Answer :
Answer:
The absolute maximum is [tex]\frac{3\sqrt 3}2[/tex] and the absolute minimum value is [tex]0.[/tex]
Step-by-step explanation:
Differentiate of [tex]f[/tex] both sides w.r.t. [tex]t[/tex],
[tex]f(t)=2 \cos t+\sin 2t[/tex]
[tex]\Rightarrow f'(t)=-2\sin t+2\cos 2t[/tex]
Now take [tex]f'(t)=0[/tex]
[tex]\Rightarrow -2\sin t+2\cos 2t=0[/tex]
[tex]\Rightarrow 2\cos 2t=2\sin t[/tex]
[tex]\Rightarrow \cos 2t=\sin t[/tex]
[tex]\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t][/tex]
[tex]\Rightarrow 2\sin ^2t+\sin t-1=0[/tex]
[tex]\Rightarrow 2\sin ^2t+2\sin t-\sin t-1=0[/tex]
[tex]\Rightarrow 2\sin t(\sin t+1)-1(\sin t+1)=0[/tex]
[tex]\Rightarrow (\sin t+1)(2\sin t-1)=0[/tex]
[tex]\Rightarrow \sin t+1=0 \;\text{and}\; 2\sin t-1=0[/tex]
[tex]\Rightarrow \sin t =-1 \;\text{and}\; \sin t =\frac 12[/tex]
In the interval [tex]0\leq t\leq \frac {\pi}2[/tex], the answer to this problem is [tex]\frac {\pi}6[/tex]
Now find the second derivative of [tex]f(t)[/tex] w.r.t. [tex]t[/tex],
[tex]f''(t)=-2\cos t-4\sin 2t[/tex]
[tex]\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3[/tex]
Thus, [tex]f(t)[/tex] is maximum at [tex]t=\frac {\pi}6[/tex] and minimum at [tex]t=0[/tex]
[tex]\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0[/tex]
Hence, the absolute maximum is [tex]\frac{3\sqrt 3}2[/tex] and the absolute minimum value is [tex]0[/tex].