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Question-22729




Question Number 22729 by selestian last updated on 22/Oct/17
Commented by math solver last updated on 22/Oct/17
0
$$\mathrm{0} \\ $$
Commented by selestian last updated on 22/Oct/17
sir please send me the solution
$${sir}\:{please}\:{send}\:{me}\:{the}\:{solution} \\ $$
Commented by math solver last updated on 22/Oct/17
i have sent. see below
$${i}\:{have}\:{sent}.\:{see}\:{below} \\ $$
Commented by math solver last updated on 22/Oct/17
hey, i don′t know if you have studied  am , gm.  don′t worry , i have another sol.too :)
$${hey},\:{i}\:{don}'{t}\:{know}\:{if}\:{you}\:{have}\:{studied} \\ $$$${am}\:,\:{gm}. \\ $$$$\left.{don}'{t}\:{worry}\:,\:{i}\:{have}\:{another}\:{sol}.{too}\::\right) \\ $$
Commented by selestian last updated on 22/Oct/17
yes i had studied am nd gm thanks
$${yes}\:{i}\:{had}\:{studied}\:{am}\:{nd}\:{gm}\:{thanks} \\ $$$$ \\ $$
Answered by math solver last updated on 22/Oct/17
Answered by $@ty@m last updated on 24/Oct/17
Let 2^x =y  ⇒sin (e^x )=y+(1/y)  ⇒y^2 −sin (e^x )y+1=0  −−−(1)  D=b^2 −4ac  =sin^2 (e^x )−4  ⇒0−4≤D≤1−4                       {∵0≤sin^2 θ≤1}  ⇒−4≤D≤−3  ⇒ (1) has no real root.
$${Let}\:\mathrm{2}^{{x}} ={y} \\ $$$$\Rightarrow\mathrm{sin}\:\left({e}^{{x}} \right)={y}+\frac{\mathrm{1}}{{y}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\mathrm{sin}\:\left({e}^{{x}} \right){y}+\mathrm{1}=\mathrm{0}\:\:−−−\left(\mathrm{1}\right) \\ $$$${D}={b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$$$=\mathrm{sin}\:^{\mathrm{2}} \left({e}^{{x}} \right)−\mathrm{4} \\ $$$$\Rightarrow\mathrm{0}−\mathrm{4}\leqslant{D}\leqslant\mathrm{1}−\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\because\mathrm{0}\leqslant\mathrm{sin}\:^{\mathrm{2}} \theta\leqslant\mathrm{1}\right\} \\ $$$$\Rightarrow−\mathrm{4}\leqslant{D}\leqslant−\mathrm{3} \\ $$$$\Rightarrow\:\left(\mathrm{1}\right)\:{has}\:{no}\:{real}\:{root}. \\ $$

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