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Question

Let f be a differentiable function on R satisfying f(t)=et(cos2tsin2t) and f(0)=1, then which of the following is/are correct ?

A
f is bounded in t(,0)
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B
number of solutions satisfying the equation f(t)=et in [0,2π] is 3
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C
limt0(f(t))1/t=1
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D
f is an even function
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Solution

The correct option is B number of solutions satisfying the equation f(t)=et in [0,2π] is 3
f(t)=et(cos2tsin2t)
Integrate both the sides with respect to t, we get
f(t)=etcos2t+C
f(0)=1C=0
f(t)=etcos2t
f(t)=etcos2t

Clearly, f is neither odd nor even.
As <t<0
0<et<1 and 0cos2t1
0etcos2t<1
f is bounded in t(,0).

Now, f(t)=et
cos2t=1
t=0,π,2π[0,2π]
Hence, f(t)=et has 3 solutions in [0,2π].

Let limt0(f(t))1/t=eL (1 form)
L=limt0(etcos2t1t) (00 form)
Applying L'Hospital rule
L=limt0(et(cos2tsin2t)1)
L=1
limt0(f(t))1/t=e

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