x)= cases(1- 2 x)/(x^2), & x 0cases] where [, R].

Question

Let $$f: \mathbf{R} \rightarrow \mathbf{R}$$ be a function given by

$$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x 0\end{cases}$$

where $$\alpha, \beta \in \mathbf{R}$$. If $f$ is continuous at $x=0$, then $$\alpha^2+\beta^2$$ is equal to :

Accepted Answer

$$f(x)=\left\{\[\begin{array}{cl} \frac{1-\cos 2 x}{x^2}, & x0 \end{array}$$\right.\]

$f(x)$ is continuous at $x=0$

$$\Rightarrow f(0)=\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)$$

$$\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\alpha \
& \lim _{x \rightarrow 0^{-}}\left(\frac{1-\cos 2 x}{x^2}\right)=\alpha \
& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 h}{x^2}=\alpha \
& \Rightarrow \lim _{h \rightarrow 0} \frac{2 \sin ^2}{h^2} h=\alpha \
& \Rightarrow \alpha=2 \
&
\end{aligned}\
\&\begin{aligned}
& \text { Also, } \lim _{x \rightarrow 0^{+}} f(x)=f(0) \
& \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\beta \sqrt{1-\cos x}}{x}=2
\end{aligned}
\end{aligned}$$

$$\Rightarrow \lim _\limits{h \rightarrow 0} \frac{\beta \sqrt{\frac{1-\cos h}{h^2}} h^2}{h}=2$$

$$\begin{aligned}
& \Rightarrow \quad \frac{\beta}{\sqrt{2}}=2 \
& \Rightarrow \quad \beta=2 \sqrt{2} \
& \Rightarrow \quad \alpha^2+\beta^2=4+8 \
& \quad=12
\end{aligned}$$

$$\therefore$$ Option (1) is correct

Solutions · 1