HL Paper 3
The function f is defined on the domain \(\left] { - \frac{\pi }{2},\frac{\pi }{2}} \right[{\text{ by }}f(x) = \ln (1 + \sin x)\) .
Show that \(f''(x) = - \frac{1}{{(1 + \sin x)}}\) .
(i) Find the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
(ii) Explain briefly why your result shows that f is neither an even function nor an odd function.
Determine the value of \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln (1 + \sin x) - x}}{{{x^2}}}\).
The integral \({I_n}\) is defined by \({I_n} = \int_{n\pi }^{(n + 1)\pi } {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x,{\text{ for }}n \in \mathbb{N}} \) .
Show that \({I_0} = \frac{1}{2}(1 + {{\text{e}}^{ - \pi }})\) .
By letting \(y = x - n\pi \) , show that \({I_n} = {{\text{e}}^{ - n\pi }}{I_0}\) .
Hence determine the exact value of \(\int_0^\infty {{{\text{e}}^{ - x}}|\sin x|{\text{d}}x} \) .
Given that \(y = \ln \left( {\frac{{1 + {{\text{e}}^{ - x}}}}{2}} \right)\), show that \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{{\text{e}}^{ - y}}}}{2} - 1\).
Hence, by repeated differentiation of the above differential equation, find the Maclaurin series for y as far as the term in \({x^3}\), showing that two of the terms are zero.
The function \(f\) is defined by \(f(x) = {{\text{e}}^{ - x}}\cos x + x - 1\).
By finding a suitable number of derivatives of \(f\), determine the first non-zero term in its Maclaurin series.
Let \(f(x) = {{\text{e}}^x}\sin x\).
Show that \(f''(x) = 2\left( {f'(x) - f(x)} \right)\).
By further differentiation of the result in part (a) , find the Maclaurin expansion of \(f(x)\), as far as the term in \({x^5}\).
Let \(f(x) = 2x + \left| x \right|\) , \(x \in \mathbb{R}\) .
Prove that f is continuous but not differentiable at the point (0, 0) .
Determine the value of \(\int_{ - a}^a {f(x){\text{d}}x} \) where \(a > 0\) .
The curves \(y = f(x)\) and \(y = g(x)\) both pass through the point \((1,{\text{ }}0)\) and are defined by the differential equations \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = x - {y^2}\) and \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = y - {x^2}\) respectively.
Show that the tangent to the curve \(y = f(x)\) at the point \((1,{\text{ }}0)\) is normal to the curve \(y = g(x)\) at the point \((1,{\text{ }}0)\).
Find \(g(x)\).
Use Euler’s method with steps of \(0.2\) to estimate \(f(2)\) to \(5\) decimal places.
Explain why \(y = f(x)\) cannot cross the isocline \(x - {y^2} = 0\), for \(x > 1\).
(i) Sketch the isoclines \(x - {y^2} = - 2,{\text{ }}0,{\text{ }}1\).
(ii) On the same set of axes, sketch the graph of \(f\).
Consider the functions \(f(x) = {(\ln x)^2},{\text{ }}x > 1\) and \(g(x) = \ln \left( {f(x)} \right),{\text{ }}x > 1\).
(i) Find \(f'(x)\).
(ii) Find \(g'(x)\).
(iii) Hence, show that \(g(x)\) is increasing on \(\left] {1,{\text{ }}\infty } \right[\).
Consider the differential equation
\[(\ln x)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{x}y = \frac{{2x - 1}}{{(\ln x)}},{\text{ }}x > 1.\]
(i) Find the general solution of the differential equation in the form \(y = h(x)\).
(ii) Show that the particular solution passing through the point with coordinates \(\left( {{\text{e, }}{{\text{e}}^2}} \right)\) is given by \(y = \frac{{{x^2} - x + {\text{e}}}}{{{{(\ln x)}^2}}}\).
(iii) Sketch the graph of your solution for \(x > 1\), clearly indicating any asymptotes and any maximum or minimum points.