Question
Let f be a real polynomial of degree n such that $f(x) = f^{'}(x)f^{''}(x)$, for all $x \in \mathbb{R}$. If $f(0) = 0$, then 36 $\left( f^{'}(2) + f^{''}(2) + \int_{0}^{2}\mspace{2mu} f(x)dx \right)$ is equal to:-
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Solution
Sol. [Key Concept: Degrees force $n = 3$; match coefficients of $f = f'f''$, then integrate.]
Let degree of polynomial be ' n '
$${n = n - 1 + n - 2 \Rightarrow n = 3 }{f(x) = {ax}^{3} + {bx}^{2} + cx }{\left( ax^{3} + bx^{2} + cx \right) = \left( 3ax^{2} + 2bx + c \right)(6ax + 2b) }{x^{3}:a = 18a^{2} \Rightarrow a = \frac{1}{18} }{x^{2}:b = 6ab + 12ab }{x:c = 4b^{2} + 6ac \Rightarrow \underline{2C} = 4b^{2} \Rightarrow b^{2} = \underline{c} }$$const : $bc = 0 \Rightarrow \text{ }b = c = 0$
$${\int_{0}^{2}\mspace{2mu} f(x)dx + f^{'}(2) + f^{''}(2) }{\int_{0}^{2}\mspace{2mu} ax^{3}dx + 12a + 12a = 4a + 24a = 28a = \frac{28}{18} = \frac{14}{9} }{36\left( \frac{14}{9} \right) = 56 }$$
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