IIT-JEE 2008 · Paper 2 · Mathematics Q13

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MathematicsAssertion and reason+3 / -1Question 13 of 66

Let a solution $y=y(x)$ of the differential equation
$x\sqrt{x^2-1}\,dy-y\sqrt{y^2-1}\,dx=0$
satisfy $y(2)=\frac2{\sqrt3}$ .

STATEMENT-1: $y(x)=\sec\left(\sec^{-1}x-\frac\pi6\right)$

and

STATEMENT-2: $y(x)$ is given by $\dfrac1y=\dfrac{2\sqrt3}{x}-\sqrt{1-\dfrac1{x^2}}$

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Answer and worked solution

Official answer: C

Separate.
$\frac{dy}{y\sqrt{y^2-1}}=\frac{dx}{x\sqrt{x^2-1}}\implies\sec^{-1}y=\sec^{-1}x+c.$
At $x=2$ , $y=\frac2{\sqrt3}$ : $\frac\pi6=\frac\pi3+c$ , so $c=-\frac\pi6$ and $y=\sec\left(\sec^{-1}x-\frac\pi6\right)$ . Statement 1 is true.

Statement 2. With $\theta=\sec^{-1}x$ ( $\cos\theta=\frac1x$ ):
$\frac1y=\cos\left(\theta-\frac\pi6\right)=\frac{\sqrt3}{2x}+\frac12\sqrt{1-\frac1{x^2}},$
not the given expression. False.

(C)

Topics:Variable separable Inverse trigonometric functions

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IIT-JEE 2008 · Paper 2, Mathematics question 13