Lizeth Herring

2022-12-27

Let f be a differentiable function on R satisfying $f\prime \left(t\right)={e}^{t}\left({\mathrm{cos}}^{2}t-sin2t\right)$ and f(0)=1, then which of the following is/are correct ?
A) f is bounded in $t\in \left(-\mathrm{\infty },0\right)$
B) number of solutions satisfying the equation $f\left(t\right)={e}^{t}$ in [0,2π] is 3
C) $\underset{t\to 0}{lim}f\left(t\right){\right)}^{1/t}=1$
D) f is an even function

Mikayla Cox

Expert

A f is bounded in $t\in \left(-\mathrm{\infty },0\right)$
B number of solutions satisfying the equation $f\left(t\right)={e}^{t}$ in [0,2π] is 3
$f\prime \left(t\right)={e}^{t}\left({\mathrm{cos}}^{2}t-\mathrm{sin}2t\right)$
Integrate both the sides with respect to t, we get
$f\left(t\right)={e}^{t}{\mathrm{cos}}^{2}t+C$
f(0)=1⇒C=0
$⇒f\left(t\right)={e}^{t}{\mathrm{cos}}^{2}t$
$⇒f\left(-t\right)={e}^{-t}{\mathrm{cos}}^{2}t$
Clearly, f is neither odd nor even.
As $-\mathrm{\infty }

$⇒O<{e}^{t}{\mathrm{cos}}^{2}t<1$
Thus, f is bounded in t∈(−∞,0).
Then, $f\left(t\right)={e}^{t}$
$⇒{\mathrm{cos}}^{2}t=1$
⇒t=0,π,2π∈[0,2π]
Therefore, $f\left(t\right)={e}^{t}$ has 3 solutions in [0,2π].
Let $\underset{t\to 0}{lim}\left(f\left(t\right){\right)}^{1/t}={e}^{L}\left({1}^{\infty }$ form)
$L=\underset{t\to 0}{lim}\left(\frac{{e}^{t}{\mathrm{cos}}^{2}t-1}{t}\right)\left(\frac{0}{0}form\right)$
Applying L'Hospital rule
$L=\underset{t\to 0}{lim}\left(\frac{{e}^{t}\left({\mathrm{cos}}^{2}t-\mathrm{sin}2t\right)}{1}\right)$
L=1
$⇒\underset{t\to 0}{lim}\left(f\left(t\right){\right)}^{1/t}=e$

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