If ∫ 1 a^2 ^2 x+b^2 ^2 x ~d x=(1)/(12) ^ -1 (3 x)+ constant, then the maximum value of a x+b x , is :

Correct answer: A. 40 aligned & ∫ ( ^2 x d x)/(a^2 ^2 x+b^2) & let x=t & ^2 d x=d t & ∫ (d t)/(a^2 t^2+b^2) & (1)/(a^2) ∫ (d t)/(t^2+((b)/(a))^2) & (1)/(a^2) (1)/((b)/(a)) ^ -1 ((t)/(b) a)+c & (1)/(a b) ^ -1 ((α)/(b) x)+c aligned on comparing a b =3 aligned & a b=12 & a=6, b=2 aligned maximum value of 6 x+2 x is 40

JEE Main 20246 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ 1 a^2 ^2 x+b^2 ^2 x ~d x=(1)/(12) ^ -1 (3 x)+ constant, then...

\int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+

constant, then the maximum value of

\mathrm{a} \sin x+\mathrm{b} \cos x

, is :

Options

A $\sqrt{40}$
B $\sqrt{41}$
C $\sqrt{39}$
D $\sqrt{42}$

Correct answer

A. $\sqrt{40}$

Step-by-step solution

\begin{aligned} & \int \frac{\sec ^2 x d x}{a^2 \tan ^2 x+b^2} \\ & \text{ let } \tan x=t \\ & \sec ^2 d x=d t \\ & \int \frac{d t}{a^2 t^2+b^2} \\ & \frac{1}{a^2} \int \frac{d t}{t^2+\left(\frac{b}{a}\right)^2} \\ & \frac{1}{a^2} \frac{1}{\frac{b}{a}} \tan ^{-1}\left(\frac{t}{b} a\right)+c \\ & \frac{1}{a b} \tan ^{-1}\left(\frac{\alpha}{b} \tan x\right)+c \end{aligned}

on comparing

\frac{\mathrm{a}}{\mathrm{b}}=3

\begin{aligned} & a b=12 \\ & a=6, b=2 \end{aligned}

maximum value of

6 \sin x+2 \cos x \text{ is } \sqrt{40}$

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