If f(x)=x^3+p x^2+q x is defined on [0,2] suc | vedclass

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(A) Given $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$.Since $f(0)=f(2)$:$f(0) = 0^3 + p(0)^2 + q(0) = 0$$f(2) = 2^3 + p(2)^2 + q(2) = 8 + 4p + 2q$Setti

Vedclass · Accepted Answer

$13$

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(A) Given $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$.Since $f(0)=f(2)$:$f(0) = 0^3 + p(0)^2 + q(0) = 0$$f(2) = 2^3 + p(2)^2 + q(2) = 8 + 4p + 2q$Setting $f(0)=f(2)$ gives $8 + 4p + 2q = 0$,which simplifies to $2p + q + 4 = 0$ (Equation $i$).Now,find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = 3x^2 + 2px + q$Given $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$:$3\left(1+\frac{1}{\sqrt{3}}\right)^2 + 2p\left(1+\frac{1}{\sqrt{3}}\right) + q = 0$$3\left(1 + \frac{1}{3} + \frac{2}{\sqrt{3}}\right) + 2p + \frac{2p}{\sqrt{3}} + q = 0$$3\left(\frac{4}{3} + \frac{2}{\sqrt{3}}\right) + 2p + \frac{2p}{\sqrt{3}} + q = 0$$4 + 2\sqrt{3} + 2p + \frac{2p}{\sqrt{3}} + q = 0$ (Equation $ii$).Subtracting Equation $i$ $(2p + q = -4)$ from Equation $ii$:$(4 + 2\sqrt{3} + 2p + \frac{2p}{\sqrt{3}} + q) - (2p + q) = 0 - (-4)$$4 + 2\sqrt{3} + \frac{2p}{\sqrt{3}} = 4$$2\sqrt{3} + \frac{2p}{\sqrt{3}} = 0$$2\sqrt{3} = -\frac{2p}{\sqrt{3}}$$2p = -2(3) = -6 \Rightarrow p = -3$.Substitute $p = -3$ into Equation $i$:$2(-3) + q + 4 = 0$$-6 + q + 4 = 0 \Rightarrow q = 2$.Therefore,$p^2 + q^2 = (-3)^2 + (2)^2 = 9 + 4 = 13$.Thus,option $A$ is correct.

If $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$ such that $f(0)=f(2)$ and $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$,then $p^2+q^2=$

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