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

### Question

The function f is defined by $$f(x) = \frac{2}{x} + 3x^2 – 3, x \neq 0$$.
(a) Find f'(x)
(b) Find the equation of the normal to the curve y = f(x) at (1, 2) in the form ax + by + d = 0, where $$a, b, d\epsilon \mathbb{Z}$$.

Ans:

(a) $$f'(x) = -2x^{-2}+6x$$ OR $$f'(x) = -\frac{2}{x^2}+6x$$
(b) finding gradient at x = 1
$$\left.\begin{matrix} \frac{dy}{dx} \end{matrix}\right|_{x=1}=4$$
$$m_{\perp} = -\frac{1}{4}$$
$$2=-\frac{1}{4}(1)+c$$  OR  $$y-2 = -\frac{1}{4}(x-1)$$
x + 4y – 9 = 0

## 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)

## 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.

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.

## 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 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]

a.

Sketch the curve for $$– 2 \leqslant x \leqslant 4$$ on the axes below.

[3]

b.

## 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.