If $$f(x) = \frac{(1+x)^{\frac{1}{2}} - 1}{1-(1-x)^{\frac{1}{2}}},$$ find (i) \(\lim_{x \to 0} f(x)\), and (ii) \(\lim_{x \to 0} \frac{df}{dx}\).
If \(y = (x^2-1)^n\), where \(n\) is a positive integer, prove that $$(1-x^2)\frac{dy}{dx} + 2nxy = 0.$$ By differentiating this equation \((n+1)\) times and using Leibniz' theorem, or otherwise, show that the function \(p_n(x)\), defined by $$p_n(x) = \frac{d^n}{dx^n}(x^2-1)^n,$$ satisfies the equation $$(1-x^2)\frac{d^2p_n}{dx^2} - 2x\frac{dp_n}{dx} + n(n+1)p_n = 0.$$ Show also that $$p_n(1) = (-1)^n p_n(-1) = 2^n n!$$
Find the minimum distance between the origin and the branch of the curve $$y = \frac{x}{x+c} \quad (c \neq 0)$$ that does not pass through the origin. Does your result remain true when \(c = 0\)?
(i) Evaluate $$\int_a^b \sqrt{[(x-a)(b-x)]} \, dx$$ where \(a\) and \(b\) (\(> a\)) are constants. (ii) If $$B(x) = \int_0^x e^{-t^4} dt,$$ prove that $$(2n+1)\int_0^x t^{2n}B(t) dt = x^{2n+1}B(x) - \frac{1}{4}n!\left[1-e^{-x^4}\sum_{r=0}^{\infty}\frac{x^{2r}}{r!}\right],$$ where \(n\) is a positive integer.
A family of plane curves has the property that if the tangent to \(f(x,y)\) of any one of the curves intersects the \(x\)-axis in \(N\), then the distance \(ON\) is equal to \(ky^2\), where \(O\) is the origin and \(k\) is a positive constant. Find the equation of the particular curve of the family that passes through the point \((0,1)\) and sketch it.
Suppose that \(a_j, b_j\) (\(1 \leq j \leq n\)) are given real numbers and that $$1 \leq a_j \leq A, \quad 1 \leq b_j \leq B \quad (1 \leq j \leq n)$$ for some \(A, B\). Show that $$a_j b_j \geq u_j B + v_j A \quad (1 \leq j \leq n),$$ where \(u_j, v_j\) are defined by the equations $$a_j^2 = u_j + v_j A^2, \quad b_j^2 = u_j B^2 + v_j.$$ Deduce that $$\frac{(\sum a_j^2)(\sum b_j^2)}{(\sum a_j b_j)^2} \leq \left(\frac{(AB)^{\frac{1}{2}} + (AB)^{-\frac{1}{2}}}{2}\right)^2.$$
The polynomial \(P(x)\) in the single variable \(x\) has real coefficients and is non-negative for every real value of \(x\). Show that there are polynomials \(Q(x), R(x)\) with real coefficients such that $$P(x) = \{Q(x)\}^2 + \{R(x)\}^2.$$
Given that the roots of the equation $$y^8 + 3y^2 + 2y - 1 = 0$$ are the fourth powers of the roots of an equation $$x^8 + ax^2 + bx + c = 0$$ with rational coefficients \(a, b, c\), find suitable values for \(a, b, c\).
Let \(N(k,l)\) be the number of sets of integers \(a_1, \ldots, a_k\) such that $$1 \leq a_{j+1} \leq 2a_j \quad (1 \leq j < k)$$ and $$a_1 = 1, \quad a_k = l.$$ Prove that $$N(k, 2s+2) - N(k, 2s) = N(k-1, s).$$ For \(k \geq 2\), \(0 \leq v < 2^{k-2}\), show that $$N(k, 2^{k-1} - 2v) = N(k, 2^{k-1} - 2v - 1) = c(v)$$ is independent of \(k\), and that $$\sum_{v=0}^{\infty} c(v)t^v = (1-t)^{-1}\prod_{r=0}^{\infty}(1-t^{2^r})^{-1}.$$
(i) \(A, B, C, D\) are the angles of a plane quadrilateral. Show that $$4\sin(A+B)\sin(B+D)\sin(D+A) = \sin 2A + \sin 2B + \sin 2C + \sin 2D.$$ (ii) If $$\arccos a + \arccos b = \frac{1}{4}\pi,$$ show that $$a^2 - 2\sqrt{2}ab + b^2 = \frac{1}{2}.$$