{Let } f(x)=≤ft(2^{-1}≤ft(^{-1}{1-x/x})), 0 Then, [26 Aug 2021 Shift 1]

Correct Answer is : (1−x)2f′(x)+2(f(x))2=0(1-x)² f'(x)+2(f(x))²=0(1−x)2f′(x)+2(f(x))2=0

Correct Answer is : (1−x)2f′(x)+2(f(x))2=0(1-x)² f'(x)+2(f(x))²=0(1−x)2f′(x)+2(f(x))2=0

Test Index

Differentiation

Section: Mathematics

Question : 44 of 79

Marks: +1, -0

\text{Let } f(x)=\cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right), 0 < x<1

Solution:

f(x)=\cos\left(2\tan^{-1}\sin\left(\cot^{-1}\sqrt{\frac{1-x}{x}}\right)\right)

\cot^{-1}\sqrt{\frac{1-x}{x}} = \sin^{-1}\sqrt{x}

∴

f(x)=\cos\left(2\tan^{-1}\sin\sin^{-1}\sqrt{x}\right)

f(x)=\cos\left(2\tan^{-1}\sqrt{x}\right)=\cos\tan^{-1}\left(\frac{2\sqrt{x}}{1-x}\right)

f(x)=\cos\cos^{-1}\left(\frac{1-x}{1+x}\right)

f(x)=\frac{1-x}{1+x}

f'(x)=\frac{-(1+x)-(1-x)}{(1+x)^2}=\frac{-2}{(1+x)^2}

f'(x)(1-x)^2=-2\left(\frac{1-x}{1+x}\right)^2

f'(x)(1-x)^2=-2\left(\frac{1-x}{1+x}\right)^2

(1-x)^2 f'(x)+2[f(x)]^2

=-2\left(\frac{1-x}{1+x}\right)^2+2\left(\frac{1-x}{1+x}\right)^2=0

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