## IB Math AA HL Paper 3 Question Bank

The IB Math AA HL Paper 3 Question Bank is a great resource for students who are preparing for their HL Paper 3 exam. The Question Bank contains a wide range of questions, from easy to challenging, that will help students get a feel for the types of questions they may see on the exam. In addition, the Question Bank includes detailed explanations for each question, so students can understand not only how to answer the question, but also why the answer is correct.

## 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) = 6a2x + 2b

Now we need to find the x values where the second derivative is equal to zero.
6a2x + 2b = 0
x = -b/(3a2)

Our Guide is written by counselors from Cambridge University for colleges like MIT and other Ivy League colleges.

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

i) Find the derivative of y2 = 2x3 + x
y2 = 2x3 + x
y = âˆš(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))/(2âˆš(2x3 + x))

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

Next, we need to set y’ equal to zero and solve for x:
(3x(2x2 + 1))/(2âˆš(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 = âˆš(2x3 + x) with respect to x:
y” = (6x(2x2 + 1) – 3(2x2 + 1)(2x))/ (2âˆš(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, Ï€/2] by f(x) = âˆ«x0 (1 + sin(t) dt.

a)
i) Determine the exact value of f(Ï€/4).
To determine the exact value of f(Ï€/4), we need to evaluate the definite integral of (1 + sin(t))2 from t = 0 to t = Ï€/4:
f(Ï€/4) = âˆ«(Ï€/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(Ï€/4) = [(1 + sin(t))3 / 3] from t = 0 to t = Ï€/4
Plugging in the limits of integration, we get:
f(Ï€/4) = [(1 + sin(Ï€/4))3 / 3] – [(1 + sin(0))3 / 3]
Since sin(0) = 0 and sin(Ï€/4) = âˆš2 / 2, we can simplify further:
f(Ï€/4) = [(1 + âˆš2 / 2)3/ 3] – (13 / 3)

ii) What is f(Ï€/8)?
To find f(Ï€/8), we need to evaluate the integral of f(x) from 0 to Ï€/8:
f(Ï€/8) = âˆ«(Ï€/8)0 (1 + sin(t)) dt
= [t – cos(t)] from 0 to Ï€/8
= (Ï€/8 – cos(Ï€/8)) – (0 – cos(0))
We can use a calculator to find that cos(Ï€/8) is approximately 0.92388, so we get:
f(Ï€/8) = Ï€/8 – 0.92388
Therefore, f(Ï€/8) â‰ˆ 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 [âˆ«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 â‰¤ x â‰¤ Ï€/2.
To show that f'(x) = (1 + sin(x))2 for 0 â‰¤ x â‰¤ Ï€/2, we need to differentiate the function f(x) with respect to x using the fundamental theorem of calculus:
f'(x) = d/dx âˆ«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 Ï€/2 from the left?
As x approaches Ï€/2 from the left, the integral âˆ«(0 to x) (1 + sin(t)) dt approaches the value of the integral from 0 to Ï€/2, which is f(Ï€/2) = Ï€/2. Therefore, we can conclude that the behavior of f(x) as x approaches Ï€/2 from the left is increasing and approaches the value Ï€/2.

ii) Determine the x-values where the function f(x) has local maximum or local minimum values on the interval [0, Ï€/2]. Justify your answer.
To determine the x-values where f(x) has local maximum or local minimum values on the interval [0, Ï€/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, Ï€/2] where f'(x) = 0.
Next, we check the endpoints of the interval [0, Ï€/2]:
f(0) = âˆ«00 (1 + sin(t))2 dt = 0
f(Ï€/2) = âˆ«(Ï€/2)2 (1 + sin(t))2 dt
Since f(0) = 0 and f(Ï€/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, Ï€/2].

d) Find the x-values where the function f(x) has points of inflection on the interval [0, Ï€/2]. Justify your answer.
To find the x-values where f(x) has points of inflection on the interval [0, Ï€/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 = Ï€/2, the points (Ï€/2, f(Ï€/2)) and (Ï€/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, Ï€/2]. Since f”(x) involves multiplication of trigonometric functions, it is defined for all x in the interval [0, Ï€/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, Ï€/2].
To determine the area between the graph of f(x) and the x-axis on the interval [0, Ï€/2], we need to integrate the absolute value of f(x) with respect to x on this interval.
âˆ«(0 to Ï€/2) |f(x)| dx = âˆ«(0 to Ï€/2) |âˆ«(0 to x) (1 + sin(t))2 dt| dx
We can use the properties of definite integrals to simplify this expression:
âˆ«(0 to Ï€/2) |âˆ«(0 to x) (1 + sin(t))2 dt| dx = âˆ«(0 to Ï€/2) |(1 + sin(x))2 * (x – 0)| dx
= âˆ«(0 to Ï€/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, Ï€/2]. We can use appropriate calculus techniques to evaluate this integral and find the exact area.

f) Evaluate the following definite integral: âˆ«20(x2 – 3x + 2) dx + âˆ«52(2x – 1) dx
First, let’s evaluate the integral on the interval [0, 2]:
âˆ«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]:
âˆ«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:
âˆ«20(x2 – 3x + 2) dx +Â  âˆ«52(2x – 1) dx = 2 + 18 = 20
Therefore, the value of the given definite integral is 20.

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