The function x^5 - 5 x^4 + 5 x^3 - 1 is - Vidyadip

MCQ

The function ${x^5} - 5{x^4} + 5{x^3} - 1$ is

A
Maximum at $x = 3$ and minimum at $x = 1$
B
Minimum at $x = 1$
✓
Neither maximum nor minimum at $x = 0$
D
Maximum at $x = 0$

✓

Answer

Correct option: C.

Neither maximum nor minimum at $x = 0$

c
(c) Let $f(x) = {x^5} - 5{x^4} + 5{x^3} - 1$

==> $f'(x) = 5{x^4} - 20{x^3} + 15{x^2} = 0$

$\therefore (x - 3)(x - 1) = 0$ or $x = 3,1$

Now $f''(x) = 20{x^3} - 60{x^2} + 30x$

Put $x = 3$ and $ 1$ , we get $f'''(3) = + ve$ and $f''(1) = - ve$ and $f''(0) = 0$.

Hence $f(x)$ neither maximum nor minimum at $x = 0$.

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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 6. Application of derivatives→STD 12 - 6. Application of derivatives — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $f(x)=\sin x+\left(x^3-3 x^2+4 x-2\right) \cos x$ for $x \in(0,1)$ Consider the following statements

$I.$ $f$ has a zero in $(0,1)$

$II.$ $f$ is monotone in $(0,1)$ Then,

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, where $|\vec{a}|=4,|\vec{b}|=3 \quad \theta \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$. Then $|(\vec{a}-\vec{b}) \times(\vec{a}+\vec{b})|^{2}+4(\vec{a} \cdot \vec{b})^{2}$ is equal to

$\int_0^\infty {\frac{{x\,dx}}{{(1 + x)(1 + {x^2})}}} = $

The least number of times a fair coin must be tossed so that the probability of getting at least one head is at least $0.8,$ is :

The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. What is the probability that out of $5$ such bulbs none will fuse after $150$ days of use

The unit vector in the direction of $\overrightarrow{\text{a}}$ is:

If $4i + 7j + 8k,\,\,\,2i + 3j + 4k\,$ and $2i + 5j + 7k$ are the position vectors of the vertices $ A, B$ and $C$ respectively of triangle $ABC$ . The position vector of the point where the bisector of angle $A$ meets $BC$ is

Solution of the equation $\cos x\cos y\frac{{dy}}{{dx}} = - \sin x\sin y$ is

If $y = {\tan ^{ - 1}}\left( {{{a\cos x - b\sin x} \over {b\cos x + a\sin x}}} \right)$ then ${{dy} \over {dx}} = $

If $tan^{-1} (x^2 + 3|x|-4 )= tan^{-1} (4 \pi + sin^{-1}(sin14))$, then the value of $cos^{-1}(cos3|x|)$ is equal to