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Question Number 94078 Answers: 0 Comments: 0

Given the function f defined by f(x) = ((∣x−2∣)/(1−∣x∣)) (i) state the domain of f. (ii) show that f(x) = { ((((2−x)/(1+x)) , x < 0)),((((2−x)/(1−x)), 0 ≤ x < 2)),((((x−2)/(1−x)) , x ≥ 2)) :} (iii) Investigate the continuity of f at x = 2.

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function}\:{f}\:\mathrm{defined}\:\mathrm{by}\:{f}\left({x}\right)\:=\:\frac{\mid{x}−\mathrm{2}\mid}{\mathrm{1}−\mid{x}\mid} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{state}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:{f}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:{f}\left({x}\right)\:=\:\begin{cases}{\frac{\mathrm{2}−{x}}{\mathrm{1}+{x}}\:,\:{x}\:<\:\mathrm{0}}\\{\frac{\mathrm{2}−{x}}{\mathrm{1}−{x}},\:\mathrm{0}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{\frac{{x}−\mathrm{2}}{\mathrm{1}−{x}}\:,\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{Investigate}\:\mathrm{the}\:\mathrm{continuity}\:\mathrm{of}\:{f}\:\mathrm{at}\:{x}\:=\:\mathrm{2}. \\ $$

Question Number 93787 Answers: 0 Comments: 4

Question Number 93730 Answers: 0 Comments: 3

what are the reasons for not using x = ((2c)/(−b ±(√(b^2 −4ac)))) as the quadratic formula? i proved it.

$$\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{reasons}\:\mathrm{for}\:\mathrm{not}\:\mathrm{using}\: \\ $$$$\:{x}\:=\:\frac{\mathrm{2}{c}}{−{b}\:\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\:\:\mathrm{as}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{formula}?\: \\ $$$$\mathrm{i}\:\mathrm{proved}\:\mathrm{it}. \\ $$

Question Number 93691 Answers: 1 Comments: 3

If Σ_(i = 1) ^(10) (x_i +4) = 60 then find the value of x^ .

$$\boldsymbol{\mathrm{If}}\:\underset{\boldsymbol{{i}}\:=\:\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\left(\boldsymbol{\mathrm{x}}_{\mathrm{i}} +\mathrm{4}\right)\:=\:\mathrm{60}\:\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\bar {\boldsymbol{{x}}}. \\ $$

Question Number 93483 Answers: 0 Comments: 0

prove that the equation of the normal to the rectangular hyperbola xy = c^2 at the point P(ct, c/t) is t^3 x −ty = c(t^4 −1). the normal to P on the hyperbola meets the x−axis at Q and the tangent to P meets the yaxis at R. show that the locus of the midpoint oc QR, as P varies is 2c^2 xy + y^4 = c^4 .

Question Number 93474 Answers: 2 Comments: 0

Q. Prove by mathematical induction that Σ_(r=1) ^n (4r + 5) = 2n^2 + 7n

$$\mathrm{Q}.\:\mathrm{Prove}\:\mathrm{by}\:\mathrm{mathematical}\:\mathrm{induction}\:\mathrm{that}\: \\ $$$$\:\:\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left(\mathrm{4}{r}\:+\:\mathrm{5}\right)\:=\:\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{7}{n}\: \\ $$

Question Number 93397 Answers: 0 Comments: 2

prove that ((1−tan^3 θ)/(1 + tan^3 θ)) = 1−2sin^3 θ

$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{3}} \theta}{\mathrm{1}\:+\:\mathrm{tan}^{\mathrm{3}} \theta}\:=\:\mathrm{1}−\mathrm{2sin}^{\mathrm{3}} \theta\: \\ $$

Question Number 93368 Answers: 1 Comments: 0

given that f(r)= sin (1 + 2r)θ show that f(r)−f(r−1) = 2 cos 2r θ sin θ hence show that Σ_(r=1) ^n cos 2r θ sin θ = cos (n +1)θ sin nθ

$$\mathrm{given}\:\mathrm{that}\:{f}\left({r}\right)=\:\:\mathrm{sin}\:\left(\mathrm{1}\:+\:\mathrm{2}{r}\right)\theta \\ $$$$\mathrm{show}\:\mathrm{that}\:{f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{r}\:\theta\:\mathrm{sin}\:\theta \\ $$$$\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\mathrm{2}{r}\:\theta\:\mathrm{sin}\:\theta\:\:=\:\mathrm{cos}\:\left({n}\:+\mathrm{1}\right)\theta\:\mathrm{sin}\:{n}\theta \\ $$

Question Number 93367 Answers: 0 Comments: 2

show using demoives theorem that sin^2 θ cos^2 4θ =(1/8)(1−cos4θ)

$$\mathrm{show}\:\mathrm{using}\:\mathrm{demoives}\:\mathrm{theorem}\:\mathrm{that}\: \\ $$$$\mathrm{sin}^{\mathrm{2}} \theta\:\mathrm{cos}^{\mathrm{2}} \mathrm{4}\theta\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}−\mathrm{cos4}\theta\right) \\ $$

Question Number 93366 Answers: 1 Comments: 0

solve the equation (z−2)^3 = (1/2)−i((√3)/2)

$$\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\left({z}−\mathrm{2}\right)^{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

Question Number 93365 Answers: 0 Comments: 3

given that α is a real number, use mathematical induction or otherwise to show that cos ((α/2))cos((α/2^2 ))cos((α/2^3 )) ...cos((α/2^n )) = ((sin α)/(2^n sin((α/2^n )))) hence find the lim_(n→∞) cos((α/2))cos((α/2^2 ))cos((α/2^3 )) ... cos((α/2^n ))

Question Number 93220 Answers: 0 Comments: 5

Please in an arithmetic mean a, A_1 , A_2 , A_3 , ... , A_n , b where A_1 , A_2 , A_3 , ... , A_n are nth arithmetic mean why is b = (n + 2)th term: like T_(n + 2) Please

Question Number 93184 Answers: 0 Comments: 0

in solving the linear congruence ax ≡ b (mod n) ⇒ n∣(ax − b) ⇒ ax −b = kn ⇔ ax −kn = b ⇒ solving the linear diophantine equation ax −kn = b what are the general solution to the equation ax−kn = b

$$\mathrm{in}\:\mathrm{solving}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{congruence} \\ $$$${ax}\:\equiv\:{b}\:\left(\mathrm{mod}\:{n}\right)\:\Rightarrow\:{n}\mid\left({ax}\:−\:{b}\right)\:\Rightarrow\:{ax}\:−{b}\:=\:{kn}\:\Leftrightarrow\:{ax}\:−{kn}\:=\:{b} \\ $$$$\Rightarrow\:\mathrm{solving}\:\mathrm{the}\:\mathrm{linear}\:\mathrm{diophantine}\:\mathrm{equation}\:{ax}\:−{kn}\:=\:{b} \\ $$$$\:\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:{ax}−{kn}\:=\:{b} \\ $$$$\: \\ $$$$ \\ $$

Question Number 93118 Answers: 1 Comments: 0

sin(x)=a −1≤a≤1

$${sin}\left({x}\right)={a} \\ $$$$−\mathrm{1}\leqslant{a}\leqslant\mathrm{1} \\ $$

Question Number 93044 Answers: 1 Comments: 2

is (8/9) is the multiplicative inerse of −1 (1/8)? why or why not?

$$\mathrm{is}\:\frac{\mathrm{8}}{\mathrm{9}}\:\mathrm{is}\:\mathrm{the}\:\mathrm{multiplicative}\:\mathrm{inerse}\:\mathrm{of}\:−\mathrm{1}\:\frac{\mathrm{1}}{\mathrm{8}}? \\ $$$$\mathrm{why}\:\mathrm{or}\:\mathrm{why}\:\mathrm{not}? \\ $$

Question Number 93081 Answers: 0 Comments: 3

Solve x^y =y^x x, y ∈ N

$$\mathrm{Solve}\:\mathrm{x}^{\mathrm{y}} =\mathrm{y}^{\mathrm{x}} \:\:\:\mathrm{x},\:\mathrm{y}\:\in\:\mathbb{N} \\ $$

Question Number 92910 Answers: 2 Comments: 2

∫(dt/(3sint+4cost))

$$\int\frac{\mathrm{dt}}{\mathrm{3sint}+\mathrm{4cost}} \\ $$

Question Number 92888 Answers: 1 Comments: 3

∫((5−t)/(1+(√((t−4)))))dt

$$\int\frac{\mathrm{5}−\mathrm{t}}{\mathrm{1}+\sqrt{\left(\mathrm{t}−\mathrm{4}\right)}}\mathrm{dt} \\ $$$$ \\ $$

Question Number 92898 Answers: 0 Comments: 0

Find a,b,c ∈ Z that satisfy (7a + 15b + 0c) mod 26 = 8 (5a + 16b + 6c) mod 26 = 21 (6a + 3b + 20c) mod 26 = 14

$$\mathrm{Find}\:{a},{b},{c}\:\in\:\mathbb{Z}\:\mathrm{that}\:\mathrm{satisfy} \\ $$$$\left(\mathrm{7}{a}\:+\:\mathrm{15}{b}\:+\:\mathrm{0}{c}\right)\:\mathrm{mod}\:\mathrm{26}\:=\:\mathrm{8} \\ $$$$\left(\mathrm{5}{a}\:+\:\mathrm{16}{b}\:+\:\mathrm{6}{c}\right)\:\mathrm{mod}\:\mathrm{26}\:=\:\mathrm{21} \\ $$$$\left(\mathrm{6}{a}\:+\:\mathrm{3}{b}\:+\:\mathrm{20}{c}\right)\:\mathrm{mod}\:\mathrm{26}\:=\:\mathrm{14} \\ $$

Question Number 92788 Answers: 0 Comments: 0

Question Number 92708 Answers: 1 Comments: 0

solve the differential equations. (a) (x + 3y^2 )(d^2 y/dx^2 ) + 6y ((dy/dx))^2 + 2(dy/dx) + 2 = 0 (b) (2y−x)(d^2 y/dx^2 ) + 2((dy/dx))^2 −2 (dy/dx) + 2 = 0

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equations}. \\ $$$$\:\left(\mathrm{a}\right)\:\left({x}\:+\:\mathrm{3}{y}^{\mathrm{2}} \right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{6}{y}\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:+\:\mathrm{2}\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{b}\right)\:\left(\mathrm{2}{y}−{x}\right)\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{2}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \:−\mathrm{2}\:\frac{{dy}}{{dx}}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$

Question Number 92712 Answers: 0 Comments: 2

find m for fix function f(x)=(((m−1)x+3)/(x−1))

$${find}\:\:\boldsymbol{{m}}\:{for}\:{fix}\:{function} \\ $$$${f}\left({x}\right)=\frac{\left(\boldsymbol{{m}}−\mathrm{1}\right){x}+\mathrm{3}}{{x}−\mathrm{1}} \\ $$

Question Number 92605 Answers: 1 Comments: 4

If ((1+x)/(1+(√(1+x)))) +((1−x)/(1−(√(1−x)))) =1 find x

$${If}\:\frac{\mathrm{1}+{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}}}\:+\frac{\mathrm{1}−{x}}{\mathrm{1}−\sqrt{\mathrm{1}−{x}}}\:=\mathrm{1} \\ $$$${find}\:{x} \\ $$

Question Number 92593 Answers: 0 Comments: 0

using the squeeze theorem show that lim_(x→a) (√x) = (√a)

$$\mathrm{using}\:\mathrm{the}\:\mathrm{squeeze}\:\mathrm{theorem}\: \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\sqrt{{x}}\:=\:\sqrt{{a}}\: \\ $$

Question Number 92555 Answers: 0 Comments: 5

Σ_(n=1) ^∞ 1/n^2 =π^2 /6

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{1}/{n}^{\mathrm{2}} =\pi^{\mathrm{2}} /\mathrm{6} \\ $$

Question Number 92529 Answers: 0 Comments: 0

two smooth spheres of masses 2m and 3m have velocites (−12i + 8j) u ms^(−1) and (5i + 12j)u , respectively where u is a constant. The spheres collide with thier line of centre of parallel to j. Given that the coefficient of restitution between the spheres is (1/4), find the loss in kinetic energy due to impact.

$$\mathrm{two}\:\mathrm{smooth}\:\mathrm{spheres}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}{m}\:\mathrm{and}\:\mathrm{3}{m}\:\mathrm{have}\:\mathrm{velocites} \\ $$$$\left(−\mathrm{12}\boldsymbol{\mathrm{i}}\:+\:\mathrm{8}\boldsymbol{\mathrm{j}}\right)\:{u}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{and}\:\left(\mathrm{5}\boldsymbol{\mathrm{i}}\:+\:\mathrm{12}\boldsymbol{\mathrm{j}}\right){u}\:,\:\mathrm{respectively}\:\mathrm{where}\:{u}\: \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}.\:\mathrm{The}\:\mathrm{spheres}\:\mathrm{collide}\:\mathrm{with}\:\mathrm{thier}\:\mathrm{line}\:\mathrm{of}\:\mathrm{centre}\:\mathrm{of} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\boldsymbol{\mathrm{j}}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{restitution}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{spheres}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{find}\:\mathrm{the}\:\mathrm{loss}\:\mathrm{in}\:\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{due}\:\mathrm{to}\:\mathrm{impact}. \\ $$

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