MCQ
Function $f(x) = \frac{{{x^2} - 2}}{{\sqrt {1 + {x^2}} }}$
- Ais always increasing
- Bis always decreasing
- ✓has exactly one point of minima
- Dhas exactly one point of maxima
✓
Answer
Correct option: C.
has exactly one point of minima
c
$f^{\prime}(x)=\frac{x\left(x^{2}+4\right)}{\left(1+x^{2}\right)^{3 / 2}},$
$f^{\prime}(x)=\frac{x\left(x^{2}+4\right)}{\left(1+x^{2}\right)^{3 / 2}},$
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
If the lines $\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2k}} = \frac{{z - 3}}{2}$, $\frac{{x - 1}}{{3k}} = \frac{{y - 5}}{1} = \frac{{z - 6}}{{ - 5}}$ are at right angles, then $k =$
→Solution of differential equation $\frac{{dy}}{{dx}} = 2xy$ is
→For a non - zero, real $a, b$ and $c$ $\left| {\begin{array}{*{20}{c}}{\frac{{{a^2} + {b^2}}}{c}}&c&c\\a&{\frac{{{b^2} + {c^2}}}{a}}&a\\b&b&{\frac{{{c^2} + {a^2}}}{b}} \end{array}} \right|$ $= \alpha \, abc$, then the values of $\alpha$ is
→If $\int\limits^1_0\text{f}(\text{x})\text{dx}=1,\int\limits^1_0\text{x}\text{f}(\text{x})\text{dx}=\text{a},\int\limits^1_0\text{x}^2\text{f}(\text{x})\text{dx}=\text{a}^2,$ then $\int\limits^1_0(\text{a}-\text{x})^2\text{f(x)}\text{dx}$ equals :
→Let $f (x + y) = f (x) + f (y)$ for all $x , y \in R.$ Then :
→The period of the function $f\left( x \right)$ = $\frac{x}{3} - \left[ {\frac{x}{3} - 5} \right] + \frac{x}{4} - \left[ {\frac{x}{4} - 5} \right] + \frac{x}{5} - \left[ {\frac{x}{5} - 5} \right]$ is (where $[.]$ is $G.I.F.$ )
→If the position vectors of $P, Q$ are respectively $5a + 4b$ and $3a - 2b$ then $\vec{\text{QP}}=$
→The corner points of the feasible region in the graphical representation of a linear programming problem are $(2,72),(15,20)$ and $(40,15)$. If $z=18 x+9 y$ be the objective function, then:
→If the curve a $y + x^2 = 7$ and $x^3 = y$ cut orthogonally at $(1, 1),$ then a is equal to:
→If $f(x)=\left\{\begin{array}{ccc}\frac{1}{|x|} & ; & |x| \geq 1 \\ a x^{2}+b & ; & |x|<1\end{array}\right.$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively
→