Home / IBDP Maths AI: Topic: SL 2.5: Direct/inverse variation: IB style Questions HL Paper 1

IBDP Maths AI: Topic: SL 2.5: Direct/inverse variation: IB style Questions HL Paper 1

Question

The following curves are sketches of the graphs of the functions given below, but in a different order. Using your graphic display calculator, match the equations to the curves, writing your answers in the table below.

(the diagrams are not to scale)

▶️Answer/Explanation

Markscheme

(i) B     (A1)

(ii) D     (A1) 

(iii) A     (A1) 

(iv) E     (A1)

(v) C     (A1) 

(vi) F     (A1)     (C6)[6 marks]

Question

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

a.Find the gradient of L.[2]

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

Write down \(f ′(x)\).[1]

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

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

▶️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

a.Consider the curve \(y = 1 + \frac{1}{{2x}},\,\,x \ne 0.\)

For this curve, write down

i)     the value of the \(x\)-intercept;

ii)    the equation of the vertical asymptote.[3]

b.Sketch the curve for \( – 2 \leqslant x \leqslant 4\) on the axes below.

[3]

 
▶️Answer/Explanation

Markscheme

i)     \(\left( {x = } \right) – 0.5\,\,\left( { – \frac{1}{2}} \right)\)       (A1)

ii)    \(x = 0\)        (A1)(A1)    (C3)

Note: Award (A1) for “\(x = \)” and (A1) for “\(0\)” seen as part of an equation.

a.

(A1)(ft)(A1)(ft)(A1)    (C3)

Note: Award (A1)(ft) for correct \(x\)-intercept, (A1)(ft) for asymptotic behaviour at \(y\)-axis, (A1) for approximately correct shape (cannot intersect the horizontal asymptote of \(y = 1\)). Follow through from part (a).

b.

Question

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

a.Write down the derivative of \(f\).[3]

b.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]

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.
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