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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 19637    Answers: 1   Comments: 0

Determine the number of five-digit integers (37abc) in base 10 such that each of the numbers (37abc), (37bca) and 37cab is divisible by 37.

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{five}-\mathrm{digit} \\ $$$$\mathrm{integers}\:\left(\mathrm{37}{abc}\right)\:\mathrm{in}\:\mathrm{base}\:\mathrm{10}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}\:\left(\mathrm{37}{abc}\right),\:\left(\mathrm{37}{bca}\right) \\ $$$$\mathrm{and}\:\mathrm{37}{cab}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{37}. \\ $$

Question Number 19634    Answers: 1   Comments: 0

How many ordered triplets (x, y, z) of positive integer satisfy lcm(x, y) = 72, lcm(x, z) = 600 and lcm(y, z) = 900?

$$\mathrm{How}\:\mathrm{many}\:\mathrm{ordered}\:\mathrm{triplets}\:\left({x},\:{y},\:{z}\right)\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{integer}\:\mathrm{satisfy}\:\mathrm{lcm}\left({x},\:{y}\right)\:=\:\mathrm{72}, \\ $$$$\mathrm{lcm}\left({x},\:{z}\right)\:=\:\mathrm{600}\:\mathrm{and}\:\mathrm{lcm}\left({y},\:{z}\right)\:=\:\mathrm{900}? \\ $$

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 19631    Answers: 1   Comments: 0

Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product then a number x is obtained which is a multiple of 17. Find the sum of digits of number x.

$$\mathrm{Two}\:\mathrm{different}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{between} \\ $$$$\mathrm{4}\:\mathrm{and}\:\mathrm{18}\:\mathrm{are}\:\mathrm{chosen}.\:\mathrm{When}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is} \\ $$$$\mathrm{subtracted}\:\mathrm{from}\:\mathrm{their}\:\mathrm{product}\:\mathrm{then}\:\mathrm{a} \\ $$$$\mathrm{number}\:{x}\:\mathrm{is}\:\mathrm{obtained}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{multiple}\:\mathrm{of}\:\mathrm{17}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{of} \\ $$$$\mathrm{number}\:{x}. \\ $$

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 19615    Answers: 1   Comments: 0

Question Number 19610    Answers: 1   Comments: 2

Prove that the radius of a circle passing through the midpoints of the sides of a triangle ABC is half the radius of a circle circum- scribed about the triangle.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{passing}\:\mathrm{through}\:\mathrm{the}\:\mathrm{midpoints} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is} \\ $$$$\mathrm{half}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{circum}- \\ $$$$\mathrm{scribed}\:\mathrm{about}\:\mathrm{the}\:\mathrm{triangle}. \\ $$

Question Number 19609    Answers: 1   Comments: 0

Question Number 19604    Answers: 1   Comments: 1

Question Number 19595    Answers: 1   Comments: 1

For x ∈ R, solve the equation below! (2^x − 4)^3 + (4^x − 2)^3 = (4^x + 2^x − 6)^3

$$\mathrm{For}\:{x}\:\in\:\mathrm{R},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{below}! \\ $$$$\left(\mathrm{2}^{{x}} \:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left(\mathrm{4}^{{x}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:=\:\left(\mathrm{4}^{{x}} \:+\:\mathrm{2}^{{x}} \:−\:\mathrm{6}\right)^{\mathrm{3}} \\ $$

Question Number 19592    Answers: 0   Comments: 2

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?

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

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.

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

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