Question

The function \[f\left( x \right) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\] is of the following types:

विकल्प

• even and increasing

• odd and increasing

• even and decreasing

• odd and decreasing

Answer 1

odd and increasing

\[f(x) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]

\[ \Rightarrow f( - x) = \log_e \left( - x^3 + \sqrt{x^6 + 1} \right)\]

\[ = \log_e \left\{ \frac{\left( - x^3 + \sqrt{x^6 + 1} \right)\left( x^3 + \sqrt{x^6 + 1} \right)}{x^3 + \sqrt{x^6 + 1}} \right\}\]

\[ = \log_e \left( \frac{x^6 + 1 - x^6}{x^3 + \sqrt{x^6 + 1}} \right)\]

\[ = \log_e \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right)\]

\[ = - \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]

\[ = - f(x) \]

\[\text { Hence,} f( - x) = - f(x)\]

\[\text { Therefore, it is an odd function } .\]

\[f(x) = \log_e \left( x^3 + \sqrt{x^6 + 1} \right)\]

\[\frac{d}{dx}\left\{ f(x) \right\} = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left( 3 x^2 + \frac{1}{2\sqrt{x^6 + 1}} \times 6 x^5 \right)\]

\[ = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left( \frac{6 x^2 \sqrt{x^6 + 1} + 6 x^5}{2\sqrt{x^6 + 1}} \right)\]

\[ = \left( \frac{1}{x^3 + \sqrt{x^6 + 1}} \right) \times \left\{ \frac{6 x^2 \left( \sqrt{x^6 + 1} + x^3 \right)}{2\sqrt{x^6 + 1}} \right\}\]

\[ = \left( \frac{6 x^2}{2\sqrt{x^6 + 1}} \right) > 0\]

\[\text { Therefore, given function is an increasing function } .\]