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Question Number 21874    Answers: 2   Comments: 0

Find the remainder if 2^(2006) is divided by 17

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{if}\:\:\:\mathrm{2}^{\mathrm{2006}} \:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\:\mathrm{17} \\ $$

Question Number 21867    Answers: 0   Comments: 0

The number of points in the cartesian plane with integral coordinates satisfying the inequalities ∣x∣ ≤ 4, ∣y∣ ≤ 4 and ∣x − y∣ ≤ 4 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{in}\:\mathrm{the}\:\mathrm{cartesian} \\ $$$$\mathrm{plane}\:\mathrm{with}\:\mathrm{integral}\:\mathrm{coordinates} \\ $$$$\mathrm{satisfying}\:\mathrm{the}\:\mathrm{inequalities}\:\mid{x}\mid\:\leqslant\:\mathrm{4},\:\mid{y}\mid\:\leqslant \\ $$$$\mathrm{4}\:\mathrm{and}\:\mid{x}\:−\:{y}\mid\:\leqslant\:\mathrm{4}\:\mathrm{is} \\ $$

Question Number 21795    Answers: 1   Comments: 3

help x∈N determine x where 7 divise 2^x +3^x

$${help} \\ $$$${x}\in{N} \\ $$$${determine}\:{x}\:{where}\:\mathrm{7}\:{divise}\:\mathrm{2}^{{x}} +\mathrm{3}^{{x}} \\ $$$$ \\ $$

Question Number 21733    Answers: 1   Comments: 0

If p is one of roots from x^2 − 2x + 6 = 0 then p^4 + 16p is equal to ...

$$\mathrm{If}\:{p}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\:\mathrm{roots}\:\mathrm{from}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{6}\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:{p}^{\mathrm{4}} \:+\:\mathrm{16}{p}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:... \\ $$

Question Number 21654    Answers: 0   Comments: 0

Find the minimum value of Q that satisfy: ∣xy(x^2 − y^2 ) + yz(y^2 − z^2 ) + zx(z^2 − x^2 )∣ ≤ Q(x^2 + y^2 + z^2 )^2

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{Q}\:\mathrm{that}\:\mathrm{satisfy}: \\ $$$$\mid{xy}\left({x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \right)\:+\:{yz}\left({y}^{\mathrm{2}} \:−\:{z}^{\mathrm{2}} \right)\:+\:{zx}\left({z}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} \right)\mid\:\leqslant\:{Q}\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$

Question Number 21653    Answers: 0   Comments: 0

Find all pair of solutions (x,y) that satisfy the equation: ((7x^2 − 13xy + 7y^2 ))^(1/3) = ∣x − y∣ + 1

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{solutions}\:\left({x},{y}\right)\:\mathrm{that}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}^{\mathrm{2}} \:−\:\mathrm{13}{xy}\:+\:\mathrm{7}{y}^{\mathrm{2}} }\:=\:\mid{x}\:−\:{y}\mid\:+\:\mathrm{1} \\ $$

Question Number 21650    Answers: 0   Comments: 0

The three distinct successive terms of an A.P are the first,second and fourth terms of a G.P. If the sum to infinity of a G.P is 3+(√5) , find the first term.

$${The}\:{three}\:{distinct}\:{successive}\:{terms}\:{of}\:{an}\:{A}.{P}\:{are} \\ $$$${the}\:{first},{second}\:{and}\:{fourth}\:{terms}\:{of}\:{a}\:{G}.{P}.\:{If}\:{the}\: \\ $$$${sum}\:{to}\:{infinity}\:{of}\:{a}\:{G}.{P}\:{is}\:\mathrm{3}+\sqrt{\mathrm{5}}\:,\:{find}\: \\ $$$${the}\:{first}\:{term}. \\ $$$$ \\ $$

Question Number 21622    Answers: 2   Comments: 0

If sec x + tan x = 2012 then 2011(cosec x + cot x) is equal to (A) 2011 (B) 2012 (C) 2013 (D) ((2011)/(2013)) (E) ((2013)/(2012))

$$\mathrm{If}\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:=\:\mathrm{2012} \\ $$$$\mathrm{then}\:\mathrm{2011}\left(\mathrm{cosec}\:{x}\:+\:\mathrm{cot}\:{x}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({A}\right)\:\mathrm{2011} \\ $$$$\left({B}\right)\:\mathrm{2012} \\ $$$$\left({C}\right)\:\mathrm{2013} \\ $$$$\left({D}\right)\:\frac{\mathrm{2011}}{\mathrm{2013}} \\ $$$$\left({E}\right)\:\frac{\mathrm{2013}}{\mathrm{2012}} \\ $$

Question Number 21643    Answers: 0   Comments: 1

(1/(1 + 1^2 + 1^4 )) + (2/(1 + 2^2 + 2^4 )) + (3/(1 + 3^2 + 3^4 )) + ... + ((2012)/(1 + 2012^2 + 2012^4 ))

$$\frac{\mathrm{1}}{\mathrm{1}\:+\:\mathrm{1}^{\mathrm{2}} \:+\:\mathrm{1}^{\mathrm{4}} }\:+\:\frac{\mathrm{2}}{\mathrm{1}\:+\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{2}^{\mathrm{4}} }\:+\:\frac{\mathrm{3}}{\mathrm{1}\:+\:\mathrm{3}^{\mathrm{2}} \:+\:\mathrm{3}^{\mathrm{4}} }\:+\:...\:+\:\frac{\mathrm{2012}}{\mathrm{1}\:+\:\mathrm{2012}^{\mathrm{2}} \:+\:\mathrm{2012}^{\mathrm{4}} } \\ $$

Question Number 21578    Answers: 0   Comments: 1

Find the whole part of A? A=(1/(√2))+(1/(√3))+(1/(√4))+......+(1/(√(9999)))+(1/(√(10000))).

$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{whole}}\:\:\boldsymbol{\mathrm{part}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{A}}? \\ $$$$\boldsymbol{\mathrm{A}}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{4}}}+......+\frac{\mathrm{1}}{\sqrt{\mathrm{9999}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{10000}}}. \\ $$

Question Number 21541    Answers: 0   Comments: 0

Question Number 21531    Answers: 1   Comments: 2

Factorise the equation by factor theorem 12x^ 3 + 4x^ 2−3x−1

$${Factorise}\:{the}\:{equation}\:{by}\:{factor}\:{theorem} \\ $$$$\mathrm{12}\hat {{x}}\mathrm{3}\:+\:\mathrm{4}\hat {{x}}\mathrm{2}−\mathrm{3}{x}−\mathrm{1} \\ $$

Question Number 21526    Answers: 1   Comments: 3

Question Number 21498    Answers: 0   Comments: 0

a^(−2 ) + b^3 + c^(−4) = ((433)/(499)) Find a + b + c

$${a}^{−\mathrm{2}\:} +\:{b}^{\mathrm{3}} \:+\:{c}^{−\mathrm{4}} \:=\:\frac{\mathrm{433}}{\mathrm{499}} \\ $$$$\mathrm{Find}\:{a}\:+\:{b}\:+\:{c} \\ $$

Question Number 21493    Answers: 1   Comments: 0

Is always a÷b = (a/b) ?

$$\mathrm{Is}\:\mathrm{always}\:\mathrm{a}\boldsymbol{\div}\mathrm{b}\:=\:\frac{\mathrm{a}}{\mathrm{b}}\:? \\ $$

Question Number 21471    Answers: 2   Comments: 0

If a + b + c = 0, then (((a + b)(b + c)(a + c))/(abc)) is equal to ...

$$\mathrm{If}\:\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\frac{\left({a}\:+\:{b}\right)\left({b}\:+\:{c}\right)\left({a}\:+\:{c}\right)}{{abc}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:... \\ $$

Question Number 21470    Answers: 1   Comments: 1

2^x = 3^y = 6^(−z) Find the value of (((2017)/x) + ((2017)/y) + ((2017)/z))^(2017)

$$\mathrm{2}^{{x}} \:=\:\mathrm{3}^{{y}} \:=\:\mathrm{6}^{−{z}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\left(\frac{\mathrm{2017}}{{x}}\:+\:\frac{\mathrm{2017}}{{y}}\:+\:\frac{\mathrm{2017}}{{z}}\right)^{\mathrm{2017}} \\ $$

Question Number 21463    Answers: 0   Comments: 0

Let A be the collection of functions f : [0, 1] → R which have an infinite number of derivatives. Let A_0 ⊂ A be the subcollection of those functions f with f(0) = 0. Define D : A_0 → A by D(f) = df/dx. Use the mean value theorem to show that D is injective. Use the fundamental theorem of calculus to show that D is surjective.

$$\mathrm{Let}\:{A}\:\mathrm{be}\:\mathrm{the}\:\mathrm{collection}\:\mathrm{of}\:\mathrm{functions} \\ $$$${f}\::\:\left[\mathrm{0},\:\mathrm{1}\right]\:\rightarrow\:\mathbb{R}\:\mathrm{which}\:\mathrm{have}\:\mathrm{an}\:\mathrm{infinite} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{derivatives}.\:\mathrm{Let}\:{A}_{\mathrm{0}} \:\subset\:{A} \\ $$$$\mathrm{be}\:\mathrm{the}\:\mathrm{subcollection}\:\mathrm{of}\:\mathrm{those}\:\mathrm{functions} \\ $$$${f}\:\mathrm{with}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{0}.\:\mathrm{Define}\:{D}\::\:{A}_{\mathrm{0}} \:\rightarrow\:{A} \\ $$$$\mathrm{by}\:{D}\left({f}\right)\:=\:{df}/{dx}.\:\mathrm{Use}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{value} \\ $$$$\mathrm{theorem}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:{D}\:\mathrm{is}\:\mathrm{injective}. \\ $$$$\mathrm{Use}\:\mathrm{the}\:\mathrm{fundamental}\:\mathrm{theorem}\:\mathrm{of} \\ $$$$\mathrm{calculus}\:\mathrm{to}\:\mathrm{show}\:\mathrm{that}\:{D}\:\mathrm{is}\:\mathrm{surjective}. \\ $$

Question Number 21422    Answers: 1   Comments: 3

Find all integer values of a such that the quadratic expression (x + a)(x + 1991) + 1 can be factored as a product (x + b)(x + c) where b and c are integers.

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{values}\:\mathrm{of}\:{a}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{expression} \\ $$$$\left({x}\:+\:{a}\right)\left({x}\:+\:\mathrm{1991}\right)\:+\:\mathrm{1}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factored} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{product}\:\left({x}\:+\:{b}\right)\left({x}\:+\:{c}\right)\:\mathrm{where}\:{b}\:\mathrm{and} \\ $$$${c}\:\mathrm{are}\:\mathrm{integers}. \\ $$

Question Number 21357    Answers: 1   Comments: 0

Solve : log_(2x+3) x^2 < 1

$$\mathrm{Solve}\::\:\mathrm{log}_{\mathrm{2}{x}+\mathrm{3}} {x}^{\mathrm{2}} \:<\:\mathrm{1} \\ $$

Question Number 21356    Answers: 1   Comments: 0

Solve : (2^((3x−1)/(x−1)) )^(1/3) < 8^((x−3)/(3x−7))

$$\mathrm{Solve}\::\:\sqrt[{\mathrm{3}}]{\mathrm{2}^{\frac{\mathrm{3}{x}−\mathrm{1}}{{x}−\mathrm{1}}} }\:<\:\mathrm{8}^{\frac{{x}−\mathrm{3}}{\mathrm{3}{x}−\mathrm{7}}} \\ $$

Question Number 21355    Answers: 1   Comments: 0

Solve : ∣x^2 + 3x∣ + x^2 − 2 ≥ 0

$$\mathrm{Solve}\::\:\mid{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\mid\:+\:{x}^{\mathrm{2}} \:−\:\mathrm{2}\:\geqslant\:\mathrm{0} \\ $$

Question Number 21354    Answers: 0   Comments: 4

Solve : (√(2x + 5)) + (√(x − 1)) > 8

$$\mathrm{Solve}\::\:\sqrt{\mathrm{2}{x}\:+\:\mathrm{5}}\:+\:\sqrt{{x}\:−\:\mathrm{1}}\:>\:\mathrm{8} \\ $$

Question Number 21321    Answers: 1   Comments: 0

The number of real solutions of the equation 4x^(99) + 5x^(98) + 4x^(97) + 5x^(96) + ..... + 4x + 5 = 0 is

$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{4}{x}^{\mathrm{99}} \:+\:\mathrm{5}{x}^{\mathrm{98}} \:+\:\mathrm{4}{x}^{\mathrm{97}} \:+\:\mathrm{5}{x}^{\mathrm{96}} \:+ \\ $$$$.....\:+\:\mathrm{4}{x}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\mathrm{is} \\ $$

Question Number 21319    Answers: 0   Comments: 0

If x, y, z are three real numbers such that x + y + z = 4 and x^2 + y^2 + z^2 = 6, then (1) (2/3) ≤ x, y, z ≤ 2 (2) 0 ≤ x, y, z ≤ 2 (3) 1 ≤ x, y, z ≤ 3 (4) 2 ≤ x, y, z ≤ 3

$$\mathrm{If}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such} \\ $$$$\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{4}\:\mathrm{and}\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{6}, \\ $$$$\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\frac{\mathrm{2}}{\mathrm{3}}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{0}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{2}\:\leqslant\:{x},\:{y},\:{z}\:\leqslant\:\mathrm{3} \\ $$

Question Number 21316    Answers: 0   Comments: 0

Let p = (x_1 − x_2 )^2 + (x_1 − x_3 )^2 + .... + (x_1 − x_6 )^2 + (x_2 − x_3 )^2 + (x_2 − x_4 )^2 + .... + (x_2 − x_6 )^2 + .... + (x_5 − x_6 )^2 = Σ_(1≤i<j≤6) ^6 (x_i − x_j )^2 . Then the maximum value of p if each x_i (i = 1, 2, ....., 6) has the value 0 and 1 is

$$\mathrm{Let}\:{p}\:=\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:....\:+ \\ $$$$\left({x}_{\mathrm{1}} \:−\:{x}_{\mathrm{6}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{3}} \right)^{\mathrm{2}} \:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{4}} \right)^{\mathrm{2}} \:+ \\ $$$$....\:+\:\left({x}_{\mathrm{2}} \:−\:{x}_{\mathrm{6}} \right)^{\mathrm{2}} \:+\:....\:+\:\left({x}_{\mathrm{5}} \:−\:{x}_{\mathrm{6}} \right)^{\mathrm{2}} \:= \\ $$$$\underset{\mathrm{1}\leqslant{i}<{j}\leqslant\mathrm{6}} {\overset{\mathrm{6}} {\sum}}\left({x}_{{i}} \:−\:{x}_{{j}} \right)^{\mathrm{2}} . \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:{p}\:\mathrm{if}\:\mathrm{each} \\ $$$${x}_{{i}} \:\left({i}\:=\:\mathrm{1},\:\mathrm{2},\:.....,\:\mathrm{6}\right)\:\mathrm{has}\:\mathrm{the}\:\mathrm{value}\:\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{1}\:\mathrm{is} \\ $$

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