IB Math AA HL Paper 3 Question Bank

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)

d) Determine the concavity of the function y = f(x) for any non-negative values of n.
To determine the concavity of the function, we need to analyze the second derivative of y = f(x).
First, find the first derivative of y = f(x) using the power rule:
dy/dx = n(ax³ + bx² + cx + d)^(n-1) * (3ax² + 2bx + c)
Next, find the second derivative by differentiating the first derivative with respect to x:
d²y/dx² = (n-1)(ax³ + bx² + cx + d)^(n-2) * (3ax² + 2bx + c)² + n(ax³ + bx² + cx + d)^(n-1) * (6ax + 2b)
To determine the concavity of the function, we need to evaluate the sign of the second derivative at different points.
If d²y/dx² > 0, the function is concave up (convex) at that point.
If d²y/dx² < 0, the function is concave down (concave) at that point.
If d²y/dx² = 0, further investigation is needed to determine the concavity.
By analyzing the sign of the second derivative at various points, we can determine the concavity of the function y = f(x) for any non-negative values of n.

2.) The function f(x) is defined on the interval [0, π/2] by f(x) = ∫^x₀ (1 + sin(t) dt.

a)
i) Determine the exact value of f(π/4).
To determine the exact value of f(π/4), we need to evaluate the definite integral of (1 + sin(t))² from t = 0 to t = π/4:
f(π/4) = ∫^(π/4)₀(1 + sin(t))² dt
Using the power rule for integration and the fact that sin(t) is a constant when integrating with respect to t, we get:
f(π/4) = [(1 + sin(t))³ / 3] from t = 0 to t = π/4
Plugging in the limits of integration, we get:
f(π/4) = [(1 + sin(π/4))³ / 3] – [(1 + sin(0))3 / 3]
Since sin(0) = 0 and sin(π/4) = √2 / 2, we can simplify further:
f(π/4) = [(1 + √2 / 2)³/ 3] – (13 / 3)

ii) What is f(π/8)?
To find f(π/8), we need to evaluate the integral of f(x) from 0 to π/8:
f(π/8) = ∫^(π/8)₀ (1 + sin(t)) dt
= [t – cos(t)] from 0 to π/8
= (π/8 – cos(π/8)) – (0 – cos(0))
We can use a calculator to find that cos(π/8) is approximately 0.92388, so we get:
f(π/8) = π/8 – 0.92388
Therefore, f(π/8) ≈ 0.0292.

b)
i)What is f'(x)?
To find the derivative of f(x), we can use the fundamental theorem of calculus and differentiate the integral with respect to x:
f'(x) = d/dx [∫^x₀(1 + sin(t)) dt]
= (1 + sin(x))
Therefore, f'(x) = 1 + sin(x).

ii) Show that f'(x) = (1 + sin(x))2 for 0 ≤ x ≤ π/2.
To show that f'(x) = (1 + sin(x))² for 0 ≤ x ≤ π/2, we need to differentiate the function f(x) with respect to x using the fundamental theorem of calculus:
f'(x) = d/dx ∫^x₀(1 + sin(t))² dt
Applying the fundamental theorem of calculus, we get:
f'(x) = (1 + sin(x))²

c)
i) What is the behavior of f(x) as x approaches π/2 from the left?
As x approaches π/2 from the left, the integral ∫(0 to x) (1 + sin(t)) dt approaches the value of the integral from 0 to π/2, which is f(π/2) = π/2. Therefore, we can conclude that the behavior of f(x) as x approaches π/2 from the left is increasing and approaches the value π/2.

ii) Determine the x-values where the function f(x) has local maximum or local minimum values on the interval [0, π/2]. Justify your answer.
To determine the x-values where f(x) has local maximum or local minimum values on the interval [0, π/2], we need to find the critical points of f(x) and test their nature using the first and second derivative tests.
First, we find the critical points by setting f'(x) = 0:
(1 + sin(x))² = 0
Since (1 + sin(x))² is always non-negative, there are no critical points on the interval [0, π/2] where f'(x) = 0.
Next, we check the endpoints of the interval [0, π/2]:
f(0) = ∫⁰₀ (1 + sin(t))2 dt = 0
f(π/2) = ∫^(π/2)₂ (1 + sin(t))² dt
Since f(0) = 0 and f(π/2) is positive (since (1 + sin(t))2 is always positive), there are no local maximum or local minimum values of f(x) on the interval [0, π/2].

d) Find the x-values where the function f(x) has points of inflection on the interval [0, π/2]. Justify your answer.
To find the x-values where f(x) has points of inflection on the interval [0, π/2], we need to find the values of x where f”(x) = 0 or is undefined.
First, we find the second derivative of f(x):
f”(x) = d/dx [(1 + sin(x))² ] = 2(1 + sin(x))cos(x)
Setting f”(x) = 0, we get:
2(1 + sin(x))cos(x) = 0
This equation is satisfied when cos(x) = 0 or sin(x) = -1. Since cos(x) = 0 has solutions at x = π/2, the points (π/2, f(π/2)) and (π/2, f(0)) are potential points of inflection.
Next, we need to check if f”(x) is undefined for any x in the interval [0, π/2]. Since f”(x) involves multiplication of trigonometric functions, it is defined for all x in the interval [0, π/2]. Therefore, there are no points of inflection where f”(x) is undefined.

e) Determine the area between the graph of f(x) and the x-axis on the interval [0, π/2].
To determine the area between the graph of f(x) and the x-axis on the interval [0, π/2], we need to integrate the absolute value of f(x) with respect to x on this interval.
∫(0 to π/2) |f(x)| dx = ∫(0 to π/2) |∫(0 to x) (1 + sin(t))² dt| dx
We can use the properties of definite integrals to simplify this expression:
∫(0 to π/2) |∫(0 to x) (1 + sin(t))² dt| dx = ∫(0 to π/2) |(1 + sin(x))² * (x – 0)| dx
= ∫(0 to π/2) (1 + sin(x))² * x dx
This is the integral that represents the area between the graph of f(x) and the x-axis on the interval [0, π/2]. We can use appropriate calculus techniques to evaluate this integral and find the exact area.

f) Evaluate the following definite integral: ∫²₀(x² – 3x + 2) dx + ∫⁵₂(2x – 1) dx
First, let’s evaluate the integral on the interval [0, 2]:
∫²₀(x² – 3x + 2) dx
To find the antiderivative of the integrand, we can apply the power rule:
= (1/3)x² – (3/2)x² + 2x | [0, 2]
Now, substitute the upper and lower limits of integration:
= [(1/3)(2)³ – (3/2)(2)² + 2(2)] – [(1/3)(0)³ – (3/2)(0)² + 2(0)]
Simplifying further:
= (8/3 – 6 + 4) – (0 – 0 + 0)
= 6/3
= 2
Next, let’s evaluate the integral on the interval [2, 5]:
∫⁵₂(2x – 1) dx
To find the antiderivative of the integrand, we can apply the power rule:
= x² – x | [2, 5]
Now, substitute the upper and lower limits of integration:
= (5² – 5) – (2² – 2)
= (25 – 5) – (4 – 2)
= 20 – 2
= 18
Finally, add the results from the two integrals:
∫²₀(x² – 3x + 2) dx +  ∫⁵₂(2x – 1) dx = 2 + 18 = 20
Therefore, the value of the given definite integral is 20.