IB Questionbank

Intro derivatives [44 marks]

1a.

Let . The following diagram shows part of the graph of .

There are -intercepts at and at . There is a maximum at A where , and a point of inflexion at B where .

Find the value of .

Markscheme

evidence of valid approach (M1)

eg

2.73205

A1 N2

[2 marks]

f(x) = −0.5x

4

+ 3x

2

+ 2x f

x x = 0 x = p x = a x = b

p

f(x) = 0, y = 0

p = 2.73

1b.

Write down the coordinates of A.

Markscheme

1.87938, 8.11721

A2 N2

[2 marks]

(1.88, 8.12)

1c.

Write down the rate of change of at A.

Markscheme

rate of change is 0 (do not accept decimals) A1 N1

[1 marks]

f

1d.

Find the coordinates of B.

[2 marks]

[1 mark]

[4 marks]

Markscheme

METHOD 1 (using GDC)

valid approach M1

eg , max/min on

sketch of either or , with max/min or root (respectively) (A1)

A1 N1

Substituting their value into (M1)

eg

A1 N1

METHOD 2 (analytical)

A1

setting (M1)

A1 N1

substituting their value into (M1)

eg

A1 N1

[4 marks]

f

′′

= 0 f

′

, x = −1

f

′

f

′′

x = 1

x f

f(1)

y = 4.5

f

′′

= −6x

2

+ 6

f

′′

= 0

x = 1

x f

f(1)

y = 4.5

1e.

Find the the rate of change of at B.

Markscheme

recognizing rate of change is (M1)

eg

rate of change is 6 A1 N2

[3 marks]

f

′

y

′

, f

′

(1)

1f.

Let be the region enclosed by the graph of , the -axis, the line and the line . The region is rotated 360° about

the -axis. Find the volume of the solid formed.

Markscheme

attempt to substitute either limits or the function into formula (M1)

involving (accept absence of and/or )

eg

128.890

A2 N3

[3 marks]

R f x x = b x = a R

x

f

2

π dx

π ∫ (−0.5x

4

+ 3x

2

+ 2x)

2

dx, ∫

1.88

1

f

2

volume = 129

2a.

Let , for .

Write down the equation of the horizontal asymptote of the graph of .

f(x) = + 2

1

x−

1

x > 1

f

[3 marks]

[2 marks]

Markscheme

(correct equation only) A2 N2

[2 marks]

y = 2

2b.

Find .

Markscheme

valid approach (M1)

eg

A1 N2

[2 marks]

f

′

(x)

(x − 1)

−

1

+ 2, f

′

(x) =

0(x−

1

)−

1

(x−

1

)

2

−(x − 1)

−

2

, f

′

(x) =

−

1

(x−

1

)

2

2c.

Let , for . The graphs of and have the same horizontal asymptote.

Write down the value of .

Markscheme

correct equation for the asymptote of

eg (A1)

A1 N2

[2 marks]

g(x) = ae

−x

+ b x ⩾ 1 f g

b

g

y = b

b = 2

2d.

Given that , find the value of .

Markscheme

correct derivative of g (seen anywhere) (A2)

eg

correct equation (A1)

eg

7.38905

A1 N2

[4 marks]

g

′

(1) = −e a

g

′

(x) = −ae

−x

−e = −ae

−

1

a = e

2

(exact), 7.39

2e.

There is a value of , for

, for which the graphs of and have the same gradient. Find this gradient.

x

1 < x < 4 f g

[2 marks]

[4 marks]

Markscheme

attempt to equate their derivatives (M1)

eg

valid attempt to solve their equation (M1)

eg correct value outside the domain of such as 0.522 or 4.51,

correct solution (may be seen in sketch) (A1)

eg

gradient is A1 N3

[4 marks]

f

′

(x) = g

′

(x), = −ae

−x

−

1

(x−

1

)

2

f

x = 2, (2, − 1)

−1

3a.

Let

.

Expand

.

Markscheme

attempt to expand (M1)

A1 N2

[2 marks]

f(x) = x

3

− 4x + 1

(x + h)

3

(x + h)

3

= x

3

+ 3x

2

h + 3xh

2

+ h

3

3b.

Use the formula

to show that the derivative of

is

.

f

′

(x) =

lim

h→0

f(x+h)−f(x)

h

f(x)

3x

2

− 4

[2 marks]

[4 marks]

Markscheme

evidence of substituting

(M1)

correct substitution A1

e.g.

simplifying A1

e.g.

factoring out h A1

e.g.

AG N0

[4 marks]

x + h

f

′

(x) =

lim

h→0

(x+h)

3

−

4

(x+h)+

1

−(x

3

−

4

x+

1

)

h

(x

3

+3x

2

h+3xh

2

+h

3

−

4

x−

4

h+

1

−x

3

+

4

x−

1

)

h

h(3x

2

+3xh+h

2

−

4

)

h

f

′

(x) = 3x

2

− 4

3c.

The tangent to the curve of f at the point

is parallel to the tangent at a point Q. Find the coordinates of Q.

Markscheme

(A1)

setting up an appropriate equation M1

e.g.

at Q,

(Q is

) A1 A1

[4 marks]

P(1, − 2)

f

′

(1) = −1

3x

2

− 4 = −1

x = −1,y = 4

(−1, 4)

3d.

The graph of f is decreasing for

. Find the value of p and of q.

Markscheme

recognizing that f is decreasing when

R1

correct values for p and q (but do not accept

) A1A1 N1N1

e.g.

;

; an interval such as

[3 marks]

p < x < q

f

′

(x) < 0

p = 1.15, q = −1.15

p = −1.15, q = 1.15

±

2

√

3

−1.15 ≤ x ≤ 1.15

3e.

Write down the range of values for the gradient of

.f

[4 marks]

[3 marks]

[2 marks]

Printed for GEMS INTERNATIONAL SCHOOL AL KHAIL

International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markscheme

,

A2 N2

[2 marks]

f

′

(x) ≥ −4

y ≥ −4

[−4,∞[