# IBDP Maths AI: Topic SL 5.1: Derivative interpreted as gradient function: IB style Questions SL Paper 1

## Question

The figure below shows the graphs of functions $$f_1 (x) = x$$ and $$f_2 (x) = 5 – x^2$$. (i) Differentiate $$f_1 (x)$$ with respect to x.

(ii) Differentiate $$f_2 (x)$$ with respect to x.

a.

Calculate the value of x for which the gradient of the two graphs is the same.

b.

Draw the tangent to the curved graph for this value of x on the figure, showing clearly the property in part (b).

c.

## Markscheme

(i) $$f_1 ‘ (x) = 1$$     (A1)

(ii) $$f_2 ‘ (x) = – 2x$$     (A1)(A1)

(A1) for correct differentiation of each term.     (C3)[3 marks]

a.

$$1 = – 2x$$     (M1)

$$x = – \frac{1}{2}$$     (A1)(ft)     (C2)[2 marks]

b.

(A1) is for the tangent drawn at $$x = \frac{1}{2}$$ and reasonably parallel to the line $$f_1$$ as shown. (A1)     (C1)[1 mark]

c.

## Question

The table given below describes the behaviour of f ′(x), the derivative function of f (x), in the domain −4 < x < 2. State whether f (0) is greater than, less than or equal to f (−2). Give a reason for your answer.

a.

The point P(−2, 3) lies on the graph of f (x).

Write down the equation of the tangent to the graph of f (x) at the point P.

b.

The point P(−2, 3) lies on the graph of f (x).

From the information given about f ′(x), state whether the point (−2, 3) is a maximum, a minimum or neither. Give a reason for your answer.

c.

## Markscheme

greater than     (A1)

Gradient between x = −2 and x = 0 is positive.     (R1)

OR

The function is increased between these points or equivalent.     (R1)     (C2)

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

a.

y = 3     (A1)(A1)     (C2)

Note: Award (A1) for y = a constant, (A1) for 3.[2 marks]

b.

minimum     (A1)

Gradient is negative to the left and positive to the right or equivalent.     (R1)     (C2)

Note: Accept a sketch. Do not award (A1)(R0).[2 marks]

c.

## Question

Consider the curve $$y = {x^2} + \frac{a}{x} – 1,{\text{ }}x \ne 0$$.

Find $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.

a.

The gradient of the tangent to the curve is $$– 14$$ when $$x = 1$$.

Find the value of $$a$$.

b.

## Markscheme

$$2x – \frac{a}{{{x^2}}}$$     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for $$2x$$, (A1) for $$– a$$ and (A1) for $${x^{ – 2}}$$.

Award at most (A1)(A1)(A0) if extra terms are present.

a.

$$2(1) – \frac{a}{{{1^2}}} = – 14$$     (M1)(M1)

Note: Award (M1) for substituting $$1$$ into their gradient function, (M1) for equating their gradient function to $$– 14$$.

Award (M0)(M0)(A0) if the original function is used instead of the gradient function.

$$a = 16$$     (A1)(ft)     (C3)

b.

## Question

A function $$f$$ is given by $$f(x) = 4{x^3} + \frac{3}{{{x^2}}} – 3,{\text{ }}x \ne 0$$.

Write down the derivative of $$f$$.

a.

Find the point on the graph of $$f$$ at which the gradient of the tangent is equal to 6.

b.

## Markscheme

$$12{x^2} – \frac{6}{{{x^3}}}$$ or equivalent     (A1)(A1)(A1)     (C3)

Note:     Award (A1) for $$12{x^2}$$, (A1) for $$– 6$$ and (A1) for $$\frac{1}{{{x^3}}}$$ or $${x^{ – 3}}$$. Award at most (A1)(A1)(A0) if additional terms seen.[3 marks]

a.

$$12{x^2} – \frac{6}{{{x^3}}} = 6$$     (M1)

Note:     Award (M1) for equating their derivative to 6.

$$(1,{\text{ }}4)$$$$\,\,\,$$OR$$\,\,\,$$$$x = 1,{\text{ }}y = 4$$     (A1)(ft)(A1)(ft)     (C3)

Note:     A frequent wrong answer seen in scripts is $$(1,{\text{ }}6)$$ for this answer with correct working award (M1)(A0)(A1) and if there is no working award (C1).[3 marks]

b.

## Question

The point A has coordinates (4 , −8) and the point B has coordinates (−2 , 4).

The point D has coordinates (−3 , 1).

Write down the coordinates of C, the midpoint of line segment AB.

a.

Find the gradient of the line DC.

b.

Find the equation of the line DC. Write your answer in the form ax + by + d = 0 where a , b and d are integers.

c.

## Markscheme

(1, −2)    (A1)(A1) (C2)
Note: Award (A1) for 1 and (A1) for −2, seen as a coordinate pair.

Accept x = 1, y = −2. Award (A1)(A0) if x and y coordinates are reversed.[2 marks]

a.

$$\frac{{1 – \left( { – 2} \right)}}{{ – 3 – 1}}$$    (M1)

Note: Award (M1) for correct substitution, of their part (a), into gradient formula.

$$= – \frac{3}{4}\,\,\,\left( { – 0.75} \right)$$     (A1)(ft)  (C2)

Note: Follow through from part (a).[2 marks]

b.

$$y – 1 = – \frac{3}{4}\left( {x + 3} \right)$$  OR  $$y + 2 = – \frac{3}{4}\left( {x – 1} \right)$$  OR  $$y = – \frac{3}{4}x – \frac{5}{4}$$      (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

OR

$$1 = – \frac{3}{4} \times – 3 + c$$  OR  $$– 2 = – \frac{3}{4} \times 1 + c$$     (M1)

Note: Award (M1) for correct substitution of their part (b) and a given point.

$$3x + 4y + 5 = 0$$  (accept any integer multiple, including negative multiples)    (A1)(ft) (C2)

Note: Follow through from parts (a) and (b). Where the gradient in part (b) is found to be $$\frac{5}{0}$$, award at most (M1)(A0) for either $$x = – 3$$ or $$x + 3 = 0$$.[2 marks]

c.

## Question

Consider the function $$f\left( x \right) = \frac{{{x^4}}}{4}$$.

Find f’(x)

a.

Find the gradient of the graph of f at $$x = – \frac{1}{2}$$.

b.

Find the x-coordinate of the point at which the normal to the graph of f has gradient $${ – \frac{1}{8}}$$.

c.

## Markscheme

x3     (A1) (C1)

Note: Award (A0) for $$\frac{{4{x^3}}}{4}$$ and not simplified to x3.[1 mark]

a.

$${\left( { – \frac{1}{2}} \right)^3}$$     (M1)

Note: Award (M1) for correct substitution of $${ – \frac{1}{2}}$$ into their derivative.

$${ – \frac{1}{8}}$$  (−0.125)     (A1)(ft) (C2)

Note: Follow through from their part (a).[2 marks]

b.

x3 = 8     (A1)(M1)

Note: Award (A1) for 8 seen maybe seen as part of an equation y = 8x + c(M1) for equating their derivative to 8.

(x =) 2     (A1) (C3)

Note: Do not accept (2, 4).[3 marks]

c.

## Question

Consider the graph of the function $$y = f(x)$$ defined below. Write down all the labelled points on the curve

that are local maximum points;

a.

where the function attains its least value;

b.

where the function attains its greatest value;

c.

where the gradient of the tangent to the curve is positive;

d.

where $$f(x) > 0$$ and $$f'(x) < 0$$ .

e.

B, F     (C1)

a.

H     (C1)

b.

F     (C1)

c.

A, E     (C1)

d.

C     (C2)

e.

## Question

Consider the curve $$y = {x^2}$$ .

Write down $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.

a.

The point $${\text{P}}(3{\text{, }}9)$$ lies on the curve $$y = {x^2}$$ . Find the gradient of the tangent to the curve at P .

b.

The point $${\text{P}}(3{\text{, }}9)$$ lies on the curve $$y = {x^2}$$ . Find the equation of the normal to the curve at P . Give your answer in the form $$y = mx + c$$ .

c.

## Markscheme

$$2x$$     (A1)     (C1)

a.

$$2 \times 3$$     (M1)
$$= 6$$     (A1)     (C2)

b.

$$m({\text{perp}}) = – \frac{1}{6}$$     (A1)(ft)

Equation $$(y – 9) = – \frac{1}{6}(x – 3)$$     (M1)

Note: Award (M1) for correct substitution in any formula for equation of a line.

$$y = – \frac{1}{6}x + 9\frac{1}{2}$$     (A1)(ft)     (C3)

Note: Follow through from correct substitution of their gradient of the normal.
Note: There are no extra marks awarded for rearranging the equation to the form $$y = mx + c$$ .

c.

[MAI 5.1] THE CONCEPT OF THE LIMIT-lala

### Question

[Maximum mark: 5]
Part of the graph of $$f(x)=\frac{e^{x}-1}{x}$$ is shown below. The function is not defined at x = 0 . (a) Complete the values of $$f$$ , correct to 4 decimal places, on the tables below: (b) Deduce the value of the limit:          $$\lim_{x\to 0}\frac{e^{x}-1}{x}$$

Ans. (b)   $$\lim_{x \to 0}\frac{e^{x}-1}{x}=1$$

### Question

[Maximum mark: 7]
Part of the graph of $$f(x)=\frac{e^{x-2}-x+1}{(x-2)^{2}}$$ is shown below: (a) Write down, correct to 4 s.f. , the value of f (0)
(b) Explain why $$f (2)$$ is not defined.
(c) Complete the values of $$f$$ , correct to 4 s.f. on the tables below: (d) Deduce the value of the limit:       $$\lim_{x\to 2}\frac{e^{x-2}-x+1}{(x-2)^{2}}$$

Ans.

(a) f (1) = 0.2838

(b) because the denominator is 0.

(c) (d)  $$\lim_{x \to 2}\frac{e^{x-2}-x+1}{(x-2)^{2}}=0.5$$

### Question

[Maximum mark: 7]
Let $$P=\frac{4Q-4\;ln(1+Q)}{Q^{2}}$$,  Q ≠ 0 , – 0.5 ≤ Q ≤ 4,

(a) Write down the values of P for
(i) Q = – 0.5                    (ii) Q = 1,                           (iii) Q = 4 .
(b) On the following diagram, sketch the graph of P vs Q, for – 0.5 ≤ Q ≤ 4
(c) P is not defined for Q = 0 . By investigating the values of P corresponding to
values of Q near zero, deduce the value of $$\lim_{Q \to 0}P$$. Ans.

(a)   (i) 3.09                (ii) 1.23                (ii) 0.598

(b) (c) $$\lim_{Q \to 0}P=2$$.

### Question

[Maximum mark: 8]
Let $$f(x)=\frac{3x+2}{x+5}$$

(a) Complete the values of $$f$$ , correct to 6 s.f. on the table below: (b) Deduce the value of the limit:      $$\lim_{x \to \infty }\frac{3x+2}{x+5}$$

(c) Complete the values of $$f$$ , correct to 6 s.f. on the table below: (d) Deduce the value of the limit:      $$\lim_{x \to -\infty }\frac{3x+2}{x+5}$$.

(e) By using a similar rationale deduce the value of the limits:  $$\lim_{x \to \pm\infty }\frac{3x+2}{2x+5}$$.

Ans.

(a) (b) $$\lim_{x \to \infty}\frac{3x+2}{x+5}=3$$

(c) (d) $$\lim_{x \to -\infty}\frac{3x+2}{x+5}=3$$

(e) $$\lim_{x \to \pm\infty}\frac{3x+2}{2x+5}=1.5$$

### Question

[Maximum mark: 7]
(a) By observing the graph of the function $$f(x)=\frac{x+3}{x-2}$$on your GDC, or otherwise
find the values of the following limits

(i)   $$\lim_{x \to 3}\frac{x+3}{x-2}$$           (ii)  $$\lim_{x \to +\infty }\frac{x+3}{x-2}$$            (iii) $$\lim_{x \to -\infty }\frac{x+3}{x-2}$$

(b) Investigate whether each of the following side limits is +∞ or -∞:

(i) $$\lim_{x \to 2^{+}}\frac{x+3}{x-2}$$                                 (ii) $$\lim_{x \to 2^{-}}\frac{x+3}{x-2}$$

(iii) $$\lim_{x \to 2^{+}}\frac{x-3}{x-2}$$                                    (iv)  $$\lim_{x \to 2^{-}}\frac{x-3}{x-2}$$

Ans.

(a)    (i)   $$\lim_{x \to 3}\frac{x+3}{x-2}=6$$          (ii)  $$\lim_{x \to +\infty }\frac{x+3}{x-2}=1$$           (iii) $$\lim_{x \to -\infty }\frac{x+3}{x-2}=1$$

(b)   (i)  $$\lim_{x \to 2^{+}}\frac{x+3}{x-2}=+\infty$$                   (ii) $$\lim_{x \to 2^{-}}\frac{x+3}{x-2}=-\infty$$

(iii)  $$\lim_{x \to 2^{+}}\frac{x-3}{x-2}=-\infty$$                     (iv)  $$\lim_{x \to 2^{-}}\frac{x-3}{x-2}=+\infty$$

### Question

[Maximum mark: 8]
The gradient of the curve $$f(x)=x^{2}$$ at point x = a is defined by the limit

$$m_{a}=\lim_{h\to0}\frac{(a+h)^{2}-a^{2}}{h}$$

For example, the gradient at x = 1 is $$m_{1}=2$$ since  $$\lim_{h\to0}\frac{(1+h)^{2}-1^{1}}{h}=2$$.

(a) By using the graph mode on your GDC, find

(i) $$\lim_{h\to 0}\frac{(2+h)^{2}-2^{2}}{h}$$ and hence the gradient $$m_{2}$$.

(ii) $$\lim_{h\to 0}\frac{(3+h)^{2}-3^{2}}{h}$$ and hence the gradient $$m_{3}$$.

(b) Find the gradient of the curve
(i)   at $$x$$ = 5.               (ii)$$x$$ = – 5

(c) Deduce the value of the gradient $$m_{a}$$ in terms of $$a$$ in general.

Ans.

(a)    (i) $$\lim_{h\to 0}\frac{(2+h)^{2}-2^{2}}{h}= 4$$,        $$m_{2}= 4$$

(ii) $$\lim_{h\to 0}\frac{(3+h)^{2}-3^{2}}{h}= 6$$,          $$m_{3}= 6$$

(b)    (i) $$m_{5}= 10$$,                             (ii) $$m_{-5}= – 10$$

(c)  $$m_{a} = 2a$$

### Question

[Maximum mark: 8]
The gradient of the curve $$f(x)=ln\;x$$ at point $$x = a$$ is defined by the limit

$$m_{a}=\lim_{h \to 0}\frac{ln(a+h)-ln\;a}{h}$$

For example, the gradient at x = 1 is $$m_{1}$$ since $$\lim_{h \to 0}\frac{ln(1+h)-ln\;1}{h}=1$$.

(a) By using the graph mode on your GDC, find

(i) $$\lim_{h \to 0}\frac{ln(2+h)-ln\;2}{h}$$ and hence the gradient $$m_{2}$$.

(ii)  $$\lim_{h \to 0}\frac{ln(5+h)-ln\;5}{h}$$ and hence the gradient $$m_{5}$$.

(b) Find the gradient $$m_{10}$$ of the curve at $$x$$ = 10 .

(c) Deduce the value of the gradient $$m_{a}$$ in terms of a in general.

Ans.

(a)     (i) $$\lim_{h \to 0}\frac{ln(2+h)-ln\;2}{h}= 0.5$$,                            $$m_{2}=0.5=\frac{1}{2}$$

(ii) $$\lim_{h \to 0}\frac{ln(5+h)-ln\;5}{h}= 0.2$$,                            $$m_{3}=0.2=\frac{1}{5}$$

(b) $$m_{5}=0.1=\frac{1}{10}$$

(c)  $$m_{a}=\frac{1}{a}$$

### Question

[Maximum mark: 8]
(a) Write down the value of e2 correct to 5 s.f.

(b) Write down, correct to 5 s.f., the values of the expression

$$\left ( 1+\frac{2}{n} \right )^{n}$$

for the values of $$n$$ shown on the following table: (c) Hence, guess the value of $$\lim_{n \to +\infty }\left ( 1+\frac{2}{n} \right )^{n}$$

(d) By following a similar rationale guess the value of $$\lim_{n \to +\infty }\left ( 1+\frac{3}{n} \right )^{n}$$

Ans.

(a) e2 ≅ 7.3891

(b) (c) $$\lim_{n \to +\infty }\left ( 1+\frac{2}{n} \right )^{n}=e^{2}$$

(d) $$\lim_{n \to +\infty }\left ( 1+\frac{3}{n} \right )^{n}=e^{3}$$

### Question

[Maximum mark: 15]
The graph of the function $$f$$ is shown below  Ans. ### Question

[Maximum mark: 14]
Let    