JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q11

MathematicsSingle correct+4 / -1Question 11 of 90

If $\dfrac{dy}{dx}+\dfrac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0$ , $x,y>0$ , $y(1)=1$ , then $y(2)$ is equal to :

2+\log_23

2+\log_32

2-\log_32

2-\log_23

Answer and worked solution

Official answer: D

Separate. $\dfrac{2^y\,dy}{2^y-1}=-\dfrac{2^x\,dx}{2^x-1}$ , giving $(2^y-1)(2^x-1)=K$ .

Constant. At $(1,1)$ , $K=1$ .

At $x=2$ . $3(2^y-1)=1$ , so $2^y=\frac43$ and $y=2-\log_23$ .

(D)

Topics:Variable separable

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 11