Question

Let $f$ be a differentiable function on $R$ satisfying $f^{\prime }\left(t\right)=e^t({\cos }^{2}t-\sin 2t)$ and $f\left(0\right)=1,$ then which of the following is/are correct ?

• A. $f$ is bounded in $t\in (-\infty ,0)$
• B. number of solutions satisfying the equation $f\left(t\right)=e^t$ in $[0,2\pi ]$ is $3$
• C. $\underset{t\to 0}{lim}(f\left(t\right){)}^{1/t}=1$
• D. $f$ is an even function

Answer 1

Answer: (A), (B)

The correct options are
A $f$ is bounded in $t\in (-\infty ,0)$
B number of solutions satisfying the equation $f\left(t\right)=e^t$ in $[0,2\pi ]$ is $3$

$f^{\prime }\left(t\right)=e^t({\cos }^{2}t-\sin 2t)$
Integrate both the sides with respect to $t,$ we get
$f\left(t\right)=e^t{\cos }^{2}t+C$
$f\left(0\right)=1\Rightarrow C=0$
$\Rightarrow f\left(t\right)=e^t{\cos }^{2}t$
$\Rightarrow f(-t)=e^{-t}{\cos }^{2}t$

Clearly, $f$ is neither odd nor even.
As $-\infty <t<0$
$\Rightarrow 0<e^t<1$ and $0\le {\cos }^{2}t\le 1$
$\Rightarrow 0\le e^t{\cos }^{2}t<1$
$\therefore f$ is bounded in $t\in (-\infty ,0).$

Now, $f\left(t\right)=e^t$
$\Rightarrow {\cos }^{2}t=1$
$\Rightarrow t=0,\pi ,2\pi \in [0,2\pi ]$
Hence, $f\left(t\right)=e^t$ has $3$ solutions in $[0,2\pi ].$

Let $\underset{t\to 0}{lim}(f\left(t\right){)}^{1/t}=e^L\left(1^{\infty }\text{form}\right)$
$L=\underset{t\to 0}{lim}\left(\frac{e^t{\cos }^{2}t-1}{t}\right)\left(\frac{0}{0}\text{form}\right)$
Applying L'Hospital rule
$L=\underset{t\to 0}{lim}\left(\frac{e^t({\cos }^{2}t-\sin 2t)}{1}\right)$
$L=1$
$\Rightarrow \underset{t\to 0}{lim}(f\left(t\right){)}^{1/t}=e$