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(B) Let $p(x) = ax^3 + bx^2 + cx + d$ be a polynomial of degree $3$.Then,the derivatives are:$p^{\prime}(x) = 3ax^2 + 2bx + c$$p^{\prime \prime}(x) =

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$-6$

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(B) Let $p(x) = ax^3 + bx^2 + cx + d$ be a polynomial of degree $3$.Then,the derivatives are:$p^{\prime}(x) = 3ax^2 + 2bx + c$$p^{\prime \prime}(x) = 6ax + 2b$$p^{\prime \prime \prime}(x) = 6a$Given $p^{\prime \prime \prime}(1) = 6$,we have $6a = 6$,which implies $a = 1$.Given $p^{\prime \prime}(1) = 0$,we substitute $a = 1$ into $p^{\prime \prime}(1) = 6a(1) + 2b = 0$:$6(1) + 2b = 0 \Rightarrow 2b = -6 \Rightarrow b = -3$.Now,we need to find $p^{\prime \prime}(0)$:$p^{\prime \prime}(0) = 6a(0) + 2b = 2b$.Substituting $b = -3$,we get $p^{\prime \prime}(0) = 2(-3) = -6$.

If $p(x)$ is a polynomial of degree $3$ which satisfies $p^{\prime \prime}(1)=0$ and $p^{\prime \prime \prime}(1)=6$,then $p^{\prime \prime}(0)$ is equal to

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