Let P be a non-zero polynomial such that P(1+ | vedclass

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

Question

(B) Given that $P(x)$ is a non-zero polynomial satisfying $P(1+x)=P(1-x)$ for all real $x$ and $P(1)=0$. Differentiating both sides with respect to $x

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given that $P(x)$ is a non-zero polynomial satisfying $P(1+x)=P(1-x)$ for all real $x$ and $P(1)=0$. Differentiating both sides with respect to $x$,we get: $P'(1+x) = -P'(1-x)$. Setting $x=0$,we obtain: $P'(1) = -P'(1) \implies 2P'(1) = 0 \implies P'(1) = 0$. Since $P(1)=0$ and $P'(1)=0$,by the Factor Theorem,$(x-1)^2$ must be a factor of $P(x)$. To check if $m$ can be larger,consider $P(x) = (x-1)^2$. This satisfies $P(1+x) = (1+x-1)^2 = x^2$ and $P(1-x) = (1-x-1)^2 = (-x)^2 = x^2$. Thus,$P(1+x)=P(1-x)$ holds. Since $P(x) = (x-1)^2$ is a valid polynomial,the largest integer $m$ such that $(x-1)^m$ divides $P(x)$ for all such $P(x)$ is $2$.

Let $P$ be a non-zero polynomial such that $P(1+x)=P(1-x)$ for all real $x$ and $P(1)=0$. Let $m$ be the largest integer such that $(x-1)^m$ divides $P(x)$ for all such $P(x)$. Then,$m$ equals

Similar Questions

If $y = \tan^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$, then at $x = \frac{\pi}{6}$, the value of $\frac{dy}{dx}$ is

Find the derivative of $\frac{2}{x+1}-\frac{x^{2}}{3x-1}$.

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

At $x=\frac{\pi^2}{4}$,find the value of $\frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)$.

Let $3f(x) - 2f(1/x) = x,$ then $f'(2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module