Let fx-x-tan-1x- gx-x-1-x2x-0 Then

The correct option is B f(x)>g(x), x>0 Let ϕ (x)= ( x+tan^ 1 x ) x1+x^2 ∴ #160; ϕ'(x)=1+1/1+x^2 1 x^2/(1+x^2)^2=1+2x^2/(1+x^2)^2 gt;0 ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

Let fx=x+ta...

Question

Let $f (x) = x + {tan}^{- 1} x, g (x) = \frac{x}{1 + x^{2}} (x > 0)$ Then

f (x) < g (x), x > 0

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f (x) > g (x), x > 0

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f (x) < g (x)

[1, \infty)

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None of these

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Solution

The correct option is B $f (x) > g (x), x > 0$
Let $ϕ (x) = (x + t a n^{- 1} x) - \frac{x}{1 + x^{2}}$
$∴$ $ϕ^{'} (x) = 1 + \frac{1}{1 + x^{2}} - \frac{1 - x^{2}}{(1 + x^{2})^{2}} = 1 + \frac{2 x^{2}}{(1 + x^{2})^{2}} > 0 f o r$ all $x$
$∴$ f(x) is increasing in the domain
$ϕ (0) = 0$
$∴$ $x + t a n^{- 1} x - \frac{x}{1 + x^{2}} > 0, x > 0$
$\Rightarrow f (x) > g (x), x > 0$

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Let f(x)=x+tan−1x,g(x)=x1+x2(x>0) Then

The correct option is B f(x)>g(x),x>0Let ϕ(x)=(x+tan−1x)−x1+x2∴ ϕ′(x)=1+11+x2−1−x2(1+x2)2=1+2x2(1+x2)2>0 for all x∴ f(x) is increasing in the domain ϕ(0)=0∴ x+tan−1x−x1+x2>0,x>0 ⇒f(x)>g(x),x>0

Let $f (x) = x + {tan}^{- 1} x, g (x) = \frac{x}{1 + x^{2}} (x > 0)$ Then