Question
Consider \(f(x) = x\ln (4 – {x^2})\) , for \( – 2 < x < 2\) . The graph of f is given below.
Let P and Q be points on the curve of f where the tangent to the graph of f is parallel to the x-axis.
(i) Find the x-coordinate of P and of Q.
(ii) Consider \(f(x) = k\) . Write down all values of k for which there are exactly two solutions.
Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .
Show that \(g'(x) = \frac{{ – 2{x^4}}}{{4 – {x^2}}} + 3{x^2}\ln (4 – {x^2})\) .
Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .
Sketch the graph of \(g’\) .
Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .
Consider \(g'(x) = w\) . Write down all values of w for which there are exactly two solutions.
Answer/Explanation
Markscheme
(i) \( – 1.15{\text{, }}1.15\) A1A1 N2
(ii) recognizing that it occurs at P and Q (M1)
e.g. \(x = – 1.15\) , \(x = 1.15\)
\(k = – 1.13\) , \(k = 1.13\) A1A1 N3
[5 marks]
evidence of choosing the product rule (M1)
e.g. \(uv’ + vu’\)
derivative of \({x^3}\) is \(3{x^2}\) (A1)
derivative of \(\ln (4 – {x^2})\) is \(\frac{{ – 2x}}{{4 – {x^2}}}\) (A1)
correct substitution A1
e.g. \({x^3} \times \frac{{ – 2x}}{{4 – {x^2}}} + \ln (4 – {x^2}) \times 3{x^2}\)
\(g'(x) = \frac{{ – 2{x^4}}}{{4 – {x^2}}} + 3{x^2}\ln (4 – {x^2})\) AG N0
[4 marks]
A1A1 N2
[2 marks]
\(w = 2.69\) , \(w < 0\) A1A2 N2
[3 marks]