Let be a polynomial function of degree four having extreme values at x = 4 ad x = 5. If , then f(2) is equal to : [2025]

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Solution

Q.
Let $f : R \to R$ be a polynomial function of degree four having extreme values at x = 4 and x = 5. If $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$ , then f(2) is equal to : [2025]

1 12

2 14

3 8

4 10

Ans.
(4)

We have, $\lim_{x \to 0} \frac{f (x)}{x^{2}} = 5$

Let $f (x) = a x^{4} + b x^{3} + c x^{2} + d x + e$

$\Rightarrow \lim_{x \to 0} \frac{(a x^{4} + b x^{3} + c x^{2} + d x + e)}{x^{2}} = 5$

$\Rightarrow c = 5 and d = e = 0$

$∴ f (x) = a x^{4} + b x^{3} + 5 x^{2}$

$f' (x) = 4 a x^{3} + 3 b x^{2} + 10 x$

$= x (4 a x^{2} + 3 b x + 10)$

$∵$ f(x) has extreme values at x = 4 and x = 5, so f(4) = 0 and f(5) = 0.

Using derivative and its values, we get

$a = \frac{1}{8} and b = \frac{- 3}{2}$

Now, $f (2) = \frac{1}{8} \times 2^{4} - \frac{3}{2} \times 2^{3} + 5 \times 2^{2}$

= 2 – 12 + 20 = 10.

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