Let f(x) be a polynomial function of the seco | vedclass

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

Question

(A) Let the polynomial function be $f(x) = ax^2 + bx + c$.Then,the derivative is $f'(x) = 2ax + b$.Given $f(1) = f(-1)$,we have $a(1)^2 + b(1) + c = a

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) Let the polynomial function be $f(x) = ax^2 + bx + c$.Then,the derivative is $f'(x) = 2ax + b$.Given $f(1) = f(-1)$,we have $a(1)^2 + b(1) + c = a(-1)^2 + b(-1) + c$.This simplifies to $a + b + c = a - b + c$,which implies $2b = 0$,so $b = 0$.Thus,$f'(x) = 2ax$.Now,we evaluate the derivatives at $a_1, a_2, a_3$: $f'(a_1) = 2aa_1$,$f'(a_2) = 2aa_2$,and $f'(a_3) = 2aa_3$.Since $a_1, a_2, a_3$ are in $A.P.$,there exists a common difference $d$ such that $a_2 = a_1 + d$ and $a_3 = a_1 + 2d$.Then $f'(a_2) - f'(a_1) = 2aa_2 - 2aa_1 = 2a(a_2 - a_1) = 2ad$ and $f'(a_3) - f'(a_2) = 2aa_3 - 2aa_2 = 2a(a_3 - a_2) = 2ad$.Since the difference between consecutive terms is constant $(2ad)$,the sequence $f'(a_1), f'(a_2), f'(a_3)$ is in $A.P.$

Let $f(x)$ be a polynomial function of the second degree. If $f(1) = f(-1)$ and $a_1, a_2, a_3$ are in $A.P.$,then $f'(a_1), f'(a_2), f'(a_3)$ are in

Similar Questions

Find the derivative: $\frac{d}{dx}[\cos((1 - x^2)^2)] = ?$

If $f(x) = \frac{1}{1 + \frac{1}{x}}$ and $g(x) = \frac{1}{1 + \frac{1}{f(x)}}$, then $g^{\prime}(2)$ is equal to

If $y = x^n \log x + x(\log x)^n$,then $\frac{dy}{dx} = $

If $y = \sin^{n} x \cos nx$,then $\frac{dy}{dx}$ is

If $f(x)$ is an even function,then $f^{\prime}(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module