MCQ
$2{x^3} - 6x + 5$ is an increasing function if
  • A
    $0 < x < 1$
  • B
    $ - 1 < x < 1$
  • $x < - 1$ or $x > 1$
  • D
    $ - 1 < x < - 1/2$

Answer

Correct option: C.
$x < - 1$ or $x > 1$
c
(c) Let $f(x) = 2{x^3} - 6x + 5$ ==> $f'(x) = 6{x^2} - 6 > 0$

==> ${x^2} - 1 > 0$==> $(x - 1)$ $(x + 1) > 0$

==>$x > 1$ 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

Similar questions

$\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right| = $
The projections of a line segment on x, y and z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are:

  1. $13;\frac{12}{13},\frac{4}{13},\frac{3}{13}$

  2. $19;\frac{12}{19},\frac{4}{19},\frac{3}{19}$

  3. $11;\frac{12}{11},\frac{14}{11},\frac{3}{11}$

  4. $\text{None of these}$

Considering only the principal values of the inverse trigonometric functions, the value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$
Solution of the differential equation, $\frac{{dy}}{{dx}} = \frac{{1 - 2y - 4x}}{{1 + y + 2x}}$ is
Area bounded by lines $y = 2 + x,$ $y = 2 - x$ and $x = 2$ is
The solutions of the equation $\left| {\,\begin{array}{*{20}{c}}x&2&{ - 1}\\2&5&x\\{ - 1}&2&x\end{array}\,} \right| = 0$ are
Let $f (x) =$ $\frac{{{\text{g}}\,(x)}}{{{\text{h}}\,(x)}}$, where $g $ and $h$ are cotinuous functions on the open interval $(a, b)$. Which of the following statements is true for $a < x < b$ ?
Let $\mathrm{f}: R \rightarrow R$ be function defined as

$f ( x )=\left\{\begin{array}{cc}3\left(1-\frac{| x |}{2}\right) & \text { if }| x | \leq 2 \text { } \\ 0 & \text { if }| x |>2 \text { }\end{array}\right.$ Let $g: R \rightarrow R$ be given by $g(x)=f(x+2)-f(x-2)$. If $n$ and $m$ denote the number of points in $R$ where $\mathrm{g}$ is not continuous and not differentiable, respectively, then $\mathrm{n}+\mathrm{m}$ is equal to $....$

If A is an invertible matrix, then det (A-1) is equal to:
  1. Det (A)
  2. $\frac{1}{\text{det(A)}}$
  3. 1
  4. None of these.
The point on the curve y2 = x where tangent makes 45° angle with x-axis is:

  1. $\Big(\frac{1}{2},\frac{1}{4}\Big)$

  2. $\Big(\frac{1}{4},\frac{1}{2}\Big)$

  3. $(4,2)$

  4. $(1,1)$