Let f : R R be a function given by f ( x ) = { 1 - cos 2 x x 2 , x < 0, x = 0 , 1 - cos x x , x > 0 where If f is continuous at x = 0 , then is equal to [2024]

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Solution

Q.
Let $f : R \to R$ be a function given by $f (x)$ $= {\begin{matrix} \frac{1 - \cos 2 x}{x^{2}}, & x < 0 \\ α, & x = 0, \\ \frac{β \sqrt{1 - \cos x}}{x}, & x > 0 \end{matrix}$

where $α, β \in R .$ If $f$ is continuous at $x = 0,$ then $α^{2} + β^{2}$ is equal to [2024]

1 3

2 6

3 12

4 48

Ans.
(3)

$f (x)$ $= {\begin{matrix} \frac{1 - \cos 2 x}{x^{2}}, & x < 0 \\ α, & x = 0, \\ \frac{β \sqrt{1 - \cos x}}{x}, & x > 0 \end{matrix}$

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

$\Rightarrow f (0) = \lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{+}} f (x)$

Now, $\lim_{x \to 0^{-}} f (x) = f (0)$

$∴$ $\lim_{x \to 0^{-}} f (x) = α \Rightarrow \lim_{x \to 0^{-}} (\frac{1 - \cos 2 x}{x^{2}}) = α$

$\Rightarrow \lim_{x \to 0^{-}} \frac{2 \sin^{2} x}{x^{2}} = α \Rightarrow \lim_{h \to 0} \frac{2 \sin^{2} h}{h^{2}} = α \Rightarrow α = 2$

Also, $\lim_{x \to 0^{+}} f (x) = f (0)$

$\Rightarrow \lim_{x \to 0^{+}} \frac{β \sqrt{1 - \cos x}}{x} = 2 \Rightarrow \lim_{h \to 0} \frac{β \sqrt{1 - \cos h}}{h} = 2$

$\Rightarrow \lim_{h \to 0} \frac{β \sqrt{2} \sin \frac{h}{2}}{h} = 2 \Rightarrow \lim_{h \to 0} \frac{β \sqrt{2} \sin \frac{h}{2}}{2 \times \frac{h}{2}} = 2$

$\Rightarrow \frac{β}{\sqrt{2}} = 2 \Rightarrow β = 2 \sqrt{2}$

Hence, $α^{2} + β^{2} = 4 + 8 = 12$

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