CAIE P2 (Pure Mathematics 2) 2014 November

1 Solve the equation \(| 3 x - 1 | = | 2 x + 5 |\).

2

1. Find \(\int _ { 0 } ^ { a } \left( \mathrm { e } ^ { - x } + 6 \mathrm { e } ^ { - 3 x } \right) \mathrm { d } x\), where \(a\) is a positive constant.
2. Deduce the value of \(\int _ { 0 } ^ { \infty } \left( \mathrm { e } ^ { - x } + 6 \mathrm { e } ^ { - 3 x } \right) \mathrm { d } x\).

3 A curve has equation $$3 \ln x + 6 x y + y ^ { 2 } = 16$$ Find the equation of the normal to the curve at the point \(( 1,2 )\). Give your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.

4

1. Find the value of \(x\) satisfying the equation \(2 \ln ( x - 4 ) - \ln x = \ln 2\).
2. Use logarithms to find the smallest integer satisfying the inequality $$1.4 ^ { y } > 10 ^ { 10 }$$

5
\includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-2_554_689_1354_726} The diagram shows part of the curve $$y = 2 \cos x - \cos 2 x$$ and its maximum point \(M\). The shaded region is bounded by the curve, the axes and the line through \(M\) parallel to the \(y\)-axis.

1. Find the exact value of the \(x\)-coordinate of \(M\).
2. Find the exact value of the area of the shaded region.

6
\includegraphics[max width=\textwidth, alt={}, center]{c703565b-8aa8-424b-9684-6592d4effdf8-3_597_931_260_607} The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = x ^ { 4 } - 3 x ^ { 3 } + 3 x ^ { 2 } - 25 x + 48 .$$ The diagram shows the curve \(y = \mathrm { p } ( x )\) which crosses the \(x\)-axis at ( \(\alpha , 0\) ) and ( 3,0 ).

1. Divide \(\mathrm { p } ( x )\) by a suitable linear factor and hence show that \(\alpha\) is a root of the equation \(x = \sqrt [ 3 ] { } ( 16 - 3 x )\).
2. Use the iterative formula \(x _ { n + 1 } = \sqrt [ 3 ] { } \left( 16 - 3 x _ { n } \right)\) to find \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7

1. Express \(5 \cos \theta - 12 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\), giving the value of \(\alpha\) correct to 2 decimal places.
2. Hence solve the equation \(5 \cos \theta - 12 \sin \theta = 8\) for \(0 ^ { \circ } < \theta < 360 ^ { \circ }\).
3. Find the greatest possible value of $$7 + 5 \cos \frac { 1 } { 2 } \phi - 12 \sin \frac { 1 } { 2 } \phi$$ as \(\phi\) varies, and determine the smallest positive value of \(\phi\) for which this greatest value occurs.
   [0pt] [4]