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Question:

Show that (x) = 2+ cot-1 x + log [√(1 + x2) – x] is increasing in R.

Solution:

Given,

(x) = 2+ cot-1 x + log [√(1 + x2) – x]

Differentiating both sides w.r.t. x, we get

$f^{\prime}(x)=2-\frac{1}{1+x^{2}}+\frac{1}{\sqrt{1+x^{2}}-x} \times \frac{d}{d x}\left(\sqrt{1+x^{2}}-x\right)$

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

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

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

$=2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}}$

For increasing function, $f^{\prime}(x) \geq 0$

$2-\frac{1}{1+x^{2}}-\frac{1}{\sqrt{1+x^{2}}} \geq 0$

$\frac{2\left(1+x^{2}\right)-1+\sqrt{1+x^{2}}}{\left(1+x^{2}\right)} \geq 0 \Rightarrow 2+2 x^{2}-1+\sqrt{1+x^{2}} \geq 0$

$2 x^{2}+1+\sqrt{1+x^{2}} \geq 0 \Rightarrow 2 x^{2}+1 \geq-\sqrt{1+x^{2}}$

On squaring both the sides, we get

4x4 + 1 + 4x2 ≥ 1 + x2

4x4 + 4x2 – x2 ≥ 0

4x4 + 3x2 ≥ 0

x2(4x2 + 3) ≥ 0

The above is true for any value of x ∈ R.

Therefore, the given function is an increasing function over R.

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