Question
Consider f (x) = 4sinx + 2.5 and \(g(x) = 4sin\left ( x-\frac{3\pi }{2} \right )+ 2.5 + q,\) where x ∈ R and q > 0. The graph of g is obtained by two transformations of the graph of f .
(a) Describe these two transformations.
The y-intercept of the graph of g is at (0, r).
(b) Given that g(x) ≥ 7, find the smallest value of r.
Answer/Explanation
Ans:
(a) translation (shift) by \(\frac{3\pi }{2}\) to the right OR positive horizontal direction by \(\frac{3\pi }{2}\)
translation (shift) by q upwards OR positive vertical direction by q
Note: accept translation by \(\binom{\frac{3\pi }{2}}{q}\)
Do not accept ‘move’ for translation/shift.
(b)
minimum of \(\left ( x-\frac{3\pi }{2} \right )\) is -4 (may be seen in sketch)
-4 + 2.5 + q ≥7
q≥ 8.5 (accept q = 8.5)
substituting x = 0 and their q (= 8.5) to find r
(r =) \(4sin\left ( \frac{-3\pi }{2} \right )+2.5 + 8.5\)
4 + 2.5 + 8.5
smallest value of r is 15
Question
Consider the function f(x) = \(2^{x} – \frac{1}{2^{x}},\) x ∈ R.
(a) Show that f is an odd function.
The function g is given by g(x) = \(\frac{x-1}{x^{2}-2x-3},\) where x ∈ R, x ≠ -1 , x ≠ 3.
(b) Solve the inequality f (x) ≥ g(x).
Answer/Explanation
Ans:
(a) attempt to replace x with −x
f(-x) = \(2^{-x}-\frac{1}{2^{-x}}\)
EITHER
\(=\frac{1}{2^{x}}-2^{x} = -f(x)\)
OR
\(=-\left (2^{x} \frac{1}{2^{x}} \right )\left ( =-f(x) \right )\)
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) attempt to find at least one intersection point
x = -1.26686….., x = 0.177935…..,x = 3.06167
x = -1.27, x = 0.178, x = 3.06
-1.27 ≤ x < – 1,
0.178 ≤ x < 3,
x ≥ 3.06
Question
Consider f (x) = 4sinx + 2.5 and \(g(x) = 4sin\left ( x-\frac{3\pi }{2} \right )+ 2.5 + q,\) where x ∈ R and q > 0. The graph of g is obtained by two transformations of the graph of f .
(a) Describe these two transformations.
The y-intercept of the graph of g is at (0, r).
(b) Given that g(x) ≥ 7, find the smallest value of r.
Answer/Explanation
Ans:
(a) translation (shift) by \(\frac{3\pi }{2}\) to the right OR positive horizontal direction by \(\frac{3\pi }{2}\)
translation (shift) by q upwards OR positive vertical direction by q
Note: accept translation by \(\binom{\frac{3\pi }{2}}{q}\)
Do not accept ‘move’ for translation/shift.
(b)
minimum of \(\left ( x-\frac{3\pi }{2} \right )\) is -4 (may be seen in sketch)
-4 + 2.5 + q ≥7
q≥ 8.5 (accept q = 8.5)
substituting x = 0 and their q (= 8.5) to find r
(r =) \(4sin\left ( \frac{-3\pi }{2} \right )+2.5 + 8.5\)
4 + 2.5 + 8.5
smallest value of r is 15
Question
Consider the function \(f(x) = \frac{{\ln x}}{x}\) , \(0 < x < {{\text{e}}^2}\) .
(i) Solve the equation \(f'(x) = 0\) .
(ii) Hence show the graph of \(f\) has a local maximum.
(iii) Write down the range of the function \(f\) .[5]
Show that there is a point of inflexion on the graph and determine its coordinates.[5]
Sketch the graph of \(y = f(x)\) , indicating clearly the asymptote, x-intercept and the local maximum.[3]
Now consider the functions \(g(x) = \frac{{\ln \left| x \right|}}{x}\) and \(h(x) = \frac{{\ln \left| x \right|}}{{\left| x \right|}}\) , where \(0 < x < {{\text{e}}^2}\) .
(i) Sketch the graph of \(y = g(x)\) .
(ii) Write down the range of \(g\) .
(iii) Find the values of \(x\) such that \(h(x) > g(x)\) .[6]
Answer/Explanation
Markscheme
(i) \(f'(x) = \frac{{x\frac{1}{x} – \ln x}}{{{x^2}}}\) M1A1
\( = \frac{{1 – \ln x}}{{{x^2}}}\)
so \(f'(x) = 0\) when \(\ln x = 1\), i.e. \(x = {\text{e}}\) A1
(ii) \(f'(x) > 0\) when \(x < {\text{e}}\) and \(f'(x) < 0\) when \(x > {\text{e}}\) R1
hence local maximum AG
Note: Accept argument using correct second derivative.
(iii) \(y \leqslant \frac{1}{{\text{e}}}\) A1
[5 marks]
\(f”(x) = \frac{{{x^2}\frac{{ – 1}}{x} – \left( {1 – \ln x} \right)2x}}{{{x^4}}}\) M1
\( = \frac{{ – x – 2x + 2x\ln x}}{{{x^4}}}\)
\( = \frac{{ – 3 + 2\ln x}}{{{x^3}}}\) A1
Note: May be seen in part (a).
\(f”(x) = 0\) (M1)
\({ – 3 + 2\ln x = 0}\)
\(x = {{\text{e}}^{\frac{3}{2}}}\)
since \(f”(x) < 0\) when \(x < {{\text{e}}^{\frac{3}{2}}}\) and \(f”(x) > 0\) when \(x > {{\text{e}}^{\frac{3}{2}}}\) R1
then point of inflexion \(\left( {{{\text{e}}^{\frac{3}{2}}},\frac{3}{{2{{\text{e}}^{\frac{3}{2}}}}}} \right)\) A1
[5 marks]
A1A1A1
Note: Award A1 for the maximum and intercept, A1 for a vertical asymptote and A1 for shape (including turning concave up).
[3 marks]
(i)
A1A1
Note: Award A1 for each correct branch.
(ii) all real values A1
(iii)
(M1)(A1)
Note: Award (M1)(A1) for sketching the graph of h, ignoring any graph of g.
\( – {{\text{e}}^2} < x < – 1\) (accept \(x < – 1\) ) A1
[6 marks]
Question
The function f is given by \(f(x) = \frac{{{3^x} + 1}}{{{3^x} – {3^{ – x}}}}\), for x > 0.
Show that \(f(x) > 1\) for all x > 0.[3]
Solve the equation \(f(x) = 4\).[4]
Answer/Explanation
Markscheme
EITHER
\(f(x) – 1 = \frac{{1 + {3^{ – x}}}}{{{3^x} – {3^{ – x}}}}\) M1A1
> 0 as both numerator and denominator are positive R1
OR
\({3^x} + 1 > {3^x} > {3^x} – {3^{ – x}}\) M1A1
Note: Accept a convincing valid argument the numerator is greater than the denominator.
numerator and denominator are positive R1
hence \(f(x) > 1\) AG
[3 marks]
one line equation to solve, for example, \(4({3^x} – {3^{ – x}}) = {3^x} + 1\), or equivalent A1
\(\left( {3{y^2} – y – 4 = 0} \right)\)
attempt to solve a three-term equation M1
obtain \(y = \frac{4}{3}\) A1
\(x = {\log _3}\left( {\frac{4}{3}} \right)\) or equivalent A1
Note: Award A0 if an extra solution for x is given.
[4 marks]
Question
Solve \({\left( {{\text{ln}}\,x} \right)^2} – \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}\).
Answer/Explanation
Markscheme
\({\left( {{\text{ln}}\,x} \right)^2} – \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) – 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)\)
EITHER
\({\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}\) M1
\( = \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}\) A1
OR
\(\left( {{\text{ln}}\,x – 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)\) M1A1
THEN
\({\text{ln}}\,x = 2\,{\text{ln}}\,2\) or \( – {\text{ln}}\,2\) A1
\( \Rightarrow x = 4\) or \(x = \frac{1}{2}\) (M1)A1
Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.
solution is \(\frac{1}{2} < x < 4\) A1
[6 marks]
Question
Solve the inequality \(|\frac{x+9}{x-9}|≤2\).
Answer/Explanation
Ans
METHOD 1
The critical values occur when \(\frac{x+9}{x-9}=±2→x=3,27\)
Consider [-∞,3]: value of function at 0 is 1 which is ≤ 2.
Consider [3,27]: value of function at 12 is 7 which is not ≤ 2.
Consider [27,∞]: value of function at 36 is \(\frac{5}{3}\) which is ≤2.
The required solution set is therefore[-∞,3]∪[27,∞].
METHOD 2