Let f be a twice differentiable function defined on R such that f(0)=1,f′(0)=2 and f′(x)≠0 for all x∈R. If ∣∣∣f(x)f′(x)f′(x)f′′(x)∣∣∣=0, for all x∈R, then the value of f(1) lies in the interval
A
(9,12)
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B
(6,9)
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C
(3,6)
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D
(0,3)
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Solution
The correct option is B(6,9) Given f(x)f′′(x)−(f′(x))2=0
Let h(x)=f(x)f′(x)
Then h′(x)=0⇒h(x)=k ⇒f(x)f′(x)=k ⇒f(x)=kf′(x) ⇒f(0)=kf′(0) ⇒k=12
Now, f(x)=12f′(x) ⇒∫2dx=∫f′(x)f(x)dx ⇒2x=ln|f(x)|+C
As f(0)=1⇒C=0 ⇒2x=ln|f(x)| ⇒f(x)=±e2x
As f(0)=1⇒f(x)=e2x ∴f(1)=e2≈7.38