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Let $f(x)=a x^3+b x^2+c x+41$ be such that $f(1)=40, f^{\prime}(1)=2$ and $f^{\prime}(1)=4$. Then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to:
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Verified Answer
The correct answer is:
51
$\begin{aligned} & f(x)=a x^3+b x^2+c x+41 \\ & f^{\prime}(x)=3 a x^2+2 b x+c x \\ & \Rightarrow f^{\prime}(1)=3 a+2 b+c=2 \ldots(1)\end{aligned}$
$\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{n})=6 \mathrm{ax}+2 \mathrm{~b} \\ & \Rightarrow \mathrm{f}^{\prime \prime}(1)=6 \mathrm{a}+2 \mathrm{~b}=4 \\ & 3 \mathrm{a}+\mathrm{b}=2 \ldots \ldots \ldots .(2)\end{aligned}$
$\begin{aligned} & (1)-(2) \\ & b+c=0...(3)\end{aligned}$
$\begin{aligned}
& \mathrm{f}(1)=40 \\
& \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40
\end{aligned}$
use (3)
$a+41=40$
by (2)
$\begin{aligned}
& -3+b=2 \Rightarrow b=5 \& c=-5 \\
& a^2+b^2+c^2=1+25+25=51
\end{aligned}$
$\begin{aligned} & \mathrm{f}^{\prime \prime}(\mathrm{n})=6 \mathrm{ax}+2 \mathrm{~b} \\ & \Rightarrow \mathrm{f}^{\prime \prime}(1)=6 \mathrm{a}+2 \mathrm{~b}=4 \\ & 3 \mathrm{a}+\mathrm{b}=2 \ldots \ldots \ldots .(2)\end{aligned}$
$\begin{aligned} & (1)-(2) \\ & b+c=0...(3)\end{aligned}$
$\begin{aligned}
& \mathrm{f}(1)=40 \\
& \mathrm{a}+\mathrm{b}+\mathrm{c}+41=40
\end{aligned}$
use (3)
$a+41=40$
by (2)
$\begin{aligned}
& -3+b=2 \Rightarrow b=5 \& c=-5 \\
& a^2+b^2+c^2=1+25+25=51
\end{aligned}$
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