First slide

Maxima and minima

Question

Let $f (x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $\lim_{x \to 0} (2 + \frac{f (x)}{x^{3}}) = 4$ , then which one of the following is not true?

Moderate

A

1) x = 1 is a point of maxima and x = -1 is a point of minimum of f.
B

f(1) – 4f(-1) =4
C

x=1 is a point of minima and x = -1 is a point of maxima of f.
D

f is an odd function

Solution

Let $f (x) = a x^{5} + b x^{4} + c x^{3}$ . Then
$\lim_{x \to 0} (2 + \frac{a x^{5} + b x^{4} + c x^{3}}{x^{3}}) = 4 \Rightarrow 2 + c = 4 \Rightarrow c = 2$
Also $f' (x) = 5 a x^{4} + 4 b x^{3} + 6 x^{2}$ $= x^{2} (5 a x^{2} + 4 b x + 6)$ .
Since $x = \pm 1 a r e p o i n t s o f e x t r e m u m, w e h a v e$
$f' (1) = 0$ and $f' (- 1) = 0$
$\Rightarrow 5 a + 4 b + 6 = 0$ $a n d 5 a - 4 b + 6 = 0$
$\Rightarrow$ $b = 0$ , $a = - \frac{6}{5}$
$\Rightarrow$ $f (x) = \frac{- 6}{5} x^{5} + 2 x^{3}$
$\Rightarrow$ $f' (x) = - 6 x^{4} + 6 x^{2}$
$\Rightarrow f^{''} (x) = - 24 x^{3} + 12 x N o w f^{''} (- 1) = 12 > 0 a n d f^{''} (1) = - 12 < 0$
$\Rightarrow$ minimum at x = -1 and maximum at x = 1

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App