Let for f(x)=7 ^8 x+7 ^6 x-3 ^4 x-3 ^2 x, I _1=∫_0^ π / 4 f(x) d x and I_2=∫_0^ π / 4 x f(x) d x . Then 7 I _1+12 I _2 is equal to :

Correct answer: B. 1 aligned f(x) & =7 ^8 x+7 ^6 x-3 ^4 x-3 ^2 x & =7 ^6 x(1+ ^2 x)-3 ^2 x(1+ ^2 x) & =(7 ^6 x-3 ^2 x)(1+ ^2 x) & =(7 ^6 x-3 ^2 x) ^2 x l_1= & ∫_0^ (π)/(4) f(x) d x=∫_0^ (π)/(4) (7 ^6 x-3 ^2 x) ^2 x d x & =.((7 ^7 x)/(7)-(3 ^3 x)/(3))|_0 ^ (π)/(4) =1-1=0 aligned aligned I_2= & ∫_0^ (π)/(4) x f(x) d x=∫_0^ (π)/(4) x(7 ^6 x-3 ^2 x) ^2 x d x & =.x( ^7 x- ^3 x)|_0 ^ (π)/(4) -∫_0^ (π)/(4) 1 ·( ^7 x- ^3 x) d x & =0-∫_0^ (π)/(4) ^3 x( ^2 x-1)( ^2 x+1) d x & =∫_0^ (π)/(4) ( ^3 x- ^5 x) ^2 x d x=( ^4 x)/(4)-.( ^6 x)/(6)|_0 ^ (π)/(4) & =(1)/(12) aligned Hence 7 I_1+12 I_2=1

JEE Main 202522 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Definite Integration

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Definite Integration

Let for f(x)=7 ^8 x+7 ^6 x-3 ^4 x-3 ^2 x, I _1=∫_0^ π / 4 f(x) d x...

Let for

f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x, \quad

\mathrm{I}

_1=\int_0^{\pi / 4} f(x) \mathrm{d} x

and

\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x

. Then

7

\mathrm{I}

_1+12

\mathrm{I}

_2

is equal to :

Options

A $2$
B $1$
C $2 \pi$
D $\pi$

Correct answer

B. $1$

Step-by-step solution

\begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \\ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x \\ l_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.\left(\frac{7 \tan ^7 x}{7}-\frac{3 \tan ^3 x}{3}\right)\right|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}

\begin{aligned} I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \\ & =0-\int_0^{\frac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x \\ & =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \\ & =\frac{1}{12} \end{aligned}

Hence

7 I_1+12 I_2=1$

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