Question

If $k_1\le {\sin }^2\theta +{\cos }^4\theta \le k_2$, then the minimum value of $\cos 2\theta +4\cos \theta +4$ is
(1) k₁ + ​k₂ (2) 2k₁ (3) 2k₂ (4) ​k₂

Answer 1

Dear Student,

$$\begin{gathered} k_1\le {\sin }^2\theta +{\cos }^4\theta \le k_2 \\ \Rightarrow k_1\le 1-{\cos }^2\theta +{\cos }^4\theta \le k_2 \\ \Rightarrow k_1\le \frac{3}{4}+\frac{1}{4}-2.\frac{1}{2}{\cos }^2\theta +{\cos }^4\theta \le k_2 \\ \Rightarrow k_1\le \frac{3}{4}+({\cos }^2\theta -\frac{1}{2}{)}^2\le k_2 \\ Theminimumvalueof\frac{3}{4}+({\cos }^2\theta -\frac{1}{2}{)}^{2}iswhenthesquaretermiszero. \\ So,minimumvalueis\frac{3}{4}. \\ \Rightarrow k_1=\frac{3}{4}. \\ Themaximumvalueof\frac{3}{4}+({\cos }^2\theta -\frac{1}{2}{)}^{2}iswhen{\cos }^2\theta ismaximum. \\ Maxvalueof{\cos }^2\theta =1,Somaximumvalueof\frac{3}{4}+({\cos }^2\theta -\frac{1}{2}{)}^2=\frac{3}{4}+\frac{1}{4}=1 \\ \Rightarrow k_2=1 \\ Now, \\ \cos 2\theta +4\cos \theta +4=2{\cos }^2\theta -1+4\cos \theta +4==2.(\cos \theta +1{)}^2+1 \\ Theaboveexpressionattainsaminimumwhensquaretermiszero. \\ Therefore,theminimumvalueis1whichisequalto{k}_2. \end{gathered}$$

Regards,