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AlgebraQuestion and Answers: Page 357

Question Number 21005 Answers: 0 Comments: 0

If z_1 = a + ib and z_2 = c + id are complex numbers such that ∣z_1 ∣ = ∣z_2 ∣ = 1 and Re(z_1 z_2 ^ ) = 0, then the pair of complex numbers ω_1 = a + ic and ω_2 = b + id satisfy (1) ∣ω_1 ∣ = 1 (2) ∣ω_2 ∣ = 1 (3) Re(ω_1 ω_2 ^ ) = 0 (4) ∣ω_1 ∣ = 2∣ω_2 ∣

$$\mathrm{If}\:{z}_{\mathrm{1}} \:=\:{a}\:+\:{ib}\:\mathrm{and}\:{z}_{\mathrm{2}} \:=\:{c}\:+\:{id}\:\mathrm{are}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{1}\:\mathrm{and} \\ $$$$\mathrm{Re}\left({z}_{\mathrm{1}} \bar {{z}}_{\mathrm{2}} \right)\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:\omega_{\mathrm{1}} \:=\:{a}\:+\:{ic}\:\mathrm{and}\:\omega_{\mathrm{2}} \:=\:{b}\:+\:{id} \\ $$$$\mathrm{satisfy} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mid\omega_{\mathrm{2}} \mid\:=\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Re}\left(\omega_{\mathrm{1}} \bar {\omega}_{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mid\omega_{\mathrm{1}} \mid\:=\:\mathrm{2}\mid\omega_{\mathrm{2}} \mid \\ $$

Question Number 20983 Answers: 1 Comments: 0

Find the number of ordered triples (a, b, c) of positive integers such that abc = 108.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{triples} \\ $$$$\left({a},\:{b},\:{c}\right)\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$${abc}\:=\:\mathrm{108}. \\ $$

Question Number 20935 Answers: 1 Comments: 0

If ∣z + ω∣^2 = ∣z∣^2 + ∣ω∣^2 , where z and ω are complex numbers, then (1) (z/ω) is purely real (2) (z/ω) is purely imaginary (3) zω^ + z^ ω = 0 (4) amp((z/ω)) = (π/2)

$$\mathrm{If}\:\mid{z}\:+\:\omega\mid^{\mathrm{2}} \:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mid\omega\mid^{\mathrm{2}} ,\:\mathrm{where}\:{z}\:\mathrm{and}\:\omega \\ $$$$\mathrm{are}\:\mathrm{complex}\:\mathrm{numbers},\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{real} \\ $$$$\left(\mathrm{2}\right)\:\frac{{z}}{\omega}\:\mathrm{is}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\left(\mathrm{3}\right)\:{z}\bar {\omega}\:+\:\bar {{z}}\omega\:=\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{amp}\left(\frac{{z}}{\omega}\right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 20934 Answers: 0 Comments: 1

If z satisfies ∣z − 1∣ < ∣z + 3∣, then ω = 2z + 3 − i satisfies (1) ∣ω − 5 − i∣ < ∣ω + 3 + i∣ (2) ∣ω − 5∣ < ∣ω + 3∣ (3) Im (iω) > 1 (4) ∣arg(ω − 1)∣ < (π/2)

$$\mathrm{If}\:{z}\:\mathrm{satisfies}\:\mid{z}\:−\:\mathrm{1}\mid\:<\:\mid{z}\:+\:\mathrm{3}\mid,\:\mathrm{then}\:\omega\:= \\ $$$$\mathrm{2}{z}\:+\:\mathrm{3}\:−\:{i}\:\mathrm{satisfies} \\ $$$$\left(\mathrm{1}\right)\:\mid\omega\:−\:\mathrm{5}\:−\:{i}\mid\:<\:\mid\omega\:+\:\mathrm{3}\:+\:{i}\mid \\ $$$$\left(\mathrm{2}\right)\:\mid\omega\:−\:\mathrm{5}\mid\:<\:\mid\omega\:+\:\mathrm{3}\mid \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Im}\:\left({i}\omega\right)\:>\:\mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mid\mathrm{arg}\left(\omega\:−\:\mathrm{1}\right)\mid\:<\:\frac{\pi}{\mathrm{2}} \\ $$

Question Number 20933 Answers: 1 Comments: 0

If z is a complex number satisfying z + z^(−1) = 1, then z^n + z^(−n) , n ∈ N, has the value (1) 2(−1)^n , when n is a multiple of 3 (2) (−1)^(n−1) , when n is not a multiple of 3 (3) (−1)^(n+1) , when n is a multiple of 3 (4) 0 when n is not a multiple of 3

$$\mathrm{If}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{satisfying} \\ $$$${z}\:+\:{z}^{−\mathrm{1}} \:=\:\mathrm{1},\:\mathrm{then}\:{z}^{{n}} \:+\:{z}^{−{n}} ,\:{n}\:\in\:{N},\:\mathrm{has} \\ $$$$\mathrm{the}\:\mathrm{value} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}\left(−\mathrm{1}\right)^{{n}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of} \\ $$$$\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} ,\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{0}\:\mathrm{when}\:{n}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3} \\ $$

Question Number 20932 Answers: 0 Comments: 0

If a, b, c are real numbers and z is a complex number such that, a^2 + b^2 + c^2 = 1 and b + ic = (1 + a)z, then ((1 + iz)/(1 − iz)) equals. (1) ((b − ic)/(1 − ia)) (2) ((a + ib)/(1 + c)) (3) ((1 − c)/(a − ib)) (4) ((1 + a)/(b + ic))

$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that},\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\mathrm{and}\:{b}\:+\:{ic}\:=\:\left(\mathrm{1}\:+\:{a}\right){z},\:\mathrm{then}\:\frac{\mathrm{1}\:+\:{iz}}{\mathrm{1}\:−\:{iz}} \\ $$$$\mathrm{equals}. \\ $$$$\left(\mathrm{1}\right)\:\frac{{b}\:−\:{ic}}{\mathrm{1}\:−\:{ia}} \\ $$$$\left(\mathrm{2}\right)\:\frac{{a}\:+\:{ib}}{\mathrm{1}\:+\:{c}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}\:−\:{c}}{{a}\:−\:{ib}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}\:+\:{a}}{{b}\:+\:{ic}} \\ $$

Question Number 20914 Answers: 1 Comments: 0

The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has (1) Only purely imaginary roots (2) All real roots (3) Two real and two purely imaginary roots (4) Neither real nor purely imaginary roots

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equation}\:{p}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{with} \\ $$$$\mathrm{real}\:\mathrm{coefficients}\:\mathrm{has}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots}.\:\mathrm{Then}\:\mathrm{the}\:\mathrm{equation}\:{p}\left({p}\left({x}\right)\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Only}\:\mathrm{purely}\:\mathrm{imaginary}\:\mathrm{roots} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{All}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{Two}\:\mathrm{real}\:\mathrm{and}\:\mathrm{two}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{Neither}\:\mathrm{real}\:\mathrm{nor}\:\mathrm{purely}\:\mathrm{imaginary} \\ $$$$\mathrm{roots} \\ $$

Question Number 20882 Answers: 1 Comments: 0

Question Number 20881 Answers: 2 Comments: 0

Question Number 20853 Answers: 0 Comments: 0

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Question Number 20867 Answers: 1 Comments: 0

The number of irrational roots of the equation (x − 1)(x − 2)(3x − 2)(3x + 1) = 21 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{irrational}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation} \\ $$$$\left({x}\:−\:\mathrm{1}\right)\left({x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:−\:\mathrm{2}\right)\left(\mathrm{3}{x}\:+\:\mathrm{1}\right)\:=\:\mathrm{21}\:\mathrm{is} \\ $$

Question Number 20823 Answers: 1 Comments: 0

Question Number 20800 Answers: 2 Comments: 1

If Re(((z + 4)/(2z − 1))) = (1/2), then z is represented by a point lying on (1) A circle (2) An ellipse (3) A straight line (4) No real locus

$$\mathrm{If}\:\mathrm{Re}\left(\frac{{z}\:+\:\mathrm{4}}{\mathrm{2}{z}\:−\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{then}\:{z}\:\mathrm{is}\:\mathrm{represented} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{point}\:\mathrm{lying}\:\mathrm{on} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{A}\:\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{An}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{A}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{real}\:\mathrm{locus} \\ $$

Question Number 20799 Answers: 1 Comments: 0

If z^2 + z∣z∣ + ∣z∣^2 = 0, then locus of z is

$$\mathrm{If}\:{z}^{\mathrm{2}} \:+\:{z}\mid{z}\mid\:+\:\mid{z}\mid^{\mathrm{2}} \:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$

Question Number 20739 Answers: 1 Comments: 0

If ∣z_1 ∣ = 2, ∣z_2 ∣ = 3, ∣z_3 ∣ = 4 and ∣2z_1 + 3z_2 + 4z_3 ∣ = 4, then the expression ∣8z_2 z_3 + 27z_3 z_1 + 64z_1 z_2 ∣ equals

$$\mathrm{If}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mathrm{2},\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{3},\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{4}\:\mathrm{and} \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} \:+\:\mathrm{3}{z}_{\mathrm{2}} \:+\:\mathrm{4}{z}_{\mathrm{3}} \mid\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\mathrm{8}{z}_{\mathrm{2}} {z}_{\mathrm{3}} \:+\:\mathrm{27}{z}_{\mathrm{3}} {z}_{\mathrm{1}} \:+\:\mathrm{64}{z}_{\mathrm{1}} {z}_{\mathrm{2}} \mid\:\mathrm{equals} \\ $$

Question Number 20723 Answers: 0 Comments: 0

∫(√x) .sinx.dx

$$\int\sqrt{{x}}\:.{sinx}.{dx} \\ $$

Question Number 20684 Answers: 1 Comments: 0

If the equation x^2 + β^2 = 1 − 2βx and x^2 + α^2 = 1 − 2αx have one and only one root in common, then ∣α − β∣ is equal to

$${If}\:{the}\:{equation}\:{x}^{\mathrm{2}} \:+\:\beta^{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{2}\beta{x}\:{and} \\ $$$${x}^{\mathrm{2}} \:+\:\alpha^{\mathrm{2}} \:=\:\mathrm{1}\:−\:\mathrm{2}\alpha{x}\:{have}\:{one}\:{and}\:{only} \\ $$$${one}\:{root}\:{in}\:{common},\:{then}\:\mid\alpha\:−\:\beta\mid\:{is} \\ $$$${equal}\:{to} \\ $$

Question Number 20671 Answers: 1 Comments: 0

The total number of positive integral solution(s) of the inequation ((x^2 (3x − 4)^3 (x − 2)^4 )/((x − 5)^5 (2x − 7)^6 )) ≤ 0 is/are

$${The}\:{total}\:{number}\:{of}\:{positive}\:{integral} \\ $$$${solution}\left({s}\right)\:{of}\:{the}\:{inequation} \\ $$$$\frac{{x}^{\mathrm{2}} \left(\mathrm{3}{x}\:−\:\mathrm{4}\right)^{\mathrm{3}} \left({x}\:−\:\mathrm{2}\right)^{\mathrm{4}} }{\left({x}\:−\:\mathrm{5}\right)^{\mathrm{5}} \left(\mathrm{2}{x}\:−\:\mathrm{7}\right)^{\mathrm{6}} }\:\leqslant\:\mathrm{0}\:{is}/{are} \\ $$

Question Number 20670 Answers: 1 Comments: 0

For the equation 3x^2 + px + 3 = 0, find the value(s) of p if one root is (i) square of the other (ii) fourth power of the other.

$${For}\:{the}\:{equation}\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{px}\:+\:\mathrm{3}\:=\:\mathrm{0}, \\ $$$${find}\:{the}\:{value}\left({s}\right)\:{of}\:{p}\:{if}\:{one}\:{root}\:{is} \\ $$$$\left({i}\right)\:{square}\:{of}\:{the}\:{other} \\ $$$$\left({ii}\right)\:{fourth}\:{power}\:{of}\:{the}\:{other}. \\ $$

Question Number 20652 Answers: 0 Comments: 4

Suppose x is a positive real number such that {x}, [x] and x are in a geometric progression. Find the least positive integer n such that x^n > 100. (Here [x] denotes the integer part of x and {x} = x − [x].)

$$\mathrm{Suppose}\:{x}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{such}\:\mathrm{that}\:\left\{{x}\right\},\:\left[{x}\right]\:\mathrm{and}\:{x}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{least} \\ $$$$\mathrm{positive}\:\mathrm{integer}\:{n}\:\mathrm{such}\:\mathrm{that}\:{x}^{{n}} \:>\:\mathrm{100}. \\ $$$$\left(\mathrm{Here}\:\left[{x}\right]\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{integer}\:\mathrm{part}\:\mathrm{of}\:{x}\right. \\ $$$$\left.\mathrm{and}\:\left\{{x}\right\}\:=\:{x}\:−\:\left[{x}\right].\right) \\ $$

Question Number 20726 Answers: 1 Comments: 0

Five distinct 2-digit numbers are in a geometric progression. Find the middle term.

$$\mathrm{Five}\:\mathrm{distinct}\:\mathrm{2}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle} \\ $$$$\mathrm{term}. \\ $$

Question Number 20619 Answers: 1 Comments: 0

Solve the equation z^(n−1) = z^ (n ∈ N)

$${Solve}\:{the}\:{equation}\:{z}^{{n}−\mathrm{1}} \:=\:\bar {{z}}\:\left({n}\:\in\:{N}\right) \\ $$

Question Number 20631 Answers: 2 Comments: 0

Solve the inequality (x + 3)^5 − (x − 1)^5 ≥ 244.

$${Solve}\:{the}\:{inequality} \\ $$$$\left({x}\:+\:\mathrm{3}\right)^{\mathrm{5}} \:−\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{5}} \:\geqslant\:\mathrm{244}. \\ $$

Question Number 20552 Answers: 1 Comments: 0

The roots of the equation (3−x)^4 +(2−x)^4 =(5−2x)^4 are (a) all real (b) all imaginary (c) two real and two imaginary (d)none of the above .

$${The}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\:\left(\mathrm{3}−{x}\right)^{\mathrm{4}} +\left(\mathrm{2}−{x}\right)^{\mathrm{4}} =\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{4}} \:{are} \\ $$$$\left({a}\right)\:{all}\:{real}\:\:\:\:\left({b}\right)\:{all}\:{imaginary} \\ $$$$\left({c}\right)\:{two}\:{real}\:{and}\:{two}\:{imaginary} \\ $$$$\left({d}\right){none}\:{of}\:{the}\:{above}\:. \\ $$

Question Number 20550 Answers: 1 Comments: 0

Find the equation of circle in complex form which touches iz + z^ + 1 + i = 0 and for which the lines (1 − i)z = (1 + i)z^ and (1 + i)z + (i − 1)z^ − 4i = 0 are normals.

$${Find}\:{the}\:{equation}\:{of}\:{circle}\:{in}\:{complex} \\ $$$${form}\:{which}\:{touches}\:{iz}\:+\:\bar {{z}}\:+\:\mathrm{1}\:+\:{i}\:=\:\mathrm{0} \\ $$$${and}\:{for}\:{which}\:{the}\:{lines}\:\left(\mathrm{1}\:−\:{i}\right){z}\:= \\ $$$$\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:{and}\:\left(\mathrm{1}\:+\:{i}\right){z}\:+\:\left({i}\:−\:\mathrm{1}\right)\bar {{z}}\:−\:\mathrm{4}{i}\:=\:\mathrm{0} \\ $$$${are}\:{normals}. \\ $$

Question Number 20549 Answers: 1 Comments: 1

Show that if z_1 z_2 + z_3 z_4 = 0 and z_1 + z_2 = 0, then the complex numbers z_1 , z_2 , z_3 , z_4 are concyclic.

$${Show}\:{that}\:{if}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} \:+\:{z}_{\mathrm{3}} {z}_{\mathrm{4}} \:=\:\mathrm{0}\:{and}\:{z}_{\mathrm{1}} \:+ \\ $$$${z}_{\mathrm{2}} \:=\:\mathrm{0},\:{then}\:{the}\:{complex}\:{numbers}\:{z}_{\mathrm{1}} , \\ $$$${z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} ,\:{z}_{\mathrm{4}} \:{are}\:{concyclic}. \\ $$

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