QJEEAdv08Macb9e: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Macb9e Solution: Let a solution of the differential equation, satify STATEMENT-1 …

The given differential equation is

$x\sqrt {{x^2} - 1} dy - y\sqrt {{y^2} - 1} dx = 0$

$= \int {{{dy} \over {y\sqrt {{y^2} - 1} }} = \int {{{dx} \over {x\sqrt {{x^2} - 1} }}} }$

$\Rightarrow {\sec ^{ - 1}}y = {\sec ^{ - 1}}x + c$

$y = \sec [{\sec ^{ - 1}}x + C]$ [ $\because$ $y(2) = {2 \over {\sqrt 3 }}$ ]

${2 \over {\sqrt 3 }} = \sec ({\sec ^{ - 1}}2 + C)$

$= {\sec ^{ - 1}}{2 \over {\sqrt 3 }} - {\sec ^{ - 1}}2 = C$

$C = {\pi \over 6} - {\pi \over 3} = {{ - \pi } \over 6}$ [ $\therefore$ $y = \sec \left[ {{{\sec }^{ - 1}}x - {\pi \over 6}} \right]$ ]

Statement 1 is true

Also, ${1 \over y} = \cos \left[ {\cos - {\pi \over x} - {\pi \over 6}} \right]$

$\cos \left( {{{\cos }^{ - 1}}{1 \over x}} \right)\cos {\pi \over 6} + \sin ({\cos ^{ - 1}}{1 \over x})\sin {\pi \over 6}$

$= {1 \over y} = {{\sqrt 3 } \over {2x}} + {1 \over 2}\sqrt {1 - {1 \over {{x^2}}}}$

$\therefore$ Statement 2 is false.