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

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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.
Answer/Explanation

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.[2]

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.[2]

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.[2]

c.
Answer/Explanation

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 \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[3]

a.

The gradient of the tangent to the curve is \( – 14\) when \(x = 1\).

Find the value of \(a\).[3]

b.
Answer/Explanation

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)

Note: Follow through from their gradient function from part (a).

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.
Answer/Explanation

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.[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.
Answer/Explanation

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)[1]

a.

Find the gradient of the graph of f at \(x =  – \frac{1}{2}\).[2]

b.

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

c.
Answer/Explanation

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;[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.
Answer/Explanation

Markscheme

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 \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[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.
Answer/Explanation

Markscheme

\(2x\)     (A1)     (C1)

a.

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

b.

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

Note: Follow through from their answer to part (b).

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:          \(\lim_{x\to 0}\frac{e^{x}-1}{x}\)

Answer/Explanation

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

Answer/Explanation

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 \(\lim_{Q \to 0}P\).

Answer/Explanation

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:  \(\lim_{x \to \pm\infty }\frac{3x+2}{2x+5}\).

Answer/Explanation

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)  \(\lim_{x \to 2^{-}}\frac{x-3}{x-2}\)

Answer/Explanation

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.

Answer/Explanation

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.

Answer/Explanation

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

Answer/Explanation

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

Answer/Explanation

Ans.

 

Question

[Maximum mark: 14]
Let

Answer/Explanation

Ans.

(a) and (b) as in exercise 9

 

 

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