APPLICATION OF DERIVATIVES — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSAPPLICATION OF DERIVATIVES1 Mark
Question
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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