If the function is continuous at x = 0, then is equal to [2025]

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Solution

Q.
If the function $f (x)$ $=$ ${\begin{matrix} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x}, & x < 0 \\ 4, & x = 0 \\ \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}), & x > 0 \end{matrix}$ is continuous at x = 0, then $k_{1}^{2} + k_{2}^{2}$ is equal to [2025]

1 20

2 5

3 10

4 8

Ans.
(3)

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0} \frac{2}{x} {\sin (k_{1} + 1) x + \sin (k_{2} - 1) x} = 4$

$\Rightarrow 2 (k_{1} + 1) + 2 (k_{2} - 1) = 4$

$\Rightarrow 2 [k_{1} + 1 + k_{2} - 1] = 4$

$\Rightarrow k_{1} + k_{2} = 2$ ... (i)

$\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0} \frac{2}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}) = 4$

$\lim_{x \to 0^{+}} \frac{1}{x} \log_{e} (\frac{2 + k_{1} x}{2 + k_{2} x}) = 2$ ;

$\lim_{x \to 0^{+}} \frac{1}{x} \log_{e} (1 + \frac{(k_{1} - k_{2}) x}{2 + k_{2} x}) = 2$

$\Rightarrow \frac{k_{1} - k_{2}}{2} = 2 \Rightarrow k_{1} - k_{2} = 4$ ... (ii)

Adding (i) and (ii), we get

$k_{1} = 3 \Rightarrow k_{2} = - 1$

$∴ k_{1}^{2} + k_{2}^{2} = {(3)}^{2} + {(- 1)}^{2} = 9 + 1 = 10$ .

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