MCQ
The function $f(x)=x-\frac{1}{x}, x \in R, x \neq 0$ is
  • increasing for all $x \in R$
  • B
    decreasing for all $x \in R$
  • C
    increasing for all $x \in(0, \infty)$
  • D
    neither increasing nor decreasing

Answer

Correct option: A.
increasing for all $x \in R$
(a) : $f(x)=x-\frac{1}{x}$
$\therefore f^{\prime}(x)=1+\frac{1}{x^2}>0$ for all $x \in R, x \neq 0$
$\therefore \quad f(x)$ is increasing for all $x \in R$, where $x \neq 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Consider three vectors $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$. Let $|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3$ and $\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}$. If $\alpha \in\left[0, \frac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\overrightarrow{c}-\overrightarrow{a}|^2$ is equal to :
$\int_0^{\pi /2} {\frac{{\sin x}}{{\sin x + \cos x}}\,dx} $ equals
If $y = 1 - x + {{{x^2}} \over {2!}} - {{{x^3}} \over {3!}} + {{{x^4}} \over {4!}} - $....., then ${{{d^2}y} \over {d{x^2}}} = $
Choose the correct answer from the given four options.The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$ is:
A random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3,$ then $P(X = 0)$ is :
Value of $\left| {\begin{array}{*{20}{c}}
  {{{(b + c)}^2}}&{{a^2}}&{{a^2}} \\ 
  {{b^2}}&{{{(a + c)}^2}}&{{b^2}} \\ 
  {{c^2}}&{{c^2}}&{{{(a + b)}^2}} 
\end{array}} \right|$ is equal to
Function $\text{f}(\text{x})=\cos\text{x}-2\lambda\text{x}$ is monotonic decreasing when :
Let $f(x)$ satisfy all the conditions of mean value theorem in  $[0, 2]. $ If $ f (0) = 0 $ and $|f'(x)|\, \le {1 \over 2}$ for all  $x$ in  $[0, 2]$ then
Let $B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right]$ and $A$ be a $2 \times 2$ matrix such that $\mathrm{AB}^{-1}=\mathrm{A}^{-1}$. If $\mathrm{BCB}^{-1}=\mathrm{A}$ and $\mathrm{C}^4+\alpha \mathrm{C}^2+\beta \mathrm{I}=\mathrm{O}$, then $2 \beta-\alpha$ is equal to :
If $f(x) = |3 − x| + (3 + x),$ where $(x) $ denotes the least integer greater than or equal to $x,$ then $f(x)$ is :