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 \(a ^2+ b…

Given function \(f(x) = ax^3 + bx^2 + cx + 41\)
\({f}^{'}(x)=3a{x}^{2}+2bx+c\)
\(\Rightarrow {f}^{'}(1)=3a+2b+c=2\ldots (i)\)
\({f}^{"}(x)=6ax+2b\)
\(\Rightarrow f^{\prime \prime}(1)=6 a+2 b=4\)

\(3 a+b=2 \ldots(i i)\)

\((i)-(i i) \Rightarrow b+c=0 \ldots(i i i)\)

\(f(1)=40 \Rightarrow a+b+c+41=40\)
use (iii) use (iii) \(\Rightarrow a+41=40 \Rightarrow a=-1\)
by (ii) \(\Rightarrow-3+b=2 \Rightarrow b=5\) and \( c=-5\)
\(a^2 + b^2 + c^2 = 1 + 25 + 25 = 51\)