Question

Let $f:R\to R$ be a function such that the third derivative of $f(x)$ vanishes for all $x$. If $f(0)=1,\,f'(2)=4$ and $f''(1)=2,$ then $f(x)$ equals to

• A. ${{x}^{2}}+1$
• B. ${{x}^{2}}+2x+1$
• C. $4x+1$
• D. ${{x}^{2}}-2x+1$

Answer 1

Answer: (A)

${{x}^{2}}+1$

Answer 2

The correct answer is A:$x^2+1$
Let $f(x)=a{{x}^{2}}+bx+c$ ..(i)
Also, given $f(0)=1$
$\Rightarrow$ $1=a\,{{(0)}^{2}}+b(0)+c$
$\Rightarrow$ $c=1$ ..(ii)
and $f'(2)=4$
on differentiating E (i), w. r. t. x, we get
$f'(x)=2ax+b$ ..(iii)
$\Rightarrow$ $f'(2)=2a\,(2)+b$
$\Rightarrow$ $4=4a+b$ ..(iv)
Again, differentiating E (iii) 0, we get
$f''(x)=2a$ But $f''(1)=2$ (given)
$\therefore$ $f''(1)=2a$
$\Rightarrow$ $2=2a$
$\Rightarrow$ $a=1$
On putting $a=1$ in E (iv), we get
$4=4+b$
$\Rightarrow$ $b=0$
On putting $a=1,\,b=0$ and $c=1$
in E (i), we get
$f(x)={{x}^{2}}+1$