Let $f(x)=a x^3+b x^2+c x+41$ be such that $f(1)=40, f^{\prime}(1)=2$ and $f^{\prime \prime}(1)=4$. Then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to:

Question

Aicharya · Accepted Answer

$\begin{aligned} & f(x)=a x^3+b x^2+c x+41 \ & f^{\prime}(x)=3 a x^2+2 b x+c x \ & \Rightarrow f^{\prime}(1)=3 a+2 b+c=2 \ldots(1)\end{aligned}$ $\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{n})=6 \mathrm{ax}+2 \mathrm{~b} \ & \Rightarrow \mathrm{f}^{\prime \prime}(1)=6 \mathrm{a}+2 \mathrm{~b}=4 \ & 3 \mathrm{a}+\mathrm{b}=2 \ldots \ldots \ldots .(2)\end{aligned}$ $\begin{aligned} & (1)-(2) \ & b+c=0...(3)\end{aligned}$ $\begin{aligned} & \mathrm{f}(1)=40 \ & \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40 \end{aligned}$ use (3) $a+41=40$ by (2) $\begin{aligned} & -3+b=2 \Rightarrow b=5 \& c=-5 \ & a^2+b^2+c^2=1+25+25=51 \end{aligned}$

Let $f(x)=a x^3+b x^2+c x+41$ be such that $f(1)=40, f^{\prime}(1)=2$ and $f^{\prime \prime}(1)=4$. Then…

Solution