IBDP Maths AI: Topic: SL 5.6: Local maximum and minimum points: IB style Questions SL Paper 1

Question

A function is represented by the equation

\[f(x) = a{x^2} + \frac{4}{x} – 3\]

Find \(f ′(x)\) .[3]

a.

The function \(f (x)\) has a local maximum at the point where \(x = −1\).

Find the value of a.[3]

b.
Answer/Explanation

Markscheme

\(f(x) = a{x^2} + 4{x^{ – 1}} – 3\)

\(f'(x) = 2ax – 4{x^{ – 2}}\)     (A3)

(A1) for 2ax, (A1) for –4x –2 and (A1) for derivative of –3 being zero.     (C3)[3 marks]

a.

\(2ax – 4x^{-2} = 0\)     (M1)

\(2a( – 1) – 4{( – 1)^{ – 2}} = 0\)     (M1)

\( -2a – 4 = 0\)

\(a = -2\)     (A1)(ft)

(M1) for setting derivative function equal to 0. (M1) for inserting \(x = -1\) but do not award (M0)(M1)     (C3)[3 marks]

b.

Question

Let \(f (x) = 2x^2 + x – 6\)

Find \(f'(x)\).[3]

a.

Find the value of \(f'( – 3)\).[1]

b.

Find the value of \(x\) for which \(f'(x) = 0\).[2]

c.
Answer/Explanation

Markscheme

\(f'(x) = 4x + 1\)     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each term differentiated correctly.

Award at most (A1)(A1)(A0) if any extra terms seen.[3 marks]

a.

\(f'( – 3) =  – 11\)     (A1)(ft)     (C1)[1 mark]

b.

\(4x + 1 = 0\)     (M1)

\(x = – \frac{{1}}{{4}}\)     (A1)(ft)     (C2)[2 marks]

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

f (x) = 5x3 − 4x2 + x

Find f‘(x).[3]

a.

Find using your answer to part (a) the x-coordinate of

(i) the local maximum point;

(ii) the local minimum point.[3]

b.
Answer/Explanation

Markscheme

15x2 – 8x + 1     (A1)(A1)(A1)     (C3)

Note: Award (A1) for each correct term.[3 marks]

a.

15x2 – 8x +1 = 0     (A1)(ft)

Note: Award (A1)(ft) for setting their derivative to zero.

(i) \((x =)\frac{1}{5}(0.2)\)     (A1)(ft) 

(ii) \((x =)\frac{1}{3}(0.333)\)     (A1)(ft)     (C3)

Notes: Follow through from their answer to part (a).[3 marks]

b.

Question

Consider the function \(f (x) = ax^3 − 3x + 5\), where \(a \ne 0\).

Find \(f ‘ (x) \).[2]

a.

Write down the value of \(f ′(0)\).[1]

b.

The function has a local maximum at x = −2.

Calculate the value of a.[3]

c.
Answer/Explanation

Markscheme

\( f ‘(x) = 3ax^2 – 3\)     (A1)(A1)     (C2)

Note: Award a maximum of (A1)(A0) if any extra terms are seen.

a.

−3     (A1)(ft)     (C1)

Note: Follow through from their part (a).

b.

\(f ‘(x) = 0\)     (M1)

Note: This may be implied from line below.

\(3a(-2)^2 – 3 = 0\)     (M1)

\((a =) \frac{1}{4}\)     (A1)(ft)     (C3)

Note: Follow through from their part (a).

c.

Question

Expand the expression \(x(2{x^3} – 1)\).[2]

a.

Differentiate \(f(x) = x(2{x^3} – 1)\).[2]

b.

Find the \(x\)-coordinate of the local minimum of the curve \(y = f(x)\).[2]

c.
Answer/Explanation

Markscheme

\(2{x^4} – x\)     (A1)(A1)     (C2)

Note: Award (A1) for \(2{x^4}\), (A1) for \( – x\).[2 marks]

a.

\(8{x^3} – 1\)     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for \(8{x^3}\), (A1)(ft) for \(–1\). Follow through from their part (a).

     Award at most (A1)(A0) if extra terms are seen.[2 marks]

b.

\(8{x^3} – 1 = 0\)     (M1)

Note: Award (M1) for equating their part (b) to zero.

\((x = )\frac{1}{2}{\text{ (0.5)}}\)     (A1)(ft)     (C2)

Notes: Follow through from part (b).

     \(0.499\) is the answer from the use of trace on the GDC; award (A0)(A0).

     For an answer of \((0.5, –0.375)\), award (M1)(A0).[2 marks]

c.

Question

Consider the curve \(y = {x^3} + kx\).

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

a.

The curve has a local minimum at the point where \(x = 2\).

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

b.

The curve has a local minimum at the point where \(x = 2\).

Find the value of \(y\) at this local minimum.[2]

c.
Answer/Explanation

Markscheme

\(3{x^2} + k\)     (A1)     (C1)[1 mark]

a.

\(3{(2)^2} + k = 0\)     (A1)(ft)(M1)

Note: Award (A1)(ft) for substituting 2 in their \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\), (M1) for setting their \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\).

\(k =  – 12\)     (A1)(ft)     (C3)

Note: Follow through from their derivative in part (a).[3 marks]

b.

\({2^3} – 12 \times 2\)     (M1)

Note: Award (M1) for substituting 2 and their –12 into equation of the curve.

\( =  – 16\)     (A1)(ft)     (C12)

Note: Follow through from their value of \(k\) found in part (b).[2 marks]

c.

Question

Consider the graph of the function \(f(x) = {x^3} + 2{x^2} – 5\).

Label the local maximum as \({\text{A}}\) on the graph.[1]

a.

Label the local minimum as B on the graph.[1]

b.

Write down the interval where \(f'(x) < 0\).[1]

c.

Draw the tangent to the curve at \(x = 1\) on the graph.[1]

d.

Write down the equation of the tangent at \(x = 1\).[2]

e.
Answer/Explanation

Markscheme

 

correct label on graph     (A1)     (C1)[1 mark]

a.

correct label on graph     (A1)     (C1)[1 mark]

b.

\( – 1.33 < x < 0\)   \(\left( { – \frac{4}{3} < x < 0} \right)\)     (A1)     (C1)[1 mark]

c.

 

tangent drawn at \(x = 1\) on graph     (A1)     (C1)[1 mark]

d.

\(y = 7x – 9\)     (A1)(A1)     (C2)

Notes: Award (A1) for \(7\), (A1) for \(-9\).

If answer not given as an equation award at most (A1)(A0).[2 marks]

e.

Question

A function is given as \(f(x) = 2{x^3} – 5x + \frac{4}{x} + 3,{\text{ }} – 5 \leqslant x \leqslant 10,{\text{ }}x \ne 0\).

Write down the derivative of the function.[4]

a.

Use your graphic display calculator to find the coordinates of the local minimum point of \(f(x)\) in the given domain.[2]

b.
Answer/Explanation

Markscheme

\(6{x^2} – 5 – \frac{4}{{{x^2}}}\)     (A1)(A1)(A1)(A1)     (C4)

Note: Award (A1) for \(6{x^2}\), (A1) for \(–5\), (A1) for \(–4\), (A1) for \({x^{ – 2}}\) or \(\frac{1}{{{x^2}}}\).

     Award at most (A1)(A1)(A1)(A0) if additional terms are seen.[4 marks]

a.

\((1.15,{\text{ }} 3.77)\) \(\left( {{\text{(1.15469…, 3.76980…)}}} \right)\)     (A1)(A1)     (C2)

Notes: Award (A1)(A1) for “\(x = 1.15\) and \(y = 3.77\)”.

     Award at most (A0)(A1)(ft) if parentheses are omitted.[2 marks]

b.

Question

A cuboid has a rectangular base of width \(x\) cm and length 2\(x\) cm . The height of the cuboid is \(h\) cm . The total length of the edges of the cuboid is \(72\) cm.

The volume, \(V\), of the cuboid can be expressed as \(V = a{x^2} – 6{x^3}\).

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

a.

Find the value of \(x\) that makes the volume a maximum.[3]

b.
Answer/Explanation

Markscheme

\(72 = 12x + 4h\;\;\;\)(or equivalent)     (M1)

Note: Award (M1) for a correct equation obtained from the total length of the edges.

\(V = 2{x^2}(18 – 3x)\)     (A1)

\((a = ){\text{ }}36\)     (A1)     (C3)

a.

\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 72x – 18{x^2}\)     (A1)

\(72x – 18{x^2} = 0\;\;\;\)OR\(\;\;\;\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\)     (M1)

Notes: Award (A1) for  \( – 18{x^2}\)  seen. Award (M1) for equating derivative to zero.

\((x = ){\text{ 4}}\)     (A1)(ft)     (C3)

Note: Follow through from part (a).

OR

Sketch of \(V\) with visible maximum     (M1)

Sketch with \(x \geqslant 0,{\text{ }}V \geqslant 0\) and indication of maximum (e.g. coordinates)     (A1)(ft)

\((x = ){\text{ 4}}\)     (A1)(ft)     (C3)

Notes: Follow through from part (a).

Award (M1)(A1)(A0) for \((4,{\text{ }}192)\).

Award (C3) for \(x = 4,{\text{ }}y = 192\).

b.

Question

A quadratic function \(f\) is given by \(f(x) = a{x^2} + bx + c\). The points \((0,{\text{ }}5)\) and \(( – 4,{\text{ }}5)\) lie on the graph of \(y = f(x)\).

The \(y\)-coordinate of the minimum of the graph is 3.

Find the equation of the axis of symmetry of the graph of \(y = f(x)\).[2]

a.

Write down the value of \(c\).[1]

b.

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

c.
Answer/Explanation

Markscheme

\(x =  – 2\)     (A1)(A1)     (C2)

Note:     Award (A1) for \(x = \) (a constant) and (A1) for \( – 2\).[2 marks]

a.

\((c = ){\text{ }}5\)     (A1)     (C1)[1 mark]

b.

\( – \frac{b}{{2a}} =  – 2\)

\(a{( – 2)^2} – 2b + 5 = 3\) or equivalent

\(a{( – 4)^2} – 4b + 5 = 5\) or equivalent

\(2a( – 2) + b = 0\) or equivalent     (M1)

Note:     Award (M1) for two of the above equations.

\(a = 0.5\)     (A1)(ft)

\(b = 2\)     (A1)(ft)     (C3)

Note:     Award at most (M1)(A1)(ft)(A0) if the answers are reversed.

Follow through from parts (a) and (b).[3 marks]

c.

Question

A factory produces shirts. The cost, C, in Fijian dollars (FJD), of producing x shirts can be modelled by

C(x) = (x − 75)2 + 100.

The cost of production should not exceed 500 FJD. To do this the factory needs to produce at least 55 shirts and at most s shirts.

Find the cost of producing 70 shirts.[2]

a.

Find the value of s.[2]

b.

Find the number of shirts produced when the cost of production is lowest.[2]

c.
Answer/Explanation
Answer/Explanation

Markscheme

(70 − 75)2 + 100     (M1)

Note: Award (M1) for substituting in x = 70.

125     (A1) (C2)[2 marks]

a.

(s − 75)2 + 100 = 500     (M1)

Note: Award (M1) for equating C(x) to 500. Accept an inequality instead of =.

OR

     (M1)

Note: Award (M1) for sketching correct graph(s).

(s =) 95    (A1) (C2)[2 marks]

b.

     (M1)

Note: Award (M1) for an attempt at finding the minimum point using graph.

OR

\(\frac{{95 + 55}}{2}\)     (M1)

Note: Award (M1) for attempting to find the mid-point between their part (b) and 55.

OR

(C’(x) =) 2x − 150 = 0     (M1)

Note: Award (M1) for an attempt at differentiation that is correctly equated to zero.

75     (A1) (C2)[2 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

A small manufacturing company makes and sells \(x\) machines each month. The monthly cost \(C\) , in dollars, of making \(x\) machines is given by
\[C(x) = 2600 + 0.4{x^2}{\text{.}}\]The monthly income \(I\) , in dollars, obtained by selling \(x\) machines is given by
\[I(x) = 150x – 0.6{x^2}{\text{.}}\]\(P(x)\) is the monthly profit obtained by selling \(x\) machines.

Find \(P(x)\) .[2]

a.

Find the number of machines that should be made and sold each month to maximize \(P(x)\) .[2]

b.

Use your answer to part (b) to find the selling price of each machine in order to maximize \(P(x)\) .[2]

c.
Answer/Explanation

Markscheme

\(P(x) = I(x) – C(x)\)     (M1)
\( = – {x^2} + 150x – 2600\)     (A1)    (C2)

a.

\( – 2x + 150 = 0\)     (M1)

Note: Award (M1) for setting \(P'(x) = 0\) .

OR

Award (M1) for sketch of \(P(x)\) and maximum point identified.     (M1)
\(x = 75\)     (A1)(ft)     (C2)

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

b.

\(\frac{{7875}}{{75}}\)     (M1)

Note: Award (M1) for \(7875\) seen.

\( = 105\) (A1)(ft)     (C2)

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

c.

[MAI 5.6-5.7] RULES OF DIFFERENTIATION-lala

Question

[Maximum mark: 3 per function]
Differentiate the following functions:

Answer/Explanation

Ans.

 

 

 

 

Question

[Maximum mark: 4]
Let \(f(x)=2x^{3}+ln\;x\)
(a) Find \({f}'(x)\)
(b) Find the gradient of the curve \(y = f (x)\) at \(x = 1\).

Answer/Explanation

Ans.

(a)\({f}'(x)=6x^{2}+\frac{1}{x}\)                           (b)\({f}'(1)=6+1=7\)

Question

[Maximum mark: 6]
Let \(f(x)=\frac{x^{3}+1}{sin\;x}\)

(a) Find \({f}'(x)\)
(b) Find the gradient of the curve \(y=f(x)\)

(i) at \(x=\frac{\pi }{4}\)                         (ii) at \(x = 1\) rad.

Answer/Explanation

Ans.

(a) \({f}'(x)=\frac{3x^{2}\;sin\;x-(x^{3}+1)cos\;x}{sin^{2}x}\)

(b) Directly by GDC  (i)  \({f}'(\frac{\pi }{4})\cong 0.518\)                          (ii)\({f}'(1)\cong 2.04\)

[Notice: the exact value for (i) is \({f}'(\frac{\pi }{4})=\frac{3\pi ^{2}}{16}\sqrt{2}-\frac{\pi ^{3}+64}{64}\sqrt{2}\)]

Question

[Maximum mark: 12]
Given the following values at \(x = 1\)

Calculate the derivatives of the following functions at \(x = 1\)

(i) \(y=3f(x)-2g(x)\)                                          (ii) \(y = f (x)g(x)\)

(iii) \(y=\frac{f(x)}{g(x)}\)                                                              (iv) \(y=2x^{3}+1+5f(x)\)

Answer/Explanation

Ans.

(a) \(\frac{dy}{dx}=2{f}'(x)-3{g}'(x)\), at \(x=1\) the value is -7

(b) \(\frac{dy}{dx}={f}'(x)g(x)+f(x){g}'(x)\), at \(x=1\) the value is 22

(c)\(\frac{dy}{dx}=\frac{{f}'(x)g(x)-f(x){g}'(x)}{g(x)^{2}}\),at \(x=1\) the value is \(\frac{2}{9}\)

(d)\(\frac{dy}{dx}=6x^{2}+5{f}'(x)\), at \(x=1\) the value is 26

 

 

Question

[Maximum mark: 4]
Let \(f(x)=6\sqrt[3]{x^{2}}\). Find \({f}'(x)\).

Answer/Explanation

Ans.

\(f(x)=6x^{\frac{2}{3}}, {f}'(x)=4x^{-\frac{1}{3}}\left ( =\frac{4}{x^{\frac{1}{3}}}=\frac{4}{\sqrt[3]{x}} \right )\)

 

 

Question

[Maximum mark: 6]
Let \(h(x)=\frac{6x}{cos\;x}\). Find \({h}'(0)\)

Answer/Explanation

Ans.

METHOD 1 (quotient)

\({h}'(x)=\frac{(cos\;x)(6)-(6x)(-sin\;x)}{(cos\;x)^{2}}\)

\({h}'(0)=\frac{(cos\;0)(6)-(6 x 0)(-sin\;0)}{(cos\;0)^{2}}=6\)

METHOD 2 (product)

\(h(x)=6x\) x \((cos\;x)^{-1}\)
\({h}'(x)=(6x)(-(cos\;x)^{-2})(-sin\;x))+(6)(cos\;x)^{-1}\)
\({h}'(0)=(6 x 0)(-(cos\;0)^{-2})(-sin\;0))+(6)(cos\;0)^{-1}=6\)

 

Question

[Maximum mark: 5]
Let \(g(x)=2x\) sin \(x\).

(a) Find g′(x).
(b) Find the exact value of the gradient of the graph of \(g\) at \(x\) = π.

Answer/Explanation

Ans.

(a)\({g}'(x)=2\;sin\;x+2\;x\;cos\;x\)
(b)\({g}'(\pi )=2\;sin\;\pi +2\pi\;cos\;\pi =-2\pi\)

Question

[Maximum mark: 4]
Consider the function\(f(x)=k\;sin\;x+3x\), where \(k\) is a constant.

(a) Find \({f}'(x)\).

(b) When \(x=\frac{\pi }{3}\), the gradient of the curve of \(f (x)\) is 8. Find the value of \(k\) .

Answer/Explanation

Ans.

(a)\({f}'(x)=k\;cos\;x+3\)
(b)\(k\;cos\;\left ( \frac{\pi }{3} \right )+3=8 \Rightarrow k\left ( \frac{1}{2} \right )+3=8\Rightarrow k=10\)

 

 

 

Question

[Maximum mark: 5]
Let \(f(x)=\frac{3x^{2}}{5x-1}\).

(a) Write down the equation of the vertical asymptote of \(y = f (x)\) .
(b) Find \({f}'(x)\) . Give your answer in the form \(\frac{ax^{2}+bx}{(5x-1)^{2}}\) where \(a\) and \(b\in \mathbb{Z}\) .

Answer/Explanation

Ans.

(a)\(x=\frac{1}{5}\)

(b)\({f}'(x)=\frac{(5x-1)(6x)-(3x^{2})(5)}{(5x-1)^{2}}=\frac{30x^{2}-6x-15x^{2}}{(5x-1)^{2}}=\frac{15x^{2}-6x}{(5x-1)^{2}}\)

 

 

Question

[Maximum mark: 8]
Let \(f(x)=x\;cos\;x\), for 0 ≤ x ≤ 6.

(a) Find \({f}'(x)\).
(b) On the grid below, sketch the graph of \(y={f}'(x)\).
(c) Write down the range of the function \(y={f}'(x)\) , for 0 ≤ x ≤ 6

Answer/Explanation

Ans.

(a)\({f}'(x)=cos\;x-x\;sin\;x\)

(b)

 

(c) y ∈ [-2.38 , 5.10]

 

Question

[Maximum mark: 7]
Let \(f(x)=e^{x}\;cos\;x\).

(a) Find \({f}'(x)\).
(b) Find the gradient of the normal to the curve of \(f\) at \(x=\pi \).
(c) Find the gradient of the tangent to the curve of \(f\) at \(x=\frac{\pi }{4}\).

Answer/Explanation

Ans.

(a)\({f}'(x)=e^{x}\) x \((-sin\;x)+cos\;x \) x \(e^{x}=e^{x}\;cos\;x-e^{x}\;sin\;x\)
\({f}'(\pi)=e^{\pi }\;cos\;\pi -e^{\pi }\;sin\;\pi =-e^{\pi }\)
gradient of normal = \(\frac{1}{e^{\pi }}\)

(b)\({f}’\left ( \frac{\pi }{4} \right )=0\)

 

Question

[Maximum mark: 7]
Let \(f(x)=xe^{x}\).
(a) Find the equation of the tangent line at \(x = 0\).
(b) Find the equation of the normal line at \(x = 0\).
(c) Solve the equation \({f}'(x)=0\).

Answer/Explanation

Ans.

(a) Point (0,0), \({f}'(x)=e^{x}+xe^{x}\), \(m_{T}=1\), Tangent line \(y=x\)

(b)\(m_{N}=-1\), Normal line \(y=-x\)

(c)\(e^{x}+xe^{x}=0\Leftrightarrow e^{x}(1+x)=0\Leftrightarrow x=-1\)

Question

[Maximum mark: 3 per function]
Find the derivative of each function below. [cover 3rd column; then compare with your answer]

Answer/Explanation

Ans.

Solutions are shown in the question.

Question

[Maximum mark: 3 per function]
Differentiate the following functions:

Answer/Explanation

Ans.

 

 

Question

[Maximum mark: 10]
Find \(\frac{ds}{dt}\) for each of the following functions

Answer/Explanation

Ans.

 

 

Question

[Maximum mark: 18]
Given that \(f(1)=2\) and \({f}'(1)=4\) , find the derivatives of the following functions at \(x =1\)

(i) \(y=f(x)^{2}\)                                     (ii) \(y=f(x)^{3}\)                                         (iii) \(y=ln\;f(x)\)

(iv) \(y=f(x^{2})\)                                     (v) \(y=f(x^{3})\)                                           (vi) \(y=\sqrt{f(x)}\)

Answer/Explanation

Ans.

(i)\(y=f(x)^{2}\Rightarrow \frac{dy}{dx}=2f(x){f}'(x)\). At \(x=1,\frac{dy}{dx}=2x2x4=16\)

(ii)\(y=f(x)^{3}\Rightarrow \frac{dy}{dx}=3f(x)^{2}{f}'(x)\). At \(x=1,\frac{dy}{dx}=3×2^{2}x4=48\)

(iii)\(y=ln\;f(x)\Rightarrow \frac{dy}{dx}=\frac{{f}'(x)}{f(x)}\). At \(x=1,\frac{dy}{dx}=\frac{4}{2}=2\)

(iv)\(y=f(x^{2})\Rightarrow \frac{dy}{dx}={f}'(x^{2})x2x\). At \(x=1, \frac{dy}{dx}=2x2x1=4\)

(v)\(y=f(x^{3})\Rightarrow \frac{dy}{dx}={f}'(x^{3})x3x^{2}\). At\(x=1, \frac{dy}{dx}=2x3x1^{2}=6\)

(vi)\(y=\sqrt{f(x)}\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{f(x)}}{f}'(x)\). At \(x=1, \frac{dy}{dx}=\frac{1}{2\sqrt{2}}x4=\frac{2}{\sqrt{2}}=\sqrt{2}\)

 

Question

[Maximum mark: 10]
(a) Show that the derivative of \(y\) = tan \(x\) is \(\frac{1}{cos^{2}x}\).

(b) Hence, differentiate the functions

(i) \(y = x\; tan\; x\)                                          (ii) \(y = tan\; 3x\)

(iii) \(y=tan^{2}x\)                                                   (iv) \(y=tan^{3}x\)

Answer/Explanation

Ans.

(a) \(y=tan\;x=\frac{sin\;x}{cos\;x}\)
\(\frac{dy}{dx}=\frac{sin\;x\;sin\;x-cos\;x(-cos\;x)}{cos^{2}x}=\frac{sin^{2}x+cos^{2}x}{cos^{2}x}=\frac{1}{cos^{2}x}\)

(b)   (i)\(\frac{dy}{dx}=\frac{x}{cos^{2}x}+tan\;x\)
(ii)\(\frac{dy}{dx}=\frac{3}{cos^{2}3x}\)
(iii)\(\frac{dy}{dx}=2\;tan\;x\frac{1}{cos^{2}x}\left ( =\frac{2\;sin\;x}{cos^{3}x} \right)\)
(iv)\(\frac{dy}{dx}=3\;tan^{2}x\frac{1}{cos^{2}x}\left ( =\frac{3\;sin^{2}x}{cos^{4}x} \right)\)

 

 

Question

[Maximum mark: 4]
Differentiate each of the following with respect to \(x\).

(a) \(y=x\;sin\;3x\)
(b) \(y=\frac{ln\;x}{x}\)

Answer/Explanation

Ans.

(a) \(\frac{dy}{dx}=sin\;3x+3x\;cos\;3x\)

(b)\(\frac{dy}{dx}=\frac{xx\frac{1}{x}-ln\;x}{x^{2}}=\frac{1-ln\;x}{x^{2}}\)

 

Question

[Maximum mark: 4]
The point P \(\left ( \frac{1}{2},0 \right )\) lies on the graph of the curve of \(y=sin(2x-1)\).
Find the gradient of the tangent to the curve at P.

Answer/Explanation

Ans.

\(y = sin\; (2x – 1)\)          \(\frac{dy}{dx}=2\;cos(2x-1)\)
At \(\left ( \frac{1}{2},0 \right )\), the gradient of the tangent = 2 cos 0 = 2

 

Question

[Maximum mark: 4]
Differentiate with respect to \(x\):                          (i) \(y=(x^{2}+1)^{2}\).                               (ii) \(y=ln(3x-1)\)

Answer/Explanation

Ans.

(i)\(\frac{d}{dx}(x^{2}+1)^{2}=2(x^{2}+1)x(2x)=4x(x^{2}+1)\)
(ii)\(\frac{d}{dx}(ln(3x-1))=\frac{1}{3x-1}x(3)=\frac{3}{3x-1}\)

 

Question

[Maximum mark: 4]
Differentiate with respect to \(x\)        (i) \(\sqrt{3-4x}\)                 (ii) \(e^{sin\;x}\)

Answer/Explanation

Ans.

(i)\(\frac{dy}{dx}=\frac{-4}{2\sqrt{3-4x}}=\frac{-2}{\sqrt{3-4x}}\)
OR \(y=\sqrt{3-4x}=(3-4x)^{\frac{1}{2}}\)                 \(\frac{dy}{dx}=\frac{1}{2}(3-4x)^{-\frac{1}{2}}(-4)\)

(ii) \(y=e^{sin\;x}\)                      \( \frac{dy}{dx}=(cos\;x)(e^{sin\;x})\)

 

Question

[Maximum mark: 5]
Let \(f(x)=e^{\frac{x}{3}}+5\;cos^{2}\;x\). Find \({f}'(x)\).

Answer/Explanation

Ans.

\({f}'(x)=\frac{1}{3}e^{\frac{x}{3}}-10\;cos\;x\;sin\;x\)

 

 

Question

[Maximum mark: 6]
Let \(f(x) = cos\;2x\) and \(g(x) = ln(3x – 5)\).
(a) Find \({f}'(x)\).
(b) Find \({g}'(x)\).
(c) Let \(h(x) = f(x) × g(x)\). Find \({h}'(x)\).

Answer/Explanation

Ans.

(a) \({f}'(x)\)=\(-sin\;2x\) x 2(=\(-2\;sin\;2x\))

(b) \({g}'(x)=3x\frac{1}{3x-5}\left ( =\frac{3}{3x-5} \right )\)

(c) product rule: \({h}'(x)=(cos\;2x)\left ( \frac{3}{3x-5} \right )+ln(3x-5)(-2\;sin\;2x)\)

 

Question

[Maximum mark: 6]
(a) Let \(f(x)=e^{5x}\). Write down \({f}'(x)\).
(b) Let \(g(x)=sin\;2x\). Write down \({g}'(x)\).
(c) Let \(h(x)=e^{5x}\;sin\;2x\). Find \({h}'(x)\).

Answer/Explanation

Ans.

(a) \({f}'(x)=5e^{5x}\)
(b) \({g}'(x)=2\;cos\;2x\)
(c) \({h}’={fg}’+{gf}’=e^{5x}(2\;cos\;2x)+sin\;2x(5e^{5x})\)

 

Question

[Maximum mark: 6]
Let \(f(x)=e^{-3x}\) and \(g(x)=sin\left ( x-\frac{\pi }{3} \right )\).
(a) Write down      (i) \({f}'(x)\);                (ii)\({g}'(x)\).
(b) Let \(h(x)=e^{-3x}\;sin\left ( x-\frac{\pi }{3} \right )\). Find the exact value of \({h}’\left ( \frac{\pi }{3} \right )\).

Answer/Explanation

Ans.

(a)      (i) \(-3e^{-3x}\)                        (ii) \(cos\left ( x-\frac{\pi }{3} \right )\)

(b)\({h}'(x)=-3e^{-3x}sin\left ( x-\frac{\pi }{3} \right )+e^{-3x}cos\left ( x-\frac{\pi }{3} \right )\)

\({h}'(\frac{\pi }{3})=-3e^{-3\frac{\pi }{3}}sin\left ( \frac{\pi }{3}-\frac{\pi }{3} \right )+e^{-3\frac{\pi }{3}}cos\left ( \frac{\pi }{3}-\frac{\pi }{3} \right )=e^{-\pi }\)

 

Question

[Maximum mark: 6]
Let \(f(x)=(2x+7)^{3}\) and \(g(x)=cos^{2}(4x)\). Find     (i) \({f}'(x)\);                    (ii)\({g}'(x)\)

Answer/Explanation

Ans.

(i) \({f}'(x)=3(2x+7)^{2}x2=6(2x+7)^{2}         (=24x^{2}+168x+294)\)
(ii)\({g}'(x)=2\;cos(4x)(-sin(4x))(4)=-8\;cos(4x)sin(4x)         (=-4sin(8x))\)

 

Question

[Maximum mark: 6]
The population \(p\) of bacteria at time \(t\) is given by \(p\) = 100e0.05t. Calculate
(a) the value of \(p\) when \(t\) = 0;
(b) the rate of increase of the population when \(t\) = 10.

Answer/Explanation

Ans.

(a)  p = 100e0 = 100

(b) Rate of increase is \(\frac{dp}{dt}=0.05x100e^{0.05t}=5e^{0.05t}\)
When \(t=10\)     \(\frac{dp}{dt}=5e^{0.05(10)}=5e^{0.5}\)                 \((=8.24=5\sqrt{e})\)

 

Question

[Maximum mark: 8]
The number of bacteria, \(n\), in a dish, after \(t\) minutes is given by \(n\) = 800e013t.
(a) Find the value of \(n\) when \(t\) = 0.
(b) Find the rate at which \(n\) is increasing when \(t\) = 15.
(c) After \(k\) minutes, the rate of increase in \(n\) is greater than 10 000 bacteria per
minute. Find the least value of \(k\), where \(k\in \mathbb{Z}\).

Answer/Explanation

Ans.

(a) \(n\) = 800e0 = 800
(b) derivative: n′(15) = 731
(c) METHOD 1
setting up inequality. n′(t) > 10 000
\(k\) = 35.1226…, least value of \(k\) is 36
METHOD 2
n′(35) = 9842, and n′(36) = 11208
least value of \(k\) is 36

 

Question

[Maximum mark: 8]
Consider the curve \(y = ln(3x – 1)\). Let P be the point on the curve where \(x = 2\).
(a) Write down the gradient of the curve at P.
(b) Find the equation of the tangent to the curve at P.
(c) The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.

Answer/Explanation

Ans.

(a) gradient is 0.6
(b) y – ln 5 = 0.6 (x – 2)       OR   directly by GDC: y = 0.6 x + 0.409
(c) at x = 2,     y = ln 5 (= 1.609…)
gradient of normal = – 5/3 = –1.6666…
normal: \(y-ln\;5=-\frac{5}{3}(x-2)\)         OR directly by GDC y = – 1.66666x + 4.94277
For y = 0:   x = 2.97 (accept 2.96)
coordinates of R are (2.97, 0)

 

Question

[Maximum mark: 7]
Let \(f(x)=3x-e^{x-2}-4\), for \(-1\leq x\leq 5\).
(a) Find the x -intercepts of the graph of \(f\) .
(b) On the grid below, sketch the graph of \(f\) .
(c) Write down the gradient of the graph of \(f\) at \(x = 2\).

Answer/Explanation

Ans.

(a) (1.54, 0) (4.13, 0)       (accept x = 1.54 x = 4.13)
(b)

(c) gradient is 2

 

Question

[Maximum mark: 6]
If \(y=ln(2x-1)\) find \(\frac{d^{2}y}{dx^{2}}\).

Answer/Explanation

Ans.

\(y-ln(2x-1)\Rightarrow \frac{dy}{dx}=\frac{2}{2x-1}\Rightarrow \frac{dy}{dx}=2(2x-1)^{-1}\)
\(\Rightarrow \frac{d^{2}y}{dx^{2}}=-2(2x-1)^{-2}(2)\)
\(\Rightarrow \frac{d^{2}y}{dx^{2}}=\frac{-4}{(2x-1)^{-2}}\)

 

 

Question

[Maximum mark: 6]
Consider the function \(y\) = tan \(x\) – 8sin \(x\) .
(a) Find \(\frac{dy}{dx}\).
(b) Find the value of cos \(x\) for which \(\frac{dy}{dx}=0\).
(c) Solve the equation \(\frac{dy}{dx}=0\). for \(-\pi \leq x\leq 2\pi\).

Answer/Explanation

Ans.

(a) \(\frac{dy}{dx}=\frac{1}{cos^{2}x}-8cosx\)

(b) \(\frac{dy}{dx}=\frac{1-8cos^{3}x}{cos^{2}x}=0\Rightarrow cosx=\frac{1}{2}\)

(c) \(x=-\frac{\pi }{3},x=\frac{\pi }{3},x=\frac{5\pi }{3}\)

 

Question

[Maximum mark: 6]
Let \(y=e^{3x}\;sin(\pi x)\). Find

(a) \(\frac{dy}{dx}\).
(b) the smallest positive value of x for which \(\frac{dy}{dx}=0\).

Answer/Explanation

Ans.

y = e3x sin(πx)

(a)\(\frac{dy}{dx}=3e^{3x}sin(\pi x)+\pi e^{3x}cos(\pi x)\)

(b) x = 0.7426… (0.743 to 3 s.f.)

Question

[Maximum mark: 6]
Let \(f(x)=cos^{3}(4x+1), 0\leq x\geq 1\). Find
(a) \({f}'(x)\)
(b) the exact values of the three roots of \({f}'(x)=0\).

Answer/Explanation

Ans.

(a) \(-12cos^{2}(4x+1)sin(4x+1)\)
(b)  \(x=\frac{\pi }{8}-\frac{1}{4},x=\frac{3\pi }{8}-\frac{1}{4},x=\frac{\pi-1 }{4}\)

 

Question

[Maximum mark: 6]
Let \(f\) be a cubic polynomial function. Given that \(f (0) = 2\), \({f}'(0)=-3\), \(f (1)\) = \({f}'(1)\)
and \({f}”(-1)=6\) , find \(f (x)\) .

Answer/Explanation

Ans.

\(f(x)=ax^{3}+bx^{2}+cx+d\)
\({f}'(x)=3ax^{2}+2bx+c\)
\({f}”(x)=6ax+2b\)
\(f(0)=2=d\)
\({f}'(1)=f(1)\rightarrow a+b+c+2=3a+2b+c\)
\(2=2a+b\)
\({f}'(0)=-3=c\)
\({f}”(-1)=6=-6a+2b\)
\(b=\frac{12}{5},a=-\frac{1}{5}\)
\(f(x)=-\frac{1}{5}x^{3}+\frac{12}{5}x^{2}-3x+2 (Accept \;a=-\frac{1}{5},b=\frac{12}{5},c=-3,d=2)\)

 

Question

[Maximum mark: 3 per function]
The following table shows the values of two functions \(f\) and \(g\) and their derivatives
when \(x = 1\) and \(x = 0\) .

Find the derivatives of the following functions when \(x = 1\).

Answer/Explanation

Ans.

 

Question

[Maximum mark: 10]
Let \(f(x)=1+3\;cos(2x)\) for \(0\leq x\leq \pi \), and \(x\) is in radians.
(a) (i) Find \({f}'(x)\).
(ii) Find the values for \(x\) for which \({f}'(x)=0\) ; Give your answers in terms of π .
The function \(g(x)\) is defined as \(g(x)=f(2x)-1, 0\leq x\leq \frac{\pi }{2}\).
(b) (i) The graph of \(f\) may be transformed to the graph of \(g\) by a stretch in the
x -direction with scale factor \(\frac{1}{2}\) followed by another transformation.
Describe fully this other transformation.
(ii) Find the solution to the equation \(g(x) = f (x)\) .

Answer/Explanation

Ans.

(a)  (i)\({f}'(x)=-6\;sin\;2x\)
(ii)  EITHER \({f}'(x)=-12sin\;x\;cos\;x=0\Rightarrow sin\;x=0\) or \(cos\;x=0\)
OR \(sin\;2x=0\), for \(0\leq 2x\leq 2\pi\)
THEN
\(x=0,\frac{\pi }{2},\pi \)

 

(b) (i)  translation in the y-direction of –1
(ii)  1.11                     (1.10 from TRACE is subject to AP)

 

 

Question

[Maximum mark: 20]
The diagram shows the graph of the function \(f\) given by \(f(x)=A\;sin\left ( \frac{\pi }{2}x \right )+B\), for
0 ≤ x ≤ 5 , where A and B are constants, and \(x\) is measured in radians.

The graph includes the points (1, 3) and (5, 3), which are maximum points of the graph.
(a) Show that A = 2 , and find the value of B .
(b) Show that \({f}'(x)=\pi cos\left ( \frac{\pi }{2}x \right )\).

The line \(y=k-\pi x\) is a tangent line to the graph for 0 ≤ x ≤ 5 .
(c) Find
(i) the point where this tangent meets the curve;
(ii) the value of \(k\) .
(d) Solve the equation \(f (x) = 2\) for 0 ≤ x ≤ 5 .

Answer/Explanation

Ans.

(a) EITHER \(A\;sin\left ( \frac{\pi }{2} \right )+B=3\) and \(A\;sin\left ( \frac{3\pi }{2} \right )+B=-1\)
\(\Leftrightarrow A+B=3,-A+B=-1\)
\(\Leftrightarrow A=2,B=1\)
OR
Amplitude = \(A =\frac{3-(-1)}{2}=\frac{4}{2}=2\)
Midpoint value = \(B =\frac{3+(-1)}{2}=\frac{2}{2}=1\)

 

(b) \(f(x)=2\;sin\left ( \frac{\pi }{2}x \right )+1\)
\({f}'(x)=\left ( \frac{\pi }{2} \right )2\;cos\left ( \frac{\pi }{2}x \right )+0=\pi \;cos\left ( \frac{\pi }{2}x \right )\)

 

(c)  (i) y = k – πx is a tangent  \(\Rightarrow -\pi =\pi \;cos\left ( \frac{\pi }{2}x\right )\)
\(\Rightarrow -1=cos\left ( \frac{\pi }{2}x\right )\)
\(\Rightarrow \frac{\pi }{2}x=\pi\) or \(3\pi\) or…
\(\Rightarrow x=2\) or 6…

Since 0 ≤ x ≤ 5, we take x = 2, so the point is (2, 1)

(ii) Tangent line is: y = –π(x – 2) + 1
y = (2π + 1) – πx
k = 2π + 1

 

 

(d) \(f(x)=2\Rightarrow 2\;sin\left ( \frac{\pi }{2}x \right )+1=2\)
\(\Rightarrow sin\left ( \frac{\pi }{2}x \right )=\frac{1}{2}\)
\(\Rightarrow \frac{\pi }{2}x=\frac{\pi }{6}or\frac{5\pi }{6}or\frac{13\pi }{6}\)
\(x=\frac{1}{3}or\frac{5}{3}\) or \(\frac{13}{3}\)

 

 

Question

[Maximum mark: 14]
The following diagram shows a waterwheel with a bucket. The wheel rotates at a
constant rate in an anticlockwise (counterclockwise) direction.

The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above
the water level. After \(t\) seconds, the height of the bucket above the water level is given
by \(h = a\) sin \(bt + 2\).
(a) Show that \(a = 4\).
The wheel turns at a rate of one rotation every 30 seconds.
(b) Show that \(b=\frac{\pi }{15}\).
In the first rotation, there are two values of \(t\) when the bucket is descending at a rate of
0.5 m s–1.
(c) Find these values of \(t\).
(d) Determine whether the bucket is underwater at the second value of \(t\).

Answer/Explanation

Ans.

(a) recognizing the amplitude is the radius: \(a=\frac{8}{2}\Rightarrow a=4\)

(b) period = 30: \(b=\frac{2\pi }{30}=\frac{\pi }{15}\)

(c) recognizing \({h}'(t)=-0.5\)
\(-0.5=\frac{4\pi }{15}cos\left ( \frac{\pi }{15}t \right )\Rightarrow t-10.6,t=19.4\)

(d) h(t) < 0 so underwater; h(t) > 0 so not underwater
h(19.4) = \(4\;sin\frac{19.4\pi }{15}+2=-1.19\)

OR

solving h(t) = 0, graph showing region below x-axis, roots 17.5, 27.5
Hence, the bucket is underwater, yes

 

 

Question

[Maximum mark: 18]
Let \(f (x) = 3sinx + 4\; cos\; x\), for –2π ≤ x ≤ 2π.
(a) Sketch the graph of \(f\).
(b) Write down
(i) the amplitude;               (ii) the period;             (iii) the x-intercept between \(-\frac{\pi }{2}\) and 0.
(c) Hence write \(f (x)\) in the form \(p\) sin \((qx + r)\).
(d) Write down one value of \(x\) such that \({f}'(x)=0\).
(e) Write down the two values of \(k\) for which the equation \(f (x) = k\) has exactly two
solutions.
(f) Let \(g(x) = ln(x + 1)\), for 0 ≤ x ≤ π. There is a value of \(x\), between 0 and 1, for which
the gradient of \(f\) is equal to the gradient of \(g\). Find this value of \(x\).

Answer/Explanation

Ans.

(a)

 

(b)  (i) 5          (ii)  2π (6.28)           (iii) –0.927
(c) \(f(x)\) = 5 sin (x + 0.927)  (accept \(p\) = 5, \(q\) = 1, \(r\) = 0.927)
(d) (man or min)
one 3 s.f. value which rounds to one of –5.6, –2.5, 0.64, 3.8

(e) \(k\) = –5,  \(k\) = 5
(f) \({g}'(x)=\frac{1}{x+1}\)
\({f}'(x)=3\;cos\;x-4\;sin\;x\)        \((5\;cos(x+0.927))\)
\({g}'(x)={f}'(x)\)
\(x=0.511\)

 

Question

[Maximum mark: 15]
(a) The function \(g\) is defined by \(g(x)=\frac{e^{x}}{\sqrt{x}}\), for 0 < x ≤ 3 .

(i) Sketch the graph of \(g\) .
(ii) Find \({g}'(x)\) .
(iii) Write down an expression representing the gradient of the normal to the
curve at any point.

(b) Let P be the point \((x, y)\) on the graph of \(g\) , and Q the point (1,0).
(i) Find the gradient of (PQ) in terms of \(x\) .
(ii) Given that the line (PQ) is a normal to the graph of \(g\) at the point P, find the
minimum distance from the point Q to the graph of \(g\) .

Answer/Explanation

Ans.

(a)

(ii) \(g(x)=\frac{e^{x}}{\sqrt{x}}\)
\({g}'(x)=\frac{e^{x}\sqrt{x}-\frac{e^{x}}{2\sqrt{x}}}{x}\)
\(=\frac{(2x-1)e^{x}}{2x\sqrt{x}}\)

(iii) gradient is \(-\frac{1}{{g}'(x)}\)
\(=\frac{2x\sqrt{x}}{(1-2x)e^{x}}\)

 

(b)  (i)  \(\frac{y-0}{x-1}=\frac{e^{x}}{\sqrt{x}(x-1)}\)

(ii) EITHER
\(\frac{e^{x}}{\sqrt{x}(x-1)}=\frac{2x\sqrt{x}}{(1-2x)e^{x}}\)
\(x=0.5454…\)
OR
\(D^{2}=(x-1)^{2}+y^{2}=(x-1)^{2}+\frac{e^{2x}}{x}\)
\(\frac{dD^{2}}{dx^{2}}=2(x-1)+\frac{2e^{2x}x-e^{2x}}{x^{2}}=0\)
\(x=0.5454…\)
THEN
distance = \(\sqrt{(1-0.5454)^{2}+\left ( \frac{e^{0.5454}}{\sqrt{0.5454}} \right )^{2}}\)
=2.38

 

 

 

 

 

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