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# F(X) = 2x − Tan−1 X − Log { X + √ X 2 + 1 } is Monotonically Increasing When - CBSE (Science) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

f(x) = 2x − tan−1 x − log $\left\{ x + \sqrt{x^2 + 1} \right\}$ is monotonically increasing when

•  x > 0

• x < 0

• x ∈ R

•  x ∈ R − {0}

#### Solution

x ∈ R

$\text { Given }: f\left( x \right) = 2x - \tan^{- 1} x - \log \left( x + \sqrt{x^2 + 1} \right)$

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

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

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

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

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

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

$\text { For f(x) to be monotonically increasing,} f'\left( x \right) > 0 .$

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

$\Rightarrow 1 + 2 x^2 - \sqrt{x^2 + 1} > 0 \left[ \because \left( 1 + x^2 \right) > 0 \right]$

$\Rightarrow 1 + 2 x^2 > \sqrt{x^2 + 1}$

$\Rightarrow \left( 1 + 2 x^2 \right)^2 > x^2 + 1$

$\Rightarrow 1 + 4 x^4 + 4 x^2 > x^2 + 1$

$\Rightarrow 4 x^4 + 3 x^2 > 0$

$\text { Thus, f(x) is monotonically increasing for x } \in R .$

Is there an error in this question or solution?

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Solution F(X) = 2x − Tan−1 X − Log { X + √ X 2 + 1 } is Monotonically Increasing When Concept: Increasing and Decreasing Functions.
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