NIMCET 2017 — Mathematics PYQ

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Question

NIMCET 2017 — Mathematics PYQ

NIMCET | Mathematics | 2017

Let $f (x)$ be a polynomial of degree four, having extreme value at $x = 1$ and $x = 2$ .

If $lim_{x \to 0} [1 + \frac{f ( x )}{x ^{2}}] = 3$ , then $f (2)$ is?

Choose the correct answer:

Explanation

Concept:
To find extreme values of a function $f$ , set $f^{'} (x) = 0$ and solve.

Calculation:
Consider a polynomial $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$

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

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

$\Rightarrow lim_{x \to 0} \frac{a x ^{4} + b x ^{3} + c x ^{2} + d x + e}{x ^{2}} = 2$

$\Rightarrow lim_{x \to 0} (a x^{2} + b x + c + \frac{d}{x} + \frac{e}{x ^{2}}) = 2$

Now, let $d = e = 0$

So,
$lim_{x \to 0} (a x^{2} + b x + c) = 2$

$\Rightarrow c = 2$

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

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

Here, extreme values are $1$ and $2$ , so $f^{'} (1) = f^{'} (2) = 0$

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

Multiply above equation by $4$ , we get
$16 a + 12 b + 16 = 0 ... (1)$

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

Subtract $(1)$ from $(2)$ we get...

$16 a - 8 = 0$

$\Rightarrow a = \frac{1}{2}$

Put $a = \frac{1}{2}$ in (1), we get

$8 + 12 b + 16 = 0$

$\Rightarrow b = - 2$

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

$f (2) =$

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

$= 8 - 16 + 8$

$= 0$

Hence, option $(1)$ is correct.