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Question

# Let f:[0,2]→R be continuous on [0,2] and twice differentiable in (0,2). If f(0)=0, f(1)=1 and f(2)=1, then

A
f(c)=13 for at least one c(0,2)
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B
2f(c)+2c=3 for at least one c(0,1)
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C
2f(c)+2c=3 for at least one c(0,2)
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D
f′′(c)=1 for at least one c(0,2)
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Solution

## The correct option is D f′′(c)=−1 for at least one c∈(0,2)→ By LMVT in the interval [0,1], f(1)−f(0)1=f′(c1)=1 for some c1∈(0,1) Similarly, f(2)−f(1)1=f′(c2)=0 for some c2∈(1,2) Since 13∈(f′(c2),f′(c1)) and f′ is continuous, ⇒f′(c)=13 for at least one c∈(0,2) by IVT. → Consider a function, H(x)=2f(x)+x2 H(0)=0,H(1)=3,H(2)=6 By LMVT in [0,1], H′(c)=2f′(c)+2c=H(1)−H(0)1=3 for some c∈(0,1) → By LMVT in [0,2], H′(c)=2f′(c)+2c=H(2)−H(0)2=3 for some c∈(0,2) → Applying Rolle's theorem to H′(x) in [c1,c2] where 0≤c1<c2≤2 H′(c1)=2f′(c1)+2c1=H(1)−H(0)1=3 H′(c2)=2f′(c2)+2c2=H(2)−H(1)2−1=3 H′′(c)=0 for some c∈(0,2) ⇒2f′′(c)+2=0 ⇒f′′(c)=−1

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