(a) The table shows some values for \(y=2x^{3}-4x^{2}+3.\)
(i) Complete the table.
(ii) On the grid, draw the graph of \(y=2x^{3}-4x^{2}+3\) for \(-1\leq x\leq 2.\)
(iii) Use your graph to solve the equation \(2x^{3}-4x^{2}+3=1.5\)
(iv) The equation \(2x^{3}-4x^{2}+3=k\) has only one solution for \(-1\leq k\leq 2.\)
Write down a possible integer value of k.
(b) 
(i) On the grid, draw the tangent to the curve at x = 1.
(ii) Use your tangent to estimate the gradient of the curve at x = 1.
(iii) Write down the equation of your tangent in the form y=mx+c
▶️ Answer/Explanation
(a)(i) Ans: 3, 2.25, 1
Substitute \(x = 0.5\) into \(y = 2x^3 – 4x^2 + 3\) to get \(y = 2(0.125) – 4(0.25) + 3 = 2.25\). Similarly, for \(x = 1\), \(y = 1\).
(ii) Plot the points and draw a smooth cubic curve.
(iii) Ans: \(x \approx -0.55\), \(0.8\), \(1.75\)
Find where the curve intersects \(y = 1.5\). Approximate solutions are \(-0.55\), \(0.8\), and \(1.75\).
(iv) Ans: \(-1\) (or any integer in \(-3\) to \(0\))
For only one solution, choose \(k\) where the horizontal line \(y = k\) touches the curve once (e.g., \(k = -1\)).
(b)(i) Draw a tangent at \(x = 1\) touching the curve.
(ii) Ans: \(5\) (range \(4.4\) to \(5.6\))
Estimate gradient using \(\frac{\Delta y}{\Delta x}\) from the tangent.
(iii) Ans: \(y = 5x – 2\) (approximate)
Using point \((1, 3)\) and gradient \(5\), the equation is \(y – 3 = 5(x – 1)\).
(a) Solve the equation tan x = 11.43 for \(0\leqslant x\leqslant 360^{\circ}\)
(b) Sketch the curve \(y=x^{3}-4x\)
(c) A curve has equation \(y=x^{3}+ax+b\)
The stationary points of the curve have coordinates (2, k) and (-2, 10 – k).
Work out the value of a, the value of b and the value of k.
▶️ Answer/Explanation
(a) Ans: 85.0°, 265.0°
First, find the reference angle: \(x = \tan^{-1}(11.43) \approx 85.0°\).
Since tan is positive in 1st and 3rd quadrants, the solutions are \(x = 85.0°\) and \(x = 180° + 85.0° = 265.0°\).
(b)
The curve \(y = x^3 – 4x\) is a cubic with roots at \(x = -2, 0, 2\).
It has a local maximum at \(x = -1.15\) and a local minimum at \(x = 1.15\).
(c) Ans: a = –12, b = 5, k = –11
Find derivative: \(\frac{dy}{dx} = 3x^2 + a\). At stationary points, \(3(2)^2 + a = 0\) gives \(a = -12\).
Substitute (2, k) and (-2, 10 – k) into \(y = x^3 – 12x + b\) and solve to find \(b = 5\) and \(k = -11\).