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Question
The following diagram shows part of the graph of \(f(x) = – 2{x^3} + 5.1{x^2} + 3.6x – 0.4\).
Find the coordinates of the local minimum point.
The graph of \(f\) is translated to the graph of \(g\) by the vector \(\left( {\begin{array}{*{20}{c}} 0 \\ k \end{array}} \right)\). Find all values of \(k\) so that \(g(x) = 0\) has exactly one solution.
Answer/Explanation
Markscheme
\(( – 0.3,{\text{ }} – 0.967)\)
\(x = – 0.3\) (exact), \(y = – 0.967\) (exact) A1A1 N2
[2 marks]
\(y\)-coordinate of local maximum is \(y = 11.2\) (A1)
negating the \(y\)-coordinate of one of the max/min (M1)
eg\(\;\;\;y = 0.967,{\text{ }}y = – 11.2\)
recognizing that the solution set has two intervals R1
eg\(\;\;\;\)two answers,
\(k < – 11.2,{\text{ }}k > 0.967\) A1A1 N3N2
[5 marks]
Notes: If working shown, do not award the final mark if strict inequalities are not used.
If no working shown, award N2 for \(k \le – 11.2\) or N1 for \(k \ge 0.967\)
Total [7 marks]
Question
Let \(f(x) = {{\text{e}}^{x + 1}} + 2\), for \( – 4 \le x \le 1\).
On the following grid, sketch the graph of \(f\).
The graph of \(f\) is translated by the vector \(\left( {\begin{array}{*{20}{c}} 3 \\ { – 1} \end{array}} \right)\) to obtain the graph of a function \(g\).
Find an expression for \(g(x)\).
Answer/Explanation
Markscheme
1A1A1 N3
Note: Curve must be approximately correct exponential shape (increasing and concave up). Only if the shape is approximately correct, award the following:
A1 for right end point in circle,
A1 for \(y\)-intercept in circle,
A1 for asymptotic to \(y = 2\), (must be above \(y = 2\)).
[3 marks]
valid attempt to find \(g\) (M1)
eg\(\;\;\;f(x – 3) – 1,{\text{ }}g(x) = {{\text{e}}^{x + 1 – 3}} + 2 – 1,{\text{ }}{{\text{e}}^{x + 1 – 3}},{\text{ }}2 – 1\), sketch
\(g(x) = {{\text{e}}^{x – 2}} + 1\) A2 N3
[3 marks]
Total [6 marks]