Let f be a polynomial function such that f(3x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $f(x) = ax^n + \dots$ be a polynomial of degree $n$.Comparing the degrees on both sides of $f(3x) = f'(x) \cdot f''(x)$:$n = (n-1) + (n-2) = 2

Vedclass · Accepted Answer

$f''(2) - f'(2) = 0$

Vedclass · Answer

(B) Let $f(x) = ax^n + \dots$ be a polynomial of degree $n$.Comparing the degrees on both sides of $f(3x) = f'(x) \cdot f''(x)$:$n = (n-1) + (n-2) = 2n - 3$,which gives $n = 3$.Let $f(x) = ax^3 + bx^2 + cx + d$.Then $f(3x) = a(3x)^3 + b(3x)^2 + c(3x) + d = 27ax^3 + 9bx^2 + 3cx + d$.$f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.$f'(x) \cdot f''(x) = (3ax^2 + 2bx + c)(6ax + 2b) = 18a^2x^3 + (6ab + 12ab)x^2 + (4b^2 + 6ac)x + 2bc$.Equating coefficients of $x^3$: $27a = 18a^2 \Rightarrow a = \frac{3}{2}$ (since $a 
eq 0$).Equating coefficients of $x^2$: $9b = 18ab = 18(\frac{3}{2})b = 27b \Rightarrow b = 0$.Equating coefficients of $x$: $3c = 4b^2 + 6ac = 0 + 6(\frac{3}{2})c = 9c \Rightarrow c = 0$.Equating constant terms: $d = 2bc = 0$.Thus,$f(x) = \frac{3}{2}x^3$.Then $f'(x) = \frac{9}{2}x^2$ and $f''(x) = 9x$.At $x = 2$,$f'(2) = \frac{9}{2}(4) = 18$ and $f''(2) = 9(2) = 18$.Therefore,$f''(2) - f'(2) = 18 - 18 = 0$.

Let $f$ be a polynomial function such that $f(3x) = f'(x) \cdot f''(x)$ for all $x \in \mathbb{R}$. Then:

Similar Questions

If $x = \sin y$,then $\frac{d^2 y}{dx^2} = . . . . . .$,$(0 < x < 1)$.

Find the second order derivative of the function $\sin (\log x)$.

If ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$,then ${I_n} - n{I_{n - 1}} = $

If $x^2+y^2=1$,then which of the following is true?

If $\cos ^{-1}\left(\frac{y}{2}\right)=\log _{e}\left(\frac{x}{5}\right)^{5}$ and $|y| < 2$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module