Let f(x) = x^3 + b x^2 + cx + d,0 < b^2 < c. Then f - Vidyadip

MCQ

Let $f(x) = {x^3} + b{x^2} + cx + d,0 < {b^2} < c$. Then $ f$

A
Is bounded
B
Has a local maxima
C
Has a local minima
✓
Is strictly increasing

✓

Answer

Correct option: D.

Is strictly increasing

d
(d) Given $f(x) = {x^3} + b{x^2} + cx + d$

$\therefore$ $f'(x) = 3{x^2} + 2bx + c$

Now its discriminant $ = 4({b^2} - 3c)$

==> $4({b^2} - c) - 8c < 0,$ as ${b^2} < c$ and $c > 0$

Therefore, $f'(x) > 0$ for all $x \in R$

Hence $ f$ is strictly increasing.

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 6. Application of derivatives→6. 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

A particle moves in a straight line with a velocity given by $\frac{{dx}}{{dt}} = x + 1$($x$ is the distance described). The time taken by a particle to traverse a distance of $99$ metre is

If $\frac{\text{dy}}{\text{dx}}=3$ then y is equal to:

3x
0
3x + c
$\frac{\text{x}}{3}+\text{c}$

$\int \frac{\left(x^{2}+1\right) e^{x}}{(x+1)^{2}} d x=f(x) e^{x}+C$, Where $C$ is a constant, then $\frac{d^{3} f}{d x^{3}}$ at $x =1$ is equal to

If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$, then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is

Line passing through (3, 4, 5) and (4, 5, 6) has direction ratios:

$1,1,1$
$\sqrt{3},\sqrt{3},\sqrt{3}$
$\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$
$7,9,11$

The perpendicular distance of the point $(2, 4, -1)$ from the line $\frac{{x + 5}}{1} = \frac{{y + 3}}{4} = \frac{{z - 6}}{{ - 9}}$ is

$\int\limits_{ - \,a}^a {\,f\,(x)\,dx} $=

The area of the region enclosed between the parabolas $y ^{2}=2 x -1$ and $y ^{2}=4 x -3$ is

If $\left| {\,\begin{array}{*{20}{c}}{x - 1}&3&0\\2&{x - 3}&4\\3&5&6\end{array}\,} \right| = 0$, then $x =$

A problem in Mathematics is given to three students whose chances of solving it are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}$ respectively. If the events of their solving the problem are independent then the probability that the problem will be solved, is