Question
The function $\text{f(x)}=\frac{2\text{x}^2-1}{\text{x}^4},\text{ x}>0,$ decreases in the interval _______.
✓
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 $(1,\infty).$
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 $(1,\infty).$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The area bounded by the curve $y=\sin x^2, x=1, x=2$ and the $x$-axis will be ____________ .
→The rate of change of the area of a circle with respect to its radius at $r=6 cm$ ________
→The value of $\int \cos x^2 d x=$ ____________
→Fill in the blank.
The negative of a matrix is obtained by multiplying it by _________.
The negative of a matrix is obtained by multiplying it by _________.
Fill in the blanks.
In a LPP, the objective function is always ________.
In a LPP, the objective function is always ________.
$y=\log _{10} x$ then derivative of $y$ w.r.t. $x$ is _________
→Fill in the blanks:
The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is ______.
→The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is ______.
Fill in the blank.
The value of $\cot^{-1}(-x)$ for all $\text{x}\in\text{R}$ in terms of $\cot^{-1}x$ is $....$
→The value of $\cot^{-1}(-x)$ for all $\text{x}\in\text{R}$ in terms of $\cot^{-1}x$ is $....$
Fill in the blanks.
The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$ is _________.
→The vector equation of the line $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2}$ is _________.
Ajay cut two circular pieces of cardboard and placed one upon other as shown in figure. One of the circle represents the equation $(x - 1)^2 + y^2 = 1,$ while other circle represents the equation $x^2 + y^2 = 1.$
Based on the above information, answer the following questions.
→Based on the above information, answer the following questions.
- Both the circular pieces of cardboard meet each other at
- $\text{x}=1$
- $\text{x}=\frac{1}{2}$
- $\text{x}=\frac{1}{3}$
- $\text{x}=\frac{1}{4}$
- Graph of given two curves can be drawn as.
- None of these
- Value of $\int\limits_{0}^{\frac{1}{2}}\sqrt{1-(\text{x}-1)^2}\text{dx}$ is.
- $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$
- $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
- $\frac{\pi}{2}+\frac{\sqrt{3}}{4}$
- $\frac{\pi}{2}-\frac{\sqrt{3}}{4}$
- Value of $\int\limits_{\frac{1}{2}}^{1}\sqrt{1-\text{x}^2}\text{dx}$ is.
- $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$
- $\frac{\pi}{6}+\frac{\sqrt{3}}{8}$
- $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$
- $\frac{\pi}{2}-\frac{\sqrt{3}}{4}$
- Area of hidden portion of lower circle is.
- $\bigg(\frac{2\pi}{3}+\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$
- $\bigg(\frac{\pi}{3}-\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
- $\bigg(\frac{\pi}{3}+\frac{\sqrt{3}}{8}\bigg)\text{ sq.units}$
- $\bigg(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\bigg)\text{ sq.units}$