Problems

Filters
Clear Filters

8 problems found

2019 Paper 2 Q1
D: 1500.0 B: 1500.0
differentiation from first principles tangent polynomial cubic if and only if curve sketching product rule quadratic proof

Let \(f(x) = (x-p)g(x)\), where g is a polynomial. Show that the tangent to the curve \(y = f(x)\) at the point with \(x = a\), where \(a \neq p\), passes through the point \((p, 0)\) if and only if \(g'(a) = 0\). The curve \(C\) has equation $$y = A(x - p)(x - q)(x - r),$$ where \(p\), \(q\) and \(r\) are constants with \(p < q < r\), and \(A\) is a non-zero constant.

  1. The tangent to \(C\) at the point with \(x = a\), where \(a \neq p\), passes through the point \((p, 0)\). Show that \(2a = q + r\) and find an expression for the gradient of this tangent in terms of \(A\), \(q\) and \(r\).
  2. The tangent to \(C\) at the point with \(x = c\), where \(c \neq r\), passes through the point \((r, 0)\). Show that this tangent is parallel to the tangent in part (i) if and only if the tangent to \(C\) at the point with \(x = q\) does not meet the curve again.

Solution: The tangent to the curve \(y = f(x)\) at \(x = a\) has the equation \(\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)\). This passes through \((p,0)\) iff \begin{align*} && \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\ \Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\ \Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\ \Leftrightarrow && 0 &= g'(a) \\ \end{align*}

  1. In this case \(g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)\) and so we must have that \(g'(a) = 0\), ie \(A(2a-(q+r)) = 0 \Rightarrow 2a = q+r\) The gradient is \(g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)\)
  2. By the same reasoning, but with \(g(x) = A(x-p)(x-q)\) we have the gradient is \(A(c-p)(c-r)\). This is parallel iff \begin{align*} && (c-p)(c-r) &= (a-q)(a-r) \end{align*} The tangent at \(x = q\) is \(\frac{y-0}{x-q} = A(q-p)(q-r)\) or \( y = A(q-p)(q-r)(x-q)\)

View
2019 Paper 3 Q2
D: 1500.0 B: 1500.0
differentiation from first principles functional equation exponential function hyperbolic functions differential equation limit definition of derivative tanh

The definition of the derivative \(f'\) of a (differentiable) function f is $$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}. \quad (*)$$

  1. The function f has derivative \(f'\) and satisfies $$f(x + y) = f(x)f(y)$$ for all \(x\) and \(y\), and \(f'(0) = k\) where \(k \neq 0\). Show that \(f(0) = 1\). Using \((*)\), show that \(f'(x) = kf(x)\) and find \(f(x)\) in terms of \(x\) and \(k\).
  2. The function g has derivative \(g'\) and satisfies $$g(x + y) = \frac{g(x) + g(y)}{1 + g(x)g(y)}$$ for all \(x\) and \(y\), \(|g(x)| < 1\) for all \(x\), and \(g'(0) = k\) where \(k \neq 0\). Find \(g'(x)\) in terms of \(g(x)\) and \(k\), and hence find \(g(x)\) in terms of \(x\) and \(k\).

Solution:

  1. \(\,\) \begin{align*} && f(0+x) &= f(0)f(x) \\ \Rightarrow && f(0) &= 0, 1\\ &&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\ \Rightarrow && f(0) &= 1 \end{align*} \begin{align*} && f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\ &&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\ &&&= f(x) \cdot f'(0) \\ &&&= kf(x) \end{align*} Since \(f'(x) = kf(x)\) we must have \(\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}\) but \(f(0) = 1\) so \(f(x) = e^{kx}\)
  2. Consider \begin{align*} && g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\ \Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\ \Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\ \Rightarrow && g(0) &= -1, 0, 1 \\ \Rightarrow && g(0) &= 0 \tag{\(|g(0)| < 1\)} \end{align*} \begin{align*} && g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\ &&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\ &&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\ &&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\ &&&= (1-(g(x))^2) \cdot g'(0)\\ &&&= k (1-(g(x))^2) \\ \end{align*} Let \(y = g(x)\) so \begin{align*} && y' &= k(1-y^2) \\ \Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\ \Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\ &&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\ x = 0, y = 0: && 0 &= \ln 1 + C \\ \Rightarrow && C &= 0 \\ \Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\ \Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\ \Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\ &&&= \tanh kx \end{align*}

View
2018 Paper 3 Q5
D: 1700.0 B: 1484.0
inequality AM-GM inequality sequences differentiation from first principles proof by induction polynomial analysis stationary points optimisation

The real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\) are all positive. For each positive integer \(n\), \(A_n\) and \(G_n\) are defined by \[ A_n = \frac{a_1+a_2 + \cdots + a_n}n \ \ \ \ \ \text{and } \ \ \ \ \ G_n = \big( a_1a_2\cdots a_n\big) ^{1/n} \,. \]

  1. Show that, for any given positive integer \(k\), \[ (k+1) ( A_{k+1} - G_{k+1}) \ge k (A_k-G_k) \] if and only if \[\lambda^{k+1}_k -(k+1)\lambda_{{k}} +k \ge 0\,, \] where \( \lambda_{{k}} = \left(\dfrac{a_{k+1}}{G_{k}}\right)^{\frac1 {k+1}}\,\).
  2. Let \[ \f(x)=x^{k+1} -(k+1)x +k \,, \] where \(x > 0\) and \(k\) is a positive integer. Show that \(\f(x)\ge0\) and that \(\f(x)=0\) if and only if \(x = 1\,\).
  3. Deduce that:
    1. \(A_n \ge G_n\) for all \(n\); \\
    2. if \(A_n=G_n\) for some \(n\), then \(a_1=a_2 = \cdots = a_n\,\).

Solution:

  1. \begin{align*} && (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\ \Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\ \Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\ \Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\ \end{align*} as required.
  2. \begin{align*} && f'(x) &= (k+1)x^k - (k+1) \\ &&&= (k+1)(x^k-1) \end{align*} Therefore \(f(x)\) is strictly decreasing on \((0,1)\) and strictly increasing on \((1,\infty)\) and so the minimum will be \(f(1) = 1 - (k+1) + k = 0\), so \(f(x) \geq 0\) with equality only at \(x = 1\).
    1. We can proceed by induction to show since the inequality holds for \(n=1\) and since if it holds for \(n=k\) it will hold for \(n=k+1\) as \(A_{k+1}-G_{k+1}\) must have the same sign as \(A_k - G_k\).
    2. The only way for equality to hold is if \(\lambda_k = 1\) for \(k = 1, \cdots n\), ie \(a_{k+1} = G_k\), but this means \(a_2 = a_1, a_3 = a_1\) etc. Therefore all values are equal.

View
2014 Paper 1 Q8
D: 1500.0 B: 1484.0
differentiation from first principles envelope limits tangent-normal curve sketching line intersection parametric parabola

Let \(L_a\) denote the line joining the points \((a,0)\) and \((0, 1-a)\), where \(0< a < 1\). The line \(L_b\) is defined similarly.

  1. Determine the point of intersection of \(L_a\) and \(L_b\), where \(a\ne b\).
  2. Show that this point of intersection, in the limit as \(b\to a\), lies on the curve \(C\) given by \[ y=(1-\sqrt x)^2\, \ \ \ \ (0< x < 1)\,. \]
  3. Show that every tangent to \(C\) is of the form \(L_a\) for some \(a\).

Solution:

  1. \(L_a : \frac{y}{x-a} = \frac{1-a-0}{0-a} = \frac{a-1}{a} \Rightarrow ay+(1-a)x = a(1-a)\) \begin{align*} && ay + (1-a)x &= a(1-a) \\ && by + (1-b)x &= b(1-b) \\ \Rightarrow && aby + b(1-a)x &= ba(1-a) \\ && aby + a(1-b)x &= ab(1-b) \\ \Rightarrow && (b-a)x &= ab(b-a) \\ \Rightarrow && x &= ab \\ && y &= \frac{a-1}{a} \cdot a(b-1) \\ &&&= (1-a)(1-b) \end{align*}
  2. As \(a \to b\), \(x \to a^2, y \to 1-2a+a^2 =(1-a)^2 = (1-\sqrt{x})^2\)
  3. \(\frac{\d y}{\d x} = 2(1-\sqrt{x})\cdot \left (-\tfrac12 \frac{1}{\sqrt{x}} \right) = 1 - \frac{1}{\sqrt{x}}\). Therefore the tangent when \(x = c^2, y = (1-c)^2\) is \begin{align*} && \frac{y-(1-c)^2}{x-c^2} &= 1 - \frac{1}{c} \\ \Rightarrow && cy + (1-c)x &= c(c-1)+c(1-c)^2 \\ &&&= c(1-c) \end{align*} Which is an equation of the form \(L_c\)

View
2005 Paper 1 Q2
D: 1516.0 B: 1484.0
conics parabola tangent-normal parametric locus geometric proof differentiation from first principles

The point \(P\) has coordinates \(\l p^2 , 2p \r\) and the point \(Q\) has coordinates \(\l q^2 , 2q \r\), where \(p\) and~\(q\) are non-zero and \(p \neq q\). The curve \(C\) is given by \(y^2 = 4x\,\). The point \(R\) is the intersection of the tangent to \(C\) at \(P\) and the tangent to \(C\) at \(Q\). Show that \(R\) has coordinates \(\l pq , p+q \r\). The point \(S\) is the intersection of the normal to \(C\) at \(P\) and the normal to \(C\) at \(Q\). If \(p\) and \(q\) are such that \(\l 1 , 0 \r\) lies on the line \(PQ\), show that \(S\) has coordinates \(\l p^2 + q^2 + 1 , \, p+q \r\), and that the quadrilateral \(PSQR\) is a rectangle.

View
1992 Paper 2 Q4
D: 1600.0 B: 1500.0
differentiation from first principles polynomials product rule proof by induction linearity abstract algebra operator

\(\lozenge\) is an operation which take polynomials in \(x\) to polynomials in \(x\); that is, given a polynomial \(\mathrm{h}(x)\) there is another polynomial called \(\lozenge\mathrm{h}(x)\). It is given that, if \(\mathrm{f}(x)\) and \(\mathrm{g}(x)\) are any two polynomials in \(x\), the following are always true:

  1. \(\lozenge(\mathrm{f}(x)\mathrm{g}(x))=\mathrm{g}(x)\lozenge\mathrm{f}(x)+\mathrm{f}(x)\lozenge\mathrm{g}(x),\)
  2. \(\lozenge(\mathrm{f}(x)+\mathrm{g}(x))=\lozenge\mathrm{f}(x)+\lozenge\mathrm{g}(x),\)
  3. \(\lozenge x=1\)
  4. if \(\lambda\) is a constant then \(\lozenge(\lambda\mathrm{f}(x))=\lambda\lozenge\mathrm{f}(x).\)
Show that, if \(\mathrm{f}(x)\) is a constant (i.e., a polynomial of degree zero), then \(\lozenge\mathrm{f}(x)=0.\) Calculate \(\lozenge x^{2}\) and \(\lozenge x^{3}.\) Prove that \(\lozenge\mathrm{h}(x)=\dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{h}(x))\) for any polynomial \(\mathrm{h}(x)\).

Solution: Claim: If \(f(x) = c\) then \(\lozenge f(x) = 0\) Proof: Consider \(g(x) = x\) then \begin{align*} (1) && \lozenge(f(x)g(x)) &= g(x) \lozenge f(x) + f(x) \lozenge g(x) \\ \Rightarrow && \lozenge(c x) &= x \lozenge f(x) + c \lozenge x \\ (4) && \lozenge(c x) &= c \lozenge x \\ \Rightarrow && 0 &= x \lozenge f(x) \\ \Rightarrow && \lozenge f(x) &= 0 \end{align*} \begin{align*} (1) && \lozenge(x^2) &= x \lozenge x + x \lozenge x \\ (3) &&&= 2 x \cdot 1 \\ &&&= 2x \\ \\ (1) && \lozenge (x^3) &= x^2 \lozenge x + x \lozenge (x^2) \\ &&&= x^2 \cdot \underbrace{1}_{(3)} + x \cdot\underbrace{ 2x}_{\text{previous part}} \\ &&&= 3x^2 \end{align*} Claim: \(\lozenge h(x) = \frac{\d }{\d x} ( h(x))\) for any polynomial \(h\). Proof: (By (strong) induction on the degree of \(h\)). Base case: True, we proved this in the first part of the question. Inductive step: Assume true for all polynomials of degree less than or equal to \(k\). Then consider \(n = k+1\). We can write \(h(x) = ax^{k+1} + h_k(x)\) where \(h_k(x)\) is a polynomial of degree less than or equal to \(k\). Then notice: \begin{align*} && \lozenge (h(x)) &= \lozenge (ax^{k+1} + h_k(x)) \\ (2) &&&= \lozenge (ax^{k+1})+ \lozenge (h_k(x)) \\ &&&=\underbrace{a\lozenge (x^{k+1})}_{(4)}+ \underbrace{\frac{\d}{\d x} (h_k(x))}_{\text{inductive hypothesis}}\\ &&&= a \underbrace{\left (x \lozenge x^k + x^k \lozenge x \right)}_{(1)} + \frac{\d}{\d x} (h_k(x)) \\ &&&= a \left ( x \cdot \underbrace{k x^{k-1}}_{\text{inductive hyp.}} + x^k \cdot \underbrace{1}_{(3)} \right) + \frac{\d}{\d x} (h_k(x)) \\ &&&= (k+1)a x^k + \frac{\d}{\d x} (h_k(x)) \\ &&&= \frac{\d }{\d x} \left ( ax^{k+1} + h_k(x) \right) \\ &&&= \frac{\d }{\d x} (h(x)) \end{align*} Therefore since our statement is true for \(n=0\) and if it is true for \(n=k\) it is true for \(n=k+1\) by the principle of mathematical induction it is true for all \(n \geq 0\)

View
1990 Paper 1 Q3
D: 1516.0 B: 1484.0
differentiation from first principles curve unit normal parametric gradient polynomial vectors geometric proof

Given a curve described by \(y=\mathrm{f}(x)\), and such that \(y\geqslant0\), a push-off of the curve is a new curve obtained as follows: for each point \((x,\mathrm{f}(x))\) with position vector \(\mathbf{r}\) on the original curve, there is a point with position vector \(\mathbf{s}\) on the new curve such that \(\mathbf{s-r}=\mathrm{p}(x)\mathbf{n},\) where \(\mathrm{p}\) is a given function and \(\mathbf{n}\) is the downward-pointing unit normal to the original curve at \(\mathbf{r}\).

  1. For the curve \(y=x^{k},\) where \(x>0\) and \(k\) is a positive integer, obtain the function \(\mathrm{p}\) for which the push-off is the positive \(x\)-axis, and find the value of \(k\) such that, for all points on the original curve, \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\).
  2. Suppose that the original curve is \(y=x^{2}\) and \(\mathrm{p}\) is such that the gradient of the curves at the points with position vectors \(\mathbf{r}\) and \(\mathbf{s}\) are equal (for every point on the original curve). By writing \(\mathrm{p}(x)=\mathrm{q}(x)\sqrt{1+4x^{2}},\) where \(\mathrm{q}\) is to be determined, or otherwise, find the form of \(\mathrm{p}\).

Solution:

  1. Suppose we have \(y = x^k\), then the tangent at \((t,t^k)\) has gradient \(\frac{\d y}{\d x} = kx^{k-1} = kt^{k-1}\) and the normal has gradient \(-\frac1k x^{1-k}\). For the push-off to be the positive \(x\)-axis, we need \(p(x)\) to be the length of the line. The line will have the equation: \begin{align*} && \frac{y - t^k}{x - t} &= -\frac1k t^{1-k} \\ y = 0: && x-t &= \frac{kt^k}{t^{1-k}} \\ && x& =t + kt^{2k-1} \end{align*} The distance from \((t,t^k)\) to \((t+kt^{2k-1},0)\) is \(\sqrt{t^{2k} + k^2t^{4k-2}} = t^k \sqrt{1+k^2t^{2k-2}} = y \sqrt{1+k^2 \frac{y^2}{x^2}} = \frac{y}{x} \sqrt{x^2 + k^2y^2}\), ie \(p(x) = \frac{y}{x}\sqrt{x^2+k^2y^2}\) If \(\left|\mathbf{r}\right|=\left|\mathbf{r-s}\right|\), then we need \(\sqrt{x^2+y^2} = \frac{y}{x}\sqrt{x^2+ky^2}\), but clearly this is satisfied when \(k = 1\).
  2. The points are \((t, t^2)\) and the normal has graident \(-\frac{1}{2t}\), the normal vector is \(\frac{1}{\sqrt{1+\frac{1}{4t^2}}}\binom{1}{-\frac1{2t}} = \frac{1}{\sqrt{4t^2+1}} \binom{2t}{-1}\). If we write \(p(x) = q(x)\sqrt{4x^2+1}\) then the the new points are at \(\binom{t+q(t)2t}{t^2-\q(t)}\) an the gradient will be \(\frac{2t-q'(t)}{1+q'(t)2t+2q(t)}\). We need it to be the case that \begin{align*} && 2t &= \frac{2t-q'(t)}{1+2q(t)+2tq'(t)} \\ \Rightarrow && 4tq(t) &= -q'(t)(1+4t^2) \\ \Rightarrow && \frac{q'(t)}{q(t)} &= -\frac{4t}{1+4t^2} \\ \Rightarrow && \ln q(t) &= -\frac12 \ln(1+4t^2)+C \\ \Rightarrow && q(t) &= A(1+4t^2)^{-1/2} \\ \Rightarrow && p(x) &= A \end{align*} So the push-off's are constants.

View
1988 Paper 1 Q7
D: 1484.0 B: 1500.0
differentiation from first principles quadratic inequality optimisation polynomial triangle inequality Chebyshev polynomial

The function \(\mathrm{f}\) is defined by \[ \mathrm{f}(x)=ax^{2}+bx+c. \] Show that \[ \mathrm{f}'(x)=\mathrm{f}(1)\left(x+\tfrac{1}{2}\right)+\mathrm{f}(-1)\left(x-\tfrac{1}{2}\right)-2\mathrm{f}(0)x. \] If \(a,b\) and \(c\) are real and such that \(\left|\mathrm{f}(x)\right|\leqslant1\) for \(\left|x\right|\leqslant1\), show that \(\left|\mathrm{f}'(x)\right|\leqslant4\) for \(\left|x\right|\leqslant1\). Find particular values of \(a,b\) and \(c\) such that, for the corresponding function \(\mathrm{f}\) of the above form \(\left|\mathrm{f}(x)\right|\leqslant1\) for all \(x\) with \(\left|x\right|\leqslant1\) and \(\mathrm{f}'(x)=4\) for some \(x\) satisfying \(\left|x\right|\leqslant1\).

Solution: Let \(f(x) = ax^2 + bx + c\) then \begin{align*} f'(x) &= 2ax + b \\ f(0) &= c \\ f(1) &= a+b+c \\ f(-1) &= a-b+c \\ f(1)+f(-1) &= 2(a+c) \\ f(1)-f(-1) &= 2b \\ f'(x) &= x(f(1)+f(-1)) + \frac12 (f(1) - f(-1)) - 2f(0)x \end{align*} as required. Since \(f'(x)\) is a straight line, the maximum value is either at \(1, -1\) or it's constant and either end suffices. \begin{align*} |f'(1)| & \leq |f(1)|\frac{3}{2} + |f(-1)| \frac12 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \\ |f'(-1)| & \leq |f(1)|\frac{1}{2} + |f(-1)| \frac32 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\ &= 4 \\ \end{align*} Therefore \(|f'(x)| \leq 4\). Suppose \(|f'(x)| = 4\) for some value in \(x \in [-1,1]\), then it must be either \(-1\) or \(1\). If \(f'(1) = 4\) then \(f(1) = 1, f(-1) = 1, f(0) = -1\) so \(f(x) = 1+ k(x^2-1) \Rightarrow f(x) = 1+2(x^2-1) = 2x^2 -1\). If \(f'(-1) = 4\) then \(f(1) = -1, f(-1) = -1, f(0) = 1 \Rightarrow f(x) = -2x^2 + 1\)

View