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

The Correct Option is A

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 
\(f(x)={{x}^{2}}+1\)