# IBDP Maths analysis and approaches Topic: SL 2.2 Concept of a function, domain, range and graph HL Paper 1

### Question:

A function f is defined by f ( x ) = $$x\sqrt{1-x^{2}} where -1\leq x\leq 1.$$
The graph of y = f (x) is shown below.

(a) Show that f is an odd function.
The range of f is a ≤ y ≤ b , where a, b ∈ R.

Ans: attempts to replace x with –x
$$f(-x) = -x\sqrt{1-(-x)^{2}}$$
$$= -x\sqrt{1-(-x)^{2}} \left ( =-f(x) \right )$$

Note: Award M1A1 for an attempt to calculate both f (-x ) and – f (-x) independently, showing that they are equal.
Note: Award M1A0 for a graphical approach including evidence that either the graph is invariant after rotation by 180° about the origin or the graph is invariant after a reflection in the y-axis and then in the x-axis (or vice versa).

so f is an odd function

(b) Find the value of a and the value of b.

Ans: attempts both product rule and chain rule differentiation to find f¢(x)

Note: Award M1 for an attempt to evaluate f(x) at least at one of their f¢(x) = 0 roots.

$$a = -\frac{1}{2} and b = \frac{1}{2}$$
Note: Award A1 for $$-\frac{1}{2}\leq y\leq \frac{1}{2}.$$

### Question:

The graph of y = f (x) for -4 ≤ x ≤ 6 is shown in the following diagram.

(a)        Write down the value of

(i)       f (2) ;

Ans: f(2) = 6

(ii)      ( f o f )(2) .                                                                                                                                                             [2]

Ans: ( f o f ) 2=− 2 [2 marks]

(b)        Let g(x) = $$\frac{1}{2} f (x) +1$$ for -4 ≤ x ≤ 6 . On the axes above, sketch the graph of g .                   [3]

Ans:

### Question

Consider the function f , where $$f(x) = \arcsin (\ln x)$$.

(a)     Find the domain of f .

Ans: $$– 1 \leqslant \ln x \leqslant 1$$     (M1)

$$\Rightarrow \frac{1}{{\text{e}}} \leqslant x \leqslant {\text{e}}$$     A1A1

(b)     Find $${f^{ – 1}}(x)$$.

Ans:

$$y = \arcsin (\ln x) \Rightarrow \ln x = \sin y$$     (M1)

$$\ln y = \sin x \Rightarrow y = {{\text{e}}^{\sin x}}$$     (M1)

$$\Rightarrow {f^{ – 1}}(x) = {{\text{e}}^{\sin x}}$$     A1

[6 marks]

### Question

a. Consider the functions given below.

$f(x) = 2x + 3$$g(x) = \frac{1}{x},x \ne 0$

(i)     Find $$\left( {g \circ f} \right)\left( x \right)$$ and write down the domain of the function.

Ans: $$\left( {g \circ f} \right)\left( x \right) = \frac{1}{{2x + 3}}$$, $$x \ne – \frac{3}{2}$$ (or equivalent)     A1

(ii)     Find $$\left( {f \circ g} \right)\left( x \right)$$ and write down the domain of the function.[2]

Ans: $$\left( {f \circ g} \right)\left( x \right) = \frac{2}{x} + 3$$, $$x \ne 0$$ (or equivalent)     A1

[2 marks]

b. Find the coordinates of the point where the graph of $$y = f(x)$$ and the graph of $$y = \left( {{g^{ – 1}} \circ f \circ g} \right)(x)$$ intersect.[4]

Ans: EITHER

$$f(x) = \left( {{g^{ – 1}} \circ f \circ g} \right)(x) \Rightarrow \left( {f \circ g} \right)\left( x \right)$$     (M1)

$$\frac{1}{{2x + 3}} = \frac{2}{x} + 3$$     A1

OR

$$\left( {{g^{ – 1}} \circ f \circ g} \right)(x) = \frac{1}{{\frac{2}{x} + 3}}$$     A1

$$2x + 3 = \frac{1}{{\frac{2}{x} + 3}}$$     M1

THEN

$$6{x^2} + 12x + 6 = 0$$ (or equivalent)     A1

$$x = – 1$$, $$y = 1$$ (coordinates are (−1, 1) )     A1

[4 marks]

### Question

Consider the function $${f_n}(x) = (\cos 2x)(\cos 4x) \ldots (\cos {2^n}x),{\text{ }}n \in {\mathbb{Z}^ + }$$.

a. Determine whether $${f_n}$$ is an odd or even function, justifying your answer.[2]

Ans: even function     A1

since $$\cos kx = \cos ( – kx)$$ and $${f_n}(x)$$ is a product of even functions     R1

OR

even function     A1

since $$(\cos 2x)(\cos 4x) \ldots = \left( {\cos ( – 2x)} \right)\left( {\cos ( – 4x)} \right) \ldots$$     R1

Note:     Do not award A0R1.

[2 marks]

b. By using mathematical induction, prove that $${f_n}(x) = \frac{{\sin {2^{n + 1}}x}}{{{2^n}\sin 2x}},{\text{ }}x \ne \frac{{m\pi }}{2}$$ where $$m \in \mathbb{Z}$$.[8]

Ans: consider the case $$n = 1$$

$$\frac{{\sin 4x}}{{2\sin 2x}} = \frac{{2\sin 2x\cos 2x}}{{2\sin 2x}} = \cos 2x$$     M1

hence true for $$n = 1$$     R1

assume true for $$n = k$$, ie, $$(\cos 2x)(\cos 4x) \ldots (\cos {2^k}x) = \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}$$     M1

Note:     Do not award M1 for “let $$n = k$$” or “assume $$n = k$$” or equivalent.

consider $$n = k + 1$$:

$${f_{k + 1}}(x) = {f_k}(x)(\cos {2^{k + 1}}x)$$     (M1)

$$= \frac{{\sin {2^{k + 1}}x}}{{{2^k}\sin 2x}}\cos {2^{k + 1}}x$$     A1

$$= \frac{{2\sin {2^{k + 1}}x\cos {2^{k + 1}}x}}{{{2^{k + 1}}\sin 2x}}$$     A1

$$= \frac{{\sin {2^{k + 2}}x}}{{{2^{k + 1}}\sin 2x}}$$     A1

so $$n = 1$$ true and $$n = k$$ true $$\Rightarrow n = k + 1$$ true. Hence true for all $$n \in {\mathbb{Z}^ + }$$     R1

Note:     To obtain the final R1, all the previous M marks must have been awarded.

[8 marks]

c. Hence or otherwise, find an expression for the derivative of $${f_n}(x)$$ with respect to $$x$$.[3]

Ans: attempt to use $$f’ = \frac{{vu’ – uv’}}{{{v^2}}}$$ (or correct product rule)     M1

$${f’_n}(x) = \frac{{({2^n}\sin 2x)({2^{n + 1}}\cos {2^{n + 1}}x) – (\sin {2^{n + 1}}x)({2^{n + 1}}\cos 2x)}}{{{{({2^n}\sin 2x)}^2}}}$$     A1A1

Note:     Award A1 for correct numerator and A1 for correct denominator.

[3 marks]

d. Show that, for $$n > 1$$, the equation of the tangent to the curve $$y = {f_n}(x)$$ at $$x = \frac{\pi }{4}$$ is $$4x – 2y – \pi = 0$$.[8]

Ans: $${f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{\left( {{2^n}\sin \frac{\pi }{2}} \right)\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right) – \left( {\sin {2^{n + 1}}\frac{\pi }{4}} \right)\left( {{2^{n + 1}}\cos \frac{\pi }{2}} \right)}}{{{{\left( {{2^n}\sin \frac{\pi }{2}} \right)}^2}}}$$     (M1)(A1)

$${f’_n}\left( {\frac{\pi }{4}} \right) = \frac{{({2^n})\left( {{2^{n + 1}}\cos {2^{n + 1}}\frac{\pi }{4}} \right)}}{{{{({2^n})}^2}}}$$     (A1)

$$= 2\cos {2^{n + 1}}\frac{\pi }{4}{\text{ }}( = 2\cos {2^{n – 1}}\pi )$$     A1

$${f’_n}\left( {\frac{\pi }{4}} \right) = 2$$     A1

$${f_n}\left( {\frac{\pi }{4}} \right) = 0$$     A1

Note:     This A mark is independent from the previous marks.

$$y = 2\left( {x – \frac{\pi }{4}} \right)$$     M1A1

$$4x – 2y – \pi = 0$$     AG

[8 marks]

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