# 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.[3]

a.

Calculate the value of x for which the gradient of the two graphs is the same.[2]

b.

Draw the tangent to the curved graph for this value of x on the figure, showing clearly the property in part (b).[1]

c.

## Markscheme

(i) $$f_1 ‘ (x) = 1$$Â Â Â Â  (A1)

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

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

a.

$$1 = – 2x$$Â Â Â Â  (M1)

$$x = – \frac{1}{2}$$Â Â Â Â  (A1)(ft)Â Â Â Â  (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.

Â  Â  Â (A1)Â Â Â Â  (C1)[1 mark]

c.

## Question

The table given below describes the behaviour of f â€²(x), the derivative function of f (x), in the domain âˆ’4 < x < 2.

State whether f (0) is greater than, less than or equal to f (âˆ’2). Give a reason for your answer.[2]

a.

The point P(âˆ’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.[2]

b.

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

From the information given about f â€²(x), state whether the point (âˆ’2, 3) is a maximum, a minimum or neither. Give a reason for your answer.[2]

c.

## Markscheme

greater than Â  Â  (A1)

Gradient between x = âˆ’2 and x = 0 is positive.Â Â Â Â  (R1)

OR

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

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

a.

y = 3 Â  Â  (A1)(A1) Â  Â  (C2)

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

b.

minimumÂ Â Â Â  (A1)

Gradient is negative to the left and positive to the right or equivalent.Â Â Â Â  (R1) Â  Â  (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 $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.[3]

a.

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

Find the value of $$a$$.[3]

b.

## Markscheme

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

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

$$a = 16$$ Â  Â  (A1)(ft) Â  Â  (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 $$f$$.[3]

a.

Find the point on the graph of $$f$$ at which the gradient of the tangent is equal to 6.[3]

b.

## Markscheme

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

Note: Â  Â  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.

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

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

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

Note: Â  Â  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 , âˆ’8) and the point B has coordinates (âˆ’2 , 4).

The point D has coordinates (âˆ’3 , 1).

Write down the coordinates of C, the midpoint of line segment AB.[2]

a.

Find the gradient of the line DC.[2]

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.[2]

c.

## Markscheme

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

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

a.

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

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

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

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

b.

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

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

OR

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

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

$$3x + 4y + 5 = 0$$Â  (accept any integer multiple, including negative multiples)Â  Â Â (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 $$x =Â – 3$$Â or $$x + 3 = 0$$.[2 marks]

c.

## Question

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

Find f’(x)[1]

a.

Find the gradient of the graph of f atÂ $$x =Â – \frac{1}{2}$$.[2]

b.

Find the x-coordinate of the point at which the normal to the graph of f has gradientÂ $${ – \frac{1}{8}}$$.[3]

c.

## Markscheme

x3Â  Â  Â (A1) (C1)

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

a.

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

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

$${ – \frac{1}{8}}$$Â  (âˆ’0.125)Â  Â  Â (A1)(ft) (C2)

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

b.

x3 = 8Â  Â  Â (A1)(M1)

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

(x =) 2Â  Â  Â (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;[1]

a.

where the function attains its least value;[1]

b.

where the function attains its greatest value;[1]

c.

where the gradient of the tangent to the curve is positive;[1]

d.

where $$f(x) > 0$$ and $$f'(x) < 0$$ .[2]

e.

## Markscheme

B, FÂ Â Â  Â (C1)

a.

HÂ Â Â Â  (C1)

b.

FÂ Â Â Â  (C1)

c.

A, EÂ Â Â Â  (C1)

d.

CÂ Â Â Â  (C2)

e.

## Question

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

Write down $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$.[1]

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 .[2]

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 $$y = mx + c$$ .[3]

c.

## Markscheme

$$2x$$Â Â Â Â  (A1)Â Â Â Â  (C1)

a.

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

b.

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

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

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

$$y =Â – \frac{1}{6}x + 9\frac{1}{2}$$Â Â Â Â  (A1)(ft)Â Â Â Â  (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:Â  Â  Â  Â  Â  $$\lim_{x\to 0}\frac{e^{x}-1}{x}$$

Ans.

Â

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

Â

### 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:Â  Â  Â  Â $$\lim_{x\to 2}\frac{e^{x-2}-x+1}{(x-2)^{2}}$$

Ans.

(a) f (1) = 0.2838

(b) because the denominator is 0.

(c)

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

### Question

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

(a) Write down the values of P for
(i) Q = – 0.5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (ii) Q = 1,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (iii) Q = 4 .
(b) On the following diagram, sketch the graph of P vs Q, for – 0.5 â‰¤ Q â‰¤ 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)Â  Â (i) 3.09Â  Â  Â  Â  Â  Â  Â  Â  (ii) 1.23Â  Â  Â  Â  Â  Â  Â  Â  (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:Â  Â  Â  $$\lim_{x \to \infty }\frac{3x+2}{x+5}$$

(c) Complete the values of $$f$$ , correct to 6 s.f. on the table below:

(d) Deduce the value of the limit:Â  Â  Â  $$\lim_{x \to -\infty }\frac{3x+2}{x+5}$$.

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

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

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

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

Â

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

Ans.

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

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

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

### 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 $$m_{1}=2$$ sinceÂ  $$\lim_{h\to0}\frac{(1+h)^{2}-1^{1}}{h}=2$$.

(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)Â  Â at $$x$$ = 5.Â  Â  Â  Â  Â  Â  Â  Â (ii)$$x$$ = – 5

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

Ans.

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

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

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

(c)Â  $$m_{a} = 2a$$

Â

### 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)Â  $$\lim_{h \to 0}\frac{ln(5+h)-ln\;5}{h}$$ 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)Â  Â  Â (i) $$\lim_{h \to 0}\frac{ln(2+h)-ln\;2}{h}= 0.5$$,Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  $$m_{2}=0.5=\frac{1}{2}$$

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

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

(c)Â  $$m_{a}=\frac{1}{a}$$

### 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 â‰… 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.

Â

### Question

[Maximum mark: 14]
Let