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प्रश्न
f(x) = 2x − tan−1 x − log \[\left\{ x + \sqrt{x^2 + 1} \right\}\] is monotonically increasing when
पर्याय
x > 0
x < 0
x ∈ R
x ∈ R − {0}
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उत्तर
x ∈ R
\[\text { Given }: f\left( x \right) = 2x - \tan^{- 1} x - \log \left( x + \sqrt{x^2 + 1} \right)\]
\[f'\left( x \right) = 2 - \frac{1}{1 + x^2} - \frac{1}{x + \sqrt{x^2 + 1}}\left( 1 + \frac{1}{2\sqrt{x^2 + 1}} . 2x \right)\]
\[ = 2 - \frac{1}{1 + x^2} - \frac{1}{x + \sqrt{x^2 + 1}}\left( 1 + \frac{x}{\sqrt{x^2 + 1}} \right)\]
\[ = 2 - \frac{1}{1 + x^2} - \frac{1}{x + \sqrt{x^2 + 1}}\left( \frac{x + \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \right)\]
\[ = 2 - \frac{1}{1 + x^2} - \frac{1}{\sqrt{x^2 + 1}}\]
\[ = \frac{2 + 2 x^2 - 1 - \sqrt{x^2 + 1}}{1 + x^2}\]
\[ = \frac{1 + 2 x^2 - \sqrt{x^2 + 1}}{1 + x^2}\]
\[\text { For f(x) to be monotonically increasing,} f'\left( x \right) > 0 . \]
\[ \Rightarrow \frac{1 + 2 x^2 - \sqrt{x^2 + 1}}{1 + x^2} > 0 \]
\[ \Rightarrow 1 + 2 x^2 - \sqrt{x^2 + 1} > 0 \left[ \because \left( 1 + x^2 \right) > 0 \right]\]
\[ \Rightarrow 1 + 2 x^2 > \sqrt{x^2 + 1}\]
\[ \Rightarrow \left( 1 + 2 x^2 \right)^2 > x^2 + 1\]
\[ \Rightarrow 1 + 4 x^4 + 4 x^2 > x^2 + 1\]
\[ \Rightarrow 4 x^4 + 3 x^2 > 0\]
\[\text { Thus, f(x) is monotonically increasing for x } \in R . \]
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