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Question Number 13491 by 433 last updated on 20/May/17

Test 1. Solve equation (k^2 −1)x^2 +(k−1)x+(k+1)=0 k∈R (30) 2. Prove ((sin(x))/(1+cos(x)))=((1−cos(x))/(sin(x))) (35) 3.P(x)=−2x^3 −2x^2 −x+2409 Find P(−11) (35) Evaluate other answers and give marks I want to see how math teachers evaluate in other countries Sorry foy my english

$${Test} \\ $$$$\mathrm{1}.\:{Solve}\:{equation} \\ $$$$\left({k}^{\mathrm{2}} −\mathrm{1}\right){x}^{\mathrm{2}} +\left({k}−\mathrm{1}\right){x}+\left({k}+\mathrm{1}\right)=\mathrm{0}\:\:{k}\in\mathbb{R} \\ $$$$\left(\mathrm{30}\right) \\ $$$$\mathrm{2}.\:{Prove} \\ $$$$\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}=\frac{\mathrm{1}−{cos}\left({x}\right)}{{sin}\left({x}\right)} \\ $$$$\left(\mathrm{35}\right) \\ $$$$\mathrm{3}.{P}\left({x}\right)=−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2409} \\ $$$${Find}\:{P}\left(−\mathrm{11}\right) \\ $$$$\left(\mathrm{35}\right) \\ $$$$ \\ $$$${Evaluate}\:{other}\:{answers}\:{and}\:{give}\:{marks} \\ $$$${I}\:{want}\:{to}\:{see}\:{how}\:{math}\:{teachers}\:{evaluate}\:{in}\:{other}\:{countries} \\ $$$${Sorry}\:{foy}\:{my}\:{english} \\ $$

Answered by 433 last updated on 20/May/17

1. If k=1 2=0 If k=−1 −2x=0⇒x=0 If k≠±1 Δ=(k−1)^2 −4(k^2 −1)(k+1) =(k−1)((k−1)−4(k+1)^2 ) =(k−1)(k−1−4k^2 −8k−4) (k−1)(−4k^2 −7k−5) Δ′=7^2 −20×4=49−80<0 −4k^2 −7k−5<0 ∀k If k>1 ⇒ Δ<0 If k∈(−∞,−1)∪(−1,1) ⇒ Δ>0 x_(1,2) =((1−k±(√((k−1)(−4k^2 −7k−5))))/(2(k^2 −1))) 2. ((sin(x))/(1+cos(x)))=((1−cos(x))/(sin(x))) ⇔ sin^2 (x)=1−cos^2 (x) sin^2 (x)+cos^2 (x)=1 3. Q(x)=x+11 P(x)=Q(x)R(x)+u P(−11)=Q(−11)R(−11)+u Q(−11)=−11+11=0 P(−11)=u (−2x^3 −2x^2 −x+2409)=(x+11)(−2x^2 +20x−221)+4840 P(−11)=4840

$$\mathrm{1}.\:{If}\:{k}=\mathrm{1}\: \\ $$$$\mathrm{2}=\mathrm{0} \\ $$$${If}\:{k}=−\mathrm{1} \\ $$$$−\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\mathrm{0} \\ $$$${If}\:{k}\neq\pm\mathrm{1} \\ $$$$\Delta=\left({k}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({k}^{\mathrm{2}} −\mathrm{1}\right)\left({k}+\mathrm{1}\right) \\ $$$$=\left({k}−\mathrm{1}\right)\left(\left({k}−\mathrm{1}\right)−\mathrm{4}\left({k}+\mathrm{1}\right)^{\mathrm{2}} \right) \\ $$$$=\left({k}−\mathrm{1}\right)\left({k}−\mathrm{1}−\mathrm{4}{k}^{\mathrm{2}} −\mathrm{8}{k}−\mathrm{4}\right) \\ $$$$\left({k}−\mathrm{1}\right)\left(−\mathrm{4}{k}^{\mathrm{2}} −\mathrm{7}{k}−\mathrm{5}\right) \\ $$$$\Delta'=\mathrm{7}^{\mathrm{2}} −\mathrm{20}×\mathrm{4}=\mathrm{49}−\mathrm{80}<\mathrm{0} \\ $$$$−\mathrm{4}{k}^{\mathrm{2}} −\mathrm{7}{k}−\mathrm{5}<\mathrm{0}\:\forall{k} \\ $$$${If}\:{k}>\mathrm{1}\:\Rightarrow\:\Delta<\mathrm{0} \\ $$$${If}\:{k}\in\left(−\infty,−\mathrm{1}\right)\cup\left(−\mathrm{1},\mathrm{1}\right)\:\Rightarrow\:\Delta>\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}−{k}\pm\sqrt{\left({k}−\mathrm{1}\right)\left(−\mathrm{4}{k}^{\mathrm{2}} −\mathrm{7}{k}−\mathrm{5}\right)}}{\mathrm{2}\left({k}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$ \\ $$$$\mathrm{2}.\:\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}=\frac{\mathrm{1}−{cos}\left({x}\right)}{{sin}\left({x}\right)}\:\Leftrightarrow \\ $$$${sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right) \\ $$$${sin}^{\mathrm{2}} \left({x}\right)+{cos}^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{3}.\:{Q}\left({x}\right)={x}+\mathrm{11} \\ $$$${P}\left({x}\right)={Q}\left({x}\right){R}\left({x}\right)+{u} \\ $$$${P}\left(−\mathrm{11}\right)={Q}\left(−\mathrm{11}\right){R}\left(−\mathrm{11}\right)+{u} \\ $$$${Q}\left(−\mathrm{11}\right)=−\mathrm{11}+\mathrm{11}=\mathrm{0} \\ $$$${P}\left(−\mathrm{11}\right)={u} \\ $$$$\left(−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2409}\right)=\left({x}+\mathrm{11}\right)\left(−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{221}\right)+\mathrm{4840} \\ $$$${P}\left(−\mathrm{11}\right)=\mathrm{4840} \\ $$

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