calculus task [69 marks]

1. [Maximum mark: 16] SPM.1.SL.TZ0.8

3
3
x + x
2
− 15x + 17.

(a) Find f ′ (x). [2]

Markscheme

′
f (x) = x
2
+ 2x − 15 (M1)A1

[2 marks]

The graph of f has horizontal tangents at the points where x = a and x = b, a < b.

(b) Find the value of a and the value of b. [3]

Markscheme

correct reasoning that f ′ (x) = 0 (seen anywhere) (M1)

2
x + 2x − 15 = 0

valid approach to solve quadratic M1

(x − 3) (x + 5), quadratic formula

correct values for x

3, −5

correct values for a and b

a= −5 and b = 3 A1

[3 marks]

(c.i) Sketch the graph of y = f (x).

Markscheme
 A1

[1 mark]

(c.ii) Hence explain why the graph of f has a local maximum point at x = a. [1]

Markscheme

first derivative changes from positive to negative at x = a A1

so local maximum at x = a AG

[1 mark]

(d.i) Find f ′′ (b). [3]

Markscheme

f
′′
(x) = 2x + 2 A1

substituting their b into their second derivative (M1)

′′
f (3) = 2 × 3 + 2

f
′′
(b) = 8 (A1)
 [3 marks]

(d.ii) Hence, use your answer to part (d)(i) to show that the graph of f has a
local minimum point at x = b. [1]

Markscheme

(b) is positive so graph is concave up R1

so local minimum at x = b AG

[1 mark]

(e) The normal to the graph of f at x = a and the tangent to the graph of f at x = b

intersect at the point (p, q) .

Find the value of p and the value of q. [5]

Markscheme

normal to f at x = a is x = −5 (seen anywhere) (A1)

attempt to find y-coordinate at their value of b (M1)

f (3) = −10 (A1)

tangent at x = b has equation y = −10 (seen anywhere) A1

intersection at (−5, −10)

p= −5 and q = −10 A1

[5 marks]
2. [Maximum mark: 16] 23M.1.SL.TZ2.7
The following diagram shows part of the graph of a quadratic function f .

The vertex of the parabola is (−2, − 5) and the y-intercept is at point P.

(a) Write down the equation of the axis of symmetry. [1]

Markscheme

x = −2 (must be an equation) A1

[1 mark]

1 2
The function can be written in the form f (x) =
4
(x − h) + k, where h, k ∈ Z.

(b) Write down the values of h and k. [2]

Markscheme

h = −2, k = −5 A1A1

[2 marks]

(c) Find the y-coordinate of P. [2]

substituting x = 0 into f (x) (M1)

1 2
y = (0 + 2) − 5
4

y = −4 (accept P(0, − 4)) A1

[2 marks]

In the following diagram, the line L is normal to the graph of f at point P.

(d) Find the equation of the line L, in the form y = ax + b. [3]

Markscheme

1 1
f ′(x) =
2
(x + 2) (=
2
x + 1) (A1)

substituting x = 0 into their derivative (M1)

f ′(0) = 1

gradient of normal is −1 (may be seen in their equation) A1

y = −x − 4 (accept a = −1, b = −4) A1

Note: Award A0 for L = −x − 4 (without the y =).

The line L intersects the graph of f at a second point, Q, as shown above.

(e) Calculate the distance between P and Q. [8]

Markscheme

equating theirf (x) to their L (M1)

1 2
(x + 2) − 5 = −x − 4
4

1
+ 2x = 0 (or equivalent) (A1)
2
x
4

valid attempt to solve their quadratic (M1)

4
x(x + 8) = 0 OR x(x + 8) = 0

x = −8 A1

Note: Accept both solutions x = −8 and x = 0 here, x = −8 may be seen in working to

substituting their value of x (not x = 0) into their f (x) or their L (M1)

2
y = −(−8) − 4 OR y =
1

4
(−8 + 2) − 5

Q(−8, 4) A1

correct substitution into distance formula (A1)

2 2
√ (−8 − 0) + (4 − (−4))

distance = √ 128 (= 8√ 2) A1

[8 marks]
3. [Maximum mark: 15] 22M.1.SL.TZ2.7
The following diagram shows part of the graph of a quadratic function f .

The graph of f has its vertex at (3, 4), and it passes through point Q as shown.

(a) Write down the equation of the axis of symmetry. [1]

Markscheme

x = 3 A1

Note: Must be an equation in the form “ x = ”. Do not accept 3 or = 3.

2a

[1 mark]

2
The function can be written in the form f (x) = a(x − h) + k.

(b.i) Write down the values of h and k. [2]

Markscheme

2
h = 3, k = 4 (accept a(x − 3) + 4) A1A1
 [2 marks]

(b.ii) Point Q has coordinates (5, 12). Find the value of a. [2]

Markscheme

attempt to substitute coordinates of Q (M1)

2
12 = a(5 − 3) + 4, 4a + 4 = 12

a = 2 A1

[2 marks]

The line L is tangent to the graph of f at Q.

(c) Find the equation of L. [4]

Markscheme

recognize need to find derivative of f (M1)

f ′(x) = 4(x − 3) or f ′(x) = 4x − 12 A1

f ′(5) = 8 (may be seen as gradient in their equation) (A1)

y − 12 = 8(x − 5) or y = 8x − 28 A1

Note: Award A0 for L = 8x − 28.

[4 marks]

Now consider another function y = g(x). The derivative of g is given by g′(x) = f (x) − d

, where d ∈ R.

(d) Find the values of d for which g is an increasing function. [3]

Markscheme
 METHOD 1

Recognizing that for g to be increasing, f (x) − d > 0, or g′> 0 (M1)

The vertex must be above the x-axis, 4 − d > 0, d − 4 < 0 (R1)

d < 4 A1

METHOD 2

attempting to find discriminant of g′ (M1)

2
(−12) − 4(2)(22 − d)

recognizing discriminant must be negative (R1)

−32 + 8d < 0 OR Δ < 0

d < 4 A1

[3 marks]

(e) Find the values of x for which the graph of g is concave-up. [3]

Markscheme

recognizing that for g to be concave up, g′′> 0 (M1)

g′′> 0 when f ′> 0, 4x − 12 > 0, x − 3 > 0 (R1)

x > 3 A1

[3 marks]
4. [Maximum mark: 7] 21N.1.SL.TZ0.5
The function f is defined for all x ∈ R. The line with equation y = 6x − 1 is the tangent to the

graph of f at x = 4.

(a) Write down the value of f ′(4). [1]

Markscheme

f ′(4) = 6 A1

[1 mark]

(b) Find f (4). [1]

Markscheme

f (4) = 6 × 4 − 1 = 23 A1

[1 mark]

The function g is defined for all x ∈ R where g(x) = x

(c) Find h(4). [2]

Markscheme

h(4) = f (g(4)) (M1)

2
h(4) = f (4 − 3 × 4) = f (4)

h(4) = 23 A1

[2 marks]

(d) Hence find the equation of the tangent to the graph of h at x = 4. [3]

Markscheme
attempt to use chain rule to find h′ (M1)

f ′(g(x)) × g′(x) OR (x
2
− 3x)′×f ′(x
2
− 3x)

h′(4) = (2 × 4 − 3)f ′(4

= 30

y − 23 = 30(x − 4) OR y = 30x − 97 A1

[3 marks]
5. [Maximum mark: 7] 19N.2.SL.TZ0.S_3
Let f (x) = x − 8, g (x) = x
4
− 3 and h (x) = f (g (x)).

(a) Find h (x). [2]

Markscheme

attempt to form composite (in any order) (M1)

4
eg f (x
4
− 3), (x − 8) − 3

h (x) = x
4
− 11 A1 N2

[2 marks]

(b) Let C be a point on the graph of h. The tangent to the graph of h at C is parallel to
the graph of f .

Find the x-coordinate of C. [5]

Markscheme

recognizing that the gradient of the tangent is the derivative (M1)

eg h
′

correct derivative (seen anywhere) (A1)

′ 3
h (x) = 4x

correct value for gradient of f (seen anywhere) (A1)

f (x) = 1, m = 1
′

setting their derivative equal to 1 (M1)

3
4x = 1

0.629960

1
(exact), A1 N3
3
x = √ 0.630
4

[5 marks]
6. [Maximum mark: 8] 19M.2.SL.TZ2.T_5
Consider the function f (x) =
1

3
3
x +
3

4
2
x − x − 1.

(b) Write down the y-intercept of the graph of y = f (x). [1]

Markscheme

−1 (A1)

Note: Accept (0, −1).

[1 mark]

(c) Sketch the graph of y = f (x) for −3 ≤ x ≤ 3 and −4 ≤ y ≤ 12. [4]

Markscheme

(A1)(A1)

(A1)(A1)

Note: Award (A1) for correct window and axes labels, −3 to 3 should be indicated on the x-axis
and −4 to 12 on the y-axis.
(A1)) for smooth curve with correct cubic shape;
(A1) for x-intercepts: one close to −3, the second between −1 and 0, and third between 1 and 2;
and y-intercept at approximately −1;
(A1) for local minimum in the 4th quadrant and maximum in the 2nd quadrant, in approximately
 correct positions.
Graph paper does not need to be used. If window not given award at most (A0)(A1)(A0)(A1).

[4 marks]

The function has one local maximum at x = p and one local minimum at x = q.

(h) Determine the range of f (x) for p ≤ x ≤ q. [3]

Markscheme

61 4
−1.27 ⩽ f (x) ⩽ 1.33 (−1.27083 … ⩽ f (x) ⩽ 1.33333 … , − ⩽ f (x) ⩽ )
48 3

(A1)(ft)(A1)(ft)(A1)

Note: Award (A1) for −1.27 seen, (A1) for 1.33 seen, and (A1) for correct weak inequalities with their
endpoints in the correct order. For example, award (A0)(A0)(A0) for answers like 5 ⩽ f (x) ⩽ 2.

Accept y in place of f (x). Accept alternative correct notation such as [−1.27, 1.33].

Follow through from their p and q values from part (g) only if their f (p) and f (q) values are
between −4 and 12. Award (A0)(A0)(A0) if their values from (g) are given as the endpoints.

[3 marks]

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