Let f ( x ) = d x ( 3 + 4 x 2 ) 4 - 3 x 2 , | x | 2 3 . If f ( 0 ) = 0 and f ( 1 ) = 1 tan - 1 0 , then 2 + 2 is equal to _______ . [2023]

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Solution

Q.
Let $f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}},$ $| x |$ $< \frac{2}{\sqrt{3}}$ . If $f (0) = 0$ and $f (1) = \frac{1}{α β} \tan^{- 1} (\frac{α}{β})$ , $α, β > 0$ , then $α^{2} + β^{2}$ is equal to _______ . [2023]

Ans.
(28)

$Given, f (x) = \int \frac{d x}{(3 + 4 x^{2}) \sqrt{4 - 3 x^{2}}}$

$x = \frac{1}{t} \Rightarrow d x = \frac{- 1}{t^{2}} d t$

$\Rightarrow f (x) = \int \frac{- \frac{1}{t^{2}} d t}{\frac{(3 t^{2} + 4)}{t^{2}} \frac{\sqrt{4 t^{2} - 3}}{t}}$

$= - \int \frac{- t d t}{(3 t^{2} + 4) \sqrt{4 t^{2} - 3}}$ $Put 4 t^{2} - 3 = z^{2} \Rightarrow t d t = \frac{z}{4} d z$

$=$ $- \frac{1}{4} \int \frac{z d z}{(3 (\frac{z^{2} + 3}{4}) + 4) z} = \int \frac{- d z}{3 z^{2} + 25} = - \frac{1}{3} \int \frac{d z}{z^{2} + {(\frac{5}{\sqrt{3}})}^{2}}$

$= \int - \frac{1}{3} \frac{\sqrt{3}}{5} \tan^{- 1} (\frac{\sqrt{3} z}{5}) + C = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{4 t^{2} - 3}) + C$

$f (x) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5} \sqrt{\frac{4 - 3 x^{2}}{x^{2}}}) + C$

$∵$ $f (0) = 0$ $∴$ $C = \frac{π}{10 \sqrt{3}}$

$Now, f (1) = - \frac{1}{5 \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3}}{5}) + \frac{π}{10 \sqrt{3}}$

$\Rightarrow f (1) = \frac{1}{5 \sqrt{3}} \cot^{- 1} (\frac{\sqrt{3}}{5}) = \frac{1}{5 \sqrt{3}} \tan^{- 1} \frac{5}{\sqrt{3}}$

$So, α = 5 : β = \sqrt{3}$ $∴$ $α^{2} + β^{2} = 28$

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