MCQ
The interval in which the function $y=x^3+5 x^2-1$ is decreasing, is
  • A
    $\left(0, \frac{10}{3}\right)$
  • B
    $(0,10)$
  • $\left(\frac{-10}{3}, 0\right)$
  • D
    None of these

Answer

Correct option: C.
$\left(\frac{-10}{3}, 0\right)$
(c) : Given, $y=x^3+5 x^2-1$
$
\Rightarrow \frac{d y}{d x}=3 x^2+10 x=x(3 x+10)
$
For function to be decreasing, $\frac{d y}{d x}<0$
x(3x + 10) < 0 => $\frac{-10}{3}$< x < 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

A stone throw upward having speed equation $S =9.8 t-4.9 t^2$ where S in $m / s$. What will be the time to achieve maximum height ?
Solution of differential equation $\frac{{dy}}{{dx}} + ay = {e^{mx}}$ is
Let $\vec a = 2\hat i - \hat j + \hat k$, $\vec b = \hat i + 2\hat j - \hat k$ and $\vec c = \hat i + \hat j - 2\hat k$ be three vectors. A vector of the type $\vec b + \lambda \vec c$ for some scalar $\lambda $,  whose projection on $\vec a$ is of magnitude $\sqrt {\frac{2}{3}} $ is
Namita walks 14 metres towards west, then turns to her right and walks 14 metres and then turns to her left and walks 10 metres.Again turning to her left she walks 14 metres.What is the shortest distance (in metres) between her starting point and the present position?
Let $A = \{1, 2, 3\}. $Then, the number of relations containing $(1, 2)$ and $(1, 3)$ which are reflexive and symmetric but not transitive is:
If A and B are two independent events with $\text{P(A)}=\frac{3}{5}$ and $\text{P(B)}=\frac{4}{9},$ then $\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$ equals,
The volume of the parallelopiped whose edges are represented by $ - 12i + \alpha k,\,\,3j - k$ and $2i + j - 15k$ is $546.$  Then $\alpha = $
If $ \cos^{-1}\left (\frac {1 - \text{x}^{2}}{1 + \text{x}^{2}}\right ) + \cos^{-1}\left (\frac {1 - \text{y}^{2}}{1 + \text{y}^{2}}\right )=\frac{\pi}{2}$, where xy < 1, then:
If $f(x)\, = \,\left\{ {\begin{array}{*{20}{c}}{\frac{{\sqrt {1 + kx} - \sqrt {1 - kx} }}{x}}&{{\rm{,for}} - 1 \le x < 0}\\{2{x^2} + 3x - 2}&{{\rm{,}}\,{\rm{for\,\, }}\,0 \le \,x \le 1}\end{array}} \right.$, is continuous at $x = 0$, then $k = $
The three points $\text{ABC}$ have position vectors $(1, x, 3), (3, 4, 7)$ and $(y, -2, -5)$ are collinear then $(x, y):$