Question and Answers Forum

Relation and FunctionsQuestion and Answers: Page 1

Question Number 224360 Answers: 1 Comments: 1

Find the equation of the circle which touches the circles x² + y² - 2x + 4y = 1, x² + y² - 12x + 2y = 4 and x² + y² + 2x - 12y + 12 = 0.

Question Number 224233 Answers: 1 Comments: 0

Prove that ∀n≥2 e^(2n−1) −1 ≥ 2n(2n−1)

$$\mathrm{Prove}\:\mathrm{that}\:\forall\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{e}^{\mathrm{2n}−\mathrm{1}} −\mathrm{1}\:\geqslant\:\mathrm{2n}\left(\mathrm{2n}−\mathrm{1}\right) \\ $$

Question Number 223055 Answers: 1 Comments: 2

Question Number 221413 Answers: 7 Comments: 0

Question Number 221154 Answers: 0 Comments: 1

f(x)= (x/(∣ x ∣ + 1)) f(f(f(f(x)))) =?

$$\:{f}\left({x}\right)=\:\frac{{x}}{\mid\:{x}\:\mid\:+\:\mathrm{1}} \\ $$$$\:\:{f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)\:=? \\ $$

Question Number 221153 Answers: 1 Comments: 1

f(x)= (1/2^x ) + (1/3^x ) + (1/4^x ) + ... +(1/(4000^x )) f(2) + f(3) + f(4)+ ... =?

$$\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}^{{x}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{{x}} }\:+\:\frac{\mathrm{1}}{\mathrm{4}^{{x}} }\:+\:...\:+\frac{\mathrm{1}}{\mathrm{4000}^{{x}} } \\ $$$$\:\:{f}\left(\mathrm{2}\right)\:+\:{f}\left(\mathrm{3}\right)\:+\:{f}\left(\mathrm{4}\right)+\:...\:=? \\ $$

Question Number 220869 Answers: 1 Comments: 1

Find the maximum value of x^2 y^3 z^4 subject to the condition x+y+z=18

$${Find}\:{the}\:{maximum}\:{value}\:{of}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} \:{subject}\:{to}\:{the}\:{condition}\:{x}+{y}+{z}=\mathrm{18} \\ $$

Question Number 219733 Answers: 2 Comments: 0

Question Number 219377 Answers: 1 Comments: 0

If ((fog)^(−1) of)(x)= 3x−8 find g(5).

$$\:\mathrm{If}\:\left(\left(\mathrm{fog}\right)^{−\mathrm{1}} \mathrm{of}\right)\left(\mathrm{x}\right)=\:\mathrm{3x}−\mathrm{8} \\ $$$$\:\mathrm{find}\:\mathrm{g}\left(\mathrm{5}\right). \\ $$

Question Number 219090 Answers: 2 Comments: 0

Prove that the sequence a_n =(1/( ((n!))^(1/n) )) is decreasing.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sequence}\:{a}_{{n}} =\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{n}!}}\:\mathrm{is}\:\mathrm{decreasing}. \\ $$

Question Number 218673 Answers: 3 Comments: 1

Question Number 218311 Answers: 1 Comments: 0

Question Number 218256 Answers: 2 Comments: 0

Question Number 217725 Answers: 0 Comments: 2

Question Number 216077 Answers: 1 Comments: 0

Find the largest value of the non negative integer p for which lim_(x→1) {((− px + sin(x − 1) + p)/(x + sin(x − 1) − 1))}^((1 − x)/(1 − (√x))) = (1/4) .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{non}\:\mathrm{negative} \\ $$$$\mathrm{integer}\:{p}\:\mathrm{for}\:\mathrm{which}\: \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left\{\frac{−\:{px}\:+\:\mathrm{sin}\left({x}\:−\:\mathrm{1}\right)\:+\:{p}}{{x}\:+\:\mathrm{sin}\left({x}\:−\:\mathrm{1}\right)\:−\:\mathrm{1}}\right\}^{\frac{\mathrm{1}\:−\:{x}}{\mathrm{1}\:−\:\sqrt{{x}}}} \:=\:\frac{\mathrm{1}}{\mathrm{4}}\:. \\ $$

Question Number 215887 Answers: 0 Comments: 2

Question Number 215331 Answers: 1 Comments: 0

Question Number 214644 Answers: 1 Comments: 0

((−6)/7)/((−7)/6)

$$\frac{−\mathrm{6}}{\mathrm{7}}/\frac{−\mathrm{7}}{\mathrm{6}} \\ $$

Question Number 213291 Answers: 1 Comments: 0

Find domain of y_(213291) : y_(213291) =((3+e^((x^2 −3x+2)/(x−6)) )/(log_(3/4) (√(x^2 −(1/4)))))

$${Find}\:{domain}\:{of}\:{y}_{\mathrm{213291}} : \\ $$$${y}_{\mathrm{213291}} =\frac{\mathrm{3}+{e}^{\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}}{{x}−\mathrm{6}}} }{\mathrm{log}_{\frac{\mathrm{3}}{\mathrm{4}}} \sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$

Question Number 212518 Answers: 2 Comments: 0

⇃

$$\:\:\:\cancel{\underbrace{\downharpoonleft}\underline{}\:} \\ $$

Question Number 209707 Answers: 0 Comments: 0

a,b ∈C : ab^− + b = 0 f : z′ = az^− + b such that f(M) = M′ 1. let z_A = z and z_(A′) = z′ and f(A) = A show that 2Re(b^− z) = bb^− (A is the set of invariant points and describes a line (△) ) 2. Deduce that (△) is a line with gradient u^( →) with affix z_u^→ = ib 3. show that (z_(MM ′) /z_u ) = ((bb^− − 2Re(bz^− ))/(ibb^− )) 4. show that 2Re(b^− z_0 ) = bb^_ where z_0 = ((z + z ′)/2) 5. Deduce that for M ∉ (△) , M is a perpendicular bisector of [MM ′]

$${a},{b}\:\in\mathbb{C}\::\:{a}\overset{−} {{b}}\:+\:{b}\:=\:\mathrm{0}\:{f}\::\:{z}'\:=\:{a}\overset{−} {{z}}\:+\:{b}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:{f}\left({M}\right)\:=\:{M}' \\ $$$$\mathrm{1}.\:{let}\:{z}_{{A}} \:=\:{z}\:{and}\:{z}_{{A}'} \:=\:{z}'\:{and}\:{f}\left({A}\right)\:=\:{A} \\ $$$${show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}\right)\:=\:{b}\overset{−} {{b}} \\ $$$$\left({A}\:{is}\:{the}\:{set}\:{of}\:{invariant}\:{points}\:{and}\right. \\ $$$$\left.\:{describes}\:{a}\:{line}\:\left(\bigtriangleup\right)\:\right) \\ $$$$\mathrm{2}.\:{Deduce}\:{that}\:\left(\bigtriangleup\right)\:{is}\:{a}\:{line}\:{with}\: \\ $$$${gradient}\:\overset{\:\rightarrow} {{u}}\:{with}\:{affix}\:{z}_{\overset{\rightarrow} {{u}}} \:=\:{ib} \\ $$$$\mathrm{3}.\:{show}\:{that}\:\frac{{z}_{{MM}\:'} }{{z}_{{u}} }\:=\:\frac{{b}\overset{−} {{b}}\:−\:\mathrm{2}{Re}\left({b}\overset{−} {{z}}\right)}{{ib}\overset{−} {{b}}} \\ $$$$\mathrm{4}.\:{show}\:{that}\:\mathrm{2}{Re}\left(\overset{−} {{b}z}_{\mathrm{0}} \right)\:=\:{b}\overset{\_} {{b}}\:{where} \\ $$$$\:{z}_{\mathrm{0}} \:=\:\frac{{z}\:+\:{z}\:'}{\mathrm{2}} \\ $$$$\mathrm{5}.\:{Deduce}\:{that}\:{for}\:{M}\:\notin\:\left(\bigtriangleup\right)\:,\:{M}\:{is}\: \\ $$$${a}\:{perpendicular}\:{bisector}\:{of}\:\left[{MM}\:'\right] \\ $$

Question Number 208533 Answers: 2 Comments: 0

z′ = (1/2)(z+(1/z)) z and z′ are complex numbers show that z = 2e^(iθ) show that M′ describes a conic section

$${z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({z}+\frac{\mathrm{1}}{{z}}\right) \\ $$$${z}\:{and}\:{z}'\:{are}\:{complex}\:{numbers} \\ $$$${show}\:{that}\:{z}\:=\:\mathrm{2}{e}^{{i}\theta} \\ $$$${show}\:{that}\:{M}'\:{describes}\:{a}\:{conic}\:{section} \\ $$

Question Number 208431 Answers: 2 Comments: 0

h_a (x) = e^(−x) + ax^2 show that h_a admits a minimum in R

$${h}_{{a}} \left({x}\right)\:=\:{e}^{−{x}} \:+\:{ax}^{\mathrm{2}} \\ $$$${show}\:{that}\:{h}_{{a}} \:{admits}\:{a}\:{minimum}\:{in}\:\mathbb{R} \\ $$

Question Number 208418 Answers: 1 Comments: 0

u_(n+1) = u_n −u_n ^3 ; u_0 ∈ ]0, 1[ . show that u_n ∈ ]0, 1[ . show that u_n converges to 0 v_n = (1/u_(n+1) ^2 ) − (1/u_n ^2 ) . express v_n interms of u_n . show that v_n converges to 2 f(x) = ((2−x)/((1−x)^2 )) . show that f is increasing and deduce that v_n is decreasing . show that v_n ≥ 2

$$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} \:;\:{u}_{\mathrm{0}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$\left..\:{show}\:{that}\:{u}_{{n}} \:\in\:\right]\mathrm{0},\:\mathrm{1}\left[\right. \\ $$$$.\:{show}\:{that}\:{u}_{{n}} \:{converges}\:{to}\:\mathrm{0} \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} } \\ $$$$.\:{express}\:{v}_{{n}} \:{interms}\:{of}\:{u}_{{n}} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:{converges}\:{to}\:\mathrm{2} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$$.\:{show}\:{that}\:{f}\:{is}\:{increasing}\:{and}\:{deduce}\:{that}\: \\ $$$${v}_{{n}} \:{is}\:{decreasing} \\ $$$$.\:{show}\:{that}\:{v}_{{n}} \:\geqslant\:\mathrm{2} \\ $$

Question Number 208103 Answers: 1 Comments: 0

Find inf{(m/n) ∣ m, n ∈ N, m<n−2}

$$\mathrm{Find}\:\mathrm{inf}\left\{\frac{{m}}{{n}}\:\mid\:{m},\:{n}\:\in\:\mathbb{N},\:{m}<{n}−\mathrm{2}\right\} \\ $$

Question Number 207390 Answers: 0 Comments: 1

Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

Contact: info@tinkutara.com