Differential Calculus Practice Questions
ATAR (Australia) · ATAR Mathematics Methods · 146 free MCQs with instant results and detailed explanations.
146
Total
32
Easy
77
Medium
37
Hard
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Sample Questions from Differential Calculus
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Q1
Easy
What is the derivative of the function f(x) = 3x^2 + 5x - 7?
A.
6x + 5
B.
3x + 5
C.
6x - 5
D.
3x^2 + 5
Show Answer & Explanation
Correct Answer: A
The derivative of the function f(x) is found by applying the power rule. For each term, we multiply by the exponent and reduce the exponent by one. Thus, the derivative is 6x from 3x^2 and 5 from 5x, giving us 6x + 5.
Q2
Easy
A function f is defined as f(x) = 2x + 3. What is the slope of the tangent line at any point on this curve?
A.
2
B.
3
C.
0
D.
1
Show Answer & Explanation
Correct Answer: A
The slope of the tangent line to the curve is given by the derivative of the function. For f(x) = 2x + 3, the derivative f'(x) = 2, which means the slope is constant at 2 for all points on the curve.
Q3
Easy
If the function g(x) = x^2 + 4x + 4, what is the critical point of g(x)?
A.
-2
B.
0
C.
2
D.
4
Show Answer & Explanation
Correct Answer: A
To find the critical point, we first find the derivative g'(x) = 2x + 4 and set it to zero. Solving 2x + 4 = 0 gives x = -2 as the critical point.
Q4
Medium
If a function is defined as g(x) = x^3 - 3x^2 + 4, what is the critical point of the function?
A.
x = 0
B.
x = 2
C.
x = 1
D.
x = 3
Show Answer & Explanation
Correct Answer: B
To find the critical points, first take the derivative g'(x) = 3x^2 - 6x and set it to zero: 3x^2 - 6x = 0, which factors to 3x(x - 2) = 0. Hence, x = 0 or x = 2. The critical point we are looking for is x = 2.
Q5
Medium
A function h(x) has a minimum at x = -1. If h'(x) < 0 for x < -1 and h'(x) > 0 for x > -1, which of the following statements is true?
A.
h(x) is increasing at x = -1.
B.
h(x) is decreasing at x = -1.
C.
h(x) has an inflection point at x = -1.
D.
h(x) is constant at x = -1.
Show Answer & Explanation
Correct Answer: A
Since h'(x) changes from negative to positive as x passes through -1, this indicates that h(x) is increasing after reaching the minimum point at x = -1.
Q6
Medium
If the function p(x) = 2x^2 - 8x + 10 represents the height of an object, what is the maximum height reached by the object?
A.
2
B.
6
C.
10
D.
8
Show Answer & Explanation
Correct Answer: C
The maximum height of a parabola in the form p(x) = ax^2 + bx + c occurs at x = -b/(2a). Here, a = 2 and b = -8, so x = -(-8)/(2*2) = 2. Substituting x = 2 back into p(x) gives p(2) = 2(2^2) - 8(2) + 10 = 10.
Q7
Medium
If f(x) = x^3 - 6x^2 + 9x, what is the critical point of f?
A.
x = 0
B.
x = 3
C.
x = 2
D.
x = 1
Show Answer & Explanation
Correct Answer: B
To find critical points, we set the derivative f'(x) = 3x^2 - 12x + 9 to zero. Solving gives x = 3 as a critical point.
Q8
Hard
Consider the function f(x) = 3x^3 - 12x^2 + 9x. What is the x-coordinate of the point where the function has a local maximum?
A.
1
B.
2
C.
3
D.
4
Show Answer & Explanation
Correct Answer: B
To find the local maximum, we first find the derivative f'(x) = 9x^2 - 24x + 9. Setting f'(x) = 0 gives us the critical points. Solving 9(x^2 - 24/9x + 1) = 0 leads us to x = 2, which is a local maximum after checking the second derivative.
Q9
Hard
A particle moves along a straight line with its position given by the function s(t) = 4t^3 - 12t^2 + 9t, where s is in meters and t is in seconds. What is the acceleration of the particle at t = 2 seconds?
A.
24 m/s²
B.
0 m/s²
C.
12 m/s²
D.
36 m/s²
Show Answer & Explanation
Correct Answer: C
To find the acceleration, we first need to find the velocity by differentiating the position function s(t). The first derivative gives us v(t) = 12t^2 - 24t + 9. Differentiating this again gives us the acceleration function a(t) = 24t - 24. Plugging in t = 2 results in a(2) = 24(2) - 24 = 12 m/s².
Q10
Hard
Consider the function f(x) = x^3 - 6x^2 + 9x. What is the x-coordinate of the point where the function has a local minimum?
A.
1
B.
2
C.
3
D.
4
Show Answer & Explanation
Correct Answer: B
To find the local minimum, we first find the derivative f'(x) = 3x^2 - 12x + 9. Setting the derivative to zero gives us the critical points. Solving 3x^2 - 12x + 9 = 0 yields x = 2 as a local minimum (second derivative test confirms this).
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Differential Calculus — ATAR (Australia) ATAR Mathematics Methods Practice Questions Online
This page contains 146 practice MCQs for the chapter Differential Calculus in ATAR (Australia) ATAR Mathematics Methods. The questions are organized by difficulty — 32 easy, 77 medium, 37 hard — so you can choose the right level for your preparation.
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