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AlgebraQuestion and Answers: Page 347

Question Number 19687 Answers: 1 Comments: 0

Let z_1 , z_2 , z_3 be three vertices of an equilateral triangle circumscribing the circle ∣z∣ = (1/2). If z_1 = (1/2) + (((√3)i)/2) and z_1 , z_2 , z_3 are in anticlockwise sense then z_2 is

$$\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{three}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{circumscribing}\:\mathrm{the} \\ $$$$\mathrm{circle}\:\mid{z}\mid\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{If}\:{z}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{3}}{i}}{\mathrm{2}}\:\mathrm{and}\:{z}_{\mathrm{1}} , \\ $$$${z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{in}\:\mathrm{anticlockwise}\:\mathrm{sense}\:\mathrm{then}\:{z}_{\mathrm{2}} \:\mathrm{is} \\ $$

Question Number 19646 Answers: 0 Comments: 1

Find the sum of all possible digits that comes at ten′s place for 3^n where n is any natural number.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{digits}\:\mathrm{that} \\ $$$$\mathrm{comes}\:\mathrm{at}\:\mathrm{ten}'\mathrm{s}\:\mathrm{place}\:\mathrm{for}\:\mathrm{3}^{{n}} \:\mathrm{where}\:{n}\:\mathrm{is} \\ $$$$\mathrm{any}\:\mathrm{natural}\:\mathrm{number}. \\ $$

Question Number 19638 Answers: 1 Comments: 0

Let P(x) is a polynomial such that P(1) = 1, P(2) = 2, P(3) = 3, and P(4) = 5. Find the value of P(6).

$$\mathrm{Let}\:{P}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{such}\:\mathrm{that} \\ $$$${P}\left(\mathrm{1}\right)\:=\:\mathrm{1},\:{P}\left(\mathrm{2}\right)\:=\:\mathrm{2},\:{P}\left(\mathrm{3}\right)\:=\:\mathrm{3},\:\mathrm{and} \\ $$$${P}\left(\mathrm{4}\right)\:=\:\mathrm{5}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{P}\left(\mathrm{6}\right). \\ $$

Question Number 19643 Answers: 1 Comments: 0

Find the real solution of the equation (√(17 + 8x − 2x^2 )) + (√(4 + 12x − 3x^2 )) = x^2 − 4x + 13.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\sqrt{\mathrm{17}\:+\:\mathrm{8}{x}\:−\:\mathrm{2}{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{4}\:+\:\mathrm{12}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} }\:=\:{x}^{\mathrm{2}} \\ $$$$−\:\mathrm{4}{x}\:+\:\mathrm{13}. \\ $$

Question Number 19629 Answers: 1 Comments: 0

If ∣z∣ = 2, then the points representing the complex numbers −1 + 5z will lie on a (1) Circle (2) Straight line (3) Parabola (4) Ellipse

$$\mathrm{If}\:\mid{z}\mid\:=\:\mathrm{2},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:\mathrm{representing} \\ $$$$\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:−\mathrm{1}\:+\:\mathrm{5}{z}\:\mathrm{will}\:\mathrm{lie} \\ $$$$\mathrm{on}\:\mathrm{a} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Straight}\:\mathrm{line} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Parabola} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Ellipse} \\ $$

Question Number 19623 Answers: 1 Comments: 0

Find the locus of z if arg(((z − 2)/(z − 3))) = (π/4)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{if}\:\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{2}}{{z}\:−\:\mathrm{3}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$

Question Number 19609 Answers: 1 Comments: 0

Question Number 19604 Answers: 1 Comments: 1

Question Number 19592 Answers: 0 Comments: 2

Question Number 19574 Answers: 0 Comments: 5

Carol was given three numbers and was asked to add the largest of the three to the product of the other two. Instead, she multiplied the largest with the sum of the other two, but still got the right answer. What is the sum of the three numbers?

Question Number 19508 Answers: 0 Comments: 0

Prove that the length of perpendicular drawn from the point z_0 to the straight line α^ z + αz^ + c = 0 is p = ∣((α^ z_0 + αz_0 ^ + c)/(2 ∣α∣))∣.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{perpendicular} \\ $$$$\mathrm{drawn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:{z}_{\mathrm{0}} \:\mathrm{to}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{line}\:\bar {\alpha}{z}\:+\:\alpha\bar {{z}}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{is} \\ $$$${p}\:=\:\mid\frac{\bar {\alpha}{z}_{\mathrm{0}} \:+\:\alpha\bar {{z}}_{\mathrm{0}} \:+\:{c}}{\mathrm{2}\:\mid\alpha\mid}\mid. \\ $$

Question Number 19507 Answers: 1 Comments: 0

Prove that three points z_1 , z_2 , z_3 are collinear if determinant ((z_1 ,z_1 ^ ,1),(z_2 ,z_2 ^ ,1),(z_3 ,z_3 ^ ,1))= 0

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{three}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{collinear}\:\mathrm{if}\:\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0} \\ $$

Question Number 19506 Answers: 1 Comments: 0

Prove that the equation of the line joining the points z_1 and z_2 can be put in the form z = tz_1 + (1 − t)z_2 , where t is a parameter.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{joining}\:\mathrm{the}\:\mathrm{points}\:{z}_{\mathrm{1}} \:\mathrm{and}\:{z}_{\mathrm{2}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{put} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{z}\:=\:{tz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:−\:{t}\right){z}_{\mathrm{2}} ,\:\mathrm{where} \\ $$$${t}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parameter}. \\ $$

Question Number 19505 Answers: 1 Comments: 0

Prove that two straight lines with complex slopes μ_1 and μ_2 are parallel and perpendicular according as μ_1 = μ_2 and μ_1 + μ_2 = 0. Hence if the straight lines α^ z + αz^ + c = 0 and β^ z + βz^ + k = 0 are parallel and perpendicular according as α^ β − αβ^ = 0 and α^ β + αβ^ = 0.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{two}\:\mathrm{straight}\:\mathrm{lines}\:\mathrm{with} \\ $$$$\mathrm{complex}\:\mathrm{slopes}\:\mu_{\mathrm{1}} \:\mathrm{and}\:\mu_{\mathrm{2}} \:\mathrm{are}\:\mathrm{parallel} \\ $$$$\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according}\:\mathrm{as}\:\mu_{\mathrm{1}} \:=\:\mu_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mu_{\mathrm{1}} \:+\:\mu_{\mathrm{2}} \:=\:\mathrm{0}.\:\mathrm{Hence}\:\mathrm{if}\:\mathrm{the}\:\mathrm{straight} \\ $$$$\mathrm{lines}\:\bar {\alpha}{z}\:+\:\alpha\bar {{z}}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\beta}{z}\:+\:\beta\bar {{z}}\:+\:{k}\:=\:\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{parallel}\:\mathrm{and}\:\mathrm{perpendicular}\:\mathrm{according} \\ $$$$\mathrm{as}\:\bar {\alpha}\beta\:−\:\alpha\bar {\beta}\:=\:\mathrm{0}\:\mathrm{and}\:\bar {\alpha}\beta\:+\:\alpha\bar {\beta}\:=\:\mathrm{0}. \\ $$

Question Number 19557 Answers: 1 Comments: 3