Questions FP2 (1279 questions)

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OCR FP2 2010 June Q3
6 marks Challenging +1.2
3 Use the substitution \(t = \tan \frac { 1 } { 2 } x\) to show that $$\int _ { 0 } ^ { \frac { 1 } { 3 } \pi } \frac { 1 } { 1 - \sin x } \mathrm {~d} x = 1 + \sqrt { 3 }$$
OCR FP2 2010 June Q4
7 marks Standard +0.8
4 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-2_947_1305_986_420} The diagram shows the curve with equation $$y = \frac { a x + b } { x + c }$$ where \(a , b\) and \(c\) are constants.
  1. Given that the asymptotes of the curve are \(x = - 1\) and \(y = - 2\) and that the curve passes through \(( 3,0 )\), find the values of \(a , b\) and \(c\).
  2. Sketch the curve with equation $$y ^ { 2 } = \frac { a x + b } { x + c }$$ for the values of \(a , b\) and \(c\) found in part (i). State the coordinates of any points where the curve crosses the axes, and give the equations of any asymptotes.
OCR FP2 2010 June Q5
8 marks Challenging +1.2
5 It is given that, for \(n \geqslant 0\), $$I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } } ( 1 - 2 x ) ^ { n } \mathrm { e } ^ { x } \mathrm {~d} x$$
  1. Prove that, for \(n \geqslant 1\), $$I _ { n } = 2 n I _ { n - 1 } - 1$$
  2. Find the exact value of \(I _ { 3 }\).
OCR FP2 2010 June Q6
7 marks Standard +0.8
6
  1. Show that \(\frac { \mathrm { d } } { \mathrm { d } x } \left( \sinh ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 2 } + 1 } }\).
  2. Given that \(y = \cosh \left( a \sinh ^ { - 1 } x \right)\), where \(a\) is a constant, show that $$\left( x ^ { 2 } + 1 \right) \frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } + x \frac { \mathrm {~d} y } { \mathrm {~d} x } - a ^ { 2 } y = 0$$
OCR FP2 2010 June Q7
11 marks Standard +0.8
7 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-3_531_1065_1208_539} The line \(y = x\) and the curve \(y = 2 \ln ( 3 x - 2 )\) meet where \(x = \alpha\) and \(x = \beta\), as shown in the diagram.
  1. Use the iteration \(x _ { n + 1 } = 2 \ln \left( 3 x _ { n } - 2 \right)\), with initial value \(x _ { 1 } = 5.25\), to find the value of \(\beta\) correct to 2 decimal places. Show all your working.
  2. With the help of a 'staircase' diagram, explain why this iteration will not converge to \(\alpha\), whatever value of \(x _ { 1 }\) (other than \(\alpha\) ) is used.
  3. Show that the equation \(x = 2 \ln ( 3 x - 2 )\) can be rewritten as \(x = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right)\). Use the NewtonRaphson method, with \(\mathrm { f } ( x ) = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right) - x\) and \(x _ { 1 } = 1.2\), to find \(\alpha\) correct to 2 decimal places. Show all your working.
  4. Given that \(x _ { 1 } = \ln 36\), explain why the Newton-Raphson method would not converge to a root of \(\mathrm { f } ( x ) = 0\).
OCR FP2 2010 June Q8
10 marks Challenging +1.2
8
  1. Using the definition of \(\cosh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that $$4 \cosh ^ { 3 } x - 3 \cosh x \equiv \cosh 3 x$$
  2. Use the substitution \(u = \cosh x\) to find, in terms of \(5 ^ { \frac { 1 } { 3 } }\), the real root of the equation $$20 u ^ { 3 } - 15 u - 13 = 0 .$$
OCR FP2 2010 June Q9
13 marks Challenging +1.8
9 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-4_486_1097_696_523} The diagram shows the curve with equation \(y = \sqrt { 2 x + 1 }\) between the points \(A \left( - \frac { 1 } { 2 } , 0 \right)\) and \(B ( 4,3 )\).
  1. Find the area of the region bounded by the curve, the \(x\)-axis and the line \(x = 4\). Hence find the area of the region bounded by the curve and the lines \(O A\) and \(O B\), where \(O\) is the origin.
  2. Show that the curve between \(B\) and \(A\) can be expressed in polar coordinates as $$r = \frac { 1 } { 1 - \cos \theta } , \quad \text { where } \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) \leqslant \theta \leqslant \pi$$
  3. Deduce from parts (i) and (ii) that \(\int _ { \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) } ^ { \pi } \operatorname { cosec } ^ { 4 } \left( \frac { 1 } { 2 } \theta \right) \mathrm { d } \theta = 24\). www.ocr.org.uk after the live examination series.
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OCR FP2 2012 June Q1
5 marks Standard +0.3
1 Express sech \(2 x\) in terms of exponentials and hence, by using the substitution \(u = e ^ { 2 x }\), find \(\int \operatorname { sech } 2 x \mathrm {~d} x\).
OCR FP2 2012 June Q2
9 marks Challenging +1.2
2 A curve has polar equation \(r = \cos \theta \sin 2 \theta\), for \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\). Find
  1. the equations of the tangents at the pole,
  2. the maximum value of \(r\),
  3. a cartesian equation of the curve, in a form not involving fractions.
OCR FP2 2012 June Q3
8 marks Standard +0.3
3
  1. By quoting results given in the List of Formulae (MF1), prove that \(\tanh 2 x \equiv \frac { 2 \tanh x } { 1 + \tanh ^ { 2 } x }\).
  2. Solve the equation \(5 \tanh 2 x = 1 + 6 \tanh x\), giving your answers in logarithmic form.
OCR FP2 2012 June Q4
9 marks Standard +0.3
4 It is given that the equation \(x ^ { 4 } - 2 x - 1 = 0\) has only one positive root, \(\alpha\), and \(1.3 < \alpha < 1.5\).
  1. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-2_433_424_1119_817} The diagram shows a sketch of \(y = x\) and \(y = \sqrt [ 4 ] { 2 x + 1 }\) for \(x \geqslant 0\). Use the iteration \(x _ { n + 1 } = \sqrt [ 4 ] { 2 x _ { n } + 1 }\) with \(x _ { 1 } = 1.35\) to find \(x _ { 2 }\) and \(x _ { 3 }\), correct to 4 decimal places. On the copy of the diagram show how the iteration converges to \(\alpha\).
  2. For the same equation, the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( x _ { n } ^ { 4 } - 1 \right)\) with \(x _ { 1 } = 1.35\) gives \(x _ { 2 } = 1.1608\) and \(x _ { 3 } = 0.4077\), correct to 4 decimal places. Draw a sketch of \(y = x\) and \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\) for \(x \geqslant 0\), and show how this iteration does not converge to \(\alpha\).
  3. Find the positive root of the equation \(x ^ { 4 } - 2 x - 1 = 0\) by using the Newton-Raphson method with \(x _ { 1 } = 1.35\), giving the root correct to 4 decimal places.
OCR FP2 2012 June Q5
8 marks Challenging +1.2
5 A function is defined by \(\mathrm { f } ( x ) = \sinh ^ { - 1 } x + \sinh ^ { - 1 } \left( \frac { 1 } { x } \right)\), for \(x \neq 0\).
  1. When \(x > 0\), show that the value of \(\mathrm { f } ( x )\) for which \(\mathrm { f } ^ { \prime } ( x ) = 0\) is \(2 \ln ( 1 + \sqrt { 2 } )\).
  2. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-3_497_659_520_708} The diagram shows the graph of \(y = \mathrm { f } ( x )\) for \(x > 0\). Sketch the graph of \(y = \mathrm { f } ( x )\) for \(x < 0\) and state the range of values that \(\mathrm { f } ( x )\) can take for \(x \neq 0\).
OCR FP2 2012 June Q6
9 marks Challenging +1.2
6 It is given that, for non-negative integers \(n\), $$I _ { n } = \int _ { 0 } ^ { \pi } x ^ { n } \sin x \mathrm {~d} x$$
  1. Prove that, for \(n \geqslant 2 , I _ { n } = \pi ^ { n } - n ( n - 1 ) I _ { n - 2 }\).
  2. Find \(I _ { 5 }\) in terms of \(\pi\).
OCR FP2 2012 June Q7
12 marks Challenging +1.8
7 \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-4_782_1065_251_500} The diagram shows the curve \(y = \frac { 1 } { x }\) for \(x > 0\) and a set of \(( n - 1 )\) rectangles of unit width below the curve. These rectangles can be used to obtain an inequality of the form $$\frac { 1 } { a } + \frac { 1 } { a + 1 } + \frac { 1 } { a + 2 } + \ldots + \frac { 1 } { b } < \int _ { 1 } ^ { n } \frac { 1 } { x } \mathrm {~d} x$$ Another set of rectangles can be used similarly to obtain $$\int _ { 1 } ^ { n } \frac { 1 } { x } \mathrm {~d} x < \frac { 1 } { c } + \frac { 1 } { c + 1 } + \frac { 1 } { c + 2 } + \ldots + \frac { 1 } { d }$$
  1. Write down the values of the constants \(a\) and \(c\), and express \(b\) and \(d\) in terms of \(n\). The function f is defined by \(\mathrm { f } ( n ) = 1 + \frac { 1 } { 2 } + \frac { 1 } { 3 } + \ldots + \frac { 1 } { n } - \ln n\), for positive integers \(n\).
  2. Use your answers to part (i) to obtain upper and lower bounds for \(\mathrm { f } ( n )\).
  3. By using the first 2 terms of the Maclaurin series for \(\ln ( 1 + x )\) show that, for large \(n\), $$f ( n + 1 ) - f ( n ) \approx - \frac { n - 1 } { 2 n ^ { 2 } ( n + 1 ) } .$$
OCR FP2 2012 June Q8
12 marks Challenging +1.2
8 The curve \(C _ { 1 }\) has equation \(y = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are polynomials of degree 2 and 1 respectively. The asymptotes of the curve are \(x = - 2\) and \(y = \frac { 1 } { 2 } x + 1\), and the curve passes through the point \(\left( - 1 , \frac { 17 } { 2 } \right)\).
  1. Express the equation of \(C _ { 1 }\) in the form \(y = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\).
  2. For the curve \(C _ { 1 }\), find the range of values that \(y\) can take.
  3. For the curve \(C _ { 1 }\), find the range of values that \(y\) can take.
    Another curve, \(C _ { 2 }\), has equation \(y ^ { 2 } = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are the polynomials found in part (i).
  4. It is given that \(C _ { 2 }\) intersects the line \(y = \frac { 1 } { 2 } x + 1\) exactly once. Find the coordinates of the point of intersection. Another curve, \(C _ { 2 }\), has equation \(y ^ { 2 } = \frac { \mathrm { p } ( x ) } { \mathrm { q } ( x ) }\), where \(\mathrm { p } ( x )\) and \(\mathrm { q } ( x )\) are the polynomials found in part (i).
  5. It is given that \(C _ { 2 }\) intersects the line \(y = \frac { 1 } { 2 } x + 1\) exactly once. Find the coordinates of the point of
    intersection. \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}
OCR FP2 2013 June Q1
5 marks Standard +0.8
1 By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), find \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 1 + \cos \theta } \mathrm { d } \theta\).
OCR FP2 2013 June Q2
8 marks Standard +0.3
2
  1. Using the definitions for \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\).
  2. Hence solve the equation \(\sinh ^ { 2 } x = 5 \cosh x - 7\), giving your answers in logarithmic form.
OCR FP2 2013 June Q3
10 marks Challenging +1.2
3 It is given that \(\mathrm { f } ( x ) = \tanh ^ { - 1 } \left( \frac { 1 - x } { 3 + x } \right)\) for \(x > - 1\).
  1. Show that \(\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 1 } { 2 ( x + 1 ) ^ { 2 } }\).
  2. Hence find the Maclaurin series for \(\mathrm { f } ( x )\) up to and including the term in \(x ^ { 2 }\).
OCR FP2 2013 June Q4
8 marks Challenging +1.2
4 It is given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \cos ^ { n } x \mathrm {~d} x\) for \(n \geqslant 0\).
  1. Show that \(I _ { n } = \frac { n - 1 } { n } I _ { n - 2 }\) for \(n \geqslant 2\).
  2. Hence find \(I _ { 11 }\) as an exact fraction.
OCR FP2 2013 June Q5
8 marks Standard +0.3
5 You are given that the equation \(x ^ { 3 } + 4 x ^ { 2 } + x - 1 = 0\) has a root, \(\alpha\), where \(- 1 < \alpha < 0\).
  1. Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 4 x _ { n } ^ { 2 } + 1 } { 3 x _ { n } ^ { 2 } + 8 x _ { n } + 1 } .$$
  2. Using the initial value \(x _ { 1 } = - 0.7\), find \(x _ { 2 }\) and \(x _ { 3 }\) and find \(\alpha\) correct to 5 decimal places.
  3. The diagram shows a sketch of the curve \(y = x ^ { 3 } + 4 x ^ { 2 } + x - 1\) for \(- 1.5 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-3_602_926_749_566} Using the copy of the diagram in your answer book, explain why the initial value \(x _ { 1 } = 0\) will fail to find \(\alpha\).
OCR FP2 2013 June Q6
8 marks Challenging +1.2
6 \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-4_656_1017_251_525} The diagram shows part of the curve \(y = \ln ( \ln ( x ) )\). The region between the curve and the \(x\)-axis for \(3 \leqslant x \leqslant 6\) is shaded.
  1. By considering \(n\) rectangles of equal width, show that a lower bound, \(L\), for the area of the shaded region is \(\frac { 3 } { n } \sum _ { r = 0 } ^ { n - 1 } \ln \left( \ln \left( 3 + \frac { 3 r } { n } \right) \right)\).
  2. By considering another set of \(n\) rectangles of equal width, find a similar expression for an upper bound, \(U\), for the area of the shaded region.
  3. Find the least value of \(n\) for which \(U - L < 0.001\).
OCR FP2 2013 June Q7
14 marks Challenging +1.2
7 The equation of a curve is \(y = \frac { x ^ { 2 } + 1 } { ( x + 1 ) ( x - 7 ) }\).
  1. Write down the equations of the asymptotes.
  2. Find the coordinates of the stationary points on the curve.
  3. Find the coordinates of the point where the curve meets one of its asymptotes.
  4. Sketch the curve.
OCR FP2 2013 June Q8
11 marks Challenging +1.2
8 The equation of a curve is \(x ^ { 2 } + y ^ { 2 } - x = \sqrt { x ^ { 2 } + y ^ { 2 } }\).
  1. Find the polar equation of this curve in the form \(r = \mathrm { f } ( \theta )\).
  2. Sketch the curve.
  3. The line \(x + 2 y = 2\) divides the region enclosed by the curve into two parts. Find the ratio of the two areas.
OCR FP2 2014 June Q1
3 marks Standard +0.3
1 Find \(\int _ { 0 } ^ { 2 } \frac { 1 } { \sqrt { 4 + x ^ { 2 } } } \mathrm {~d} x\), giving your answer exactly in logarithmic form.
OCR FP2 2014 June Q2
5 marks Moderate -0.3
2 It is given that \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\).
  1. Using the standard Maclaurin expansion for \(\ln ( 1 + x )\), write down the first four terms in the expansion of \(\mathrm { f } ( x )\), stating the set of values of \(x\) for which the expansion is valid.
  2. Hence find the exact value of $$1 - \frac { 1 } { 2 } \left( \frac { 1 } { 2 } \right) ^ { 2 } + \frac { 1 } { 3 } \left( \frac { 1 } { 2 } \right) ^ { 4 } - \frac { 1 } { 4 } \left( \frac { 1 } { 2 } \right) ^ { 6 } + \ldots .$$