IB Math AA HL Paper 3 Question Bank

1.) The family of curves y = f(x) = (ax³ + bx² + cx + d)ⁿ, where a, b, c, d, and n are constants, is given.

a) Determine the following:

i) the domain and range of the function y = f(x) for any non-negative values of n.
The domain of the function y = f(x) is all real numbers because the polynomial is defined for all real values of x. The range of the function y = f(x) is all non-negative real numbers because the power of the polynomial is non-negative.

ii) Determine the x-intercepts of the function y = f(x) for any non-negative values of n.
To find the x-intercepts of the function, we set y = 0 and solve for x. Therefore, we need to find the roots of the polynomial ax³ + bx² + cx + d = 0. We can use the factor theorem or the synthetic division to find the roots of the polynomial.

iii) Determine the y-intercept of the function y = f(x) for any non-negative values of n.
The y-intercept of the function y = f(x) is (0,dⁿ) since the function is equal to dⁿ when x = 0.

b) For n = 2, determine the inflection point of the function y = f(x).
For n = 2, the function becomes y = (ax³ + bx² + cx + d)². To find the inflection point, we need to find the second derivative of the function, which is:
f”(x) = 6a^2x + 2b

Now we need to find the x values where the second derivative is equal to zero.
6a^2x + 2b = 0
x = -b/(3a²)

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c) Next, Consider the curve, y² = 2x³ + x

i) Find the derivative of y² = 2x³ + x
y² = 2x³ + x
y = √(2x³ + x)
Next, we can take the derivative of this expression using the chain rule. The chain rule states that if we have a function u(x) = f(g(x)), then the derivative of u(x) with respect to x is given by:
du/dx = df/dg * dg/dx

In this case, we have u(x) = sqrt(2x³ + x) and we have to find the derivative of this function with respect to x
du/dx = (1/2)(2x³ + x)^(-1/2)(6x² + 1) = (3x(2x² + 1))/(2√(2x³ + x))

So the derivative of y = sqrt(2x³ + x) with respect to x is (3x(2x² + 1))/(2√(2x³ + x))

ii) Hence, find the local minima or maxima of y² = 2x³ + x
To find the local maxima or minima of y² = 2x³ + x, we need to find the critical points of the function
y = √(2x³ + x).

A critical point of a function is a point where the derivative of the function is equal to zero or does not exist.

First, we need to find the derivative of y = √(2x³ + x) with respect to x:
y’ = (3x(2x² + 1))/(2√(2x³ + x))

Next, we need to set y’ equal to zero and solve for x:
(3x(2x² + 1))/(2√(2x³ + x)) = 0
3x(2x² + 1) = 0
x = 0

Now, we need to check if x = 0 is a local maximum or minimum. To do this, we need to find the second derivative of y = √(2x³ + x) with respect to x:
y” = (6x(2x² + 1) – 3(2x² + 1)(2x))/ (2√(2x³ + x))²
We need to evaluate y” at x = 0, we get
y”(0) = -3/2, which is negative and that means that x = 0 is a local minimum.

So the local minima of y² = 2x³ + x is (0, 0)

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