1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# 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)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(6,9)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(3,6)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(0,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
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) ⇒∫2 dx=∫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

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Methods of Solving First Order, First Degree Differential Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program