IB DP Maths Topic 2.2 Function graphing skills SL Paper 1

Question

The velocity v ms−1 of a particle at time t seconds, is given by \(v = 2t + \cos 2t\) , for \(0 \le t \le 2\) .

Write down the velocity of the particle when \(t = 0\) .

[1]
a.

When \(t = k\) , the acceleration is zero.

(i)     Show that \(k = \frac{\pi }{4}\) .

(ii)    Find the exact velocity when \(t = \frac{\pi }{4}\) .

[8]
b(i) and (ii).

When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)  .

Sketch a graph of v against t .

[4]
c.

Let d be the distance travelled by the particle for \(0 \le t \le 1\) .

(i)     Write down an expression for d .

(ii)    Represent d on your sketch.

[3]
d(i) and (ii).
Answer/Explanation

Markscheme

\(v = 1\)     A1     N1

[1 mark]

a.

(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\)     A1

\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = – 2\sin 2t\)     A1A1

Note: Award A1 for coefficient 2 and A1 for \( – \sin 2t\) .

evidence of considering acceleration = 0     (M1)

e.g. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(2 – 2\sin 2t = 0\)

correct manipulation     A1

e.g. \(\sin 2k = 1\) , \(\sin 2t = 1\)

\(2k = \frac{\pi }{2}\) (accept \(2t = \frac{\pi }{2}\) )     A1

\(k = \frac{\pi }{4}\)     AG     N0

(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v     (M1)

e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)

\(v = \frac{\pi }{2}\)     A1     N2

[8 marks]

b(i) and (ii).


     A1A1A2      N4

Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\) , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\) and concave up to the right of \(\frac{\pi }{4}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\) , do not award the second A1 for the zero gradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.

[4 marks]

c.

(i) correct expression     A2

e.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1\) , \(1 + \frac{{\sin 2}}{2}\) , \(\int_0^1 {v{\rm{d}}t} \)

(ii)


     A1

Note: The line at \(t = 1\) needs to be clearly after \(t = \frac{\pi }{4}\) .

[3 marks]

d(i) and (ii).

Question

Consider \(f(x) = \ln ({x^4} + 1)\) .

The second derivative is given by \(f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}\) .

The equation \(f”(x) = 0\) has only three solutions, when \(x = 0\) , \( \pm \sqrt[4]{3}\) \(( \pm 1.316 \ldots )\) .

Find the value of \(f(0)\) .

[2]
a.

Find the set of values of \(x\) for which \(f\) is increasing.

[5]
b.

(i)     Find \(f”(1)\) .

(ii)     Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .

[5]
c.

There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .

Sketch the graph of \(f\) , for \(x \ge 0\) .

[3]
d.
Answer/Explanation

Markscheme

substitute \(0\) into \(f\)     (M1)

eg   \(\ln (0 + 1)\) , \(\ln 1\)

\(f(0) = 0\)     A1 N2

[2 marks]

a.

\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere)     A1A1

Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .

 

recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere)     R1

eg   \(f'(x) > 0\) , diagram of signs

attempt to solve \(f'(x) > 0\)     (M1)

eg   \(4{x^3} = 0\) , \({x^3} > 0\)

\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) )     A1     N1

[5 marks]

b.

(i)     substituting \(x = 1\) into \(f”\)     (A1)

eg   \(\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)

\(f”(1) = 2\)     A1     N2

 

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in \(f”(x)\) , no change in concavity,

\(f’\) increasing both sides of zero

attempt to find \(f”(x)\) for \(x < 0\)     (M1)

eg   \(f”( – 1)\) , \(\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}\) , diagram of signs

correct working leading to positive value     A1

eg   \(f”( – 1) = 2\) , discussing signs of numerator and denominator

there is no point of inflexion at \(x = 0\)     AG     N0

 

[5 marks]

c.

     A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

    Only if this A1 is awarded, then award the following:

    A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.

    Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

Question

Let \(f(x) = 3\sin \left( {\frac{\pi }{2}x} \right)\), for \(0 \leqslant x \leqslant 4\).

(i)     Write down the amplitude of \(f\).

(ii)     Find the period of \(f\).

[3]
a.

On the following grid sketch the graph of \(f\).

M16/5/MATME/SP1/ENG/TZ1/03.b

[4]
b.
Answer/Explanation

Markscheme

(i)     3     A1     N1

(ii)     valid attempt to find the period     (M1)

eg\(\,\,\,\,\,\)\(\frac{{2\pi }}{b},{\text{ }}\frac{{2\pi }}{{\frac{\pi }{2}}}\)

period \( = 4\)     A1     N2

[3 marks]

a.

M16/5/MATME/SP1/ENG/TZ1/03.b/M     A1A1A1A1     N4

[4 marks]

b.

Question

The following diagram shows the graph of a function \(f\), with domain \( – 2 \leqslant x \leqslant 4\).

N17/5/MATME/SP1/ENG/TZ0/03

The points \(( – 2,{\text{ }}0)\) and \((4,{\text{ }}7)\) lie on the graph of \(f\).

Write down the range of \(f\).

[1]
a.

Write down \(f(2)\);

[1]
b.i.

Write down \({f^{ – 1}}(2)\).

[1]
b.ii.

On the grid, sketch the graph of \({f^{ – 1}}\).

[3]
c.
Answer/Explanation

Markscheme

correct range (do not accept \(0 \leqslant x \leqslant 7\))     A1     N1

eg\(\,\,\,\,\,\)\([0,{\text{ }}7],{\text{ }}0 \leqslant y \leqslant 7\)

[1 mark]

a.

\(f(2) = 3\)     A1     N1

[1 mark]

b.i.

\({f^{ – 1}}(2) = 0\)     A1     N1

[1 mark]

b.ii.

N17/5/MATME/SP1/ENG/TZ0/03.c/M     A1A1A1     N3

Notes:     Award A1 for both end points within circles,

A1 for images of \((2,{\text{ }}3)\) and \((0,{\text{ }}2)\) within circles,

A1 for approximately correct reflection in \(y = x\), concave up then concave down shape (do not accept line segments).

[3 marks]

c.

Question

Consider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph of f.

Write down the value of f (0).

[1]
a.i.

Write down the value of f −1 (1).

[1]
a.ii.

Write down the range of f −1.

[1]
b.

On the grid above, sketch the graph of f −1.

[4]
c.
Answer/Explanation

Markscheme

\(f\left( 0 \right) =  – \frac{1}{2}\)     A1 N1

[1 mark]

a.i.

f −1 (1) = 2     A1 N1

[1 mark]

a.ii.

−2 ≤ y ≤ 2, y∈ [−2, 2]  (accept −2 ≤ x ≤ 2)     A1 N1

[1 mark]

b.

A1A1A1A1  N4

Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.

(y = x does not need to be explicitly seen)

Only if this mark is awarded, award marks as follows:

A1 for both correct invariant points in circles,

A1 for the three other points in circles,

A1 for correct domain.

[4 marks]

c.

Question

The following diagram shows the graph of a function \(f\), for −4 ≤ x ≤ 2.

On the same axes, sketch the graph of \(f\left( { – x} \right)\).

[2]
a.

Another function, \(g\), can be written in the form \(g\left( x \right) = a \times f\left( {x + b} \right)\). The following diagram shows the graph of \(g\).

Write down the value of a and of b.

[4]
b.
Answer/Explanation

Markscheme

A2 N2
[2 marks]

a.

recognizing horizontal shift/translation of 1 unit      (M1)

eg  = 1, moved 1 right

recognizing vertical stretch/dilation with scale factor 2      (M1)

eg   a = 2,  ×(−2)

a = −2,  b = −1     A1A1 N2N2

[4 marks]

b.

Question

Consider a function \(f\). The line L1 with equation \(y = 3x + 1\) is a tangent to the graph of \(f\) when \(x = 2\)

Let \(g\left( x \right) = f\left( {{x^2} + 1} \right)\) and P be the point on the graph of \(g\) where \(x = 1\).

Write down \(f’\left( 2 \right)\).

[2]
a.i.

Find \(f\left( 2 \right)\).

[2]
a.ii.

Show that the graph of g has a gradient of 6 at P.

[5]
b.

Let L2 be the tangent to the graph of g at P. L1 intersects L2 at the point Q.

Find the y-coordinate of Q.

[7]
c.
Answer/Explanation

Markscheme

recognize that \(f’\left( x \right)\) is the gradient of the tangent at \(x\)     (M1)

eg   \(f’\left( x \right) = m\)

\(f’\left( 2 \right) = 3\)  (accept m = 3)     A1 N2

[2 marks]

a.i.

recognize that \(f\left( 2 \right) = y\left( 2 \right)\)     (M1)

eg  \(f\left( 2 \right) = 3 \times 2 + 1\)

\(f\left( 2 \right) = 7\)     A1 N2

[2 marks]

a.ii.

recognize that the gradient of the graph of g is \(g’\left( x \right)\)      (M1)

choosing chain rule to find \(g’\left( x \right)\)      (M1)

eg  \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u’ = 2x\)

\(g’\left( x \right) = f’\left( {{x^2} + 1} \right) \times 2x\)     A2

\(g’\left( 1 \right) = 3 \times 2\)     A1

\(g’\left( 1 \right) = 6\)     AG N0

[5 marks]

b.

 at Q, L1L2 (seen anywhere)      (M1)

recognize that the gradient of L2 is g’(1)  (seen anywhere)     (M1)
eg  m = 6

finding g (1)  (seen anywhere)      (A1)
eg  \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)

attempt to substitute gradient and/or coordinates into equation of a straight line      M1
eg  \(y – g\left( 1 \right) = 6\left( {x – 1} \right),\,\,y – 1 = g’\left( 1 \right)\left( {x – 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)

correct equation for L2 

eg  \(y – 7 = 6\left( {x – 1} \right),\,\,y = 6x + 1\)     A1

correct working to find Q       (A1)
eg   same y-intercept, \(3x = 0\)

\(y = 1\)     A1 N2

[7 marks]

c.
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