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Increasing and Decreasing Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIncreasing and Decreasing Functions1 Mark
MCQ
The function $\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$ is of the following type:
  • A
    Even and increasing.
  • ✓
    Odd and increasing.
  • C
    Even and decreasing.
  • D
    Odd and decreasing.
✓

Answer

Correct option: B.
Odd and increasing.
$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$\Rightarrow\text{f}(-\text{x})=\log_\text{e}\Big(-\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=\log_\text{e}\Bigg\{\frac{\big(-\text{x}^3+\sqrt{\text{x}^6+1}\big)\big(\text{x}^3+\sqrt{\text{x}^6+1}\big)}{\text{x}^3+\sqrt{\text{x}^6+1}}\Bigg\}$
$=\log_\text{e}\Big(\frac{\text{x}^6+1-\text{x}^6}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=\log_\text{e}\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)$
$=-\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$=-\text{f}(\text{x})$
Hence, f(-x) = -f(x)
Therefore, it is an odd function.
$\text{f}(\text{x})=\log_\text{e}\Big(\text{x}^3+\sqrt{\text{x}^6+1}\Big)$
$\frac{\text{d}}{\text{dx}}\{\text{f}(\text{x})\}=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Big(3\text{x}^2+\frac{1}{2\sqrt{\text{x}^6+1}}\times6\text{x}^5\Big)$
$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\bigg(\frac{6\text{x}^2\sqrt{\text{x}^6+1}+6\text{x}^5}{2\sqrt{\text{x}^6+1}}\bigg)$
$=\Big(\frac{1}{\text{x}^3+\sqrt{\text{x}^6+1}}\Big)\times\Bigg\{\frac{6\text{x}^2\big(\sqrt{\text{x}^6+1+\text{x}^3}\big)}{2\sqrt{\text{x}^6+1}}\Bigg\}$
$=\Big(\frac{6\text{x}^2}{2\sqrt{\text{x}^6+1}}\Big)>0$
Therefore the given function is an increasing function.

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