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Re: Nairaland Mathematics Clinic by Nobody: 4:14pm On Jun 04, 2015 |
tonywirelex:| 3-x | ≥ 1, i.e. 3-x ≥ 1 & -(3-x) ≥ 1 -x ≥ 1 - 3 & -3+x ≥ 1 x ≤ 2 & x ≥ 4 (2) solve x > x² x> x² x-x²>0 x(1-x)>0 x > 0 & 1 - x > 0 x > 0 & 1 > x or 1 > x > 0. X will be numbers between 1 and 0, e.g. 1/2 (3) find the domain f(x)= 1/√1-x² Singularities at x = 1 and x = -1 define domain. Pretty much all real numbers except 1 & -1 U ( 2 , infinity ), U ( -infinity , -2 ), 0 (4) if f(x)= x² - 1 then f(secθ) - f(tanθ) =?f(secθ) - f(tanθ) = sec2θ - 1 - (tan2θ - 1) = sec2θ - 1 - tan2θ + 1 but sec2θ = 1 + tan2θ thus, f(secθ) - f(tanθ) = 1 (5) if f(x) = logx, then find f(2sinx) + f(cosx) at x=π/4f(2sinx) + f(cosx) = log(2sinx) + log(cosx) = log(2sinxcosx) = log(sin2x), which at x=π/4, is log(Sin(π/2)) = log1 = 0 (6) period of tanx and cotx is ?cotx = 1/tanx (derivative of same curve) tan0 = 0 tanπ = 0 Period is π (7) if f(x)= x+1/x, x≠0, then f(1/x) =?f(x)= x+1/x f(1/x)= 1/x+1/(1/x) = x + 1/x if f(x) = (x+1)/x, then f(1/x) = 1/(1+x) (8 ) if f(x) = (2x+1/3x-2), then f(f(2)) =?f(2) = 5/4 f(f(2)) = f(5/4) = (2(5/4)+1)/(3(5/4)-2) =(5/2+1)/(15/4-2) = 4/2=2 (9) lim (1+2x)^1/x as x→0 =?lim (1+x)^x as x→∞ = e x = 1/2h in equation, then lim (1+1/h)^2h as h→∞ = e2, if approaching x approaching 0 from right (10) evaluate lim 1-cos2x/x²(1+cos2x) as x→0 =? use l'hopital rule to find solution. DIfferentiate top and bottom, (-2sin2x)/(x²(-2sin2x) + 2x(1+cos2x)) differentiate again (4cos2x)/(x²(-4cos2x)+2x(-2sin2x)+2(1+cos2x)+2x(-2sin2x)) at x=0 three terms in the denominator vanish lim 1-cos2x/x²(1+cos2x) as x→0 = 4/4 = 1 (11) lim (1+(4/x))^x as x→∞ =? x = 4y lim (1+(1/y))4y as x→∞ = e4 (12) lim 2sinx-sin2x/x^3 as x→0 =?lim 2sinx-sin2x/x^3 as x→0 if, lim (2sinx-sin2x)/x3 as x→0 using l'hopital rule... differentiate num and denom (2cosx - 2cos2x)/(3x2) (-2sinx + 4 sin2x)/6x (-2cosx+8cos2x)/6 with x = 0 (at the origin) 6/6 = 1 hence, lim (2sinx-sin2x)/x3 as x→0 = 1 if, lim 2sinx-sin2x)/x3 as x→0, then using l'hopital rule lim2sinx as x→0 = 0 lim sin2x/x3 as x→0 8 cos2x/6, after differentiating twice lim sin2x/x3 as x→0 = 8/6 = 4/3 (13) evaluate lim x(π/2 - arc tanx) =?Incomplete question (14) derivative of x²/1+x² with respect to x² =?f(x) = x²/(1+x²) d/d(x²)(x²/(1+x²)) d/du(u/(1+u)) = 1/(1+u)² derivative is thus 1/(1+x²)² (15) if y=x^4 - 7x^3 + 3, then d^3 y/dx^3 at x=2 is? dy/dx = 4x3-21x2 d2y/dx2 = 12x2 - 42x d3y/dx3 = 24x - 42 at x = 2, d3y/dx3 = 6 (16) d/dx (2sin²x + cos2x) is ?2.2sinxcox - 2sin2x 2.sin2x - 2sin2x = 0. Also cos2x = 1 - 2sin²x d/dx(2sin²x + 1 - 2sin²x) = 0 (17) differntiate with respect to x, the function 2arc tan√xy = 2arc tan√x tan(y/2) = √x x = tan²(y/2) dx/dy = 2.tan(y/2). sec²(y/2). 1/2 dy/dx = 1/(dx/dy) but, sec²u = 1 + tan²u tan²(y/2) = x dx/dy = 2√x(1+x).1/2 dx/dy = √x(1+x) dy/dx = 1/(√x(1+x)) (18) find d/dx f(x) where f(x) = cosh^-1 (2x) at x=2 y = cosh-1(2x) 2x = coshy 2.dx/dy = sinhy cosh²y - sinh²y = 1 cosh²y - 1 = sinh²y sinhy =√(cosh²y - 1) 2.dx/dy = √(cosh²y - 1) 2.dx/dy = √(4x² - 1) dx/dy = √(4x² - 1)/2 dy/dx = 2/√(4x² - 1) at x = 2, dy/dx = 2/√15 (19) if x²/a² + logy²/b² =1, then dy/dx =?Even though solvable, question appears wrongly worded (20) differentiate x^sinx with respect to xy = xsinx logxy = sinx Iny/Inx = sinx Iny = Inxsinx (1/y)dy/dx = d(Inxsinx)/dx (1/y)dy/dx = Inx.cosx + (sinx)/x dy/dx = y.(Inx.cosx + (sinx)/x) dy/dx = xsinx(Inx.cosx + (sinx)/x) (21) find ∫(sec²x + tan²x)dx =?sec²x = 1 + tan²x =∫(sec²x + sec²x - 1)dx =∫(2sec²x - 1)dx =2tanx - x + C (22) evaluate ∫(x²-x+1/√x) dx=x3/3 - x2/2 + 2√x + C (23) ∫ dx/√4-9x² =? =arcsin(3x/2)/3 + C make x = (2/3)sinu to solve problem dx/du = (2/3)cosu ∫ dx/√(4-9x²) = ∫ dx/√(4-4sin²u) = ∫ (2/3)cosudu/2cosu =∫ (1/3)du =(1/3)u + C =(1/3)arcsin(3x/2) + C (24) determine ∫ dx/4+x²∫ dx/(4+x²) using x = 2sinhu dx/du = 2coshu x² = 4 sinh²u ∫ dx/4+x² = (1/2)∫ (1/coshu)du = (1/2)∫sechudu using standard tables, maybe =(1/2)arctan(sinhu) + C =(1/2)arctan(x/2) + C (25) ∫ (logx)²/x dx =?∫(Inx/In10)²/x dx = (1/In10)² ∫ (Inx)²/x dx u = Inx du/dx = 1/x dx = xdu (1/In10)² ∫ (Inx)²/x dx = (1/In10)² ∫ (u)²/x .xdu (1/In10)² ∫ (u)²/x .xdu (1/In10)² (u)3/3 (1/In10)² ( (Inx)3/3) + C (26) ∫ Interesting solving these questions... 5 Likes 1 Share |
Re: Nairaland Mathematics Clinic by Emdee590(m): 5:12pm On Jun 04, 2015 |
naturalwaves:Bro ! Na like that e be . Thank you |
Re: Nairaland Mathematics Clinic by Emdee590(m): 5:28pm On Jun 04, 2015 |
goofyone:Thank you man I appreciate but the option ticked here says 192 , I don't know how come Thank you |
Re: Nairaland Mathematics Clinic by Nobody: 5:36pm On Jun 04, 2015 |
Emdee590: Sorry, I had thought you meant only 4-digit numbers. Of course, 5-digit numbers can also be created from the number set. That will happen in 5 x 4 x 3 x 2 x 1 ways, i.e. 5! ways, since the numbers cannot be repeated. Only 4- and 5-digit numbers can be created. Hence 5! = 120 120+72 = 192. Sorry for the misunderstanding. |
Re: Nairaland Mathematics Clinic by Nobody: 7:10pm On Jun 04, 2015 |
benji93:Just what I was thinking... |
Re: Nairaland Mathematics Clinic by Geofavor(m): 9:42pm On Jun 04, 2015 |
bolaji3071: |
Re: Nairaland Mathematics Clinic by naturalwaves: 11:15pm On Jun 04, 2015 |
@goofyone, weldone bro. |
Re: Nairaland Mathematics Clinic by Nobody: 1:58am On Jun 05, 2015 |
naturalwaves:Thanks 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 2:48am On Jun 05, 2015 |
bolaji3071: (1-0.1)*(1+0.15)*old cost = new cost 1.035*old cost = new cost 35% increase in food bill 2. The minimum point on the curve y=x^2-6x+5 is atdy/dx = 2x - 6 at what point is there no change in dy/dx? At the maximum or minimum point, rate of change is slowest or zero 2x-6 = 0; x = 3 3. A man runs a distance of 9km/h for the first 4km and then 2km/h for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is question not clear 4. If 25^x-1=64(5/2)^fgdghh6, then x has the value?25(x-1) = 64.(5/2)6 52(x-1) = 26.(5/2)6 52(x-1) = 56 x = 4 5. Find the area of the curved surface of a cone whose base radius is 6cm and whose height is 8cm. (take pie =22/7)a cone is made up of an sector and a circle (?) Area of the sector is φ/360.πl2, where l is length of cone side length of arc is 2πr = φ/360.2πl Area of sector part of cone = 360.2πr/2πl. 1/360. πl2 = πrl Area = πrl with height of 8cm, and base radium of 6cm, l2 = 82+62 l = 10 Area = 6π.10 = 60π 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:32pm On Jun 05, 2015 |
bolkay47:Is that your final answer .? |
Re: Nairaland Mathematics Clinic by Emdee590(m): 5:08pm On Jun 05, 2015 |
goofyone:Thank you man |
Re: Nairaland Mathematics Clinic by ositadima1(m): 6:27pm On Jun 06, 2015 |
doubleDx: Bros, longest time o. |
Re: Nairaland Mathematics Clinic by Profmaojo: 12:07pm On Jun 07, 2015 |
Let x e (AUBUC)'[ X £ ( AUBUC) X £ A and x £ B and x £ C x e A' and x e B' and x e c' x e A' n B' n C' therefore AUBUC=A'nB'nC'........e=element of, £=not an element of quote author=thankyouJesus post=33628212] let me use "ojoro" method to prove: let u = {1, 2, . . . .10} a = {1,2,3} b = {4,5,6} c = {7,8,9} AUBUC = {1,2,3. . . 9} (AUBUC)' = {10} A'nB'nC' = {10}. QED[/quote] |
Re: Nairaland Mathematics Clinic by Soneh(m): 8:56pm On Jun 07, 2015 |
Please guys I need help in the following question. 1. Find the roots of the polynomial p(x) = x4 + 4x3 + 6x2 + 4x + 5=0 given that one of the roots is x = -i 2. Each of my 39 friends has either a dog, a cat or a rabbit. 24 of the have a dog, 17 have a cat and 16 have a rabbit. The number having both dog and cat is 1 more than the number having both cat and rabbit. There are 9 who have both dog and rabbit, while two of them have all three. How many of my friends have both rabbit and cat? 3. Prove that the equation mx(x2 + 2x + 3) = x2 – 2x – 3 has exactly one real root if m = 1 and exactly 3 real roots if m= -2/3 cc Dejt4u Benbuks jackpot Arithmetic Naturalwaves Richies Amazing angel Agentofchange1
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Re: Nairaland Mathematics Clinic by naturalwaves: 10:37pm On Jun 07, 2015 |
Soneh: The answer is 5....see solution attached below.
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Re: Nairaland Mathematics Clinic by Nobody: 11:51am On Jun 08, 2015 |
Soneh: If -i is a root, i will also be a root following the complex conjugate theorem. I can prove that if you want, but i don't think it's relevant to do that here. Hence, two roots already, ± i (x+i)(x-i) = x2 -i.i = x2 + 1 x4 + 4x3 + 6x2 + 4x + 5 divided by x2 + 1 to get the other factor. Division produces x2 + 4x + 5 Solving with quadratic equation: 1/2(-4 ± √(16 - 20)) x = -2 ± i 2 Likes |
Re: Nairaland Mathematics Clinic by dejt4u(m): 11:58am On Jun 08, 2015 |
goofyone:make sense 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 1:45pm On Jun 08, 2015 |
Soneh: For this question, you can use Descartes's rule of signs. You should actually use the more detailed Sturm's extension of this rule, but i think the former could be okay here. Descartes's rule of signs says if v is the variation of sign in a polynomial p(x) and the number of positive real zeros (roots) is n, then n<=v and v-n is even. Same applies to number of negative real zeros with p(-x) With m=1, you get x3 + x2 + 5x + 3 = 0 p(x) = x3 + x2 + 5x + 3 No sign variation, so v=0. n will also be 0 so there are no positive real roots. p(-x) = -x3 + x2 - 5x + 3 Number of sign variation is 3 (- to + to - to +) n will be 1, so that 3-1 is even. So there's a negative real root. Equation thus has one negative real root and two complex roots which will be conjugates. With m= -2/3 p(x) = -(2/3)x3 - (7/3)x2 + 3 Number of sign variation is 1. n will be 1 so that 1-1=0 which is even. Hence, one positive real root. p(-x) = (2/3)x3 - (7/3)x2 + 3. Number of sign variation here is 2. n cannot be 1, cos v-n should be even. So n is 2. And 2-2=0. So we have two negative real roots. Hence, three real roots (one positive and two negative) |
Re: Nairaland Mathematics Clinic by Soneh(m): 8:12pm On Jun 08, 2015 |
goofyone:i want to say a very big thank your to all of those gurus who helped me in solving the maths problem i pasted, i appreciate all of you with a million thanks. pls i still need you help on this other question> 1. let a function f be defined by f(x)=x2-2x-3/x2+2x+3 a. determine the domain of f(x) b. find the range of f(x) c. find and expression for f(3x+1) 2. show that | 1 1 1 | ................... | x y z | = (y-x)(z-x)(z-y) ( its a matrix problem) ....................| x2 y2 z2 | waiting for your help gurus........... |
Re: Nairaland Mathematics Clinic by naturalwaves: 8:23pm On Jun 08, 2015 |
Soneh: Check the first question well, there are 2 functions there, which of them do we go with? I guess it is the second one but quickly review it or take a snapshot of the question....waiting.... |
Re: Nairaland Mathematics Clinic by Soneh(m): 8:29pm On Jun 08, 2015 |
naturalwaves:modified |
Re: Nairaland Mathematics Clinic by Richiez(m): 9:38pm On Jun 08, 2015 |
Keep the good work going...it's good to see that the thread is very much alive! 1 Like |
Re: Nairaland Mathematics Clinic by naturalwaves: 10:23pm On Jun 08, 2015 |
Soneh: 1 Like
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Re: Nairaland Mathematics Clinic by naturalwaves: 10:26pm On Jun 08, 2015 |
Soneh: 1 Like 1 Share
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Re: Nairaland Mathematics Clinic by Nobody: 10:33pm On Jun 08, 2015 |
Soneh: | 1 1 1 | | x y z | | x2 y2 z2 | = (y.z2 - z.y2) - (x.z2 - z.x2) + (x.y2 - y.x2) =zy(z-y)-zx(z-x)+xy(y-x) =z(y(z-y) - x(z-x)) + xy(y-x) =z(yz-xz + x2 - y2) + xy(y-x) =z(z(y-x)+(x-y)(x+y))+xy(y-x) =z(z(y-x)-(y-x)(x+y))+xy(y-x) =z(y-x).(z-(x+y))+xy(y-x) =(y-x)(z.(z-(x+y)+xy) =(y-x)(z2-zx-zy+xy) =(y-x)(z2-zy+xy-zx) =(y-x)(z(z-y)-x(z-y)) =(y-x)(z-x)(z-y) 1 Like |
Re: Nairaland Mathematics Clinic by naturalwaves: 10:53pm On Jun 08, 2015 |
Soneh: I would have loved to see the snap shot of this question but this is what I got attached below. Modified.... Goofyone has done justice to the factorisation, check his solution.
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Re: Nairaland Mathematics Clinic by naturalwaves: 10:56pm On Jun 08, 2015 |
PS: all solutions are subject to converse opinions and criticisms. |
Re: Nairaland Mathematics Clinic by benji93: 1:14am On Jun 09, 2015 |
goofyone: |
Re: Nairaland Mathematics Clinic by benji93: 1:16am On Jun 09, 2015 |
goofyone: |
Re: Nairaland Mathematics Clinic by benji93: 1:19am On Jun 09, 2015 |
goofyone, you are right brother but you misstated. (1 + x)^1/x = e as x approaches 0 (1 + 1/x)^x =e as x approaches infinity. so as u rightly did we substitute 1/2h for x (1+2x)^1/x = (1 + 1/h)^2h = ((1 + 1/h)^h)^2 = e^2 as h approaches infinity but as h approaches infinity x approaches 0, therefore the limit is e^2 but you have really done well by solving all these, God bless this forum 1 Like |
Re: Nairaland Mathematics Clinic by benji93: 2:05am On Jun 09, 2015 |
hello guys i am imploring all Engineering and Physics students to join Nairaland Physics gurus, as there are certain physics questions that do not suit the purpose of this thread.I am thinking of forming an Engineering group but we can start with tish |
Re: Nairaland Mathematics Clinic by Soneh(m): 6:12am On Jun 09, 2015 |
goofyone:i want to say a big thank you to all maths gurus who helped me out in solving my mathematical problem, i love you all and i really appreciate you all..... 1 Like |
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