IIT-JEE 2012 · Paper 1 · Mathematics Q55

MathematicsMultiple correct+4Question 55 of 60

If $y(x)$ satisfies the differential equation $y'-y\tan x=2x\sec x$ and $y(0)=0$ , then

y\left(\dfrac\pi4\right)=\dfrac{\pi^2}{8\sqrt2}

y'\left(\dfrac\pi4\right)=\dfrac{\pi^2}{18}

y\left(\dfrac\pi3\right)=\dfrac{\pi^2}9

y'\left(\dfrac\pi3\right)=\dfrac{4\pi}3+\dfrac{2\pi^2}{3\sqrt3}

Answer and worked solution

Official answer: A, D

Integrating factor $e^{-\int\tan x\,dx}=\cos x$ :
$(y\cos x)'=2x\implies y\cos x=x^2\implies y=x^2\sec x$
(using $y(0)=0$ ).

Derivative. $y'=2x\sec x+x^2\sec x\tan x$ .

(A) $y(\frac\pi4)=\frac{\pi^2}{16}\sqrt2=\frac{\pi^2}{8\sqrt2}$ . True.
(B) $y'(\frac\pi4)=\frac{\pi\sqrt2}2+\frac{\pi^2\sqrt2}{16}$ , not $\frac{\pi^2}{18}$ . False.
(C) $y(\frac\pi3)=\frac{2\pi^2}9$ . False.
(D) $y'(\frac\pi3)=\frac{4\pi}3+\frac{\pi^2}9\cdot2\sqrt3=\frac{4\pi}3+\frac{2\pi^2}{3\sqrt3}$ . True.

(A), (D)

Topics:Linear first-order equations

IIT-JEE 2012 · Paper 1, Mathematics question 55