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Question

# Let f:[0,2]→R a function which is continuous on [0,2] and is differentiable on (0,2) with f(0)=1. Let F(x)=x2∫0f(√t)dt, for x∈[0,2], if F′(x)=f′(x),∀x∈(0,2), then F(2) equals to

A
e21
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B
e41
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C
e1
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D
e4
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Solution

## The correct option is A e4−1From Newton-Leibnitz's formula,ddx[∫ψ(x)ϕ(x)f(t)dt]=f{ψ(x)}{ddxψ(x)}−f{ϕ(x)}{ddxϕ(x)}Given that,F(x)=∫x20f(√t)dt∴F′(x)=f(x)⋅ddx(x2)−f(0)⋅ddx(0)⇒F′(x)=2xf(x)...(i)Also given that, F′(x)=f′(x)⇒2xf(x)=f′(x) [ from (i)]⇒f′(x)f(x)=2xIntegrating both sides w.r.t. x,⇒∫f′(x)f(x)dx=∫2xdx⇒ln{f(x)}=x2+c⇒f(x)=ex2+c⇒f(x)=kex2Now, f(0)=1⇒k=1Hence,f(x)=ex2∴F(2)=∫40etdt=[et]40=e4−1

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