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TS EAMCET 5 May 2018 Shift 2 Solved Paper

Section: Mathematics

Question : 66 of 160

Marks: +1, -0

y= (\sin^{-1} 2 x)^{2}+ (\cos^{-1} 2 x)^{2}

(1-4 x^{2}) y_{2}-4 x y_{1}=

Solution:

It is given that ,

y= (\sin^{-1} 2 x)^{2}+ (\cos^{-1} 2 x)^{2}

Differentiate both side with respect to

x

\frac{dy}{dx}=2 \sin^{-1} 2 x \frac{1}{\sqrt{1-(2x)^{2}}}

+2 \cos^{-1} 2 x \frac{-1}{\sqrt{1-(2x)^{2}}}

\sqrt{1-4x^{2}} \frac{dy}{dx}=4 (\sin^{-1} 2x+\cos^{-1} 2x)

Differentiate with respect to

x

\sqrt{1-4x^{2}} y_{2}+ \frac{1}{2\sqrt{1-4x^{2}}}(-8x) y_{1}

=4 \left( \frac{1}{\sqrt{1-4x^{2}}} \cdot 2 + \frac{1}{\sqrt{1-4x^{2}}} \cdot 2 \right)

\sqrt{1-4x^{2}} y_{2} - \frac{4x}{y_{1}\sqrt{1-4x^{2}}}

=8 \left( 2 \sqrt{1-4x^{2}} \right)

(1-4x^{2}) y_{2} - 4x y_{1} = 16

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