CREDZ TO CHILLICHAMPZ
Q1:
y=2x(3x-1)^{5}
a) find dy/dx
dy/dx = 2(3x-1)^{4}(18x-1)
b) Hence find the set of values for x for which dy/dx \leq 0
x = \frac {1}{3} , or x \leq \frac{1}{18}
Q2:
a) Simplify f(x) = \frac{6}{2x+5} + \frac{2}{2x-5} + \frac{60}{4x^2-25} , x > 4
Answer: \frac {8}{2x-5}
b) find f^-1(x) and state its domain
Answer: f^-1(x) = \frac{8+5x}{2x} , domain 0 < x < \frac{8}{3}
Q3:
V = 16000e^-kt + A . When t = 0, V = 17500 and when t=2, V = 13500
a) A = 1500 , because 17500 = 16000 + A
b) show that k = ln(\frac {2}{\sqrt{3}})
c) when V = 6000, t = 8.82 years (3 significant figures)
Q4:
The curve C has the equation y = e^-2x + x^2 - 3 . The point A is where the curve C crosses the y-axis.
a) Find an equation for the normal at the point A in the form y = mx + c
dy/dx = -2e^-2x + 2x . substitute x = 0 , dy/dx = -2 . Therefore gradient of the normal was \frac{1}{2} .
substitute x = 0 into y to get y-coordinate of A, which was -2. A(0, -2) so equation of line was y = \frac{1}{2} x - 2
b) The normal then intersects C again at the point B. Show that the x coordinate of B is a solution of x = \sqrt{1 + \frac{1}{2} x - e^-2x} .
c) Use this as an iterative formula with x_1 = 1
x_2 = 1.168 and x_3 = 1.220 (both to 3 decimal places)
Q5:
Graph of y = 2|5-x| + 3
a) Find the set of values of k such that f(x) = k , where k is a constant, has exactly one root.
Answer : k = 3 or k > 13
b) Solve the equation f(x) = \frac{1}{2} x + 10
Answers, x = \frac{6}{5} or x = \frac{34}{3}
c) The graph f(x) is transformed into the graph 4f(x-1) . The new coordinates of the vertex are (p, q) . State the values of p and q.
Answer: The original vertex of f(x) was (5, 3) . Shift one right and then stretch vertically factor 4. So p = 6 , q = 12
Q6:
y = \frac{ln(x^2 + 1)}{(x^2 + 1)}
a) find dy/dx
dy/dx = \frac{2x(1- ln(x^2 + 1)}{(x^2 + 1)^2}
b) Hence find the the stationary points
Answer: 2x = 0 , therefore x = 0 . When x = 0, y = 0 so (0, 0) is a stationary point.
Also, ln(x^2 + 1) = 1 . This simplifies to x^2 = e -1 . So x = \sqrt{e-1} or x = -\sqrt{e-1}
Both these points give a y coordinate of \frac{1}{e}
(0, 0) , (\sqrt{e-1}, \frac{1}{e}) , (-\sqrt{e-1}, \frac{1}{e})
Q7:
a) Solve the equation \frac{tan2x + tan32}{1-tan2xtan32} = 5
Answers, x = -66.65 , x = 23.35 (2 decimal places)
b) Show that tan(3\theta + 45) = \frac{tan3\theta + 1}{1-tan3\theta}
c) Solve the equation (1- tan3\theta)(tan\theta + 28) = (tan3\theta + 1)
Answers, \theta = 36.5 , \theta = 126.5
Q8:
a) Show that d/d\theta (sec\theta) = sec\theta tan\theta
b) Find dy/dx in terms of x
dy/dx = \frac{1}{x\sqrt{(lnx)^4 - (lnx)^2}}
Q9:
a) find Rsin(x-\alpha)
Answer : \sqrt{5} sin(x-1.107)
b) Minimum value for M(\theta) = 85
Occurred when \theta = 2.678 (3 decimal places) Exact value was \frac{\pi}{2} + arctan(2)
c) Maximum value for N(\theta ) = \frac{30}{5} = 6
Occured when \theta= 5.266 (3 decimal places) Exact value was \frac{3\pi + arctan(2)}{2}