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Show that f(x) = 2x + cot–1x + log(1+x2-x) is increasing in R - Mathematics

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प्रश्न

Show that f(x) = 2x + cot–1x + `log(sqrt(1 + x^2) - x)` is increasing in R

योग
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उत्तर

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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अध्याय 6: Application Of Derivatives - Exercise [पृष्ठ १३६]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 6 Application Of Derivatives
Exercise | Q 20 | पृष्ठ १३६

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