Question

The third derivative of a function f(x) vanishes for all x. If f(0)=1, f'1(1)=2 and f”(1) = – 1, then f(x) is equal to

• A. $(-\frac{3}{2})x^2+3x+9$
• B. $(-\frac{1}{2})x^2-3x+1$
• C. $(-\frac{1}{2})x^2+3x+1$
• D. $(-\frac{3}{2})x^2-7x+2$

Answer 1

The correct option is C $(-\frac{1}{2})x^2+3x+1$

Given, f’’’(x) = 0
On integrating
$f^{\prime \prime }\left(x\right)=c\Rightarrow f^{\prime \prime }\left(1\right)=c\Rightarrow -1=c$ (given)
$\therefore$ f’’(x) = – 1
f’(x) = – x + d
$\because$ f’(1) = – 1 + d = 2
$\therefore$ d = 3
f’(x) = 3 – x
or $f\left(x\right)=-\frac{x^2}{2}+3x+e\because f\left(0\right)=0+0+e=1\therefore e=1$
Hence, $f\left(x\right)=-\frac{x^2}{2}+3x+1$