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Question
Show that f(x) = 2x + cot–1x + `log(sqrt(1 + x^2) - x)` is increasing in R
Solution
Given that f(x) = 2x + cot–1x + `log(sqrt(1 + x^2) - x)`
Differentiating both sides w.r.t. x, we get
f'(x) = `2 - 1/(1 + x^2) + 1/(sqrt(1 + x^2) - x) xx "d"/"dx" (sqrt(1 + x^2) - x)`
= `2 - 1/(1 + x^2) + ((1/(2sqrt(1 + x^2)) xx (2x - 1)))/(sqrt(1 + x^2) - x)`
= `2 - 1/(1 + x^2) + (x - sqrt(1 + x^2))/(sqrt(1 + x^2) (sqrt(1 + x^2 - x))`
= `2 - 1/(1 + x^2) - ((sqrt(1 + x^2) - x))/(sqrt(1 + x^2) (sqrt(1 + x^2) - x))`
= `2 - 1/(1 + x^2) - 1/sqrt(1 + x^2)`
For increasing function, f '(x) ≥ 0
∴ `2 - 1/(1 + x^2) - 1/sqrt(1 + x^2) ≥ 0`
⇒ `(2(1 + x^2) - 1 + sqrt(1 + x^2))/((1 + x^2)) ≥ 0`
⇒ `2 + 2x^2 - 1 + sqrt(1 + x^2) ≥ 0`
⇒ `2x^2 + 1 + sqrt(1 + x^2) ≥ 0`
⇒ `2x^2 + 1 ≥ - sqrt(1 + x^2)`
Squaring both sides, we get 4x4 + 1 + 4x2 ≥ 1 + x2
⇒ 4x4 + 4x2 – x2 ≥ 0
⇒ 4x4 + 3x2 ≥ 0
⇒ x2(4x2 + 3) ≥ 0
Which is true for any value of x ∈ R.
Hence, the given function is an increasing function over R.
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