# IB DP Math AA : Topic: SL 2.2 : Concept of a function and graph: IB Style Questions-HL Paper 2

### Question

The functions f and g are defined for $$x\in \mathbb{R}\; by \; f(x)=6x^2-12x+1\;and\;g(x)=-x+c,\;where\; c\in \mathbb{R}$$
(a) Find the range of f . [2]
(b) Given that (g o f ) (x) ≤ 0 for all $$x\in \mathbb{R}$$, determine the set of possible values for c . [4]

Ans:

(a) By completing the square for $f\left(x\right)=6x^2-12x+1$, we have $f\left(x\right)=6\left(x-1\right)^2-5$. Thus, $\text{R}_f=\left[-5,\infty\right)$.
(b) $$\begin{eqnarray} \left(g\circ f\right)\left(x\right)=c-f\left(x\right) \leq 0. \end{eqnarray}$$ By considering the graph of $y=-f\left(x\right)$, in order for $c-f\left(x\right)\leq 0$ we must have $c\leq -5$.

### Question

Consider the function $$g$$ , where $$g(x) = \frac{{3x}}{{5 + {x^2}}}$$ .

(a)     Given that the domain of $$g$$ is $$x \geqslant a$$ , find the least value of $$a$$ such that $$g$$ has an inverse function.

(b)     On the same set of axes, sketch

(i)     the graph of $$g$$ for this value of $$a$$ ;

(ii)     the corresponding inverse, $${g^{ – 1}}$$ .

(c)     Find an expression for $${g^{ – 1}}(x)$$ .

### Markscheme

(a)     $$a = 2.24$$     $$\sqrt 5$$     A1

(b)     (i)

A2

Note: Award A1 for end point

A1 for its asymptote.

(ii)     sketch of $${g^{ – 1}}$$ (see above)     A2

Note: Award A1 for end point

A1 for its asymptote.

(c)     $$y = \frac{{3x}}{{5 + {x^2}}} \Rightarrow y{x^2} – 3x + 5y = 0$$     M1

$$\Rightarrow x = \frac{{3 \pm \sqrt {9 – 20{y^2}} }}{{2y}}$$     A1

$${g^{ – 1}}(x) = \frac{{3 \pm \sqrt {9 – 20{x^2}} }}{{2x}}$$     A1

[8 marks]

### Question

The following graph represents a function $$y = f(x)$$, where $$– 3 \le x \le 5$$.

The function has a maximum at $$(3,{\text{ }}1)$$ and a minimum at $$( – 1,{\text{ }} – 1)$$.

a.The functions $$u$$ and $$v$$ are defined as $$u(x) = x – 3,{\text{ }}v(x) = 2x$$ where $$x \in \mathbb{R}$$.

(i)     State the range of the function $$u \circ f$$.

(ii)     State the range of the function $$u \circ v \circ f$$.

(iii)     Find the largest possible domain of the function $$f \circ v \circ u$$.[7]

b.

(i)     Explain why $$f$$ does not have an inverse.

(ii)     The domain of $$f$$ is restricted to define a function $$g$$ so that it has an inverse $${g^{ – 1}}$$.

State the largest possible domain of $$g$$.

(iii)     Sketch a graph of $$y = {g^{ – 1}}(x)$$, showing clearly the $$y$$-intercept and stating the coordinates of the endpoints.[6]

c.

Consider the function defined by $$h(x) = \frac{{2x – 5}}{{x + d}}$$, $$x \ne – d$$ and $$d \in \mathbb{R}$$.

(i)     Find an expression for the inverse function $${h^{ – 1}}(x)$$.

(ii)     Find the value of $$d$$ such that $$h$$ is a self-inverse function.

For this value of $$d$$, there is a function $$k$$ such that $$h \circ k(x) = \frac{{2x}}{{x + 1}},{\text{ }}x \ne – 1$$.

(iii)     Find $$k(x)$$.[8]

### Markscheme

Note:     For Q12(a) (i) – (iii) and (b) (ii), award A1 for correct endpoints and, if correct, award A1 for a closed interval.

Further, award A1A0 for one correct endpoint and a closed interval.

(i)     $$– 4 \le y \le – 2$$     A1A1

(ii)     $$– 5 \le y \le – 1$$     A1A1

(iii)     $$– 3 \le 2x – 6 \le 5$$     (M1)

Note:     Award M1 for $$f(2x – 6)$$.

$$3 \le 2x \le 11$$

$$\frac{3}{2} \le x \le \frac{{11}}{2}$$     A1A1

[7 marks]

a.

(i)     any valid argument eg $$f$$ is not one to one, $$f$$ is many to one, fails horizontal line test, not injective     R1

(ii)     largest domain for the function $$g(x)$$ to have an inverse is $$[ – 1,{\text{ }}3]$$     A1A1

(iii)

$$y$$-intercept indicated (coordinates not required)     A1

correct shape     A1

coordinates of end points $$(1,{\text{ }}3)$$ and $$( – 1,{\text{ }} – 1)$$     A1

Note:     Do not award any of the above marks for a graph that is not one to one.

[6 marks]

b.

(i)     $$y = \frac{{2x – 5}}{{x + d}}$$

$$(x + d)y = 2x – 5$$     M1

Note:     Award M1 for attempting to rearrange $$x$$ and $$y$$ in a linear expression.

$$x(y – 2) = – dy – 5$$     (A1)

$$x = \frac{{ – dy – 5}}{{y – 2}}$$     (A1)

Note:     $$x$$ and $$y$$ can be interchanged at any stage

$${h^{ – 1}}(x) = \frac{{ – dx – 5}}{{x – 2}}$$     A1

Note:     Award A1 only if $${h^{ – 1}}(x)$$ is seen.

(ii)     self Inverse $$\Rightarrow h(x) = {h^{ – 1}}(x)$$

$$\frac{{2x – 5}}{{x + d}} \equiv \frac{{ – dx – 5}}{{x – 2}}$$     (M1)

$$d = – 2$$     A1

(iii)     METHOD 1

$$\frac{{2k(x) – 5}}{{k(x) – 2}} = \frac{{2x}}{{x + 1}}$$     (M1)

$$k(x) = \frac{{x + 5}}{2}$$     A1

METHOD 2

$${h^{ – 1}}\left( {\frac{{2x}}{{x + 1}}} \right) = \frac{{2\left( {\frac{{2x}}{{x + 1}}} \right) – 5}}{{\frac{{2x}}{{x + 1}} – 2}}$$     (M1)

$$k(x) = \frac{{x + 5}}{2}$$     A1

[8 marks]

Total [21 marks]

c.
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