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

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

If $y = \sec(\tan^{-1} x)$,then $\frac{dy}{dx}$ at $x = 1$ is

Let $f$ and $g$ be differentiable functions on $R$ such that $f \circ g$ is the identity function. If for some $a, b \in R$,$g^{\prime}(a) = 5$ and $g(a) = b$,then $f^{\prime}(b)$ is equal to

Differentiate the following with respect to $x:$ $\cos^{-1}(e^x)$

If $y=\sqrt{2 x+\cos ^2\left(2 x+\frac{\pi}{4}\right)}$,then find $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module