C2-Algebra, Coordinate geometry, Logarithms, Differentiation, Integration

If 2x+5, 6x-10 and 8x+20 are consecutive terms in a geometric series find the possible values of x and of the terms.

(6x-10)/(2x+5) = (8x+20)/(6x-10) (6x-10)^2 = (8x+20)(2x+5) 36x^2 - 120x + 100 = 16x^2 + 80x + 100 20x^2 - 200x = 0 20x(x-10)=0 So x=0 or 10 So terms are 5, -10, -20 or 25, 50, 100

Find the remainder when x^3 - 4x^2 + 2x - 3 is divided by (x+1).

(ax-b): a=1, b=-1 as x--1=x+1 f(-1)=-1-4-2-3=-10 Remainder=-10

Find the remainder when 4x^4 - x^2 - 8x + 6 is divided by (2x-1).

(ax-b): a=2, b=1 f(1/2)=1/4-1/4-4+6=2 Remainder=2

Find the equation of the tangent to the circle x^2 + y^2 + 10x + 2y + 13=0 at the point (-3, 2).

(x+5)^2 + (y+1)^2 = 13 Centre = (-5,-1), radius = root13 ml=-1-2/-5+3=-3/-2 so mt=-2/3 y-2=-2/3(x+3) 3y-6=-2x-6 So 2x+3y=0

Find the equation of a circle with centre (4,-2) and radius 3.

(x-4)^2 + (y+2)^2 = 9

Solve: 6 log 3 (x) + 2 log 3 (y) = 7 2 log 3 (x) + 4 log 3 (y) = 9

12 log 3 (x) + 4 log 3 (y) = 14 [1] 2 log 3 (x) + 4 log 3 (y) = 9 [2] 10 log 3 (x) = 5 So 3^5=x^10 So x=root3 Into [2]: log 3 (3) + 4 log 3 (y) = 9 4 log 3 (y)=8 log 3 (y)=2 y=9

Solve 2^(2x+1)+1=3(2^x)

2(2^2x)-3(2^x)+1=0 2(2^x)^2-3(2^x)+1=0 Let y=2^x 2y^2-3y+1=0 (2y-1)(y-1)=0 So y=1/2 or 1 x=-1 or 0

Solve 2x^3 - 5x^2 - 10 = 23x.

2x^3 - 5x^2 -23x - 10 = 0 If (x-p) is a factor p could be -1, 10, -10, 1, 2, -5, -2, 5 f(-2)=0 so (x+2) is a factor Divide 2x^3 - 5x^2 -23x - 10 by (x+2): =2x^2 - 9x - 5 =(2x+1)(x-5) So f(x)=(x+2)(2x+1)(x-5)=0 So x=-2, -1/2, 5

Solve 3 + log 2 (2x-5)=log 2 (5x+4).

3=log 2 (5x+4) - log 2 (2x-5) 3=log 2 (5x+4) / (2x-5) 2^3=(5x+4) / (2x-5) 8(2x-5)=5x+4 16x-40=5x+4 11x=44 x=4

Solve 3^(2x-1)-4^(x+3)=0

3^(2x-1)=4^(x+3) log (3^(2x-1))=log (4^(x+3)) (2x-1)log (3)=(x+3)log (4) 2x log (3) - log (3)= x log (4) + 3 log (4) 2x log (3) - x log 4= 3 log (4) + log (3) x (2 log (3) - log (4))= 3 log (4) + log (3) x=3 log (4) + log (3) / 2 log (3) - log (4)

Solve 5^x * 5^2x+1 > 2000

5^3x+1 > 2000 log 5^3x+1 > log 2000 (3x+1) log 5 > log 2000 3x log 5 > log 400 x> log 400 / 3 log 5

Solve log 5x = log 2x+1.

5x=2x+1 x=1/3

Simplify log (81) / log (27).

= log (3^4) / log (3^3) = 4 log (3) / 3 log (3) = 4/3

Divide x^3 + 9x^2 + 17x - 15 by (x + 5).

= x^2 + 4x -3

Divide x^3 + 5x - 6 by (x - 1).

= x^2 + x + 6

(2,-1) S 4x^3 - 3 dx...

=(2,-1) [x^4 - 3x + c] =(10 + c) - (4 + c) =6

Evaluate (2, 0) S x^-1/2 dx.

=(2,t) 2[x^1/2] =2root2 -2roott As t->0, roott -> 0 So = 2root2

(3,1) S 3x^2 dx...

=(3,1) [x^3] =27 - 1 =26

Evaluate (infinity, 2) S x^-1/2 dx.

=(t,2) 2[rootx] =2roott -2root2 As t->infinity roott-> inifinity So = infinity (undefined)

log(3)1/9...

=-2 as 3^-2=1/9

Evaluate (infinity, 0) S e^-x dx.

=-[e^-t -1] =1 - e^-t As t-> 0, 1/e^t -> 0 so = 1

log(5)1...

=0 as 5^0=1

log (10)...

=1 as 10^1=10

5!...

=1*2*3*4*5 =120

log(9)3...

=1/2 as 9^1/2=3

Find the approximate area of (7,3) S root(x-1) dx, taking strips of 1 unit.

=1/2[root2 + 2(root3 + root4 + root5) + root6] =7.9 units to 1 d.p.

log3(9)...

=2 as 3^2=9

log (1000)

=3 as 10^3=1000

log2(8)...

=3 as 2^3=8

(3+2x)^4...

=4C0 (3)^4 (2x)^0 +4C1 (3)^3 (2x)^1 +4C2 (3)^2 (2x)^2 +4C3 (3)^1 (2x)^3 +4C4 (3)^0 (2x)^4 =81 + 216x + 216x^2 + 96x^3 + 16x^4

log2(32)...

=5 as 2^5=32

5C2...

=5!/((5-2)!*2!) =10

Expand (2-3x)^5 up to the x^3 term.

=5C0 (2)^5 (3x)^0 +5C1 (2)^4 (3x)^1 +5C2 (2)^3 (3x)^2 +5C3 (2)^2 (3x)^3 =32 - 240x + 720x^2 - 1080x^3 +...

Find the term in x^5 in the expansion of (5-2x)^8.

=8C5 (5)^3 (-2x)^5 =-224000x^5

Simplify log (5) + 3/5 log (32) - log 4.

=log (5) + log (32^3/5) - log (4) =log (5) + log (8) - log (4) = log (40/4)= log (10) = 1

Simplify 3 log 4 (2) - log 4 (6) + log 4 (12).

=log 4 (2^3) - log 4 (6) + log 4 (12) =log 4 (8/6) + log 4 (12) =log 4 (4/3 *12) =log 4 (16) =2

2log(5)+log(4)...

=log(5^2)+log(4) =log(100) =2

log(x^3 y^2 / z)...

=log(x^3) + log(y^2) - log(z) =3log(x) + 2log(y) - log(z)

log 100...

=log10(100) =2 as 10^2=100

log2(72)-log2(9)...

=log2(72/9) =log2(8) =3

log3(4)+log3(5)...

=log3(20)

Divide x^3 - 5x^2 + 8x - 4 by (x - 2).

=x^2 - 3x + 2

Find the area bounded by the curve y=(x+1)(x-2), the line x=3 and the x axis.

A= (-1, 2) S x^2 - x - 2 dx + (3, 2) S x^2 - x - 2 dx =(2,-1) [1/3 x^3 - 1/2 x^2 -2x] + (3,2) [1/3 x^3 - 1/2 x^2 -2x] =[8/3 - 2 -4 +1/3 +1/2 -2] + [9 - 9/2 - 6 -8/3 + 2 + 4] =-(-9/2) + 11/6 as total área =8/3 sq units

Find the area bounded by the curve y=x(x+2) and the x axis.

A=(-2, 0) S x^2 + 2x dx =(-2,0) [1/3 x^3 + x^2] =4/3 - 0 =4/3 sq units

Find the area bounded by the curve y=x^2+1,  the lines x=-1 and x=3 and the x axis.

Area=(3,-1) S x^2 + 1 dx =(3,-1) [1/3 x^3 + x] =12+4/3 =40/3 sq units

If x^3 + 2x^2 + ax + b has factors (x+1) and (x-2), find a and b.

As (x+1) is a factor f(-1)=0 So -1+2-a+b=0 so 1-a+b=0 [1] As (x-2) is a factor f(2)=0 So 8+8+2a+b=0 so 16+2a+b=0 [2] [1]-[2]: -15-3a=0 so a=-5 Into [1]: 1+5+b=0 so b=-6

Find the area bounded by the curve y=x^2 and the line y=x.

At Ps of intersection: x=x^2 x(x-1)=0 x=0, y=0 x=1, y=1 A=(1,0) S x^2 - x dx =(1,0) [1/3 x^3 - 1/2 x^2] =-1/6= 1/6 as total area

A man invests £500 at the start of each year in an account paying 3.5% interest per year. At the start of each year he puts in another £500. After how many year has he accumulated more than £20000?

End of year 1=a=500*1.035 End of year 2=(500+ 500*1.035)1.035=500*1.035 + 500*(1.035^2)=a + ar So r= 1.035 Sn=500*1.035(1.035^n - 1) / (1.035- 1) >=20000 1.035^n >= 20000*0.035 / 500*1.035 + 1 >= 2.3526... >= log 2.3526... / log 1.035 >= 25 years to nearest year

Factorise 2x^3 - 3x^2 - 11x + 6.

Factor is (x-p) p could be: 1, -1, 2, -2, 3, -3, 6, -6 Try f(p) f(-2)=0 so (x--2) is factor= (x+2) Divide 2x^3 - 3x^2 - 11x + 6 by (x-2) to get other factor =2x^2 - 7x^2 + 3 =(2x-1)(x-3) So f(x)=(x+2)(2x-1)(x-3)

Show that (x-1) is a factor of 2x^3 - 3x^2 - x + 2.

If (x-1) is a factor f(1)=0 So f(1)=2-3-1+2=0 Therefore by the factor theorem (x-1) is a factor

Explain how S(infinity)= a/(1 - r).

If -1<r<1 and n is large Sn=S(inifinity)=a(1-r^n)/1-r But as r^n -> 0 and n ->; infinity, S(infinity)=a/(1 - r)

A stationary point...

Is a turning point At stationary point gradient and dy/dx=0

Prove that log a (x) + log a (y) = log a (xy).

Let log a (x)= p; a^p=x Let log a (y)= q; a^q=y So xy= a^p a^q= a^(p+q) So log a (xy)= p+q So log a (xy)= log a (x) + log a (y)

Solve 3(5^2x)-4(5^x)+1=0.

Let y=5^x 3y^2-4y+1=0 (3y-1)(y-1)=0 y=1/3 or 1 So 5^x=1/3 or 1 x=log 5 (1/3) or 0

Is (x + 2) a factor of x^4 - 3x^2?

Remainder 4 Therefore not a factor

Prove that the sum of terms of a geometric series Sn=a(r^n -1) / r - 1.

Sn = a + ar + ar^2....+ar^n-1 [1] rSn = ar + ar^2 + ar^3...+ar^n [2] [1]-[2]= Sn-rSn=a - ar^n Sn(1-r)=a(1- r^n) So Sn=a(1- r^n)/(1-r)=a(1 - r^n)/(1-r) * -1/-1 =a(r^n -1) / r-1 So Sn=a(r^n -1) / r-1

Expand root(x^2 / 5 y^2) in terms of log x and log y.

Taking logs: log (root(x^2 / 5 y^2))=1/2 log (x^2 / 5 y^2) =1/2 (log (x^2) - log (y^2) - log (5)) =1/2 (2 log (x) - 2 log (y) - log (5))

If a polynomial f(x) has a remainder...

The remainder is equal to f(x) where x=b/a; and (ax-b) will be a factor of f(x)

If a polynomial f(x) is divided by (ax-b)...

The remainder is the value of f(b/a)

When d^2 y / dx^2 &lt; 0...

The stationary point is a maximum

When d^2 y / dx^2 &gt; 0...

The stationary point is a minimum

When dy/dx is positive (or negative) on both sides of the stationary point...

The stationary point is a point of inflexion

If f(x) is a polynomial, and f(p)=0...

Then (x-p) is a factor of f(x)

If f(x) is a polynomial, and (x-p) is a factor of f(x)...

Then f(p)=0

An&nbsp;open rectangular box has a total surface area of 486 cm². If the length of the box is twice the width. Find the maximum volume and prove that this volume is a maximum.

V=2xh [1] 486=2x^2 + 2xh + 4xh 243=x^2 + 3xh h=(243-x^2)/3x [2] [2] into [1]: V=486x/3 -2x^3 / 3 V=162x - 2x^3 / 3 dV/dx= 162 - 2x^2 = 0 at stationary points x^2=81 so x=9 as x has to be > 0 for max volume So V=972 cm^3 d^2 V / dx^2 = -4x = -36 so maximum as d^2 V / dx^2 <0

In the sequence 4, 12, 36, 108 find the 6th term and the sum of the first ten terms.

a=4 r=3 ar^5=243*4=972 S10=a(r^10 -1) / r-1 = 4(3^10 -1) / 2= 118096

The second term of a geometric series is 5 and its S(infinity)=20. Find the common ratio and first term.

ar=5 a/(1-r)=20 a=20(1-r) So 20(1-r)r=5 4(1-r)r=1 4r-4r^2=1 4r^2 - 4r + 1=0 (2r-1)(2r-1)=0 r=1/2 so a=10

If the 6th term of a geometric series is 972 and the 9th term is 26244 find the 1st term, common ratio and sum of the first 10 terms.

ar^5=972 ar^8=26244 ar^5 / ar^8 = r^3=27 r=3 972/3^5 = a = 4 S10=a(r^10 -1) / r-1 = 4(3^10 - 1) / 2

Find the range of values of x for which y is increasing: y=x^3 + 5x^2 - 8x +1

dy/dx = 3x^2 + 10x - 8 y is increasing when dy/dx &gt; 0 So 3x^2 + 10x - 8 &gt; 0 (3x-2)(x+4)>0 CVs=2/3 or -4 y is increasing when x>2/3 and <-4

Find the coordinates of the stationary points on the curve y=12x^5 + 5x^6 + 1 and determine the nature of the stationary points.

dy/dx = 60x^4 + 30x^5 = 0 30x^4(2 + x)=0 x=0 or -2 y=1 or -63 d^2 y / dx^2 = 240x^3 + 150x^4 = 480 for (-2, 63) so minimum = 0 for (0,1) dy/dx is increasing either side of 0,1 so point of inflexion

Find the stationary points on the curve: y = x^3 + 3x^2 - 9x - 4

dy/dx= 3x^2 + 6x - 9 (3x-3)(x+3)=0 x=1 or -3 y=-9 or 23 P's= (1,-9) and (-3, 23)

When x^3 + ax^2 + bx -1 is divided by (x-1) the remainder is 3 and when divided by (x+2) the remainder is -27. Find a and b.

f(1)=3=1+a+b-1=3=a+b; so 6=2a+2b [1] f(-2)=-27=-8+4a-2b-1=-27=-9+4a-2b [2] [1]+[2]: -21=-9+6a -12=6a so a=-2 3=-2+b so b=5

Solve: log x - log 2 = 2 log y x-5y+2=0

log (x/2)=log (y^2) So x/2=y^2 So 2y^2-5y+2=0 (2y-1)(y-2)=0 y=1/2 or 2 x=1/2 or 8

Show log 3 (9) = 2 by working in logs to base 10.

log 3 (9) = 2 as 3^2=9 log 3 (9) = log 9 / log 3 = 0.95.../0.48... =2

Find the value of log 4 (8) by changing to base 2.

log 4 (8)= log 2 (8) / log 2 (4) =3/2

Express log 5 (3) in terms of a log in base 3.

log 5 (3)= log 3 (3) / log 3 (5) =1/log 3 (5)

Solve log 9 (4) + log 3 (x)=3.

log 9 (4) = log 3 (4) / log 3 (9) So log 3 (4) / log 3 (9) + log 3 (x)=3 log 3 (4) + 2 log 3 (x)=6 log 3 (4x^2)=6 So 3^6=4x^2 So x=root(3^6 / 4)=3^3 / 2=27/2

Prove that log a (N)=log b (N) / log b (a).

log a (N) = x So a^x= N [1] Taking logs to base b: log b (a^x) = x log b (a) So x= log b (a^x) / log b (a) [2] So [1] into [2] gives: x= log b (N) / log b (a) So log a (N) = log b (N) / log b (a)

Prove that log a (x/y) = log a (x) - log a (y).

log a (x)=p so a^p=x log a (y)=q so a^q=y x/y=a^p / a^q=a^(p-q) So log a (x/y)= p-q So log a (x/y) = log a (x) - log a (y)

Prove that log a (x^n) = n log a (x).

log a (x^n)= log a (x1) + log a (x2) +... log a (xn) =n lots of log a (x) =n log a (x)

Solve: 2 log x (y)=1 xy=64

log x (y^2)=1 So y^2=x^1=x So y^3=64 y=4 x=16

Find the coefficient of 1/x^3 in the expansion of (2-3/2x)^7.

term in 1/x^3 = 7C3 (2)^4 (-3/2x)^3 =-1890/x^3 So coefficient=-1890

Solve 0.92^x &lt; 0.1

x > log (0.1) / log (0.92)

Solve 3^x=12.

x= log 3 (12)

Solve log 3 (x)=2.

x=3^2=9

Find the centre and radius of the circle x^2 + y^2 + 10x - 12y + 9 = 0.

x^2 + 10x + y^2 - 12y + 9 = 0 (x+5)^2 + (y-6)^2 = 52 Centre= (-5, 6) Radius=root52=2root13

Solve log x (27)=3

x^3=27 x=^3root(27)=3