If α=∫_ (1)/(2) ^2 ^ -1 x 2 x^2-3 x+2 d x then the value of 7 ( 2 α 7 π ) is ____________. (Here, the inverse trigonometric function ^ -1 x assumes values in (-(π)/(2), (π)/(2)) .)

Correct answer: 0 α=∫_ (1)/(2) ^2 ^ -1 x 2 x^2-3 x+2 d x ....(i) aligned & Let x=(1)/(t) & d x=-(1)/(t^2) d t & α=∫_2^ (1)/(2) ^ -1 ((1)/(t)) (2)/(t^2)-(3)/(t)+2 ((-1)/(t^2)) d t aligned α=∫_ (1)/(2) ^2 ^ -1 t 2 t^2-3 t+2 dt ....(ii) Now by (i) + (ii) aligned & 2 α=∫_ (1)/(2) ^2 ((π)/(2))/(2 x^2-3 x+2) d x & α=(π)/(8) ∫_ (1)/(2) ^2 (d x)/(x^2-(3 x)/(2)+1) & α=(π)/(8) ∫_ (1)/(2) ^2(x-(3)/(4))^2+(7)/(16) & α= π 8 × 7 4 [ ^ -1 ( x-(3)/(4) 7 4 )]_ (1)/(2) ^2 aligned aligned & α= π 2 7 [ ^ -1 4 x-3 7 ]_ (1)/(2) ^2 & α= π 2 7 [ ^ -1 5 7 - ^ -1 (- 1 7 )] & α= π 2 7 ^ -1 ( 5 7 + 1 7 ) 1-(5)/(7) & α= π 2 7 ^ -1 (3 7 ) aligned Now 7 ( 2 7 α π ) aligned & 7 × ( ^ -1 (3 7 )) & 7 × 3 7 & =21 aligned

JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Mathematics
Chapter: Definite Integration

JEE Advanced 2025 · Paper 2 · Mathematics · Definite Integration

If α=∫_ (1)/(2) ^2 ^ -1 x 2 x^2-3 x+2 d x then the value of 7 ( 2 α...

\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x then the value of

\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)

is ____________. (Here, the

inverse trigonometric function\tan ^{-1} x

assumes values in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

Correct answer

Step-by-step solution

\alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x

....(i)

\begin{aligned} & \text{ Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t\end{aligned}

\alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} \mathrm{t}}{2 \mathrm{t}^2-3 \mathrm{t}+2} \mathrm{dt}

....(ii) Now by (i) + (ii)

\begin{aligned} & 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2\left(x-\frac{3}{4}\right)^2+\frac{7}{16} \\ & \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2 \end{aligned}

$\begin{aligned} & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\ & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7}) \end{aligned}

Now

\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)

\begin{aligned} & \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\ & \sqrt{7} \times 3 \sqrt{7} \\ & =21 \end{aligned}

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