Download App

Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
$2 x^3-6 x+5$ is an increasing function, if
  • A
    $0 < x < 1$
  • B
    $-1 < x < 1$
  • ✓
    $x < -1$ or $x > 1$
  • D
    $-1 < x < -\frac{1}{2}$
✓

Answer

Correct option: C.
$x < -1$ or $x > 1$
Let $f(x)=2 x^3-6 x+5$
On differentiating $\ce{w.r.t. x}$, we get $f^{\prime}(x)=6 x^2-6$
Since, it is increasing function.
$\Rightarrow 6 x^2-6>0 $
$\Rightarrow(x-1)(x+1)>0$
$\Rightarrow x >1 \text { or } x<-1$

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

Explore more

More "M.C.Q (1 Marks)" from Application of Derivatives→Application of Derivatives — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $ \alpha _1, \alpha _2$ are two values of $\alpha $ for which the system $2 \alpha x + y = 5, x - 6y = \alpha $ and $x + y = 2$ is consistent, then $ |2(\alpha _1 + \alpha _2)| $ is -
→
Solution of the differential equation $\frac{{dx}}{x} + \frac{{dy}}{y} = 0$ is
→
If the area of the triangle with vertices $(-3,0),(3,0)$ and $(0, k)$ is $9$ sq units, then the value/s of $k$ is will be
→
A bag $X$ contains $2$ white and $3$ black balls and another bag $Y$ contains $4$ white and $2$ black balls. One bag is selected at random and a ball is drawn from it. Then, the probability chosen to be white is,
→
A rifle man is firing at a distant target and has only $10\%$ chance of hitting it. The minimum number of rounds he must fire in order to have $50\%$ chance of hitting it at least once is
→
Let $f(x) = \int\limits_0^x {{{\cos t} \over t}dt,\,\,x > 0} $ then $f(x)$ has
→
If $\int\frac{\cos8\text{x}+1}{\tan2\text{x}-\cot2\text{x}}\text{ dx}=\text{a}\cos8\text{x}+\text{C},$ then $a =$
→
Let ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}{{\alpha _1}}&{{\beta _1}}&{{\gamma _1}}\\{{\alpha _2}}&{{\beta _2}}&{{\gamma _2}}\\{{\alpha _3}}&{{\beta _3}}&{{\gamma _3}}\end{array}\,} \right|$, then ${\Delta _1} \times {\Delta _2}$ can be expressed as the sum of how many determinants
→
The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14 l^2$ is equal to.................
→
length of the $1^{\mathrm{er}}$ from the point $(0, -1, 3)$ to the plane $2x + y - 2z + 1 = 0$ is:
→