For a , b > 0 , let f ( x ) = { tan ( ( a + 1 ) x ) + b tan x x , x < 0 3 , x = 0 a x + b 2 x 2 - a x b a x x , x > 0 be a continuous function at x = 0 . Then b a is equal to [2024]

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Solution

Q.
For $a, b > 0,$ let $f (x)$ $= {\begin{matrix} \frac{\tan ((a + 1) x) + b \tan x}{x}, & x < 0 \\ 3, & x = 0 \\ \frac{\sqrt{a x + b^{2} x^{2}} - \sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{matrix}$ be a continuous function at $x = 0 .$ Then $\frac{b}{a}$ is equal to [2024]

1 5

2 6

3 8

4 4

Ans.
(2)

$We have, f (x) is continuous at x = 0 .$

$∴$ $f (0^{-}) = f (0) = f (0^{+})$

$∴$ $\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} \frac{\tan [(a + 1) x] + b \tan x}{x}$

$= \lim_{x \to 0^{-}} \frac{\tan [(a + 1) x]}{(a + 1) x} \times (a + 1) + b \lim_{x \to 0^{-}} \frac{\tan x}{x}$

$= a + 1 + b = 3$ $[∵ f (0) = 3]$

$and \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} \frac{\sqrt{a x + b^{2} x^{2}} - \sqrt{a x}}{b \sqrt{a} x \sqrt{x}}$

$= \lim_{x \to 0^{+}} \frac{\sqrt{a + b^{2} x} - \sqrt{a}}{b \sqrt{a} x} \times \frac{\sqrt{a + b^{2} x} + \sqrt{a}}{\sqrt{a + b^{2} x} + \sqrt{a}}$

$= \lim_{x \to 0^{+}} \frac{b^{2} x}{b \sqrt{a} x (\sqrt{a + b^{2} x} + \sqrt{a})}$

$= \lim_{x \to 0^{+}} \frac{b^{2}}{b \sqrt{a} (\sqrt{a + b^{2} x} + \sqrt{a})} = \frac{b^{2}}{b \sqrt{a} (\sqrt{a + 0} + \sqrt{a})} = \frac{b}{2 a} = 3$

$∴$ $a + 1 + b = 3 and \frac{b}{2 a} = 3 \Rightarrow \frac{b}{a} = 6$

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