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

(A) $x=1$ is a point of minima and $x=-1$ is a point of maxima of $f$.

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

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Application of Derivatives

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?

(A) $x = 1$ is a point of minima and $x = - 1$ is a point of maxima of $f$ .✓ Correct
(B) $x = 1$ is a point of maxima and $x = - 1$ is a point of minimum of $f$
(C)f is an odd function
(D) $f (1) - 4 f (- 1) = 4$

Answer: (A)

Since

lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4

, we need

lim_{x \to 0} \frac{f ( x )}{x ^{3}} = 2

, which forces

f (0) = 0

and, by repeated use of L'Hopital, pins down the low-order coefficients. With

x = \pm 1

as critical points of this degree-five polynomial,

f^{'} (x) = a (x - 1) (x + 1) (p x^{2} + q x + r)

; the finite limit kills

r

and then

q

and leaves

- \frac{a p}{3} = 2

, so

f^{'} (x) = - 6 x^{2} (x - 1) (x + 1) = - 6 x^{2} (x^{2} - 1)

. The sign of

f^{'}

runs

-, +, +, -

across

- 1, 0, 1

, so

x = - 1

is a local minimum and

x = 1

a local maximum, while

x = 0

is only an inflection. Integrating,

f (x) = 2 x^{3} - \frac{6}{5} x^{5}

, which is odd, with

f (1) = \frac{4}{5}

f (- 1) = - \frac{4}{5}

and

f (1) - 4 f (- 1) = 4

. The claim that

x = 1

is a minimum and

x = - 1

a maximum reverses the true picture, so that is the statement which is not true.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Single correct option