Question

If f : $R\to R$ is a twice differentiable function such that $|f"\left(x\right)|\le 1$; and f(0)=0=f'(0). Then which of the following CANNOT be true.

• A. $f\left(1\right)=\frac{1}{3}$
  
• B. f(-3)=-1
  
• C. $f\left(\frac{1}{2}\right)=-\frac{1}{9}$
  
• D. f(-5)=-13

Answer 1

The correct option is D f(-5)=-13

$-1\le f"\left(x\right)\le 1$
Integrate both sides of inequality from 0 to x
$–x\le f\left(x\right)\le x$ [using f ¢(0) = 0]
Again integrate on both sides from 0 to x
$-\frac{x^2}{2}\le f\left(x\right)\le \frac{x^2}{2}$ [using f(0)=0]
Only (D) does not lie in required range as \(–12.5 \leq f(–5) \leq 12.5\)