Question
The figure below shows the graphs of functions \(f_1 (x) = x\) and \(f_2 (x) = 5 – x^2\).
a.(i) Differentiate \(f_1 (x) \) with respect to x.
(ii) Differentiate \(f_2 (x) \) with respect to x.[3]
▶️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]
\(1 = – 2x\) (M1)
\(x = – \frac{1}{2}\) (A1)(ft) (C2)[2 marks]
(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]
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]
Write down the equation of the tangent to the graph of f (x) at the point P.[2]
c.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]
▶️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]
y = 3 (A1)(A1) (C2)
Note: Award (A1) for y = a constant, (A1) for 3.[2 marks]
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]
Question
Consider the curve \(y = {x^2} + \frac{a}{x} – 1,{\text{ }}x \ne 0\).
a.Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[3]
The gradient of the tangent to the curve is \( – 14\) when \(x = 1\).
b.Find the value of \(a\).[3]
▶️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.
\(2(1) – \frac{a}{{{1^2}}} = – 14\) (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).
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]
Find the point on the graph of \(f\) at which the gradient of the tangent is equal to 6.[3]
▶️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]
\(12{x^2} – \frac{6}{{{x^3}}} = 6\) (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]
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]
Find the gradient of the line DC.[2]
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]
▶️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]
\(\frac{{1 – \left( { – 2} \right)}}{{ – 3 – 1}}\) (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]
\(y – 1 = – \frac{3}{4}\left( {x + 3} \right)\) 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]
Question
Consider the function \(f\left( x \right) = \frac{{{x^4}}}{4}\).
Find f’(x)[1]
Find the gradient of the graph of f at \(x = – \frac{1}{2}\).[2]
Find the x-coordinate of the point at which the normal to the graph of f has gradient \({ – \frac{1}{8}}\).[3]
▶️Answer/Explanation
Markscheme
x3 (A1) (C1)
Note: Award (A0) for \(\frac{{4{x^3}}}{4}\) and not simplified to x3.[1 mark]
\({\left( { – \frac{1}{2}} \right)^3}\) (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]
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]
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]
where the function attains its least value;[1]
where the function attains its greatest value;[1]
where the gradient of the tangent to the curve is positive;[1]
where \(f(x) > 0\) and \(f'(x) < 0\) .[2]
▶️Answer/Explanation
Markscheme
B, F (C1)
H (C1)
F (C1)
A, E (C1)
C (C2)
Question
Consider the curve \(y = {x^2}\) .
Write down \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[1]
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]
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]
▶️Answer/Explanation
Markscheme
\(2x\) (A1) (C1)
\(2 \times 3\) (M1)
\( = 6\) (A1) (C2)
\(m({\text{perp}}) = – \frac{1}{6}\) (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 \(y = mx + c\) .
[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: 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
