Let f be a differentiable function on R satisfying f ′ ( t ) =...

Lizeth Herring

Lizeth Herring

Answered

2022-12-27

Let f be a differentiable function on R satisfying f ( t ) = e t ( cos 2 t s i n 2 t ) and f(0)=1, then which of the following is/are correct ?
A) f is bounded in t ( , 0 )
B) number of solutions satisfying the equation f ( t ) = e t in [0,2π] is 3
C) lim t 0 f ( t ) ) 1 / t = 1
D) f is an even function

Answer & Explanation

Mikayla Cox

Mikayla Cox

Expert

2022-12-28Added 6 answers

The correct answers are
A f is bounded in t ( , 0 )
B number of solutions satisfying the equation f ( t ) = e t in [0,2π] is 3
f ( t ) = e t ( cos 2 t sin 2 t )
Integrate both the sides with respect to t, we get
f ( t ) = e t cos 2 t + C
f(0)=1⇒C=0
f ( t ) = e t cos 2 t
f ( t ) = e t cos 2 t
Clearly, f is neither odd nor even.
As < t < O
O < e t < 1 a n d   O cos 2 t 1
O < e t cos 2 t < 1
Thus, f is bounded in t∈(−∞,0).
Then, f ( t ) = e t
cos 2 t = 1
⇒t=0,π,2π∈[0,2π]
Therefore, f ( t ) = e t has 3 solutions in [0,2π].
Let lim t 0 ( f ( t ) ) 1 / t = e L ( 1 form)
L = lim t 0 ( e t cos 2 t 1 t ) ( 0 0 f o r m )
Applying L'Hospital rule
L = lim t 0 ( e t ( cos 2 t sin 2 t ) 1 )
L=1
lim t 0 ( f ( t ) ) 1 / t = e

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