Regent College Maths Department

Regent College Maths Department

Core Mathematics 1

Calculus (Differentiation and

Integration)

vasanthakumaran.kand

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vasanthakumaran.kand

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September 2011 C1 Note

Copyright www.pgmaths.co.uk - For AS, A2 notes and IGCSE / GCSE worksheets 33

Differentiation

The derivative of f(x) as the gradient of the tangent to the graph of y = f (x) at a point; the gradient of the tangent as a limit; interpretation as a rate of change.

For example, knowledge thatd

d

y

x is the rate of change of y with respect to x. Knowledge of the chain

rule is not required. Finding the gradient of the curve Consider the curve shown below.

How do we find the gradient to the curve at the point P? At GCSE we would have found the gradient by drawing a tangent to the curve at that point and then measuring the gradient of that tangent. This was only estimate – we now want to find the exact value of this gradient. If P is a point on the curve f ( )y x then the gradient of the curve at P is the gradient of the tangent to the curve at P. If Q is another point on the curve then, as Q moves closer and closer to P, the gradient of the chord PQ get closer and closer to the gradient of the tangent to the curve at P.

The gradient of PQ is

f f f fx h x x h x

x h x h

.

( , f ( ))P x x

( , f ( ))Q x h x h

f ( )y x



The gradient of the curve at P is the gradient of the tangent to the curve at P. This is obtained by moving Q closer and closer to P (i.e. by making h tend to 0).

The gradient of the curve at P is denoted by f ( )x or d

d

y

x.

So 0

f ( ) f ( )f ( ) lim

h

x h xx

h

where “0

limh

” means that we make h tend to 0.

For example if 2f ( )x x then 2 2 2f ( ) ( ) 2x h x h x hx h . So we have that

0

2 2 2

0

2

0

0

f ( ) f ( )f ( ) lim

2lim

2lim

lim 2

2

h

h

h

h

x h xx

h

x hx h x

h

hx h

h

x h

x

Thus we have proved that if 2f ( )x x then f ( ) 2x x .

Alternatively we could write that if 2xy then d

2d

yx

x or simply say “ 2x differentiates to x2 ”.

This means that the gradient at any point on the curve 2y x is 2x . So, for example, at the point (3, 9) the gradient is 6.

Since ddy

x is the gradient of the curve, we see that

x

y

d

d represents the rate of change of y with respect

to x.



The notation f(x) may be used. Differentiation of xn, and related sums and differences.

E.g., for 1n , the ability to differentiate expressions such as 2 5 1x x and 12

2 5 3x x

x

is

expected. Applications of differentiation to gradients, tangents and normals. Use of differentiation to find equations of tangents and normals at specific points on a curve.

We have just proved that if 2xy then d

2d

yx

x .

We could have proved that if 3y x then 2d3

d

yx

x .

We already know that the gradient of 1y is 0 (it is a horizontal line) and that the gradient of the line y x is 1. We could write this more formally by saying:

If 1y then d

0d

y

x . If y x then

d1

d

y

x

So we see the following:

y d

d

y

x

1 0 x 1

2x 2x 3x 23x

We see a pattern here which continues. The pattern is that that if ny x then 1d

dny

nxx

. This

holds for all values of n – positive and negative, fractions and whole numbers.

So, for example if 6y x then 5d6

d

yx

x .

The general result is as follows:

If ny kx then 1d

dny

nkxx

.

So, for example if 52y x then 4d10

d

yx

x .



Example 1

Find x

y

d

d when 2 5 1y x x

We first need to multiply out the bracket to get 22 3 5y x x .

We now differentiate using the rule on the previous page to get d

4 3d

yx

x .

Example 2

Find x

y

d

d when 1

2

2 5 3x xy

x

We first need to rewrite y in the form 3 1 1

2 2 25 3y x x x , using the laws of indices.

We now differentiate using the rule on the previous page to get 1 1 1

2 2 2d 3 5 3

d 2 2 2

yx x x

x

.

Example 3

If 22 3 1y x x then find the gradient of the curve at the point (2, 15).

First we find d

4 3d

yx

x .

We then need to replace x in this expression with the x coordinate of the point we are looking at.

The x coordinate of the point (2, 15) is 2 so we substitute 2x into the equation d

4 3d

yx

x to get

d11

d

y

x .



Some more examples...

y d

d

y

x

22 3 5 6x x x x 2 5x

213

3x

x xx

21 3x

1 1

2 21

2

2 6 2 62 6

x xx x

xx

1 3

2 23x x

213 4 7 12

7 12x x x x

x xx x

21 12x

2 3 1 122 2 2

1

2

2 4 4 12 2

222

x x xx x x

xx

1 1 3

2 2 23

4x x x



Tangents

The gradient of a tangent to the curve at any point is the gradient of the curve at that point.

Example 4

Find the equation of the tangent to the curve 3 23 1y x x x at the point (2, -1).

2d3 6 1

d

yx x

x .

We want to know x

y

d

d at the point (2, -1) so we substitute 2x into the equation 2d

3 6 1d

yx x

x

to get d

1d

y

x .

So the gradient of the tangent at (2, -1) is 1.

Using the earlier work on straight lines we know that the equation is 1

12

y

x

.

Rearranging gives 3y x .



Normals

The gradient of a normal to the curve at any point is the negative reciprocal of the gradient of the

curve at that point.

So..

if d

3d

y

x at a point on the curve, the gradient of the normal is

1

3 .

if d

4d

y


1

4.

if d 3

d 5

y


5

3 .

if d

1d

y

x at a point on the curve, the gradient of the normal is -1.

Example 5

Find the equation of the normal to the curve 3 23 1y x x x at the point (1, 0).

2d3 6 1

d

yx x

x .

We want to know x

y

d

d at the point (1, 0) so we substitute 1x into the equation 2d

3 6 1d

yx x

x to

get d

2d

y

x .

So the gradient of the normal at (1, 0) is 1

2.

Using the earlier work on straight lines we know that the equation is 0 1

1 2

y

x

.

Rearranging gives 2 1 0y x .



Example 6

A curve C has equation y = x3 – 5x2 + 5x + 2.

(a) Find x

y

d

d in terms of x.

The points P and Q lie on C. The gradient of C at both P and Q is 2. The x-coordinate of P is 3. (b) Find the x-coordinate of Q. (c) Find an equation for the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants. (d) Find an equation for the normal to C at P, giving your answer in the form 0ax by c , where m and c are constants.

Edexcel GCE Pure Mathematics P1 January 2002

(a) 2d3 10 5

d

yx x

x

(b) At P and Q , 2d3 10 5 2

d

yx x

x so 23 10 3 0x x

This gives 3 1 3 0x x and so 1

3 or 3

x x . So x coordinate of Q is 1

3.

(c) At P, 3x , so 3 23 5 3 5 3 2 1y . Also at P d

2d

y

x

Hence the equation is 1

23

y

x

. That is 2 7y x .

(d) At Q, 3x , and 1y . Also at P d

2d

y

x . So gradient of normal is

1

2 .

Hence the equation is 1 1

3 2

y

x

. That is 2 2 3y x . Hence

2 1 0y x . Second order derivatives.

Consider the curve 3 23 9 11y x x x .

We see that 2d3 6 9

d

yx x

x .

The “gradient of the gradient” is denoted by 2

2

d

d

y

x. We see that in this example

2

2

d6 6

d

yx

x



Indefinite Integration Indefinite integration as the reverse of differentiation. Students should be aware that a constant of integration is required.

Integration of nx . For example, the ability to integrate expressions such as 1221

2 3x x ,

12

22x

x

is

expected. Given f ( )x and a point on the curve, students should be able to find an equation of the curve in the form f( )y x .

We know that if 32 5y x x then 2d6 5

d

yx

x .

However we also that

if 32 5 2y x x then 2d6 5

d

yx

x .

or if 32 5 3y x x then 2d6 5

d

yx

x .

So if we were given 2d6 5

d

yx

x and asked to find y we cannot get a definite answer.

As we have seen y could be 32 5x x or it could be 32 5 2x x or it could be 32 5 3x x .

All we can say is that if 2d6 5

d

yx

x then 32 5y x x c where c can be any constant.

To find a particular solution of 2d6 5

d

yx

x we need to have more information.

Example 1

If 2d6 5

d

yx

x and the curve passes through the point (1, 13) then find y.

We see that 32 5y x x c

Since the curve passes through the point (1, 13), we have that 213 2 1 5 1 c and so 6c . Hence see that the solution is 32 5 6y x x .



Notation

The “indefinite integral” of f ( )x is denoted by f ( ) dx x .

The general rule we use is 1

d ( 1)1

pp x

kx x k c pp

Here are some examples

31

2 2

2

83

2 12

3 d 3

55 d

2

4 d 4 d

6d 6 d 6

x x c

x x x c

x x x x x c

x x x x cx

NB We can only integrate an expression made up of terms of the form pkx . Example 2

Find 2 1 2 3 dx x x .

We first of all need to express 2 1 2 3x x as 24 8 3x x

We now use the above rule to see that

2

3 2

2 1 2 3 d

4 8 3 d

44 3

3

x x x

x x x

x x x c

Example 3

Find 28

dx x

xx

.

We first of all need to express 28x x

x

as

312 2

12

288

x xx x

x

We now use the above rule to see that

312 2

3 52 2

2

16 23 5

8d

8 d

x xx

x

x x x

x x c



Some more examples...

y dy x

22 3 5 6x x x x 3 21 56

3 2x x x c

2 33

33

xx x

x

1 23

2x x c

1 1

2 21

2

2 6 2 62 6

x xx x

xx

3 1

2 24

123

x x c

3 1 122 2 2

1

2

3 4 7 127 12

x x x xx x x

xx

5 3 1

2 2 22 14

245 3

x x x c

1. (i) Given that y = 5x3 + 7x + 3, find

(a) xy

dd ,

(3)

(b) 2

2

dd

xy .

(1)

(ii) Find ⌡⌠

−√+ 2

131x

x dx.

(4) Q2, Jan 2005

2. The curve C has equation y = 4x2 + x

x−5 , x ≠ 0. The point P on C has x-coordinate 1.

(a) Show that the value of xy

dd at P is 3.

(5)

(b) Find an equation of the tangent to C at P. (3)

This tangent meets the x-axis at the point (k, 0). (c) Find the value of k.

(2) Q7, Jan 2005

3. The gradient of the curve C is given by

xy

dd = (3x – 1)2.

The point P(1, 4) lies on C. (a) Find an equation of the normal to C at P.

(4)

(b) Find an equation for the curve C in the form y = f(x). (5)

(c) Using xy

dd = (3x – 1)2, show that there is no point on C at which the tangent is parallel to the

line y = 1 – 2x. (2)

Q9, Jan 2005

Core Maths 1 Calculus Page 1

4. Given that y = 6x – 2

4x

, x ≠ 0,

(a) find xy

dd ,

(2)

(b) find y⌡⌠ dx.

(3) Q2, May 2005

5. (a) Show that xx

√√− 2)3( can be written as 2

121

69 xx +−− .

(2)

Given that xy

dd =

xx

√√− 2)3( , x > 0, and that y = 3

2 at x = 1,

(b) find y in terms of x.

(6) Q7, May 2005

6. The curve C has equation y = 31 x3 – 4x2 + 8x + 3.

The point P has coordinates (3, 0).

(a) Show that P lies on C.

(1)

(b) Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

(5)

Another point Q also lies on C. The tangent to C at Q is parallel to the tangent to C at P.

(c) Find the coordinates of Q. (5)

Q10, May 2005

7. Given that y = 2x2 – 3

6x

, x ≠ 0,

(a) find xy

dd , (2)

(b) find ⌡⌠ xy d . (3)

Q4, Jan 2006


8. The curve with equation y = f(x) passes through the point (1, 6). Given that

f ′(x) = 3 + 21

25 2

x

x + , x > 0,

find f(x) and simplify your answer. (7) Q8, Jan 2006

9. Figure 2

Figure 2 shows part of the curve C with equation

y = (x – 1)(x2 – 4).

The curve cuts the x-axis at the points P, (1, 0) and Q, as shown in Figure 2. (a) Write down the x-coordinate of P and the x-coordinate of Q.

(2)

(b) Show that xy

dd = 3x2 – 2x – 4.

(3)

(c) Show that y = x + 7 is an equation of the tangent to C at the point (–1, 6). (2)

The tangent to C at the point R is parallel to the tangent at the point (–1, 6). (d) Find the exact coordinates of R.

(5) Q9, Jan 2006

y

C

x P O 1 Q

4


10. Find xxx d)26( 212

⌡⌠ ++ − , giving each term in its simplest form.

(4)

Q1, May 2006 11. Differentiate with respect to x (a) x4 + 6√x,

(3)

(b) x

x 2)4( + . (4)

Q5, May 2006

12. The curve C with equation y = f(x), x ≠ 0, passes through the point (3, 7 2

1 ).

Given that f ′(x) = 2x + 2

3x

,

(a) find f(x).

(5)

(b) Verify that f(–2) = 5. (1)

(c) Find an equation for the tangent to C at the point (–2, 5), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4) Q10, May 2006

13. Given that

y = 4x3 – 1 + 2 21

x , x > 0,

find xy

dd .

(4)

Q1, Jan 2007 14. (a) Show that (4 + 3√x)2 can be written as 16 + k√x + 9x, where k is a constant to be found.

(2)

(b) Find ⌡⌠ √+ xx d)34( 2 .

(3)

Q6, Jan 2007


15. The curve C has equation y = f(x), x ≠ 0, and the point P(2, 1) lies on C. Given that

f ′(x) = 3x2 – 6 – 2

8x

,

(a) find f(x).

(5)

(b) Find an equation for the tangent to C at the point P, giving your answer in the form y = mx + c, where m and c are integers. (4)

Q7, Jan 2007

16. The curve C has equation y = 4x + 23

3x – 2x2, x > 0.

(a) Find an expression for xy

dd .

(3)

(b) Show that the point P(4, 8) lies on C. (1)

(c) Show that an equation of the normal to C at the point P is

3y = x + 20. (4)

The normal to C at P cuts the x-axis at the point Q. (d) Find the length PQ, giving your answer in a simplified surd form.

(3) Q8, Jan 2007

17. Given that y = 3x2 + 4√x, x > 0, find

(a) xy

dd ,

(2)

(b) 2

2

dd

xy ,

(2)

(c) ⌡⌠ xy d .

(3)

Q3, May 2007


18. The curve C with equation y = f(x) passes through the point (5, 65). Given that f ′(x) = 6x2 – 10x – 12, (a) use integration to find f(x).

(4)

(b) Hence show that f(x) = x(2x + 3)(x – 4). (2)

(c) Sketch C, showing the coordinates of the points where C crosses the x-axis. (3)

Q9, May 2007

19. The curve C has equation y = x2(x – 6) +x4 , x > 0.

The points P and Q lie on C and have x-coordinates 1 and 2 respectively. (a) Show that the length of PQ is √170.

(4)

(b) Show that the tangents to C at P and Q are parallel. (5)

(c) Find an equation for the normal to C at P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4)

Q10, May 2007

20. Find ⌡⌠ −+ xxx d)743( 52 .

(4)

Q1, Jan 2008

21. (a) Write xx 32 +√ in the form 2x

p + 3x q, where p and q are constants.

(2)

Given that y = 5x – 7 + xx 32 +√ , x > 0,

(b) find xy

dd , simplifying the coefficient of each term. (4)

Q5, Jan 2008


22. The curve C has equation y = f(x), x > 0, and f ′(x) = 4x – 6√x + 2

8x

.

Given that the point P(4, 1) lies on C, (a) find f(x) and simplify your answer.

(6)

(b) Find an equation of the normal to C at the point P(4, 1).

(4) Q9, Jan 2008

23. The curve C has equation

y = (x + 3)(x – 1)2.

(a) Sketch C, showing clearly the coordinates of the points where the curve meets the coordinate axes.

(4)

(b) Show that the equation of C can be written in the form

y = x3 + x2 – 5x + k,

where k is a positive integer, and state the value of k. (2)

There are two points on C where the gradient of the tangent to C is equal to 3. (c) Find the x-coordinates of these two points.

(6) Q10, Jan 2008

24. Find ⌡⌠ + xx d)52( 2 .

(3)

Q1, June 2008

25. f(x) = 3x + x3, x > 0. (a) Differentiate to find f ′(x).

(2)

Given that f ′(x) = 15, (b) find the value of x.

(3)

Q4, June 2008


26. The curve C has equation y = kx3 – x2 + x – 5, where k is a constant.

(a) Find xy

dd .

(2)

The point A with x-coordinate – 21 lies on C. The tangent to C at A is parallel to the line with

equation 2y – 7x + 1 = 0. Find (b) the value of k,

(4)

(c) the value of the y-coordinate of A. (2)

Q9, June 2008

27. The gradient of a curve C is given by xy

dd = 2

22 )3(x

x + , x ≠ 0.

(a) Show that xy

dd = x2 + 6 + 9x–2.

(2)

The point (3, 20) lies on C. (b) Find an equation for the curve C in the form y = f(x).

(6)

Q11, June 2008

28. Find ⌡⌠ +− xxx d)3812( 35 , giving each term in its simplest form.

(4)

Q2, Jan 2009

29. A curve has equation y = f(x) and passes through the point (4, 22).

Given that

f ′(x) = 3x2 – 21

3x – 7,

use integration to find f(x), giving each term in its simplest form. (5)

Q4, Jan 2009


30. Given that x

xx√− 2

322 can be written in the form 2x

p – xq,

(a) write down the value of p and the value of q.

(2)

Given that y = 5x4 – 3 + x

xx√− 2

322 ,

(b) find xy

dd , simplifying the coefficient of each term.

(4) Q6, Jan 2009


y = 9 – 4x – x8 , x > 0.

The point P on C has x-coordinate equal to 2.

(a) Show that the equation of the tangent to C at the point P is y = 1 – 2x.

(6)

(b) Find an equation of the normal to C at the point P. (3)

The tangent at P meets the x-axis at A and the normal at P meets the x-axis at B. (c) Find the area of the triangle APB.

(4) Q11, Jan 2009

32. Given that y = 2x3 + 2

3x

, x≠ 0, find

(a) xy

dd ,

(3)

(b) ⌡⌠ xy d , simplifying each term.

(3)

Q3, June 2009


33. f(x) =x

x√√− 2)43( , x > 0.

(a) Show that f(x) = BAxx ++− 2

121

9 , where A and B are constants to be found. (3)

(b) Find f' (x). (3)

(c) Evaluate f' (9). (2)

Q9, June 2009 34. The curve C has equation

y = x3 – 2x2 – x + 9, x > 0. The point P has coordinates (2, 7). (a) Show that P lies on C.

(1)

(b) Find the equation of the tangent to C at P, giving your answer in the form y = mx + c, where m and c are constants.

(5)

The point Q also lies on C. Given that the tangent to C at Q is perpendicular to the tangent to C at P,

(c) show that the x-coordinate of Q is 31 (2 + √6).

(5)

Q11, June 2009

35. Given that y = x4 + 31

x + 3, find xy

dd .

(3) Q1, Jan 2010

36. xy

dd = 2

1

5−

x + x√x, x > 0.

Given that y = 35 at x = 4, find y in terms of x, giving each term in its simplest form.

(7)

Q4, Jan 2010



xxxy )8)(3( −+

= , x > 0.

(a) Find xy

dd in its simplest form.

(4)

(b) Find an equation of the tangent to C at the point where x = 2.

(4)

Q6, Jan 2010 38. Find

⌡

⌠−+ xxx d)568( 2

13 ,

giving each term in its simplest form.

(4) Q2, May 2010

39. Given that

y = 8x3 – 4√x + x

x 23 2 + , x > 0,

find xy

dd .

(6) Q7, May 2010

40. The curve C has equation y = f(x), x > 0, where

xy

dd = 3x –

x√5 – 2.

Given that the point P (4, 5) lies on C, find (a) f(x),

(5)

(b) an equation of the tangent to C at the point P, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

(4) Q11, May 2010


41. Find

⌡

⌠+− xxxx d)4312( 3

125 ,


(5)

Q2, Jan 2011 42. The curve with equation y = f(x) passes through the point (−1, 0).

Given that

f ′(x) = 12x2 − 8x + 1,

find f(x). (5)

Q7, Jan 2011 43. The curve C has equation

y = 3

21 x – 2

3

9x + x8 + 30, x > 0.

(a) Find xy

dd .

(4)

(b) Show that the point P(4, –8) lies on C. (2)

(c) Find an equation of the normal to C at the point P, giving your answer in the form ax + by + c = 0 , where a, b and c are integers.

(6)

(5) Q7, Jan 2011

44. Given that y = 2x5 + 7 + 3

1x

, x ≠ 0, find, in their simplest form,

(a) xy

dd ,

(3)

(b) ⌡⌠ xy d .

(4) Q2, May 2011


45. Given that x

xx√+ 2

5

36 can be written in the form 6x p + 3xq,

(a) write down the value of p and the value of q.

(2)

Given that xy

dd =

xxx

√+ 2

5

36 and that y = 90 when x = 4,

(b) find y in terms of x, simplifying the coefficient of each term.

(5) Q6, May 2011


y = (x + 1)(x + 3)2.

(a) Sketch C, showing the coordinates of the points at which C meets the axes. (4)

(b) Show that xy

dd = 3x2 + 14x + 15.

(3) The point A, with x-coordinate –5, lies on C. (c) Find the equation of the tangent to C at A, giving your answer in the form y = mx + c, where

m and c are constants. (4)

Another point B also lies on C. The tangents to C at A and B are parallel. (d) Find the x-coordinate of B. (3)

Q10, May 2011

47. Given that y = x4 + 21

6x , find in their simplest form

(a) xy

dd ,

(3)

(b) ⌡⌠ xy d .

(3) Q1, Jan 2012


48. A curve with equation y = f(x) passes through the point (2, 10). Given that

f ′(x) = 3x2 − 3x + 5, find the value of f(1).

(5) Q7, Jan 2012

49. The curve C1 has equation

y = x2(x + 2).

(a) Find xy

dd .

(2)

(b) Sketch C1, showing the coordinates of the points where C1 meets the x-axis. (3)

(c) Find the gradient of C1 at each point where C1 meets the x-axis. (2)

The curve C2 has equation

y = (x − k)2(x − k + 2), where k is a constant and k > 2. (d) Sketch C2, showing the coordinates of the points where C2 meets the x and y axes.

(3) Q8, Jan 2012

50.

Figure 2


Figure 2 shows a sketch of the curve C with equation

y = 2 − x1 , x ≠ 0.

The curve crosses the x-axis at the point A. (a) Find the coordinates of A.

(1)

(b) Show that the equation of the normal to C at A can be written as

2x + 8y −1 = 0. (6)

The normal to C at A meets C again at the point B, as shown in Figure 2. (c) Find the coordinates of B.

(4) Q10, Jan 2012

51. Find

⌡⌠

++ x

xx d526 2

2 ,


(4) Q1, May 2012

52. y = 5x3 − 34

6x + 2x − 3.

(a) Find xy

dd , giving each term in its simplest form.

(4)

(b) Find 2

2

dd

xy .

(2) Q4, May 2012

53. The point P (4, –1) lies on the curve C with equation y = f(x), x > 0, and f(x), x > 0, and

f '(x) = x21 –

x√6 + 3.


(a) Find the equation of the tangent to C at the point P, giving your answer in the form

y = mx + c, where m and c are integers. (4)

(b) Find f(x). (4)

Q7, May 2012

54. xy

dd = –x3 + 32

54x

x − , x ≠ 0.


(6)

Q8, Jan 2013


y = 2x − 8√x + 5, x ≥ 0.

(a) Find xy

dd , giving each term in its simplest form.

(3)

The point P on C has x-coordinate equal to 41 .

(b) Find the equation of the tangent to C at the point P, giving your answer in the form

y = ax + b, where a and b are constants. (4)

The tangent to C at the point Q is parallel to the line with equation 2x − 3y + 18 = 0. (c) Find the coordinates of Q.

(5)

Q8, Jan 2013 56. Find

xx

xx d3410 4⌡⌠

√−− ,


(4)

Q2, May 2013


57. f'(x) = 2

22 )3(xx− , x ≠ 0.

(a) Show that f'(x) = 9x–2 + A + Bx2, where A and B are constants to be found.

(3)

(b) Find f"(x). (2)

Given that the point (–3, 10) lies on the curve with equation y = f(x), (c) find f(x).

(5)

Q9, May 2013 58.

Figure 2

Figure 2 shows a sketch of the curve H with equation y = x3 + 4 , x ≠ 0.

(a) Give the coordinates of the point where H crosses the x-axis.

(1)

(b) Give the equations of the asymptotes to H. (2)

(c) Find an equation for the normal to H at the point P(–3, 3). (5)

This normal crosses the x-axis at A and the y-axis at B. (d) Find the length of the line segment AB. Give your answer as a surd.

(3)

Q11, May 2013


59. Find

xx

x d43 22

⌡⌠

− ,


(4)

Q3, May 2013_R 60. A curve has equation y = f(x). The point P with coordinates (9, 0) lies on the curve. Given that

f'(x) = x

x√+ 9 , x > 0,

(a) find f(x).

(6)

(b) Find the x-coordinates of the two points on y = f(x) where the gradient of the curve is equal to 10.

(4) Q10, May 2013_R

61. Find⌡⌠ + xx d)48( 3 , giving each term in its simplest form.

(3) Q1, May 2014

62. Differentiate with respect to x, giving each answer in its simplest form, (a) (1 – 2x)2,

(3)

(b) 2

5

26

xxx + .

(4) Q7, May 2014

63. A curve with equation y = f(x) passes through the point (4, 25).

Given that f ′(x) = 83 x2 – 2

1

10−

x + 1, x > 0,

(a) find f(x), simplifying each term. (5) (b) Find an equation of the normal to the curve at the point (4, 25). Give your answer in the

form ax + by + c = 0, where a, b and c are integers to be found. (5)

Q10, May 2014


64. Given that 5 62y xx

= +√

, x > 0, find in their simplest form

(a) ddyx

(3)

(b) dy x∫ (3)

Q4, May 2014_R

65. 12d 6

dy x x xx

−= + √ , x > 0


(7) Q8, May 2014_R

66. The curve C has equation 21 83

y x= + .

The line L has equation y = 3x + k, where k is a positive constant. (a) Sketch C and L on separate diagrams, showing the coordinates of the points at which

C and L cut the axes. (4)

Given that line L is a tangent to C, (b) find the value of k.

(5)

Q9, May 2014_R


67.

Figure 3

A sketch of part of the curve C with equation

1820 4y xx

= − − , x > 0

is shown in Figure 3. Point A lies on C and has an x coordinate equal to 2. (a) Show that the equation of the normal to C at A is y = –2x + 7.

(6) The normal to C at A meets C again at the point B, as shown in Figure 3. (b) Use algebra to find the coordinates of B.

(5) Q11, May 2014_R


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