Let f (x) = ax^3 + bx^2 + ex + 41 be such that f(1) = 40, f'(1) = 2 and f''(1) = 4.

Question

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

Then \(a^{2}+b^{2}+c^{2}\) is equal to :

(1) 62

(2) 73

(3) 54

(4) 51

Answer 1

Correct option is : (4) 51 

\( f(x)=a x^{3}+b x^{2}+c x+41\)

\(f^{\prime}(x)=3 a x^{2}+2 b x+c x\)

\(\Rightarrow \mathrm{f}^{\prime}(1)=3 \mathrm{a}+2 \mathrm{b}+\mathrm{c}=2\)  ..........(1)

\(f^{\prime \prime}(n)=6 a x+2 b\)

\(\Rightarrow f^{\prime \prime}(1)=6 a+2 b=4\)

\(3 a+b=2\) .........(2)

(1) - (2)

\(\mathrm{b}+\mathrm{c}=0\)  ..........(3)

\(\mathrm{f}(1)=40\)

\(a+b+c+41=40\)

use (3)

\(a+41=40\)

by (2)

\(-3+b=2 \Rightarrow b=5 \ \&\ c=-5\)

\(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1+25+25=51\)