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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES1 Mark
Question
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The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval _______.
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Answer

The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval $(1,\infty).$
Solution:
We have, $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4}$
$\Rightarrow\ \text{f}'(\text{x})=\frac{\text{x}^4\cdot4\text{x}-(2\text{x}^2-1)\cdot4\text{x}^3}{\text{x}^8}$ $\bigg[\because\Big(\frac{\text{f}}{\text{g}}\Big)'=\frac{\text{gf}'-\text{fg}'}{\text{g}^2}\bigg]$
$=\frac{4\text{x}^5-8\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{-4\text{x}^5+4\text{x}^3}{\text{x}^8}$
$=\frac{4\text{x}^3(-\text{x}^2+1)}{\text{x}^8}$
For decreasing, f'(x) < 0
$\Rightarrow\ \frac{4\text{x}^3(1-\text{x}^2)}{\text{x}^8}<0$
$\Rightarrow\ 1-\text{x}^2<0$
$\Rightarrow\ \text{x}^2>1$
$\Rightarrow\ \text{x}>\pm1$
Since, x > 0
Hence, $\text{x}\in(1,\infty)$
Therefore, the function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},$ where x > 0, decreases in the interval $(1,\infty).$

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