Question Number 39464 by nishant last updated on 06/Jul/18
![Domain of the explicit form of the function y represented implicitly by the equation (1+x)cosy−x^2 =0 is (a) (−1,1] (b) (−1, 1−(√)5/2] (c) [1−(√)5/2, 1+(√)5/2] (d) [0, 1+(√)5/2]](https://www.tinkutara.com/question/Q39464.png)
$${Domain}\:\:{of}\:\:{the}\:\:{explicit}\:\:{form}\:\:{of} \\ $$$${the}\:\:{function}\:\:\:{y}\:\:\:{represented}\: \\ $$$${implicitly}\:\:\:{by}\:\:{the}\:\:{equation}\: \\ $$$$\left(\mathrm{1}+{x}\right){cosy}−{x}^{\mathrm{2}} =\mathrm{0}\:\:{is} \\ $$$$\left({a}\right)\:\:\left(−\mathrm{1},\mathrm{1}\right]\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:\:\:\:\left(−\mathrm{1},\:\mathrm{1}−\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$$$\left({c}\right)\:\:\:\left[\mathrm{1}−\sqrt{}\mathrm{5}/\mathrm{2},\:\mathrm{1}+\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$$$\left({d}\right)\:\:\left[\mathrm{0},\:\mathrm{1}+\sqrt{}\mathrm{5}/\mathrm{2}\right] \\ $$
Answered by MJS last updated on 06/Jul/18
![cos y =(x^2 /(x+1)) ⇒ x≠−1 ∧ (x^2 /(∣x+1∣))≤1 x^2 ≤∣x+1∣ f(x)=x^2 −x−1=0 ⇒ x_1 =(1/2)−((√5)/2); x_2 =(1/2)+((√5)/2) f(x)=x^2 +x+1=0 ⇒ no real solution ((1/2)−((√5)/2)≤x≤(1/2)+((√5)/2)) domain is [(1/2)−((√5)/2); (1/2)+((√5)/2)]](https://www.tinkutara.com/question/Q39502.png)
$$\mathrm{cos}\:{y}\:=\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{1}}\:\Rightarrow\:{x}\neq−\mathrm{1}\:\wedge\:\frac{{x}^{\mathrm{2}} }{\mid{x}+\mathrm{1}\mid}\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} \leqslant\mid{x}+\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}};\:{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$$\:\:\:\:\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$\mathrm{domain}\:\mathrm{is}\:\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}};\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right] \\ $$