Question

# If f : R→R is a twice differentiable function such that |f"(x)|≤1; and f(0)=0=f'(0). Then which of the following CANNOT be true.

A
f(1)=13
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B
f(-3)=-1
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C
f(12)=19
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D
f(-5)=-13
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Solution

## The correct option is A f(-5)=-13−1≤f"(x)≤1 Integrate both sides of inequality from 0 to x –x≤f(x)≤x [using f ¢(0) = 0] Again integrate on both sides from 0 to x −x22≤f(x)≤x22 [using f(0)=0] Only (D) does not lie in required range as $$–12.5 \leq f(–5) \leq 12.5$$

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