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1951 Paper 2 Q310
D: 1500.0 B: 1500.0

A particle \(P\) of mass \(m\) is attached by a light elastic string, of unstretched length \(l\) and modulus of elasticity \(\lambda m\), to a point \(O\) on a smooth horizontal plane. Initially the particle is at rest on the plane and \(OP\) is of length \(l\). The particle is then given an initial velocity \(V\) on the plane in a direction perpendicular to \(OP\). Prove that, if \(3V^2 < 4\lambda l\), the length of \(OP\) in the ensuing motion never exceeds \(2l\).

1951 Paper 2 Q401
D: 1500.0 B: 1500.0

Show that if \(a, b, c\) are real numbers different from \(\pm 1\) and such that \[ a^2+b^2+c^2+2abc=1, \] then all three or none of them lie between \(-1\) and \(+1\).

1951 Paper 2 Q402
D: 1500.0 B: 1500.0

Show that in any algebraic equation \[ x^n - p_1x^{n-1} + p_2x^{n-2} - \dots + (-1)^n p_n = 0 \] the coefficient \(p_r\) is the sum of all the products formed by taking the roots together \(r\) at a time. If the roots are \(x_1, x_2, \dots, x_n\), prove that \[ (1-p_2+p_4-\dots)^2 + (p_1-p_3+p_5-\dots)^2 = (1+x_1^2)(1+x_2^2)\dots(1+x_n^2). \]

1951 Paper 2 Q403
D: 1500.0 B: 1500.0

If a sequence of quantities \(x_0, x_1, x_2, \dots\) satisfy the recurrence relation \[ x_{n+2} - 2x_{n+1} + 2x_n = 0 \] and \(x_0=1, x_1=2\), show that \(x_{4n}=(-4)^n\), and find \(x_{4n-1}, x_{4n-2}\) and \(x_{4n-3}\). Prove also that \[ \sum_{r=1}^{4n} x_r^2 = \frac{8}{5}(2^{4n}-1). \]

1951 Paper 2 Q404
D: 1500.0 B: 1500.0

A pack of 52 playing cards is shuffled and dealt to four players. One person finds he has 5 cards of a particular suit. Prove that the chance that his partner holds all the remaining 8 cards of the same suit is rather greater than 1 in 50,000.

1951 Paper 2 Q405
D: 1500.0 B: 1500.0

Justify Newton's method for approximating to a root of the equation \(f(x)=0\), namely, that if \(a\) is a first approximation, \(a_1 = a - \frac{f(a)}{f'(a)}\) is in general a better approximation. Illustrate as simply as you can, graphically or otherwise, the general circumstances in which (i) \(a_1\) is nearer to the actual root than \(a\), (ii) the actual root lies between \(a\) and \(a_1\). Consider the positive root between 1 and 2 of the equation \(3\sin x = 2x\) by taking \(a = \frac{\pi}{2}\). Find the next approximation and state which of the two cases mentioned above it illustrates.

1951 Paper 2 Q406
D: 1500.0 B: 1500.0

State Leibnitz's theorem for the \(n\)th differential coefficient of the product of two functions. If \[ y = (1-x^2)^{\frac{1}{2}m} \frac{d^m u}{dx^m}, \] where \(m\) is a positive integer, and \(u\) satisfies the equation \[ (1-x^2) \frac{d^2u}{dx^2} - 2x \frac{du}{dx} + n(n+1)u=0, \] prove that \(y\) satisfies the equation \[ (1-x^2) \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + \left\{ (n-m)(n+m+1) - \frac{m^2}{1-x^2} \right\} y = 0. \]

1951 Paper 2 Q407
D: 1500.0 B: 1500.0

Give sufficient conditions for the truth of Rolle's theorem, that if \(f(x)\) has equal values for \(x=a\) and \(x=b\), then \(f'(x)\) will be zero for some value of \(x\) between \(a\) and \(b\). By considering the function \[ f(x) = \begin{vmatrix} 1 & \phi(x) & \psi(x) \\ 1 & \phi(a) & \psi(a) \\ 1 & \phi(b) & \psi(b) \end{vmatrix}, \] prove that if \(\phi'(x)\) and \(\psi'(x)\) exist for \(a \le x \le b\), and \(\psi'(x) \ne 0\), then there is a value \(c\) such that \(a < c < b\), and \[ \frac{\phi(b)-\phi(a)}{\psi(b)-\psi(a)} = \frac{\phi'(c)}{\psi'(c)}. \] Evaluate \[ \lim_{x \to 0} \frac{e^x-1-\log(1+x)}{1-\cos x}. \]

1951 Paper 2 Q408
D: 1500.0 B: 1500.0

Prove that the equation \(x^3-3px^2+4q=0\) will have three real roots if \(p\) and \(q\) are the same sign and \(p^6>q^2\). Show that two roots will be positive or negative as the sign of \(p\) and \(q\) is positive or negative. (It may be assumed that neither \(p\) nor \(q\) vanish.) Find out by these results as much as possible about the roots of the equation \[ x^3 - 6x^2 + 16 = 0. \]

1951 Paper 2 Q409
D: 1500.0 B: 1500.0

If \[ I(p,q) = \int_0^{\log(1+\sqrt{2})} \sinh^p x \cosh^q x \, dx, \] where \(p>1\), prove that \[ (p+q)I(p,q) = 2^{\frac{q-1}{2}} + (q-1)I(p,q-2) = 2^{\frac{q+1}{2}} - (p-1)I(p-2,q). \]