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Question

(B) Given that $P(x)$ is a polynomial with real coefficients and $P^{\prime \prime}(x) 
eq 0$ for all $x$. Since $P^{\prime \prime}(x)$ is a constant

Vedclass · Accepted Answer

$P(1) 
eq 0$

Vedclass · Answer

(B) Given that $P(x)$ is a polynomial with real coefficients and $P^{\prime \prime}(x) 
eq 0$ for all $x$. Since $P^{\prime \prime}(x)$ is a constant for a quadratic polynomial,let us consider the simplest case where $P(x)$ is a quadratic polynomial: $P(x) = ax^2 + bx + c$,where $a 
eq 0$. Then,$P^{\prime}(x) = 2ax + b$ and $P^{\prime \prime}(x) = 2a$. Given $P(0) = 1$,we have $c = 1$. Given $P^{\prime}(0) = -1$,we have $b = -1$. Thus,$P(x) = ax^2 - x + 1$. Evaluating at $x = 1$: $P(1) = a(1)^2 - (1) + 1 = a$. Since $P^{\prime \prime}(x) = 2a 
eq 0$,it follows that $a 
eq 0$. Therefore,$P(1) = a 
eq 0$. This property holds for higher-degree polynomials as well,as $P(1)$ cannot be zero under the constraint $P^{\prime \prime}(x) 
eq 0$ given the initial conditions. Thus,the correct option is $B$.

$A$ polynomial $P(x)$ with real coefficients has the property that $P^{\prime \prime}(x) \neq 0$ for all $x$. Suppose $P(0) = 1$ and $P^{\prime}(0) = -1$. What can you say about $P(1)$?

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