CAIE P2 (Pure Mathematics 2) 2015 June

1. Use logarithms to solve the equation \(2^x = 20^5\), giving the answer correct to 3 significant figures. [2]
2. Hence determine the number of integers \(n\) satisfying $$20^{-5} < 2^n < 20^5.$$ [2]

1. Given that \((x + 2)\) is a factor of $$4x^3 + ax^2 - (a + 1)x - 18,$$ find the value of the constant \(a\). [3]
2. When \(a\) has this value, factorise \(4x^3 + ax^2 - (a + 1)x - 18\) completely. [3]

It is given that \(\theta\) is an acute angle measured in degrees such that $$2\sec^2\theta + 3\tan\theta = 22.$$

1. Find the value of \(\tan\theta\). [3]
2. Use an appropriate formula to find the exact value of \(\tan(\theta + 135°)\). [3]

\includegraphics{figure_4} The diagram shows the curve \(y = e^x + 4e^{-2x}\) and its minimum point \(M\).

1. Show that the \(x\)-coordinate of \(M\) is \(\ln 2\). [3]
2. The region shaded in the diagram is enclosed by the curve and the lines \(x = 0\), \(x = \ln 2\) and \(y = 0\). Use integration to show that the area of the shaded region is \(\frac{5}{2}\). [4]

1. By sketching a suitable pair of graphs, show that the equation $$|3x| = 16 - x^4$$ has two real roots. [3]
2. Use the iterative formula \(x_{n+1} = \sqrt[4]{16 - 3x_n}\) to find one of the real roots correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
3. Hence find the coordinates of each of the points of intersection of the graphs \(y = |3x|\) and \(y = 16 - x^4\), giving your answers correct to 3 decimal places. [2]

\includegraphics{figure_6} The diagram shows part of the curve with equation $$y = 4\sin^2 x + 8\sin x + 3$$ and its point of intersection \(P\) with the \(x\)-axis.

1. Find the exact \(x\)-coordinate of \(P\). [3]
2. Show that the equation of the curve can be written $$y = 5 + 8\sin x - 2\cos 2x,$$ and use integration to find the exact area of the shaded region enclosed by the curve and the axes. [6]

1. Find the gradient of the curve $$3\ln x + 4\ln y + 6xy = 6$$ at the point \((1, 1)\). [4]
2. The parametric equations of a curve are $$x = \frac{10}{t} - t, \quad y = \sqrt{2t - 1}.$$ Find the gradient of the curve at the point \((-3, 3)\). [6]