Question Number 32220 by rahul 19 last updated on 21/Mar/18
$${Find}\:{the}\:{value}\:{of}\:{a}\:{for}\:{which}\:{the}\:{equation} \\ $$$$\mathrm{sin}\:^{\mathrm{4}} {x}+{a}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{1}=\mathrm{0}\:{will}\:{have}\:{a}\:{solution}. \\ $$
Commented by rahul 19 last updated on 21/Mar/18
Commented by rahul 19 last updated on 21/Mar/18
![pls explain this: Hence,t^2 +at+1=0 should have at least one solution in [0,1]..........must have exactly one root in [0,1].](https://www.tinkutara.com/question/Q32222.png)
$${pls}\:{explain}\:{this}: \\ $$$$\boldsymbol{{H}}{ence},{t}^{\mathrm{2}} +{at}+\mathrm{1}=\mathrm{0}\:{should}\:{have}\:{at}\:{least} \\ $$$${one}\:{solution}\:{in}\:\left[\mathrm{0},\mathrm{1}\right]……….{must}\:{have} \\ $$$${exactly}\:{one}\:{root}\:{in}\:\left[\mathrm{0},\mathrm{1}\right]. \\ $$
Commented by MJS last updated on 21/Mar/18
![because t=sin^2 x it should be clear that t∈[0;1] ⇒ t≧0 t^2 +at+1=0 (t−t_1 )(t−t_2 )=0 t^2 +(−t_1 −t_2 )t+t_1 t_2 =0 ⇒ a=−t_1 −t_2 (but we don′t need this) 1=t_1 t_2 ∧t≧0 (from above) ⇒ ⇒ t_1 =t_2 ^(−1) (but we don′t need this) so if there′s one root in [0;1], there′s also a change of sign, so (f(0)<0∧f(1)>0)∨(f(0)>0∧f(1)<0) ⇒ ⇒ f(0)×f(1)<0](https://www.tinkutara.com/question/Q32228.png)
$$\mathrm{because}\:{t}=\mathrm{sin}^{\mathrm{2}} {x}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be} \\ $$$$\mathrm{clear}\:\mathrm{that}\:{t}\in\left[\mathrm{0};\mathrm{1}\right]\:\Rightarrow\:{t}\geqq\mathrm{0} \\ $$$${t}^{\mathrm{2}} +{at}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)=\mathrm{0} \\ $$$${t}^{\mathrm{2}} +\left(−{t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right){t}+{t}_{\mathrm{1}} {t}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{a}=−{t}_{\mathrm{1}} −{t}_{\mathrm{2}} \:\left(\mathrm{but}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{this}\right) \\ $$$$\mathrm{1}={t}_{\mathrm{1}} {t}_{\mathrm{2}} \wedge{t}\geqq\mathrm{0}\:\left(\mathrm{from}\:\mathrm{above}\right)\:\Rightarrow \\ $$$$\Rightarrow\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} ^{−\mathrm{1}} \:\left(\mathrm{but}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{this}\right) \\ $$$$\mathrm{so}\:\mathrm{if}\:\mathrm{there}'\mathrm{s}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{1}\right], \\ $$$$\mathrm{there}'\mathrm{s}\:\mathrm{also}\:\mathrm{a}\:\mathrm{change}\:\mathrm{of}\:\mathrm{sign},\:\mathrm{so} \\ $$$$\left({f}\left(\mathrm{0}\right)<\mathrm{0}\wedge{f}\left(\mathrm{1}\right)>\mathrm{0}\right)\vee\left({f}\left(\mathrm{0}\right)>\mathrm{0}\wedge{f}\left(\mathrm{1}\right)<\mathrm{0}\right)\:\Rightarrow \\ $$$$\Rightarrow\:{f}\left(\mathrm{0}\right)×{f}\left(\mathrm{1}\right)<\mathrm{0} \\ $$