The correct option is B number of solutions satisfying the equation f(t)=et in [0,2π] is 3
f′(t)=et(cos2t−sin2t)
Integrate both the sides with respect to t, we get
f(t)=etcos2t+C
f(0)=1⇒C=0
⇒f(t)=etcos2t
⇒f(−t)=e−tcos2t
Clearly, f is neither odd nor even.
As −∞<t<0
⇒0<et<1 and 0≤cos2t≤1
⇒0≤etcos2t<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 limt→0(f(t))1/t=eL (1∞ form)
L=limt→0(etcos2t−1t) (00 form)
Applying L'Hospital rule
L=limt→0(et(cos2t−sin2t)1)
L=1
⇒limt→0(f(t))1/t=e