IBDP Maths AI: Topic SL 5.3: The derivative of functions: 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.
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

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

Consider the function \(f(x) = 2{x^3} – 5{x^2} + 3x + 1\).

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

a.

Write down the value of \(f'(2)\).[1]

b.

Find the equation of the tangent to the curve of \(y = f(x)\) at the point \((2{\text{, }}3)\).[2]

c.
Answer/Explanation

Markscheme

\(f'(x) = 6{x^2} – 10x + 3\)     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for each correct term and no extra terms.
Award (A1)(A1)(A0) if each term correct and extra term seen.
Award(A1)(A0)(A0) if two terms correct and extra term seen.
Award (A0) otherwise.[3 marks]

a.

\(f'(2) = 7\)     (A1)(ft)     (C1)[1 mark]

b.

\(y = 7x – 11\) or equivalent     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) on their (b) for \(7x\) (must have \(x\)), (A1)(ft) for \( – 11\). Accept \(y – 3 = 7(x – 2)\) .[2 marks]

c.

Question

Consider the function \(f(x) = \frac{1}{2}{x^3} – 2{x^2} + 3\).

Find \(f'(x)\).[2]

a.

Find \(f”(x)\).[2]

b.

Find the equation of the tangent to the curve of \(f\) at the point \((1{\text{, }}1.5)\).[2]

c.
Answer/Explanation

Markscheme

\(\frac{{3{x^2}}}{2} – 4x\)     (A1)(A1)     (C2)

Note: Award (A1) for each correct term and no extra terms; award (A1)(A0) for both terms correct and extra terms; (A0) otherwise.[2 marks]

a.

\(3x – 4\)     (A1)(ft)(A1)(ft)     (C2)

Note: accept \(3{x^1} – {4^0}\)[2 marks]

b.

\(y = – 2.5x + 4\) or equivalent     (A1)(ft)(A1)     (C2)

Note: Award (A1)(ft) on their (a) for \( – 2.5x\) (must have \(x\)), (A1) for \(4\) or equivalent correct answer only.
Accept \(y – 1.5 = – 2.5(x – 1)\)[2 marks]

c.

Question

The straight line, L, has equation \(2y – 27x – 9 = 0\).

Find the gradient of L.[2]

a.

Sarah wishes to draw the tangent to \(f (x) = x^4\) parallel to L.

Write down \(f ′(x)\).[1]

b.

Find the x coordinate of the point at which the tangent must be drawn.[2]

c, i.

Write down the value of \(f (x)\) at this point.[1]

c, ii.
Answer/Explanation

Markscheme

y = 13.5x + 4.5     (M1)

Note: Award (M1) for 13.5x seen.

gradient = 13.5     (A1)     (C2)[2 marks]

a.

4x3     (A1)     (C1)[1 mark]

b.

4x3 = 13.5     (M1)

Note: Award (M1) for equating their answers to (a) and (b).

x = 1.5     (A1)(ft)[2 marks]

c, i.

\(\frac{{81}}{{16}}\)   (5.0625, 5.06)     (A1)(ft)     (C3)

Note: Award (A1)(ft) for substitution of their (c)(i) into x4 with working seen.[1 mark]

c, ii.

Question

Consider \(f:x \mapsto {x^2} – 4\).

Find \(f ′(x)\).[1]

a.

Let L be the line with equation y = 3x + 2.

Write down the gradient of a line parallel to L.[1]

b.

Let L be the line with equation y = 3x + 2.

Let P be a point on the curve of f. At P, the tangent to the curve is parallel to L. Find the coordinates of P.[4]

c.
Answer/Explanation

Markscheme

\(2x\)     (A1)     (C1)[1 mark]

a.

3     (A1)     (C1)[1 mark]

b.

\(2x = 3\)     (M1)

Note: (M1) for equating their (a) to their (b).

\(x =1.5\)     (A1)(ft)

\(y = (1.5)^2 – 4\)     (M1)

Note: (M1) for substituting their x in f (x).

(1.5, −1.75)   (accept x = 1.5, y = −1.75)     (A1)(ft)     (C4)

Note: Missing coordinate brackets receive (A0) if this is the first time it occurs.[4 marks]

c.

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 figure shows the graphs of the functions \(f(x) = \frac{1}{4}{x^2} – 2\) and \(g(x) = x\) .

Differentiate \(f(x)\) with respect to \(x\) .[1]

a.

Differentiate \(g(x)\) with respect to \(x\) .[1]

b.

Calculate the value of \(x\) for which the gradients of the two graphs are the same.[2]

c.

Draw the tangent to the parabola at the point with the value of \(x\) found in part (c).[2]

d.
Answer/Explanation

Markscheme

\(\frac{1}{2}x{\text{ }}\left( {\frac{2}{4}x} \right)\)     (A1)     (C1)

Note: Accept an equivalent, unsimplified expression (i.e. \(2 \times \frac{1}{4}x\)).[1 mark]

a.

\(1\)     (A1)     (C1)[1 mark]

b.

\(\frac{1}{2}x = 1\)     (M1)

\(x = 2\)     (A1)(ft)     (C2)

Notes: Award (M1)(A0) for coordinate pair \((2{\text{, }} – 1)\) seen with or without working. Follow through from their answers to parts (a) and (b).[2 marks]

c.

tangent drawn to the parabola at the \(x\)-coordinate found in part (c)     (A1)(ft)

candidate’s attempted tangent drawn parallel to the graph of \(g(x)\)     (A1)(ft)     (C2)[2 marks]

d.

Question

\[f(x) = \frac{1}{3}{x^3} + 2{x^2} – 12x + 3\]

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

a.

Find the interval of \(x\) for which \(f(x)\) is decreasing.[3]

b.
Answer/Explanation

Markscheme

\(f'(x) = {x^2} + 4x – 12\)     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for each term. Award at most (A1)(A1)(A0) if other terms are seen.[3 marks]

a.

\( – 6 \leqslant x \leqslant 2\)     OR     \( – 6 < x < 2\)     (A1)(ft)(A1)(ft)(A1)     (C3)

Notes: Award (A1)(ft) for \( – 6\), (A1)(ft) for \(2\), (A1) for consistent use of strict (\(<\)) or weak (\(\leqslant\)) inequalities. Final (A1) for correct interval notation (accept alternative forms). This can only be awarded when the left hand side of the inequality is less than the right hand side of the inequality. Follow through from their solutions to their \(f'(x) = 0\) only if working seen.[3 marks]

b.

Question

The equation of a curve is given as \(y = 2x^{2} – 5x + 4\).

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]

a.

The equation of the line L is \(6x + 2y = -1\).

Find the x-coordinate of the point on the curve \(y = 2x^2 – 5x + 4\) where the tangent is parallel to L.[4]

b.
Answer/Explanation

Markscheme

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4x – 5\)     (A1)(A1)     (C2)

Notes: Award (A1) for each correct term. Award (A1)(A0) if any other terms are given.[2 marks]

a.

\(y = – 3x – \frac{1}{2}\)     (M1)

Note: Award (M1) for rearrangement of equation

gradient of line is –3     (A1)

\(4x – 5 = -3\)     (M1)

Notes: Award (M1) for equating their gradient to their derivative from part (a). If \(4x – 5 = -3\) is seen with no working award (M1)(A1)(M1).

\(x = \frac{1}{2}\)     (A1)(ft)     (C4)

Note: Follow through from their part (a). If answer is given as (0.5, 2) with no working award the final (A1) only.[4 marks]

b.

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

A curve is described by the function \(f (x) = 3x – \frac{2}{{x^2}}\), \(x \ne 0\).

Find \(f ‘ (x) \).[3]

a.

The gradient of the curve at point A is 35.

Find the x-coordinate of point A.[3]

b.
Answer/Explanation

Markscheme

\(f'(x) = 3 + \frac{4}{{{x^3}}}\)     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for 3, (A1) for + 4 and (A1) for \(\frac{1}{{{x^3}}}\)  or \(x^{-3}\). Award at most (A1)(A1)(A0) if additional terms are seen.

a.

\(3 + \frac{4}{{{x^3}}} = 35\)     (M1)

Note: Award (M1) for equating their derivative to 35 only if the derivative is not a constant.

\({x^3} = \frac{1}{8}\)     (A1)(ft)

\(\frac{1}{2}(0.5)\)     (A1)(ft)     (C3)

b.

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

Let \(f(x) = {x^4}\).

Write down \(f'(x)\).[1]

a.

Point \({\text{P}}(2,6)\) lies on the graph of \(f\).

Find the gradient of the tangent to the graph of \(y = f(x)\) at \({\text{P}}\).[2]

b.

Point \({\text{P}}(2,16)\) lies on the graph of \(f\).

Find the equation of the normal to the graph at \({\text{P}}\). Give your answer in the form \(ax + by + d = 0\), where \(a\), \(b\) and \(d\) are integers.[3]

c.
Answer/Explanation

Markscheme

\(\left( {f'(x) = } \right)\)   \(4{x^3}\)     (A1)     (C1)[1 mark]

a.

\(4 \times {2^3}\)     (M1)

Note: Award (M1) for substituting 2 into their derivative.

\( = 32\)     (A1)(ft)     (C2)

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

b.

\(y – 16 =  – \frac{1}{{32}}(x – 2)\)   or   \(y =  – \frac{1}{{32}}x + \frac{{257}}{{16}}\)     (M1)(M1)

Note: Award (M1) for their gradient of the normal seen, (M1) for point substituted into equation of a straight line in only \(x\) and \(y\) (with any constant ‘\(c\)’ eliminated).

\(x + 32y – 514 = 0\) or any integer multiple     (A1)(ft)     (C3)

Note: Follow through from their part (b).[3 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

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

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 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

Consider the function \(f(x) = a{x^2} + c\).

Find \(f'(x)\)[1]

a.

Point \({\text{A}}( – 2,\,5)\)  lies on the graph of \(y = f(x)\) . The gradient of the tangent to this graph at \({\text{A}}\) is \( – 6\) .

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

b.

Find the value of \(c\) .[2]

c.
Answer/Explanation

Markscheme

\(2ax\)      (A1)   (C1)

Note: Award (A1) for \(2ax\).  Award (A0) if other terms are seen.

a.

\(2a( – 2) =  – 6\)       (M1)(M1)

Note: Award (M1) for correct substitution of \(x =  – 2\)  in their gradient function, (M1) for equating their gradient function to \( – 6\) . Follow through from part (a).

\((a = )1.5\,\,\,\left( {\frac{3}{2}} \right)\)       (A1)(ft) (C3)

b.

\({\text{their }}1.5 \times {( – 2)^2} + c = 5\)         (M1)

Note: Award (M1) for correct substitution of their \(a\) and point \({\text{A}}\). Follow through from part (b).

\((c = ) – 1\)         (A1)(ft) (C2)

c.

Question

A company sells fruit juices in cylindrical cans, each of which has a volume of \(340\,{\text{c}}{{\text{m}}^3}\). The surface area of a can is \(A\,{\text{c}}{{\text{m}}^2}\) and is given by the formula

\(A = 2\pi {r^2} + \frac{{680}}{r}\) ,

where \(r\) is the radius of the can, in \({\text{cm}}\).

To reduce the cost of a can, its surface area must be minimized.

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\)[3]

a.

Calculate the value of \(r\) that minimizes the surface area of a can.[3]

b.
Answer/Explanation

Markscheme

\(\left( {\frac{{{\text{d}}A}}{{{\text{d}}r}}} \right) = 4\pi r – \frac{{680}}{{{r^2}}}\)        (A1)(A1)(A1)      (C3)

Note: Award (A1) for \(4\pi r\) (accept \(12.6r\)), (A1) for \( – 680\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ – 2}}\)

Award at most (A1)(A1)(A0) if additional terms are seen.

a.

\(4\pi r – \frac{{680}}{{{r^2}}} = 0\)          (M1)

Note: Award (M1) for equating their \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\) to zero.

\(4\pi {r^3} – 680 = 0\)          (M1)

Note: Award (M1) for initial correct rearrangement of the equation. This may be assumed if \({r^3} = \frac{{680}}{{4\pi }}\) or \(r = \sqrt[3]{{\frac{{680}}{{4\pi }}}}\) seen.

OR

sketch of \(A\) with some indication of minimum point         (M1)(M1)

Note: Award (M1) for sketch of \(A\), (M1) for indication of minimum point.

OR

sketch of \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\) with some indication of zero      (M1)(M1)

Note: Award (M1) for sketch of \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\), (M1) for indication of zero.

\((r = )\,\,3.78\,({\text{cm}})\,\,\,\,\,(3.78239…)\)                (A1)(ft) (C3)  

Note: Follow through from part (a).

b.

Question

Consider the function \(f(x) = {x^3} – 3{x^2} + 2x + 2\) . Part of the graph of \(f\) is shown below.

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

a.

There are two points at which the gradient of the graph of \(f\) is \(11\). Find the \(x\)-coordinates of these points.[3]

b.
Answer/Explanation

Markscheme

\((f'(x) = )\,\,3{x^2} – 6x + 2\)        (A1)(A1)(A1)     (C3)

Note: Award (A1) for \(3{x^2}\), (A1) for \( – 6x\) and (A1) for \( + 2\).
Award at most (A1)(A1)(A0) if there are extra terms present.

a.

\(11 = 3{x^2} – 6x + 2\)        (M1)

Note: Award (M1) for equating their answer from part (a) to \(11\), this may be implied from \(0 = 3{x^2} – 6x – 9\) .

\((x = )\,\, – 1\,\,,\,\,\,\,(x = )\,\,3\)        (A1)(ft)(A1)(ft)     (C3)

Note: Follow through from part (a).
If final answer is given as coordinates, award at most (M1)(A0)(A1)(ft) for \(( – 1,\,\, – 4)\) and \((3,\,\,8)\) .

b.

Question

The equation of a curve is \(y = \frac{1}{2}{x^4} – \frac{3}{2}{x^2} + 7\).

The gradient of the tangent to the curve at a point P is \( – 10\).

Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[2]

a.

Find the coordinates of P.[4]

b.
Answer/Explanation

Markscheme

\(2{x^3} – 3x\)     (A1)(A1)     (C2)

Note:     Award (A1) for \(2{x^3}\), award (A1) for \( – 3x\).

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

a.

\(2{x^3} – 3x =  – 10\)    (M1)

Note:     Award (M1) for equating their answer to part (a) to \( – 10\).

\(x =  – 2\)    (A1)(ft)

Note:     Follow through from part (a). Award (M0)(A0) for \( – 2\) seen without working.

\(y = \frac{1}{2}{( – 2)^4} – \frac{3}{2}{( – 2)^2} + 7\)    (M1)

Note:     Award (M1) substituting their \( – 2\) into the original function.

\(y = 9\)    (A1)(ft)     (C4)

Note:     Accept \(( – 2,{\text{ }}9)\).[4 marks]

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

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 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.

Question

A sketch of the function \(f(x) = 5{x^3} – 3{x^5} + 1\) is shown for \( – 1.5 \leqslant x \leqslant 1.5\) and \( – 6 \leqslant y \leqslant 6\) .

Write down \(f'(x)\) .[2]

a.

Find the equation of the tangent to the graph of \(y = f(x)\) at \((1{\text{, }}3)\) .[2]

b.

Write down the coordinates of the second point where this tangent intersects the graph of \(y = f(x)\) .[2]

c.
Answer/Explanation

Markscheme

\(f'(x) = 15{x^2} – 15{x^4}\)     (A1)(A1)     (C2)

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

a.

\(f'(1) = 0\)     (M1)

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

\(y = 3\)     (A1)(ft)     (C2)

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

b.

\(( – 1.38{\text{, }}3)\) \(( – 1.38481 \ldots {\text{, }}3)\)     (A1)(ft)(A1)(ft)     (C2)

Note: Follow through from their answer to parts (a) and (b).

Note: Accept \(x = – 1.38\), \(y = 3\) (\(x = – 1.38481 \ldots\) , \(y = 3\)) .

c.

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.

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