Question-57959

Question Number 57959 by rahul 19 last updated on 15/Apr/19

Answered by MJS last updated on 16/Apr/19

3sin^(−1) x =y ⇒ x=sin (y/3) with x∈[−1; 1] ∧ y∈[−((3π)/2); ((3π)/2)] 3sin^(−1) x increases for −1≤x≤1 sin^(−1) (−4x^3 +3x) =y ⇒ −4x^3 +3x=sin y with (−4x^3 +3x)∈[−1; 1] ∧ y∈[−(π/2); (π/2)] (−4x^3 +3x)∈[−1; 1] ⇒ x∈[−1; 1] but −4x^3 +3x has a minimum at (((−(1/2))),((−1)) ) and a maximum at (((1/2)),(1) ) ⇒ ⇒ sin^(−1) (−4x^3 +3x) decreases for −1≤x<−(1/2) and (1/2)<x≤1 and increases for −(1/2)<x<(1/2) ⇒ answer is (a)

$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:={y}\:\Rightarrow\:{x}=\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$$\mathrm{with}\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\mathrm{3}\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:\mathrm{increases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:={y}\:\Rightarrow\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}=\mathrm{sin}\:{y} \\ $$$$\mathrm{with}\:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\pi}{\mathrm{2}};\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\Rightarrow\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$$$\mathrm{but}\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\:\mathrm{has}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{1}}\end{pmatrix}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:\mathrm{decreases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}\:\mathrm{and}\:\mathrm{increases}\:\mathrm{for}\:−\frac{\mathrm{1}}{\mathrm{2}}<{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\left(\mathrm{a}\right) \\ $$

Commented by MJS last updated on 16/Apr/19

solving −4x^3 +3x=sin y x^3 −(3/4)x+(1/4)sin y =0 D=(p^3 /(27))+(q^2 /4)=(((−(3/4))^3 )/(27))+((((1/4)sin y)^2 )/4)=((−1+sin^2 y)/(64))= =−((cos^2 y)/(64)) ≤0 ⇒ 3 real solutions ⇒ we need the trigonometric method, which gives x_1 =sin (y/3) x_2 =(1/2)((√3)cos (y/3) −sin (y/3)) x_3 =−(1/2)((√3)cos (y/3) +sin (y/3)) ⇒ 3sin^(−1) x =sin^(−1) (−4x^3 +3x) in certain cases

$$\mathrm{solving}\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}=\mathrm{sin}\:{y} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{y}\:=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} }{\mathrm{27}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{y}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{−\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{y}}{\mathrm{64}}= \\ $$$$=−\frac{\mathrm{cos}^{\mathrm{2}} \:{y}}{\mathrm{64}}\:\leqslant\mathrm{0}\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{method},\:\mathrm{which}\:\mathrm{gives} \\ $$$${x}_{\mathrm{1}} =\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{{y}}{\mathrm{3}}\:−\mathrm{sin}\:\frac{{y}}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{{y}}{\mathrm{3}}\:+\mathrm{sin}\:\frac{{y}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\mathrm{3sin}^{−\mathrm{1}} \:{x}\:=\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:\mathrm{in}\:\mathrm{certain}\:\mathrm{cases} \\ $$

Commented by rahul 19 last updated on 17/Apr/19

$${Thank}\:{you}\:{sir}. \\ $$

Commented by MJS last updated on 17/Apr/19

you′re welcome. I hope my answer is helpful I′m not 100% sure if this is a legal proof...

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}.\:\mathrm{I}\:\mathrm{hope}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{helpful} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{legal}\:\mathrm{proof}… \\ $$

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