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# Solve log 2 (2 x + 1) – log 2 x = 2 PowerPoint PPT Presentation

Log tips. When solving, you can often either: Get in the form log a b = c . Then rearrange as a c = b Get in the form log a b = log a c . Then b = c. Actual exam questions:. Solve log 2 (2 x + 1) – log 2 x = 2. log 2 ( x + 1) – log 2 x = log 2 7.

Solve log 2 (2 x + 1) – log 2 x = 2

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Log tips

When solving, you can often either:

Get in the form logab = c. Then rearrange as ac = b

Get in the form logab = logac. Then b = c.

Actual exam questions:

Solve

log2 (2x + 1) – log2 x = 2

log 2 (x + 1) – log 2 x = log 2 7

Solve simultaneous equations:

a = 3b,

log3a + log3b = 2

2 log3x – log3 7x = 1

log5 (4 – x) – 2 log5x = 1

log2y = –3

log2 (11 – 6x) = 2 log2 (x – 1) + 3

logx 64 = 2

Log tips

Other types of question:

52x – 12(5x) + 35 = 0

3x = 10

Circles

The points A and B have coordinates (–2, 11) and (8, 1) respectively.

Given that AB is a diameter of the circle C,

(a) show that the centre of C has coordinates (3, 6),

(1)

(b) find an equation for C.

(4)

(c) Verify that the point (10, 7) lies on C.

(1)

(d) Find an equation of the tangent to C at the point (10, 7), giving your answer in the form y = mx + c, where m and c are constants.

(4)

Remember that you need the centre (a,b) and the radius r, which gives the equation:

(x-a)2 + (y-b)2 = r2

The tangent is perpendicular to the radius at the point of contact.

Circles

How could you tell if a line and a circle intersect:

once

twice

0 times

y = 4-x

x2 + y2 = 1

Equate the expressions then look at the discriminant:

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b2 – 4ac < 0

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b2 – 4ac > 0

b2 – 4ac = 0

Circles

8. The circle C, with centre at the point A, has equation x2 + y2 – 10x + 9 = 0.

Find

(a)the coordinates of A,

(2)

(2)

(c)the coordinates of the points at which C crosses the x-axis.

(2)

Given that the line l with gradient is a tangent to C, and that l touches C at the point T,

(d)find an equation of the line which passes through A and T.

(3)

Circles

If PR is the diameter, how would you prove that a = 13?

Circles

Circles

Factor Theorem

(a)Use the factor theorem to show that (x + 4) is a factor of

2x3+ x2 – 25x + 12.

(2)

(b)Factorise 2x3 + x2 – 25x + 12 completely.

(4)

f(x) = x3 + ax2 + bx+ 3, where a and b are constants.

Given that when f (x) is divided by (x + 2) the remainder is 7,

(a) show that 2a − b = 6

(2)

Given also that when f (x) is divided by (x −1) the remainder is 4,

(b) find the value of a and the value of b.

(4)

Trigonometric Solutions

(a) Given that sin  = 5 cos, find the value of tan .

(1)

(b) Hence, or otherwise, find the values of  in the interval0  < 360 for which

sin  = 5 cos,

(3)

(a) Find all the values of  , to 1 decimal place, in the interval 0 < 360 for which

5 sin ( + 30) = 3.

(4)

(b) Find all the values of  , to 1 decimal place, in the interval 0 < 360 for which

tan2 = 4.

(5)

Find all the solutions, in the interval 0 ≤ x < 2, of the equation

2 cos2x + 1 = 5 sin x,

giving each solution in terms of .

Trigonometric Solutions

(a) Sketch, for 0 ≤ x ≤ 2, the graph of y = sin ( x + ( /6)).

(2)

(b) Write down the exact coordinates of the points where the graph meets the coordinate axes.

(3)

(c) Solve, for 0 ≤ x ≤ 2, the equation

sin (x + ( /6)) = 0.65,

(a) Show that the equation

3 sin2 – 2 cos2 = 1

can be written as

5 sin2 = 3.

(2)

(b) Hence solve, for 0 < 360, the equation

3 sin2 – 2 cos2 = 1,

(7)

Solve, for 0 x < 360°,

(a)sin(x – 20) = 1/√2,

(4)

(b) cos 3x = –1/2 .

(6)

Trigonometric Solutions

(a) Show that the equation

4 sin2x + 9 cosx – 6 = 0

can be written as

4 cos2x – 9 cosx + 2 = 0.

(2)

(b) Hence solve, for 0 x < 720°,

4 sin2x + 9 cosx – 6 = 0,

(6)

Show that the equation

tan 2x = 5 sin 2x

can be written in the form

(1 – 5 cos 2x) sin 2x = 0

(2)

(b) Hence solve, for 0 ≤x ≤180°,

tan 2x = 5 sin 2x

(5)

(i) Find the solutions of the equation sin(3x - 15 ) = ½ for which 0 ≤ x ≤ 180

Trigonometric Solutions

Summary of tips:

• To get all your solutions when you do you inverse sin/cos/tan:

• Remember that sin(180-x) = sin(x) and cos(360-x) = cos(x)

• sin and cos repeat every 360 (i.e. you can add 360 to your solution as many times as you like).

• But tan repeats every 180.

• If you’re working in radians, then sin(pi – x) = sin(x), etc.

• Ensure your calculator is correctly set to either radians or degrees mode.

• If you have sin2, then make sure you get both positive and negative solution. Likewise for cos2 and tan2.

• Make sure you write out enough solutions before you start manipulating: if you had sin(3x) = ½ for example and had the range 0 < x < 360, then you’d initially need to write values up to 1080 since you’re going to be dividing by 3.

Areas of sector/Arc lengths/Sine and Cosine Rule

Areas of sector/Arc lengths/Sine and Cosine Rule

Areas

Only 1 in 36 candidates (across the country) got this question fully correct.

Optimisation

Optimisation

Optimisation

Optimisation

Integration

Integration

Figure 1 shows part of a curve C with equation , x > 0.

The points P and Q lie on C and have x-coordinates 1 and 4 respectively. The region R, shaded in Figure 1, is bounded by C and the straight line joining P and Q.

(a)Find the exact area of R.

(8)

(b)Use calculus to show that y is increasing for x > 2.

(4)

Integration

Integration

Examiner’s Report:

(a) A pleasing majority of the candidates were able to differentiate these fractional powers correctly, but a sizeable group left the constant term on the end. They then put the derivative equal to zero. Solving the equation which resulted caused more problems as the equation contained various fractional powers. Some tried squaring to clear away the fractional powers, but often did not deal well with the square roots afterwards. There were many who expressed 6x-1/2 = 1/(6x1/2) and tended to get in a muddle after that. Those who took out a factor x1/2 usually ended with x = 0 as well as x = 4 and if it was not discounted, they lost an accuracy mark. Those who obtained the solution x = 4 sometimes neglected to complete their solution by finding the corresponding y value. Some weaker candidates did not differentiate at all in part (a), with some integrating, and others substituting various values into y.

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Geometric Series

Geometric Series

Geometric Series

Binomial Expansion