Question Number 32563 by rahul 19 last updated on 27/Mar/18
$$\boldsymbol{{T}}{he}\:{set}\:{of}\:{values}\:{of}\:'{a}'\:{for}\:{which}\: \\ $$$${all}\:{the}\:{solutions}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{4sin}\:^{\mathrm{4}} {x}+{a}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{3}=\mathrm{0}\:{are}\:{real}\:{and}\: \\ $$$${distinct}\:{is}\:? \\ $$
Commented by rahul 19 last updated on 27/Mar/18
$${i}\:{tried}\:{doing}\:{the}\:{same}\:{procedure} \\ $$$${as}\:{in}\:{Ques}.\:{No}.\:\mathrm{32220}. \\ $$
Commented by rahul 19 last updated on 27/Mar/18
$${Correct}\:{ans}.\:{is}\:\left[−\mathrm{7},−\mathrm{4}\sqrt{\mathrm{3}}\right). \\ $$
Answered by rahul 19 last updated on 28/Mar/18
$${D}\geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} \geqslant\mathrm{48} \\ $$$$\Rightarrow{a}\geqslant\mathrm{4}\sqrt{\mathrm{3}}\:{and}\:{a}\leqslant−\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${Now}\:{let}\:{sin}^{\mathrm{2}} {x}={t} \\ $$$$\Rightarrow\mathrm{4}{t}^{\mathrm{2}} +{at}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right).{f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}×\left(\mathrm{7}+{a}\right)<\mathrm{0} \\ $$$$\Rightarrow{a}<−\mathrm{7}. \\ $$$$\boldsymbol{{W}}{hat}'{s}\:{wrong}\:{with}\:\:{my}\:{solution}? \\ $$
Commented by rahul 19 last updated on 29/Mar/18
$${what}\:{does}\:\vee,\:\wedge\:{means}\:{in}\:{your} \\ $$$${solution}? \\ $$
Commented by MJS last updated on 29/Mar/18
$$\mathrm{I}\:\mathrm{found}\:\mathrm{it}\:\mathrm{was}\:\mathrm{off}\:\mathrm{topic} \\ $$$$\mathrm{look}\:\mathrm{below} \\ $$
Commented by MJS last updated on 29/Mar/18
$$\wedge=``{and}'' \\ $$$$\vee=``{or}'' \\ $$
Answered by MJS last updated on 28/Mar/18
![s^2 +(a/4)s+(3/4)=0 exactly 1 solution: (s−s_0 )^2 =s^2 −2s_0 s+s_0 ^2 −2s_0 =(a/4) ∧ s_0 ^2 =(3/4) (a=4(√3) ∧ s=−((√3)/2)) ∨ (a=−4(√3) ∧ s=((√3)/2)) BUT: t^4 +(a/4)t^2 +(3/4)=0 ⇒ t=(√s) ⇒ s≥0 ⇒ a=−4(√3) ∧ s=((√3)/2) ∧ t=±(((3)^(1/4) (√2))/2) this shows that the equation f(x)=4sin^4 x−4(√3)sin^2 x+3 has exactly 1 solution in [0;(π/2)] with a>−4(√3) it has none with −7≤a<−4(√3) it has 2 with a<−7 it has 1 again a→−∞ ⇒ x_0 →0 but f(0)=3∀a∈R](Q32573.png)
$${s}^{\mathrm{2}} +\frac{{a}}{\mathrm{4}}{s}+\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution}: \\ $$$$\left({s}−{s}_{\mathrm{0}} \right)^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{2}{s}_{\mathrm{0}} {s}+{s}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$−\mathrm{2}{s}_{\mathrm{0}} =\frac{{a}}{\mathrm{4}}\:\wedge\:{s}_{\mathrm{0}} ^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left({a}=\mathrm{4}\sqrt{\mathrm{3}}\:\wedge\:{s}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:\vee\:\left({a}=−\mathrm{4}\sqrt{\mathrm{3}}\:\wedge\:{s}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\mathrm{BUT}: \\ $$$${t}^{\mathrm{4}} +\frac{{a}}{\mathrm{4}}{t}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow\:{t}=\sqrt{{s}}\:\Rightarrow\:{s}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{a}=−\mathrm{4}\sqrt{\mathrm{3}}\:\wedge\:{s}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\wedge\:{t}=\pm\frac{\sqrt[{\mathrm{4}}]{\mathrm{3}}\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{this}\:\mathrm{shows}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation} \\ $$$${f}\left({x}\right)=\mathrm{4sin}^{\mathrm{4}} {x}−\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{3}\:\mathrm{has} \\ $$$$\mathrm{exactly}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{in}\:\left[\mathrm{0};\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{with}\:{a}>−\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{it}\:\mathrm{has}\:\mathrm{none} \\ $$$$\mathrm{with}\:−\mathrm{7}\leqslant{a}<−\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{it}\:\mathrm{has}\:\mathrm{2} \\ $$$$\mathrm{with}\:{a}<−\mathrm{7}\:\mathrm{it}\:\mathrm{has}\:\mathrm{1}\:\mathrm{again} \\ $$$${a}\rightarrow−\infty\:\Rightarrow\:{x}_{\mathrm{0}} \rightarrow\mathrm{0}\:\mathrm{but}\:{f}\left(\mathrm{0}\right)=\mathrm{3}\forall{a}\in\mathbb{R} \\ $$
Commented by rahul 19 last updated on 28/Mar/18
$${sir}\:{pls}\:{check}\:{my}\:{method}. \\ $$
Commented by MJS last updated on 29/Mar/18
![your method is ok but the conclusions might be wrong in qu.32220 we were looking for solutions, here we′re looking for all distinct solutions, so there must be 4 in [0;π] it seems difficult to understand the nature of the connection between the functions f(x)=c_1 sin^4 x+c_2 sin^2 x+c_3 g(s)=c_1 s^4 +c_2 s^2 +c_3 h(t)=c_1 t^2 +c_2 t+c_3 you should draw them or plot them, if possible.](Q32616.png)
$$\mathrm{your}\:\mathrm{method}\:\mathrm{is}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{conclusions}\:\mathrm{might}\:\mathrm{be}\:\mathrm{wrong} \\ $$$$\mathrm{in}\:\mathrm{qu}.\mathrm{32220}\:\mathrm{we}\:\mathrm{were}\:\mathrm{looking} \\ $$$$\mathrm{for}\:\mathrm{solutions},\:\mathrm{here}\:\mathrm{we}'\mathrm{re}\:\mathrm{looking} \\ $$$$\mathrm{for}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{distinct}}\:\mathrm{solutions},\:\mathrm{so}\:\mathrm{there} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{4}\:\mathrm{in}\:\left[\mathrm{0};\pi\right] \\ $$$$\mathrm{it}\:\mathrm{seems}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand} \\ $$$$\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{connection} \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{functions} \\ $$$${f}\left({x}\right)={c}_{\mathrm{1}} \mathrm{sin}^{\mathrm{4}} {x}+{c}_{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}+{c}_{\mathrm{3}} \\ $$$${g}\left({s}\right)={c}_{\mathrm{1}} {s}^{\mathrm{4}} +{c}_{\mathrm{2}} {s}^{\mathrm{2}} +{c}_{\mathrm{3}} \\ $$$${h}\left({t}\right)={c}_{\mathrm{1}} {t}^{\mathrm{2}} +{c}_{\mathrm{2}} {t}+{c}_{\mathrm{3}} \\ $$$$\mathrm{you}\:\mathrm{should}\:\mathrm{draw}\:\mathrm{them}\:\mathrm{or}\:\mathrm{plot} \\ $$$$\mathrm{them},\:\mathrm{if}\:\mathrm{possible}. \\ $$
Commented by MJS last updated on 29/Mar/18
Commented by MJS last updated on 29/Mar/18
Commented by MJS last updated on 29/Mar/18
Commented by MJS last updated on 29/Mar/18
![we′re losing important information: 1. picture 4sin^4 x−4(√3)sin^2 x+3 4sin^4 x−7sin^2 x+3 [(π/4);((3π)/4)] 2. picture 4s^4 −4(√3)s^2 +3 4s^4 −7s^2 +3 [((√2)/2); 1.11^((∗)) ] 3. picture 4t^2 −4(√3)t+3 4t^2 −7t+3 [((√2)/2); 1.02^((∗)) ] (∗) upper borders chosen for symmetry of pics](Q32621.png)
$$\mathrm{we}'\mathrm{re}\:\mathrm{losing}\:\mathrm{important}\:\mathrm{information}: \\ $$$$\mathrm{1}.\:\mathrm{picture} \\ $$$$\mathrm{4sin}^{\mathrm{4}} {x}−\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{3} \\ $$$$\mathrm{4sin}^{\mathrm{4}} {x}−\mathrm{7sin}^{\mathrm{2}} {x}+\mathrm{3} \\ $$$$\left[\frac{\pi}{\mathrm{4}};\frac{\mathrm{3}\pi}{\mathrm{4}}\right] \\ $$$$\mathrm{2}.\:\mathrm{picture} \\ $$$$\mathrm{4}{s}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{3}}{s}^{\mathrm{2}} +\mathrm{3} \\ $$$$\mathrm{4}{s}^{\mathrm{4}} −\mathrm{7}{s}^{\mathrm{2}} +\mathrm{3} \\ $$$$\left[\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:\mathrm{1}.\mathrm{11}^{\left(\ast\right)} \right] \\ $$$$\mathrm{3}.\:\mathrm{picture} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}{t}+\mathrm{3} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} −\mathrm{7}{t}+\mathrm{3} \\ $$$$\left[\frac{\sqrt{\mathrm{2}}}{\mathrm{2}};\:\mathrm{1}.\mathrm{02}^{\left(\ast\right)} \right] \\ $$$$ \\ $$$$\left(\ast\right)\:\mathrm{upper}\:\mathrm{borders}\:\mathrm{chosen}\:\mathrm{for} \\ $$$$\mathrm{symmetry}\:\mathrm{of}\:\mathrm{pics} \\ $$