tonywirelex:
(1) solve | 3-x | ≥ 1

(2) solve x > x²

(3) find the domain f(x)= 1/√1-x²

(4) if f(x)= x² - 1 then f(secθ) - f(tanθ) =?

(5) if f(x) = logx, then find f(2sinx) + f(cosx) at x=π/4

(6) period of tanx and cotx is ?

(7) if f(x)= x+1/x, x≠0, then f(1/x) =?

(8 ) if f(x) = (2x+1/3x-2), then f(f(2)) =?

(9) lim (1+2x)^1/x as x→0 =?

(10) evaluate lim 1-cos2x/x²(1+cos2x) as x→0 =?

(11) lim (1+(4/x))^x as x→∞ =?

(12) lim 2sinx-sin2x/x^3 as x→0 =?

(13) evaluate lim x(π/2 - arc tanx) =?

(14) derivative of x²/1+x² with respect to x² =?

(15) if y=x^4 - 7x^3 + 3, then d^3 y/dx^3 at x=2 is?

(16) d/dx (2sin²x + cos2x) is ?

(17) differntiate with respect to x, the function 2arc tan√x

(18) find d/dx f(x) where f(x) = cosh^-1 (2x) at x=2

(19) if x²/a² + logy²/b² =1, then dy/dx =?

(20) differentiate x^sinx with respect to x

(21) find ∫(sec²x + tan²x)dx =?

(22) evaluate ∫(x²-x+1/√x) dx

(23) ∫ dx/√4-9x² =?

(24) determine ∫ dx/4+x²

(25) ∫ ^(logx)²/_x dx =?

(26) ∫

naturalwaves:

You omitted something in question 1. Check well and repost.

goofyone:


Question (1) missing an item

Question (2):
if number is to be greater than 4000, then it can only start with any of 4,5,6 (3 ways of starting), i.e. 3 x....
second term can have any of the other 4 numbers, once first term is fixed (no repetition), then third any of 3, then fourth any of 2.
Hence, 3 x 4 x 3 x 2 = 72 numbers

Emdee590:

Thank you man I appreciate but the option ticked here says 192 , I don't know how come

Thank you

benji93:

you right if line that defines this distribution is a straight,but in the case where it is a curve,hmm

bolaji3071:
Gurus in the house am here again o.
1. Mother reduced the quantity of food bought for the family by 10% when she found out that the cost of living had Increased by 15%. Thus the fractional Increase in the family food bill is now?

2. The minimum point on the curve y=x^2-6x+5 is at

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

4. If 25^x-1=64(5/2)^fgdghh6, then x has the value?

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)

naturalwaves:
@goofyone, weldone bro.

bolaji3071:
Gurus in the house am here again o.
1. Mother reduced the quantity of food bought for the family by 10% when she found out that the cost of living had Increased by 15%. Thus the fractional Increase in the family food bill is now?

2. The minimum point on the curve y=x^2-6x+5 is at

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

4. If 25^x-1=64(5/2)^fgdghh6, then x has the value?

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)

bolkay47:
y~n=5/2+C1{10/3}^n+C2{-7/10}^n

goofyone:


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.

doubleDx:


Wow! General benbuks, is that you? You've don't a great job here bruv....Good to see you guys keeping the thread alive! That's the spirit; keep up the good work guys. God bless y'all!

I have been kinda busy lately so I check around once in a while, but will start contributing soon!

I like this your new moniker sha, nice one!

Nice work people!

Hi chief General Ositadima1; nice to have you back! Hailings bruv!

Soneh:

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?

Soneh:

1. Find the roots of the polynomial p(x) = x⁴ + 4x³ + 6x² + 4x + 5=0 given that one of the roots is x = -i

goofyone:


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

Soneh:

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

goofyone:


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)

Soneh:

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
x² - 2x - 3

---

f(x)=
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...........

naturalwaves:


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....

Soneh:

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)

Soneh:


c. find and expression for f(3x+1)

Soneh:

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...........

Soneh:


2. show that | 1 1 1 |
................... | x y z | = (y-x)(z-x)(z-y) ( its a matrix problem)
....................| x² y² z² |

goofyone:

| 3-x | ≥ 1, i.e. 3-x ≥ 1 & -(3-x) ≥ 1
-x ≥ 1 - 3 & -3+x ≥ 1
x ≤ 2 & x ≥ 4




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



Singularities at x = 1 and x = -1 define domain. Pretty much all real numbers except 1 & -1
U ( 2 , infinity ), U ( -infinity , -2 ), 0


f(secθ) - f(tanθ) = sec²θ - 1 - (tan²θ - 1)
= sec²θ - 1 - tan²θ + 1
but sec²θ = 1 + tan²θ
thus, f(secθ) - f(tanθ) = 1



f(2sinx) + f(cosx) = log(2sinx) + log(cosx)
= log(2sinxcosx) = log(sin2x), which at x=π/4, is log(Sin(π/2)) = log1 = 0


cotx = 1/tanx (derivative of same curve)
tan0 = 0
tanπ = 0
Period is π


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)


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




lim (1+x)^x as x→∞ = e
x = 1/2h in equation, then
lim (1+h)^2h as h→∞ = e², if approaching x approaching 0 from righted

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
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




x = 4y

lim (1+(1/y))^4y as x→∞

= e⁴


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



Incomplete question



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




dy/dx = 4x³-21x²
d²y/dx² = 12x² - 42x
d³y/dx³ = 24x - 42
at x = 2, d³y/dx³ = 6


2.2sinxcox - 2sin2x
2.sin2x - 2sin2x = 0.

Also cos2x = 1 - 2sin²x
d/dx(2sin²x + 1 - 2sin²x) = 0


y = 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))




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


Even though solvable, question appears wrongly worded


y = x^sinx
log_xy = 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 = x^sinx(Inx.cosx + (sinx)/x)


sec²x = 1 + tan²x
=∫(sec²x + sec²x - 1)dx
=∫(2sec²x - 1)dx
=2tanx - x + C



=x³/3 - x²/2 + 2√x + C





=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



∫ 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



∫(^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




Interesting solving these questions...

goofyone:

| 3-x | ≥ 1, i.e. 3-x ≥ 1 & -(3-x) ≥ 1
-x ≥ 1 - 3 & -3+x ≥ 1
x ≤ 2 & x ≥ 4




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



Singularities at x = 1 and x = -1 define domain. Pretty much all real numbers except 1 & -1
U ( 2 , infinity ), U ( -infinity , -2 ), 0


f(secθ) - f(tanθ) = sec²θ - 1 - (tan²θ - 1)
= sec²θ - 1 - tan²θ + 1
but sec²θ = 1 + tan²θ
thus, f(secθ) - f(tanθ) = 1



f(2sinx) + f(cosx) = log(2sinx) + log(cosx)
= log(2sinxcosx) = log(sin2x), which at x=π/4, is log(Sin(π/2)) = log1 = 0


cotx = 1/tanx (derivative of same curve)
tan0 = 0
tanπ = 0
Period is π


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)


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




lim (1+x)^x as x→∞ = e
x = 1/2h in equation, then
lim (1+h)^2h as h→∞ = e², if approaching x approaching 0 from righted

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
use l'hopital 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




x = 4y

lim (1+(1/y))^4y as x→∞

= e⁴


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



Incomplete question



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




dy/dx = 4x³-21x²
d²y/dx² = 12x² - 42x
d³y/dx³ = 24x - 42
at x = 2, d³y/dx³ = 6


2.2sinxcox - 2sin2x
2.sin2x - 2sin2x = 0.

Also cos2x = 1 - 2sin²x
d/dx(2sin²x + 1 - 2sin²x) = 0


y = 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))




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


Even though solvable, question appears wrongly worded


y = x^sinx
log_xy = 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 = x^sinx(Inx.cosx + (sinx)/x)


sec²x = 1 + tan²x
=∫(sec²x + sec²x - 1)dx
=∫(2sec²x - 1)dx
=2tanx - x + C



=x³/3 - x²/2 + 2√x + C





=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



∫ 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



∫(^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




Interesting solving these questions...

goofyone:


| 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)