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प्रश्न
The function \[f\left( x \right) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\] is of the following types:
पर्याय
even and increasing
odd and increasing
even and decreasing
odd and decreasing
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उत्तर
odd and increasing
\[f(x) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]
\[ \Rightarrow f( - x) = \log_e \left( - x^3 + \sqrt{x^6 + 1} \right)\]
\[ = \log_e \left\{ \frac{\left( - x^3 + \sqrt{x^6 + 1} \right)\left( x^3 + \sqrt{x^6 + 1} \right)}{x^3 + \sqrt{x^6 + 1}} \right\}\]
\[ = \log_e \left( \frac{x^6 + 1 - x^6}{x^3 + \sqrt{x^6 + 1}} \right)\]
\[ = \log_e \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right)\]
\[ = - \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]
\[ = - f(x) \]
\[\text { Hence,} f( - x) = - f(x)\]
\[\text { Therefore, it is an odd function } .\]
\[f(x) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]
\[\frac{d}{dx}\left\{ f(x) \right\} = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left( 3 x^2 + \frac{1}{2\sqrt{x^6 + 1}} \times 6 x^5 \right)\]
\[ = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left( \frac{6 x^2 \sqrt{x^6 + 1} + 6 x^5}{2\sqrt{x^6 + 1}} \right)\]
\[ = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left\{ \frac{6 x^2 \left( \sqrt{x^6 + 1} + x^3 \right)}{2\sqrt{x^6 + 1}} \right\}\]
\[ = \left( \frac{6 x^2}{2\sqrt{x^6 + 1}} \right) > 0\]
\[\text { Therefore, given function is an increasing function } .\]
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