Past Exam Questions

The equation of a curve is \(y = x + \\cos 2x\). Find the \(x\)-coordinates of the stationary points of the curve for which \(0 \leq x \leq \pi\), and determine the nature of each of these stationary points.

The curve \(y = e^{-4x} \tan x\) has two stationary points in the interval \(0 \leq x < \frac{1}{2} \pi\).

(a) Obtain an expression for \(\frac{dy}{dx}\) and show it can be written in the form \(\sec^2 x (a + b \sin 2x) e^{-4x}\), where \(a\) and \(b\) are constants.

(b) Hence find the exact \(x\)-coordinates of the two stationary points.

The curve \(y = e^x + 4e^{-2x}\) has one stationary point.

(i) Find the \(x\)-coordinate of this point.

(ii) Determine whether the stationary point is a maximum or a minimum point.

The equation of a curve is \(y = 2 \cos x + \sin 2x\). Find the \(x\)-coordinates of the stationary points on the curve for which \(0 < x < \pi\), and determine the nature of each of these stationary points.

The equation of a curve is \(y = \cos^3 x \sqrt{\sin x}\). It is given that the curve has one stationary point in the interval \(0 < x < \frac{1}{2}\pi\).

Find the \(x\)-coordinate of this stationary point, giving your answer correct to 3 significant figures.

The curve with equation \(y = xe^{1-2x}\) has one stationary point.

(a) Find the coordinates of this point.

(b) Determine whether the stationary point is a maximum or a minimum.

The diagram shows the curve \(y = \frac{\ln x}{x^4}\) and its maximum point \(M\).

Find the exact coordinates of \(M\).

The equation of a curve is \(y = e^{-5x} \tan^2 x\) for \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\).

Find the \(x\)-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate.

The equation of a curve is \(y = x^{-\frac{2}{3}} \ln x\) for \(x > 0\). The curve has one stationary point.

Find the exact coordinates of the stationary point.

A curve has equation \(y = \\cos x \\sin 2x\).

Find the \(x\)-coordinate of the stationary point in the interval \(0 < x < \frac{1}{2} \pi\), giving your answer correct to 3 significant figures.

The equation of a curve is \(y = x \arctan\left(\frac{1}{2}x\right)\).

(a) Find \(\frac{dy}{dx}\).

(b) The tangent to the curve at the point where \(x = 2\) meets the y-axis at the point with coordinates \((0, p)\). Find \(p\).

A curve has equation

\(y = \frac{2 - \tan x}{1 + \tan x}\).

Find the equation of the tangent to the curve at the point for which \(x = \frac{1}{4} \pi\), giving the answer in the form \(y = mx + c\) where \(c\) is correct to 3 significant figures.

The equation of a curve is \(y = x \sin 2x\), where \(x\) is in radians. Find the equation of the tangent to the curve at the point where \(x = \frac{1}{4} \pi\).

By differentiating \(\frac{1}{\cos x}\), show that the derivative of \(\sec x\) is \(\sec x \tan x\). Hence show that if \(y = \ln(\sec x + \tan x)\) then \(\frac{dy}{dx} = \sec x\).

The diagram shows the curve \(y = \sqrt{\left( \frac{1-x}{1+x} \right)}\).

(i) By first differentiating \(\frac{1-x}{1+x}\), obtain an expression for \(\frac{dy}{dx}\) in terms of \(x\). Hence show that the gradient of the normal to the curve at the point \((x, y)\) is \((1+x)\sqrt{1-x^2}\). [5]

(ii) The gradient of the normal to the curve has its maximum value at the point \(P\) shown in the diagram. Find, by differentiation, the \(x\)-coordinate of \(P\). [4]

The polynomial \(2x^3 + ax^2 + bx - 4\), where \(a\) and \(b\) are constants, is denoted by \(p(x)\). The result of differentiating \(p(x)\) with respect to \(x\) is denoted by \(p'(x)\). It is given that \((x + 2)\) is a factor of \(p(x)\) and of \(p'(x)\).

(i) Find the values of \(a\) and \(b\).

(ii) When \(a\) and \(b\) have these values, factorise \(p(x)\) completely.

Find the exact coordinates of the points on the curve \(y = \frac{x^2}{1 - 3x}\) at which the gradient of the tangent is equal to 8.

The curve \(y = \\sin(x + \frac{1}{3}\pi) \\cos x\) has two stationary points in the interval \(0 \leq x \leq \pi\).

(i) Find \(\frac{dy}{dx}\).

(ii) By considering the formula for \(\cos(A + B)\), show that, at the stationary points on the curve, \(\cos(2x + \frac{1}{3}\pi) = 0\).

(iii) Hence find the exact \(x\)-coordinates of the stationary points.

Find \(\frac{dy}{dx}\) in each of the following cases:

1. \(y = \ln(1 + \sin 2x)\),
2. \(y = \frac{\tan x}{x}\).

The equation of a curve is \(y = \frac{1 + e^{-x}}{1 - e^{-x}}\), for \(x > 0\).

(i) Show that \(\frac{dy}{dx}\) is always negative.

(ii) The gradient of the curve is equal to \(-1\) when \(x = a\). Show that \(a\) satisfies the equation \(e^{2a} - 4e^{a} + 1 = 0\). Hence find the exact value of \(a\).