Let $\mathrm{f}(\mathrm{x})$ be a polynomial of degree four having extreme values at $\mathrm{x}=1$ and $\mathrm{x}=2$. If $\lim _{\mathrm{x} \rightarrow 0}\left[1+\frac{\mathrm{f}(\mathrm{x})}{\mathrm{x}^{2}}\right]=3$, then $f(2)$ is equal to :

JEE Main · 2015 · 04 Apr 2015 · Offline · Application of Derivatives

Let

f (x)

be a polynomial of degree four having extreme values at

x = 1

and

x = 2

. If

lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3

, then

f (2)

is equal to :

Answer: (B)

A quartic with turning points at

x = 1

and

x = 2

satisfies

f^{'} (1) = f^{'} (2) = 0

. The limit condition says

\frac{f ( x )}{x ^{2}} \to 2

x \to 0

, which is possible only if

f

carries no constant or linear term and its

x^{2}

coefficient is exactly

2

; so write

f (x) = a x^{4} + b x^{3} + 2 x^{2}

. Then

f^{'} (x) = 4 a x^{3} + 3 b x^{2} + 4 x

, and

f^{'} (1) = 4 a + 3 b + 4 = 0

together with

f^{'} (2) = 32 a + 12 b + 8 = 0

solve to

a = \frac{1}{2}

b = - 2

. Thus

f (x) = \frac{1}{2} x^{4} - 2 x^{3} + 2 x^{2}

, and

f (2) = 8 - 16 + 8 = 0

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option