Question
Consider the function \(f(x)=\frac{4x+2}{x-2}, x\neq 2\)
(a) Sketch the graph of \(y=f(x)\). On your sketch, indicate the values of any axis intercepts and label any asymptotes with their equations.
(b) Write down the range of f.
Consider the function \(g(x)=x^{2}+bx+c\). The graph of g has an axis of symmetry at \(x=2\).
The two roots of \(g(x)=0\) are \(-\frac{1}{2}\) and p, where \(p\in \mathbb{Q}\).
(c) Show that \(p=\frac{9}{2}\).
(d) Find the value of b and the value of c.
(e) Find the y-coordinate of the vertex of the graph of \(y=g(x)\).
(f)Find the product of the solutions of the equation \(f(x)=g(x)\).
▶️Answer/Explanation
Ans:
(a) Step 1] Domain of function
The denominator
, so:
Domain:
, i.e.,
Step 2] X and Y intercept
Y-intercept (when x = 0 ):
Y-intercept:
X-intercept (when f(x)=0):
X-intercept:
The denominator is zero at
As
(left) f(x)→−∞ and As
→ (right) f(x)→+∞ , hence x=2 is the vertical asymptote to the curve
(b) Range \(y\neq 4\) or y
(c) We know that,
Axis of symmetry at
= 2 (given) , for \(g(x)=x^{2}+bx+c\)
Axis of symmetry at
= 2, put a =1,we get b = -4,
Applying the product of roots, we get
\( \frac{-1}{2}\times p = \frac{c}{a}= \frac{c}{1}\) , put p = \( \frac{9}{2}\)
\(\frac{-1}{2}\times \frac{9}{2}\ = \frac{c}{a}= \frac{c}{1}\)
Therefore, c = \( \frac{-1}{2}\times \frac{9}{2} = \frac{-9}{4}\)therefore \(g(x)=x^{2}+(-4)x+\frac{-9}{4} =(x+\frac{1}{2})(x-\frac{9}{2}) \)
(e) Substituting the axis of symmetry \(x=2\) into the \(g(x)=x^{2}+(-4)x+\frac{-9}{4}\) we get the y coordinate of vertex as
\(y = 2^{2}+(-4)2+\frac({-9}{4})\)
\(y=-\frac{25}{4}\)
Solving \(f(x)=g(x)\) can be done by following ways
Question
The graph of \( y=f(\left| x\right|)\) for \(-6\leq x\leq 6\) is shown in the following diagram.
(a) On the following axes, sketch the graph of \(y=\left| f(\left| x\right|)\right|\) for \(-6\leq x\leq 6\).
It is given that f is an odd function.
(b) On the following axes, sketch the graph of \(y=f(x)\) for \(-6\leq x\leq 6\).
It is also given that \(\int_{0}^{4}f(\left| x\right|)dx=1.6\).
(c) Write down the value of
(i) \(\int_{-4}^{0}f(x)dx\);
(ii) \(\int_{-4}^{4}(f(\left| x\right|)+f(x))dx\).
▶️Answer/Explanation
Detailed Solution
(a) To find the graph for \(y=\left| f(\left| x\right|)\right|\) , the part below the -axis is taken as mirror image above the X-axis (reflection of all negative sections in x-axis), therefore the graph is as follows
(a) 
(b) To find the graph for \(y=f(x)\) for \(-6\leq x\leq 6\) , since the mod is removed from x and it is given that y=f(x) is an odd function, therefore the graph will be same for x≥0 whereas for x<0 it will be symmetric about the origin that is just same behavior as the graph of y = \(x^{3}\) as shown below
(b) 
(c) (i) From the graph in part (b), we can conclude that the area \(\int_{-4}^{0}f(x)dx\) = \(-\int_{0}^{4}f(\left| x\right|)dx\) = \(-1.6\)
(ii) Since f(x) is an odd function therefore, \(\int_{-4}^{4} f(x))dx\) = 0, and \(\int_{-4}^{4}(f(\left| x\right|) = 2 \int_{0}^{4}(f(\left| x\right|) \)=\(2\times (-1.6)\) = \(3.2\)
Question
The graph below shows \(y = f(x)\) , where \(f(x) = x + \ln x\) .
(a) On the graph below, sketch the curve \(y = {f^{ – 1}}(x)\) .
(b) Find the coordinates of the point of intersection of the graph of \(y = f(x)\) and the graph of \(y = {f^{ – 1}}(x)\) .
▶️Answer/Explanation
Markscheme
(a)
A1A1
Note: Award A1 for correct asymptote with correct behaviour and A1 for shape.
[2 marks]
(b) intersect on \(y = x\) (M1)
\(x + \ln x = x \Rightarrow \ln x = 0\) (A1)
intersect at (1, 1) A1 A1
[4 marks]
Total [6 marks]
Question
The function f is defined by \(f(x) = \frac{{2x – 1}}{{x + 2}}\), with domain \(D = \{ x: – 1 \leqslant x \leqslant 8\} \).
a. Express \(f(x)\) in the form \(A + \frac{B}{{x + 2}}\), where \(A\) and \(B \in \mathbb{Z}\).[2]
b. Hence show that \(f'(x) > 0\) on D.[2]
c. State the range of f.[2]
(i) Find an expression for \({f^{ – 1}}(x)\).
(ii) Sketch the graph of \(y = f(x)\), showing the points of intersection with both axes.
(iii) On the same diagram, sketch the graph of \(y = f'(x)\).[8]
(i) On a different diagram, sketch the graph of \(y = f(|x|)\) where \(x \in D\).
(ii) Find all solutions of the equation \(f(|x|) = – \frac{1}{4}\).[7]
▶️Answer/Explanation
Markscheme
a.by division or otherwise
\(f(x) = 2 – \frac{5}{{x + 2}}\) A1A1
[2 marks]
\(f'(x) = \frac{5}{{{{(x + 2)}^2}}}\) A1
> 0 as \({(x + 2)^2} > 0\) (on D) R1AG
Note: Do not penalise candidates who use the original form of the function to compute its derivative.
[2 marks]
\(S = \left[ { – 3,\frac{3}{2}} \right]\) A2
Note: Award A1A0 for the correct endpoints and an open interval.
[2 marks]
(i) EITHER
rearrange \(y = f(x)\) to make x the subject M1
obtain one-line equation, e.g. \(2x – 1 = xy + 2y\) A1
\(x = \frac{{2y + 1}}{{2 – y}}\) A1
OR
interchange x and y M1
obtain one-line equation, e.g. \(2y – 1 = xy + 2x\) A1
\(y = \frac{{2x + 1}}{{2 – x}}\) A1
THEN
\({f^{ – 1}}(x) = \frac{{2x + 1}}{{2 – x}}\) A1
Note: Accept \(\frac{5}{{2 – x}} – 2\)
(ii), (iii)
A1A1A1A1
[8 marks]
Note: Award A1 for correct shape of \(y = f(x)\).
Award A1 for x intercept \(\frac{1}{2}\) seen. Award A1 for y intercept \( – \frac{1}{2}\) seen.
Award A1 for the graph of \(y = {f^{ – 1}}(x)\) being the reflection of \(y = f(x)\) in the line \(y = x\). Candidates are not required to indicate the full domain, but \(y = f(x)\) should not be shown approaching \(x = – 2\). Candidates, in answering (iii), can FT on their sketch in (ii).
(i)
A1A1A1
Note: A1 for correct sketch \(x > 0\), A1 for symmetry, A1 for correct domain (from –1 to +8).
Note: Candidates can FT on their sketch in (d)(ii).
(ii) attempt to solve \(f(x) = – \frac{1}{4}\) (M1)
obtain \(x = \frac{2}{9}\) A1
use of symmetry or valid algebraic approach (M1)
obtain \(x = – \frac{2}{9}\) A1
[7 marks]