Q. $f(x)$ is a polynomial of degree 2, $f(0) = 4,f^{'}(0) = 3$ and $f^{''}(0) = 4,\ then\ f( - 1)$ is equal to
A) 3
B) $- 2$
C) 2
D) $- 3$
Explanation
(a)
Let $f(x) = ax^{2} + bx + c$ ….(i)
$\Rightarrow \ \ f^{'}(x) = 2ax + b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$
and ${\ f}^{''}(x) = 2a$
Given, $f(0) = 4,\ f^{'}(0) = 3\ and\ f^{''}(0) = 4,\ we\ get\$
$c = 4,\ \ \ \ b = 3,\ \ \ \ a = 2$
$\therefore\ f(x) = 2x^{2} + 3x + 4\$
$\therefore\ f( - 1) = 2( - 1)^{2} + 3( - 1) + 4 = 3\ \ \ \ \ \ \ \$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.