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Question Number 19785 Answers: 1 Comments: 0

If x^((x^4 )) = 4, what is the value of x^((x^2 )) + x^((x^8 )) ?

$$\mathrm{If}\:{x}^{\left({x}^{\mathrm{4}} \right)} \:=\:\mathrm{4},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${x}^{\left({x}^{\mathrm{2}} \right)} \:+\:{x}^{\left({x}^{\mathrm{8}} \right)} ? \\ $$

Question Number 19688 Answers: 1 Comments: 0

The vertices of a square are z_1 , z_2 , z_3 and z_4 taken in the anticlockwise order, then z_3 = (1) −iz_1 + (1 + i)z_2 (2) iz_1 + (1 + i)z_2 (3) z_1 + (1 + i)z_2 (4) (1 + i)z_1 + z_2

$$\mathrm{The}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{square}\:\mathrm{are}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \\ $$$$\mathrm{and}\:{z}_{\mathrm{4}} \:\mathrm{taken}\:\mathrm{in}\:\mathrm{the}\:\mathrm{anticlockwise}\:\mathrm{order}, \\ $$$$\mathrm{then}\:{z}_{\mathrm{3}} \:= \\ $$$$\left(\mathrm{1}\right)\:−{iz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{iz}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:{z}_{\mathrm{1}} \:+\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{1}\:+\:{i}\right){z}_{\mathrm{1}} \:+\:{z}_{\mathrm{2}} \\ $$

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 19683 Answers: 1 Comments: 0

In an A.P; the common difference is −2 and the largest term exceeds the middle term by 58. Find the smallest term of the A.P.

$$\mathrm{In}\:\mathrm{an}\:\mathrm{A}.\mathrm{P};\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{is}\:−\mathrm{2}\:\mathrm{and}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{term}\:\:\mathrm{exceeds}\:\mathrm{the}\:\mathrm{middle}\:\mathrm{term}\:\mathrm{by}\:\mathrm{58}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{A}.\mathrm{P}. \\ $$

Question Number 19679 Answers: 0 Comments: 1

Question Number 19675 Answers: 1 Comments: 1

Question Number 19668 Answers: 0 Comments: 2

Question Number 19665 Answers: 0 Comments: 0

Question Number 19666 Answers: 1 Comments: 0

∫ x^4 (√(x^2 + 1)) dx

$$\int\:{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\:{dx} \\ $$

Question Number 19659 Answers: 1 Comments: 0

((1+secθ)/(secθ))=((sin^(2 ) θ)/(1−cosθ))

$$\frac{\mathrm{1}+{sec}\theta}{{sec}\theta}=\frac{{sin}^{\mathrm{2}\:} \theta}{\mathrm{1}−{cos}\theta} \\ $$$$ \\ $$

Question Number 19658 Answers: 0 Comments: 0

=((sec x−1)/(sec x+1)) =((sec x+(1−1)−1)/(sec x+1)) =(((sec x+1)−2)/(sec x+1)) =((sec x+1)/(sec x+1))−(2/(sec x+1)) =1−(2/(sec x+1)) =1−2(sec x+1)^(−1) (d/dx)(1−2(sec x+1)^(−1) ) =−2(−1(sec x+1)^(−2) )(sec x tan x) =((2 sec x tan x)/((sec x+1)^2 ))

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 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

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

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