JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q11

← JEE Main 2022 · 28 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 11 of 90

Let the slope of the tangent to a curve $y=f(x)$ at $(x,y)$ be given by $2\tan x(\cos x-y)$ . If the curve passes through the point $\left(\frac\pi4,0\right)$ , then the value of $\displaystyle\int_0^{\pi/2}y\,dx$ is equal to :

\left(2-\sqrt2\right)+\dfrac\pi{\sqrt2}

2-\dfrac\pi{\sqrt2}

\left(2+\sqrt2\right)+\dfrac\pi{\sqrt2}

2+\dfrac\pi{\sqrt2}

Answer and worked solution

Official answer: B

Linear. $y'+2\tan x\,y=2\sin x$ ; the integrating factor is $\sec^2x$ : $y\sec^2x=2\sec x+C$ .

Point. $\left(\frac\pi4,0\right)$ gives $C=-2\sqrt2$ , so $y=2\cos x-2\sqrt2\cos^2x$ .

Integral. $2-2\sqrt2\cdot\frac\pi4=2-\dfrac\pi{\sqrt2}$ .

(B)

Topics:Linear first-order equations Properties of definite integrals

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JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 11