Let f(0)=0 and 2∫0f′(2t)ef(2t)dt=5, then the value of f(4) is:
A
2ln3
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B
ln10
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C
ln11
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D
2ln2
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Solution
The correct option is Cln11 We have 2∫0f′(2t)ef(2t)dt=5
Put ef(2t)=y⇒2f′(2t)ef(2t)dt=dy
Now 12ef(4)∫ef(0)dy=5⇒ef(4)∫ef(0)dy=10 ⇒ef(4)−ef(0)=10⇒ef(4)=10+e0=11
Hence f(4)=ln11.