Let f be a real polynomial of degree n such that , for all . If f(0) = 0, then is equal to: [2026]

Question

Let f be a real polynomial of degree n such that $f (x) = f' (x) f'' (x)$ , for all $x \in R$ . If f(0) = 0, then $36 (f' (2) + f'' (2) + \int_{0}^{2} f (x) d x)$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

We have, $f (x) = f' (x) \cdot f'' (x)$

From balancing degree,

n = n – 1 + n – 2

$\Rightarrow$ n = 3

$∴$ f(x) is a cubic polynomial,

Also, f(0) = 0

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

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

$f'' (x) = 6 a x + 2 b$

Now, $f (x) = f' (x) \cdot f'' (x)$

$a x^{3} + b x^{2} + c x = (3 a x^{2} + 2 b x + c) (6 a x + 2 b)$

$\Rightarrow a x^{3} + b x^{2} + c x = 18 a^{2} x^{3} + (6 a b + 12 a b) x^{2} + (4 b^{2} + 6 a c) x + 2 b c$

Comparing coefficients, we get

$a = 18 a^{2}$

$\Rightarrow a = \frac{1}{18}$ ( $∵ a \neq 0$ )

b = 18 ab

$\Rightarrow b = b$ $(Using a = \frac{1}{18})$

$c = 4 b^{2} + 6 a c$

$c = 4 b^{2} + \frac{c}{3}$ $(∵ a = \frac{1}{18})$

$\Rightarrow \frac{2 c}{3} = 4 b^{2} \Rightarrow c = 6 b^{2}$

2bc = 0

$\Rightarrow b = 0 or c = 0$

If $b = 0 \Rightarrow c & if c = 0 \Rightarrow b = 0$

$∴ f (x) = \frac{1}{18} x^{3} \Rightarrow f' (x) = \frac{x^{2}}{6}$

$\Rightarrow f'' (x) = \frac{x}{3}$

Now, $36 (f' (2) + f'' (2) + \int_{0}^{2} f (x) d x)$

$= 36 (\frac{4}{6} + \frac{2}{3} + \int_{0}^{2} \frac{x^{3}}{19} d x)$

$= 24 + 24 + \frac{36}{18} (\frac{x^{4}}{4}) |_{0}^{2}$

$= 48 + \frac{2}{4} (2^{4}) = 48 + 8 = 56$

Solution

Q.
Let f be a real polynomial of degree n such that $f (x) = f' (x) f'' (x)$ , for all $x \in R$ . If f(0) = 0, then $36 (f' (2) + f'' (2) + \int_{0}^{2} f (x) d x)$ is equal to: [2026]

1 42

2 46

3 56

4 66

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