A quadratic polynomial P(x) satisfies the con | vedclass

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(B) Since $P(x)$ is a quadratic polynomial with roots at $x = 0$ and $x = 1$,we can write it as $P(x) = a(x)(x - 1) = a(x^2 - x)$.Given the condition

Vedclass · Accepted Answer

$-6$

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(B) Since $P(x)$ is a quadratic polynomial with roots at $x = 0$ and $x = 1$,we can write it as $P(x) = a(x)(x - 1) = a(x^2 - x)$.Given the condition $\int_{0}^{1} P(x) dx = 1$,we substitute $P(x)$ into the integral:$\int_{0}^{1} a(x^2 - x) dx = 1$$a \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} = 1$$a \left( \frac{1}{3} - \frac{1}{2} \right) = 1$$a \left( \frac{2 - 3}{6} \right) = 1$$a \left( -\frac{1}{6} \right) = 1$$a = -6$.Thus,the leading coefficient is $-6$.

$A$ quadratic polynomial $P(x)$ satisfies the conditions $P(0) = 0$,$P(1) = 0$,and $\int_{0}^{1} P(x) dx = 1$. The leading coefficient of the quadratic polynomial is:

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