y=f(x) is a function which satisfies (i)f(0)=0 (ii) f′′(x)=f′(x) and (iii) f′(0)=1 then the area bounded by the graph of y=f(x) the lines x=0,x−1=0 and y+1=0 is
A
e
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B
e−2
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C
e−1
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D
e+1
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Solution
The correct option is Be−1 If f′(x)=x+c⇒f′(x)=kex ⇒f′(x)=ex{f′(0)=1⇒k=1} ⇒f(x)=ex+λ ⇒f(x)=ex−1{f(0)=0⇒λ=−1} A=∫10(ex−1+1)dx=e−1