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Question Number 146876 by gsk2684 last updated on 16/Jul/21
if the maximum value of 4sin^2 x+3cos^2 x+sin (x/2)+cos (x/2)+3 is a+(√b) then find a+b
$${if}\:{the}\:{maximum}\:{value}\:{of}\: \\ $$$$\mathrm{4sin}\:^{\mathrm{2}} {x}+\mathrm{3cos}\:^{\mathrm{2}} {x}+\mathrm{sin}\:\frac{{x}}{\mathrm{2}}+\mathrm{cos}\:\frac{{x}}{\mathrm{2}}+\mathrm{3} \\ $$$${is}\:{a}+\sqrt{{b}}\:{then}\:{find}\:{a}+{b} \\ $$
Answered by liberty last updated on 16/Jul/21
f(x)=4sin^2 x+3−3sin^2 x+3+sin ((x/2))+cos ((x/2)) f(x)=sin^2 x+sin ((x/2))+cos ((x/2))+6 f(x)=sin^2 x+sin ((x/2))+cos ((x/2))+6 f ′(x)=sin 2x+(1/2)cos ((x/2))−(1/2)sin ((x/2))=0 sin 2x = (1/2)sin ((x/2))−(1/2)cos ((x/2)) 2sin 2x −sin ((x/2))+cos ((x/2))=0 let (x/2)=u →x=2u 2sin 4u+(cos u−sin u)=0 4sin 2u (cos^2 u−sin^2 u)+(cos u−sin u)=0 (cos u−sin u){4sin 2u(cos u+sin u)+1}=0 (•) cos u−sin u=0 cos u=sin u ⇒tan ((x/2))=1 x=(π/2) f((π/2))= 1+6 +((√2)/2)+((√2)/2)=7+(√2) f_(max) = 7+(√2) ≈ 8.414213 then { ((a=7)),((b=2)) :} ⇒a+b=9
$${f}\left({x}\right)=\mathrm{4sin}\:^{\mathrm{2}} {x}+\mathrm{3}−\mathrm{3sin}\:^{\mathrm{2}} {x}+\mathrm{3}+\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)=\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{6} \\ $$$${f}\left({x}\right)=\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{6} \\ $$$${f}\:'\left({x}\right)=\mathrm{sin}\:\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{2sin}\:\mathrm{2}{x}\:−\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${let}\:\frac{{x}}{\mathrm{2}}={u}\:\rightarrow{x}=\mathrm{2}{u} \\ $$$$\mathrm{2sin}\:\mathrm{4}{u}+\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)=\mathrm{0} \\ $$$$\mathrm{4sin}\:\mathrm{2}{u}\:\left(\mathrm{cos}\:^{\mathrm{2}} {u}−\mathrm{sin}\:^{\mathrm{2}} {u}\right)+\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:{u}−\mathrm{sin}\:{u}\right)\left\{\mathrm{4sin}\:\mathrm{2}{u}\left(\mathrm{cos}\:{u}+\mathrm{sin}\:{u}\right)+\mathrm{1}\right\}=\mathrm{0} \\ $$$$\left(\bullet\right)\:\mathrm{cos}\:{u}−\mathrm{sin}\:{u}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}\:{u}=\mathrm{sin}\:{u}\:\Rightarrow\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{x}=\frac{\pi}{\mathrm{2}} \\ $$$$\:{f}\left(\frac{\pi}{\mathrm{2}}\right)=\:\mathrm{1}+\mathrm{6}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{7}+\sqrt{\mathrm{2}} \\ $$$${f}_{{max}} =\:\mathrm{7}+\sqrt{\mathrm{2}}\:\approx\:\mathrm{8}.\mathrm{414213} \\ $$$$\:{then}\:\begin{cases}{{a}=\mathrm{7}}\\{{b}=\mathrm{2}}\end{cases}\:\Rightarrow{a}+{b}=\mathrm{9} \\ $$
Commented by gsk2684 last updated on 16/Jul/21
thank you
$${thank}\:{you}\: \\ $$
Commented by liberty last updated on 16/Jul/21

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