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Question Number 19589 Answers: 1 Comments: 0

Let A and B is 3×3 matrix of equal number where A=symmetric matrix ....B=skew symmetric matrix and the relation... (A+B)(A−B)=(A−B)(A+B) then..the value of.. ... k (AB)^T =(−1)^k (AB) (a) −1 (c) 2 (b) 1 (d) 3

$${Let}\:{A}\:{and}\:{B}\:{is}\:\mathrm{3}×\mathrm{3}\:{matrix}\:{of}\:{equal}\:{number} \\ $$$${where}\:{A}={symmetric}\:{matrix}\: \\ $$$$....{B}={skew}\:{symmetric}\:{matrix} \\ $$$${and}\:{the}\:{relation}...\:\left({A}+{B}\right)\left({A}−{B}\right)=\left({A}−{B}\right)\left({A}+{B}\right) \\ $$$${then}..{the}\:{value}\:{of}..\:...\:{k} \\ $$$$\:\:\:\:\left({AB}\right)^{{T}} =\left(−\mathrm{1}\right)^{{k}} \left({AB}\right) \\ $$$$\left({a}\right)\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:\mathrm{2} \\ $$$$ \\ $$$$\left({b}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\:\mathrm{3} \\ $$

Question Number 19588 Answers: 0 Comments: 0

Question Number 19586 Answers: 0 Comments: 4

Given in an isosceles triangle a lateral side b and the base angle α. Compute the distance from the centre of the inscribed circle to the centre of the circumscribed circle.

$$\mathrm{Given}\:\mathrm{in}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{a} \\ $$$$\mathrm{lateral}\:\mathrm{side}\:\mathrm{b}\:\mathrm{and}\:\mathrm{the}\:\mathrm{base}\:\mathrm{angle}\:\alpha. \\ $$$$\mathrm{Compute}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{inscribed}\:\mathrm{circle}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumscribed}\:\mathrm{circle}. \\ $$

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 19657 Answers: 0 Comments: 2

differentiate the function with respect to x 1. ((secx−1)/(secx+1))

$${differentiate}\:{the}\:{function}\:{with}\:{respect}\: \\ $$$${to}\:{x} \\ $$$$\mathrm{1}.\:\:\:\:\frac{{secx}−\mathrm{1}}{{secx}+\mathrm{1}} \\ $$

Question Number 19547 Answers: 0 Comments: 2

A matrix has N rows and 2k−1 columns. Each column is filled with M ones and N−M zeros. A given row j is “cool” if and only if Σ_(i=1) ^(2k−1) a_(ji) ≥ k. Find the minimum and the maximum number of cool rows for given N, k and M.

$$\mathrm{A}\:\mathrm{matrix}\:\mathrm{has}\:{N}\:\mathrm{rows}\:\mathrm{and}\:\mathrm{2}{k}−\mathrm{1}\: \\ $$$$\mathrm{columns}.\:\mathrm{Each}\:\mathrm{column}\:\mathrm{is}\:\mathrm{filled}\:\mathrm{with} \\ $$$${M}\:\mathrm{ones}\:\mathrm{and}\:{N}−{M}\:\mathrm{zeros}. \\ $$$$\mathrm{A}\:\mathrm{given}\:\mathrm{row}\:{j}\:\mathrm{is}\:``{cool}''\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{2}{k}−\mathrm{1}} {\sum}}{a}_{{ji}} \:\geqslant\:{k}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{number}\:\mathrm{of}\:\mathrm{cool}\:\mathrm{rows} \\ $$$$\mathrm{for}\:\mathrm{given}\:{N},\:{k}\:\mathrm{and}\:{M}. \\ $$

Question Number 19516 Answers: 1 Comments: 0

Let Akbar and Birbal together have n marbles, where n > 0. Akbar says to Birbal, “If I give you some marbles then you will have twice as many marbles as I will have.” Birbal says to Akbar, “If I give you some marbles then you will have thrice as many marbles as I will have.” What is the minimum possible value of n for which the above statements are true?

$$\mathrm{Let}\:\mathrm{Akbar}\:\mathrm{and}\:\mathrm{Birbal}\:\mathrm{together}\:\mathrm{have}\:{n} \\ $$$$\mathrm{marbles},\:\mathrm{where}\:{n}\:>\:\mathrm{0}. \\ $$$$\mathrm{Akbar}\:\mathrm{says}\:\mathrm{to}\:\mathrm{Birbal},\:``\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{twice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.''\:\mathrm{Birbal} \\ $$$$\mathrm{says}\:\mathrm{to}\:\mathrm{Akbar},\:``\mathrm{If}\:\mathrm{I}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some} \\ $$$$\mathrm{marbles}\:\mathrm{then}\:\mathrm{you}\:\mathrm{will}\:\mathrm{have}\:\mathrm{thrice}\:\mathrm{as} \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{as}\:\mathrm{I}\:\mathrm{will}\:\mathrm{have}.'' \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of} \\ $$$${n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{above}\:\mathrm{statements}\:\mathrm{are} \\ $$$$\mathrm{true}? \\ $$

Question Number 19511 Answers: 1 Comments: 1

In the arrangement shown, the wedge is smooth and has a mass M. The sphere has a mass m. The system is released from rest from the position shown. There is no friction anywhere. Find the contact force between the wall and the sphere.

$$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown},\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{is}\:\mathrm{smooth}\:\mathrm{and}\:\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{M}.\:\mathrm{The}\:\mathrm{sphere} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{mass}\:{m}.\:\mathrm{The}\:\mathrm{system}\:\mathrm{is}\:\mathrm{released} \\ $$$$\mathrm{from}\:\mathrm{rest}\:\mathrm{from}\:\mathrm{the}\:\mathrm{position}\:\mathrm{shown}. \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{friction}\:\mathrm{anywhere}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{contact}\:\mathrm{force}\:\mathrm{between}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{sphere}. \\ $$

Question Number 19509 Answers: 1 Comments: 1

Blocks P and R starts from rest and moves to the right with acceleration a_P = 12t m/s^2 and a_R = 3 m/s^2 . Here t is in seconds. The time when block Q again comes to rest is

$$\mathrm{Blocks}\:{P}\:\mathrm{and}\:{R}\:\mathrm{starts}\:\mathrm{from}\:\mathrm{rest}\:\mathrm{and} \\ $$$$\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}_{{P}} \:=\:\mathrm{12}{t}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \:\mathrm{and}\:{a}_{{R}} \:=\:\mathrm{3}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} .\:\mathrm{Here}\:{t} \\ $$$$\mathrm{is}\:\mathrm{in}\:\mathrm{seconds}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{when}\:\mathrm{block}\:{Q} \\ $$$$\mathrm{again}\:\mathrm{comes}\:\mathrm{to}\:\mathrm{rest}\:\mathrm{is} \\ $$

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 19531 Answers: 1 Comments: 0

Question Number 19557 Answers: 1 Comments: 3

Question Number 19556 Answers: 0 Comments: 0

Question Number 19499 Answers: 1 Comments: 0

Find the last digit of 2^(253)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{last}\:\mathrm{digit}\:\mathrm{of}\:\:\mathrm{2}^{\mathrm{253}} \\ $$

Question Number 19554 Answers: 0 Comments: 0

proof (a^− +b^− ).c^− =a^− .c^− +b^− .c^−

$${proof}\:\left(\overset{−} {{a}}+\overset{−} {{b}}\right).\overset{−} {{c}}=\overset{−} {{a}}.\overset{−} {{c}}+\overset{−} {{b}}.\overset{−} {{c}} \\ $$

Question Number 19472 Answers: 0 Comments: 3

A fishing boat is anchored 9 km away from the nearest point on the shore. A messanger must be sent from the fishing boat to a camp, 15 km from the point on shore closest to boat. If the messanger can walk at a speed of 5 km per hour and can row at 4 km/h, determine the distance of that point (in km) from the shore, where he must land so as to reach the shore in least possible time.

$$\mathrm{A}\:\mathrm{fishing}\:\mathrm{boat}\:\mathrm{is}\:\mathrm{anchored}\:\mathrm{9}\:\mathrm{km}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{shore}.\:\mathrm{A} \\ $$$$\mathrm{messanger}\:\mathrm{must}\:\mathrm{be}\:\mathrm{sent}\:\mathrm{from}\:\mathrm{the}\:\mathrm{fishing} \\ $$$$\mathrm{boat}\:\mathrm{to}\:\mathrm{a}\:\mathrm{camp},\:\mathrm{15}\:\mathrm{km}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{shore}\:\mathrm{closest}\:\mathrm{to}\:\mathrm{boat}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{messanger} \\ $$$$\mathrm{can}\:\mathrm{walk}\:\mathrm{at}\:\mathrm{a}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{5}\:\mathrm{km}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{and}\:\mathrm{can}\:\mathrm{row}\:\mathrm{at}\:\mathrm{4}\:\mathrm{km}/\mathrm{h},\:\mathrm{determine}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{of}\:\mathrm{that}\:\mathrm{point}\:\left(\mathrm{in}\:\mathrm{km}\right)\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{shore},\:\mathrm{where}\:\mathrm{he}\:\mathrm{must}\:\mathrm{land}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to} \\ $$$$\mathrm{reach}\:\mathrm{the}\:\mathrm{shore}\:\mathrm{in}\:\mathrm{least}\:\mathrm{possible}\:\mathrm{time}. \\ $$

Question Number 19455 Answers: 1 Comments: 0

Prove that if z = cos 6° + i sin 6°, then (1/(z^2 + 1)) − ((iz)/(z^4 − 1)) + ((iz^3 )/(z^8 − 1)) + ((iz^7 )/(z^(16) − 1)) = 0.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{z}\:=\:\mathrm{cos}\:\mathrm{6}°\:+\:{i}\:\mathrm{sin}\:\mathrm{6}°,\:\mathrm{then} \\ $$$$\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\frac{{iz}}{{z}^{\mathrm{4}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{3}} }{{z}^{\mathrm{8}} \:−\:\mathrm{1}}\:+\:\frac{{iz}^{\mathrm{7}} }{{z}^{\mathrm{16}} \:−\:\mathrm{1}}\:=\:\mathrm{0}. \\ $$

Question Number 19435 Answers: 1 Comments: 0

If α = cos ((2π)/5) + i sin ((2π)/5) , then find the value of α + α^2 + α^3 + α^4 .

$$\mathrm{If}\:\alpha\:=\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:+\:{i}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:,\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\alpha\:+\:\alpha^{\mathrm{2}} \:+\:\alpha^{\mathrm{3}} \:+\:\alpha^{\mathrm{4}} . \\ $$

Question Number 19434 Answers: 1 Comments: 0

If (((1 + i(√3))/(1 − i(√3))))^n is an integer, then n is

$$\mathrm{If}\:\left(\frac{\mathrm{1}\:+\:{i}\sqrt{\mathrm{3}}}{\mathrm{1}\:−\:{i}\sqrt{\mathrm{3}}}\right)^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{n}\:\mathrm{is} \\ $$

Question Number 19433 Answers: 1 Comments: 0

(((√(5 + 12i)) + (√(5 − 12i)))/((√(5 + 12i)) − (√(5 − 12i)))) =

$$\frac{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:+\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}{\sqrt{\mathrm{5}\:+\:\mathrm{12}{i}}\:−\:\sqrt{\mathrm{5}\:−\:\mathrm{12}{i}}}\:= \\ $$

Question Number 19419 Answers: 1 Comments: 1

sin z=200 find z

$$\mathrm{sin}\:\boldsymbol{{z}}=\mathrm{200} \\ $$$$ \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{z}} \\ $$

Question Number 19415 Answers: 1 Comments: 0

PS is a line segment of length 4 and O is the midpoint of PS. A semicircular arc is drawn with PS as diameter. Let X be the midpoint of this arc. Q and R are points on the arc PXS such that QR is parallel to PS and the semicircular arc drawn with QR as diameter is tangent to PS. What is the area of the region QXROQ bounded by the two semicircular arcs?

$${PS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{of}\:\mathrm{length}\:\mathrm{4}\:\mathrm{and}\:{O} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PS}.\:\mathrm{A}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:{PS}\:\mathrm{as}\:\mathrm{diameter}.\:\mathrm{Let} \\ $$$${X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{this}\:\mathrm{arc}.\:{Q}\:\mathrm{and}\:{R} \\ $$$$\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{arc}\:{PXS}\:\mathrm{such}\:\mathrm{that}\:{QR} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{PS}\:\mathrm{and}\:\mathrm{the}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{drawn}\:\mathrm{with}\:{QR}\:\mathrm{as}\:\mathrm{diameter}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:{PS}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{region}\:{QXROQ}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{semicircular}\:\mathrm{arcs}? \\ $$

Question Number 19413 Answers: 1 Comments: 1

How many integer pairs (x, y) satisfy x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{integer}\:\mathrm{pairs}\:\left({x},\:{y}\right)\:\mathrm{satisfy} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{4}{y}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:−\:\mathrm{2}{x}\:−\:\mathrm{4}{y}\:−\:\mathrm{8}\:=\:\mathrm{0}? \\ $$

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