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Question 52 of 71

Q.Prove that tan⁡−1x<x\tan^{-1}x < x, for all x>0x>0.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 6mImportance★★★★★
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Define f(x)=x−tan⁡−1xf(x)=x-\tan^{-1}x and show it is strictly increasing with f(0)=0f(0)=0, forcing f(x)>0f(x)>0 for x>0x>0.

1. Define the auxiliary function.

f(x)=x−tan⁡−1x,x≥0f(x)=x-\tan^{-1}x,\qquad x\ge 0

2. Evaluate at x=0x=0.

f(0)=0−tan⁡−10=0−0=0f(0)=0-\tan^{-1}0=0-0=0

3. Differentiate.

f′(x)=1−11+x2=(1+x2)−11+x2=x21+x2f'(x)=1-\frac{1}{1+x^2}=\frac{(1+x^2)-1}{1+x^2}=\frac{x^2}{1+x^2}

4. Sign of f′(x)f'(x) for x>0x>0.

Since x2>0x^2>0 and 1+x2>01+x^2>0 for every x>0x>0, we get f′(x)>0f'(x)>0 for all x>0x>0. Hence ff is strictly increasing on (0,∞)(0,\infty).

5. Conclude. …

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