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 $p^2+q^2=$

• Correct answer
  A

  13
• B

  5
• C

  $2+\frac{1}{\sqrt{3}}$
• D

  1

Solution

Given, $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$ $$ \begin{array}{ll} f(0)=f(2) & \\ & 0=2^3+p(2)^2+q(2) \\ \Rightarrow & 4 p+2 q+8=0 \end{array} $$
$$ \begin{aligned} \text { Now } f^{\prime}(x) & =\frac{d}{d x}\left(x^3+p x^2+q x\right) \\ f^{\prime}(x) & =3 x^2+2 p x+q \end{aligned} $$ Given, $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$ $$ \begin{aligned} & 3\left(1+\frac{1}{\sqrt{3}}\right)^2+2 p\left(1+\frac{1}{\sqrt{3}}\right)+q=0 \\ & \therefore 2 p\left(1+\frac{1}{\sqrt{3}}\right)+q+3\left(1+\frac{1}{3}+\frac{2}{\sqrt{3}}\right)=0 \end{aligned} $$
Eq. (ii) Subtracting by Eq. (i), we get $$ \begin{aligned} & \Rightarrow \quad \frac{2 p}{\sqrt{3}}+\frac{6}{\sqrt{3}}=0 \\ & \Rightarrow \quad 2 P+6=0 \Rightarrow P=-3 \end{aligned} $$ Substituting $P=-3$ in Eq. (i) $$ \begin{aligned} & \quad 2(-3)+q+4=0 \\ & \Rightarrow \quad q=2 \\ & \therefore p^2+q^2=(-3)^2+(2)^2=9+4 \\ & p^2+q^2=13 \end{aligned} $$ Hence, option (a) is correct.

Hence, the correct answer is option: 'A'.