Lizeth Herring

Answered

2022-12-27

Let f be a differentiable function on R satisfying $f\prime (t)={e}^{t}({\mathrm{cos}}^{2}t-sin2t)$ and f(0)=1, then which of the following is/are correct ?

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

C) $\underset{t\to 0}{lim}f(t){)}^{1/t}=1$

D) f is an even function

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

C) $\underset{t\to 0}{lim}f(t){)}^{1/t}=1$

D) f is an even function

Answer & Explanation

Mikayla Cox

Expert

2022-12-28Added 6 answers

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

$f\prime (t)={e}^{t}({\mathrm{cos}}^{2}t-\mathrm{sin}2t)$

Integrate both the sides with respect to t, we get

$f(t)={e}^{t}{\mathrm{cos}}^{2}t+C$

f(0)=1⇒C=0

$\Rightarrow f(t)={e}^{t}{\mathrm{cos}}^{2}t$

$\Rightarrow f(-t)={e}^{-t}{\mathrm{cos}}^{2}t$

Clearly, f is neither odd nor even.

As $-\mathrm{\infty}<t<O$

$\Rightarrow O<{e}^{t}<1and\text{}O\le {\mathrm{cos}}^{2}t\le 1$

$\Rightarrow O<{e}^{t}{\mathrm{cos}}^{2}t<1$

Thus, f is bounded in t∈(−∞,0).

Then, $f(t)={e}^{t}$

$\Rightarrow {\mathrm{cos}}^{2}t=1$

⇒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)

$L=\underset{t\to 0}{lim}(\frac{{e}^{t}{\mathrm{cos}}^{2}t-1}{t})(\frac{0}{0}form)$

Applying L'Hospital rule

$L=\underset{t\to 0}{lim}(\frac{{e}^{t}({\mathrm{cos}}^{2}t-\mathrm{sin}2t)}{1})$

L=1

$\Rightarrow \underset{t\to 0}{lim}(f(t){)}^{1/t}=e$

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

$f\prime (t)={e}^{t}({\mathrm{cos}}^{2}t-\mathrm{sin}2t)$

Integrate both the sides with respect to t, we get

$f(t)={e}^{t}{\mathrm{cos}}^{2}t+C$

f(0)=1⇒C=0

$\Rightarrow f(t)={e}^{t}{\mathrm{cos}}^{2}t$

$\Rightarrow f(-t)={e}^{-t}{\mathrm{cos}}^{2}t$

Clearly, f is neither odd nor even.

As $-\mathrm{\infty}<t<O$

$\Rightarrow O<{e}^{t}<1and\text{}O\le {\mathrm{cos}}^{2}t\le 1$

$\Rightarrow O<{e}^{t}{\mathrm{cos}}^{2}t<1$

Thus, f is bounded in t∈(−∞,0).

Then, $f(t)={e}^{t}$

$\Rightarrow {\mathrm{cos}}^{2}t=1$

⇒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)

$L=\underset{t\to 0}{lim}(\frac{{e}^{t}{\mathrm{cos}}^{2}t-1}{t})(\frac{0}{0}form)$

Applying L'Hospital rule

$L=\underset{t\to 0}{lim}(\frac{{e}^{t}({\mathrm{cos}}^{2}t-\mathrm{sin}2t)}{1})$

L=1

$\Rightarrow \underset{t\to 0}{lim}(f(t){)}^{1/t}=e$

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