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

The correct answers are
A f is bounded in $t\in (-\mathrm{\infty },0)$
B number of solutions satisfying the equation $f(t)=e^t$ in [0,2π] is 3
$f\prime (t)=e^t({\cos }^{2}t-\sin 2t)$
Integrate both the sides with respect to t, we get
$f(t)=e^t{\cos }^{2}t+C$
f(0)=1⇒C=0
$\Rightarrow f(t)=e^t{\cos }^{2}t$
$\Rightarrow f(-t)=e^{-t}{\cos }^{2}t$
Clearly, f is neither odd nor even.
As $-\mathrm{\infty }<t<O$
$\Rightarrow O<e^t<1andO\le {\cos }^{2}t\le 1$
$\Rightarrow O<e^t{\cos }^{2}t<1$
Thus, f is bounded in t∈(−∞,0).
Then, $f(t)=e^t$
$\Rightarrow {\cos }^{2}t=1$
⇒t=0,π,2π∈[0,2π]
Therefore, $f(t)=e^t$ has 3 solutions in [0,2π].
Let $\underset{t\to 0}{lim}(f(t){)}^{1/t}=e^L(1^{\infty }$ form)
$L=\underset{t\to 0}{lim}(\frac{e^t{\cos }^{2}t-1}{t})(\frac{0}{0}form)$
Applying L'Hospital rule
$L=\underset{t\to 0}{lim}(\frac{e^t({\cos }^{2}t-\sin 2t)}{1})$
L=1
$\Rightarrow \underset{t\to 0}{lim}(f(t){)}^{1/t}=e$