If y = {2}{{a²-b²} ^{-1}≤ft[{a-b/a+b} x/2] , then ≤ft.d²y/dx²|ₓ=π/2} =

Correct Answer is : ba2{b}{a²}a2b

Correct Answer is : ba2{b}{a²}a2b

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AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

Section: Mathematics

Question : 67 of 160

Marks: +1, -0

If y =

\frac{2}{\sqrt{a^2-b^2}}

\tan^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right]

\left.\frac{d^2y}{dx^2}\right|_{x=\pi/2}

Solution:

As the given expression is given as

y = \frac{2}{\sqrt{a^{2}-b^{2}}} \tan^{-1} \left[\sqrt{\frac{a-b}{a+b}} \tan \frac{x}{2}\right]

\frac{dy}{dx} = \left[ \frac{2}{\sqrt{a^{2}-b^{2}}} \cdot \frac{1}{1+\left(\frac{a-b}{a+b}\right)\tan^{2}\frac{x}{2}} \cdot \sqrt{\frac{a-b}{a+b}} \cdot \sec^{2}\frac{x}{2} \cdot \frac{1}{2} \right]

= \frac{\frac{\sec^{2} x}{2}}{a+b} \cdot \frac{a+b}{(a+b)+(a-b)\tan^{2} \frac{x}{2}}

= \frac{\frac{\sec^{2} x}{2}}{a+b} \cdot \frac{a+b}{a \tan^{2} \frac{x}{2} - b \tan^{2} \frac{x}{2}}

= \frac{\frac{\sec^{2} x}{2}}{1+\tan^{2} \frac{x}{2}}

\left[ a+b \left( \frac{1-\tan^{2} \frac{x}{2}}{1+\tan^{2} \frac{x}{2}} \right) \right]

Now the expression will become

\frac{dy}{dx} = \frac{1}{a+b\cos x}

Differentiate the above expression

\frac{d^{2}y}{dx^{2}} = \frac{-1}{(a+b\cos x)^{2}} \cdot (-b\sin x)

\frac{d^{2}y}{dx^{2}} = \frac{b\sin x}{(a+b\cos x)^{2}}

Afler substitute the values

\left.\frac{d^{2}y}{dx^{2}}\right|_{x=\pi/2} = \frac{b \sin \frac{\pi}{2}}{a+b\cos \frac{\pi}{2}}

= \frac{b}{a^{2}}

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