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NCEA Level 3 Calculus study materials
NCEA Level 3 Calculus study resources

NCEA Level 3 Calculus practice questions

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Question 1Differentiation and integration (calculus methods)

Differentiate f(x) = 3x^4 - 5x^2 + 7x - 2 with respect to x.

  1. 12x^3 - 10x
  2. 12x^4 - 10x^2 + 7x
  3. 12x^3 - 10x + 7 Correct answer.
  4. 3x^3 - 5x + 7

Explanation

Apply the power rule term by term: the derivative of 3x^4 is 12x^3, of -5x^2 is -10x, of 7x is 7, and the constant -2 differentiates to zero. Combining these gives f'(x) = 12x^3 - 10x + 7. Constant terms always vanish under differentiation, which is why the -2 does not appear.

Question 2Differentiation and integration (calculus methods)

Find the gradient of the curve y = x^3 - 4x at the point where x = 2.

  1. 0
  2. 8 Correct answer.
  3. 12
  4. 4

Explanation

Differentiate to get dy/dx = 3x^2 - 4, then substitute x = 2 to obtain 3(4) - 4 = 8. The gradient of the curve at a point equals the value of the derivative there, so the answer is 8.

Question 3Differentiation and integration (calculus methods)

Evaluate the integral of (6x^2 + 4x) dx.

  1. 2x^3 + 2x^2 + C Correct answer.
  2. 2x^3 + 4x^2 + C
  3. 6x^3 + 4x^2 + C
  4. 12x + 4 + C

Explanation

Integrate term by term using the rule that the integral of x^n is x^(n+1)/(n+1). Thus 6x^2 integrates to 2x^3 and 4x integrates to 2x^2, giving 2x^3 + 2x^2 + C. The constant of integration C must be included for an indefinite integral.

Question 4Differentiation and integration (calculus methods)

What is the derivative of y = sin(3x) with respect to x?

  1. 3cos(3x) Correct answer.
  2. -3sin(3x)
  3. -3cos(3x)
  4. cos(3x)

Explanation

The chain rule applies because the sine function has an inner function 3x. The derivative of sin(u) is cos(u) multiplied by du/dx, so d/dx sin(3x) = cos(3x) times 3 = 3cos(3x).

Question 5Differentiation and integration (calculus methods)

Differentiate y = e^(5x).

  1. 5xe^(5x)
  2. 5e^(5x) Correct answer.
  3. e^(5x)
  4. e^(5x)/5

Explanation

For an exponential with a linear inner function, the chain rule gives d/dx e^(kx) = k e^(kx). Here k = 5, so the derivative is 5e^(5x). The exponential itself is unchanged; only the inner derivative multiplies the front.

Question 6Differentiation and integration (calculus methods)

The derivative of ln(x) with respect to x, for x > 0, is:

  1. 1/ln(x)
  2. 1/x Correct answer.
  3. ln(x)/x
  4. x

Explanation

The natural logarithm has the standard derivative d/dx ln(x) = 1/x for x > 0. This follows from the inverse relationship with e^x, whose derivative is itself, giving the reciprocal result.

Question 7Differentiation and integration (calculus methods)

Find the value of the definite integral of 2x dx from x = 1 to x = 3.

  1. 9
  2. 4
  3. 8 Correct answer.
  4. 6

Explanation

The antiderivative of 2x is x^2. Evaluating from 1 to 3 gives 3^2 - 1^2 = 9 - 1 = 8. The definite integral equals the difference of the antiderivative at the upper and lower limits.

Question 8Differentiation and integration (calculus methods)

A curve has derivative dy/dx = 3x^2 - 2. Given that the curve passes through (1, 4), find the equation of the curve.

  1. y = x^3 - 2x - 5
  2. y = x^3 - 2x + 5 Correct answer.
  3. y = x^3 - 2x + 4
  4. y = 6x + C

Explanation

Integrating 3x^2 - 2 gives y = x^3 - 2x + C. Substituting the point (1, 4) gives 4 = 1 - 2 + C, so C = 5. The curve is therefore y = x^3 - 2x + 5.

Question 9Differentiation and integration (calculus methods)

Differentiate y = (2x + 1)^4 using the chain rule.

  1. 8(2x + 1)^3 Correct answer.
  2. (2x + 1)^3
  3. 4(2x + 1)^3
  4. 8(2x + 1)^4

Explanation

The chain rule gives d/dx (2x + 1)^4 = 4(2x + 1)^3 times the derivative of the inner function 2x + 1, which is 2. Multiplying gives 8(2x + 1)^3.

Question 10Differentiation and integration (calculus methods)

Find the equation of the tangent to the curve y = x^2 - 3x + 1 at the point where x = 2.

  1. y = -x + 3
  2. y = 4x - 7
  3. y = x + 1
  4. y = x - 3 Correct answer.

Explanation

At x = 2, y = 4 - 6 + 1 = -1, and dy/dx = 2x - 3 gives a gradient of 1. Using point-slope form: y - (-1) = 1(x - 2), so y = x - 3. The tangent line touches the curve at (2, -1) with slope 1.

Question 11Differentiation and integration (calculus methods)

Differentiate y = x^2 ln(x) with respect to x, for x > 0.

  1. 2x + 1/x
  2. x + ln(x)
  3. 2x ln(x)
  4. 2x ln(x) + x Correct answer.

Explanation

Use the product rule with u = x^2 and v = ln(x). Then u' = 2x and v' = 1/x, so dy/dx = 2x ln(x) + x^2 times (1/x) = 2x ln(x) + x. Simplifying the second term gives x.

Question 12Differentiation and integration (calculus methods)

Use the quotient rule to differentiate y = (3x + 1)/(x - 2), for x ≠ 2.

  1. (3x - 7)/(x - 2)^2
  2. 7/(x - 2)^2
  3. -7/(x - 2)
  4. -7/(x - 2)^2 Correct answer.

Explanation

The quotient rule gives dy/dx = [(3)(x - 2) - (3x + 1)(1)]/(x - 2)^2. Expanding the numerator: 3x - 6 - 3x - 1 = -7, so dy/dx = -7/(x - 2)^2. The negative sign arises because the denominator grows faster than the numerator.

Question 13Differentiation and integration (calculus methods)

The curve y = x^3 - 3x has a local maximum and a local minimum. At which x-values do these stationary points occur?

  1. x = 3 and x = -3
  2. x = -1 and x = 1 Correct answer.
  3. x = 1 only
  4. x = 0 and x = 2

Explanation

Set dy/dx = 3x^2 - 3 = 0, giving x^2 = 1, so x = -1 or x = 1. The second derivative is 6x, which is negative at x = -1 (local maximum) and positive at x = 1 (local minimum). Both stationary points are required.

Question 14Differentiation and integration (calculus methods)

Determine the nature of the stationary point of y = x^4 at x = 0.

  1. It is a local minimum. Correct answer.
  2. It is a local maximum.
  3. It is both a maximum and a minimum.
  4. It is a point of inflection.

Explanation

The derivative 4x^3 is zero at x = 0, and the second derivative 12x^2 is also zero there, so the second-derivative test is inconclusive. Checking the first derivative sign: 4x^3 is negative for x < 0 and positive for x > 0, so the function decreases then increases, giving a local minimum at the origin.

Question 15Differentiation and integration (calculus methods)

Find the derivative of y = cos(x^2).

  1. -sin(x^2)
  2. -2x sin(x^2) Correct answer.
  3. 2x sin(x^2)
  4. -2x cos(x^2)

Explanation

The chain rule applies with outer function cosine and inner function x^2. The derivative of cos(u) is -sin(u) times du/dx, so d/dx cos(x^2) = -sin(x^2) times 2x = -2x sin(x^2).

Question 16Differentiation and integration (calculus methods)

Evaluate the definite integral of (3x^2 + 1) dx from x = 0 to x = 2.

  1. 10 Correct answer.
  2. 8
  3. 6
  4. 12

Explanation

The antiderivative of 3x^2 + 1 is x^3 + x. Evaluating from 0 to 2 gives (8 + 2) - (0 + 0) = 10. The definite integral equals the antiderivative at the upper limit minus its value at the lower limit.

Question 17Differentiation and integration (calculus methods)

The gradient function of a curve is dy/dx = 6x - 4. Given the curve passes through (2, 5), find y.

  1. y = 3x^2 - 4x + 1 Correct answer.
  2. y = 6x^2 - 4x + 1
  3. y = 3x^2 - 4x - 1
  4. y = 3x^2 - 4x + 5

Explanation

Integrating 6x - 4 gives y = 3x^2 - 4x + C. Substituting (2, 5): 5 = 12 - 8 + C = 4 + C, so C = 1. The curve is y = 3x^2 - 4x + 1.

Question 18Differentiation and integration (calculus methods)

Find the x-coordinate of the point on the curve y = x^2 - 4x + 7 where the tangent is parallel to the line y = 2x + 3.

  1. x = 4
  2. x = 2
  3. x = 3 Correct answer.
  4. x = 1

Explanation

Parallel lines have equal gradients, and the line y = 2x + 3 has gradient 2. Setting dy/dx = 2x - 4 = 2 gives 2x = 6, so x = 3. At this x-value the tangent has the same slope as the given line.

Question 19Differentiation and integration (calculus methods)

Differentiate y = (x^2 + 1)^3 (x - 1) with respect to x.

  1. 6x(x^2 + 1)^2
  2. 6x(x^2 + 1)^2 (x - 1) + (x^2 + 1)^3 Correct answer.
  3. 3(x^2 + 1)^2 (x - 1) + (x^2 + 1)^3
  4. 3(x^2 + 1)^2 (x - 1)

Explanation

Apply the product rule with u = (x^2 + 1)^3 and v = x - 1. Then u' = 3(x^2 + 1)^2 times 2x = 6x(x^2 + 1)^2 by the chain rule, and v' = 1. The derivative is u'v + uv' = 6x(x^2 + 1)^2 (x - 1) + (x^2 + 1)^3.

Question 20Differentiation and integration (calculus methods)

Find the area enclosed between the curve y = x^2 and the line y = 4.

  1. 8 square units
  2. 32/3 square units Correct answer.
  3. 64/3 square units
  4. 16/3 square units

Explanation

The curves meet where x^2 = 4, i.e. x = -2 and x = 2. The enclosed area is the integral from -2 to 2 of (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = (8 - 8/3) - (-8 + 8/3) = 32/3. This counts the region between the line and the parabola.

Question 21Differentiation and integration (calculus methods)

A curve is defined implicitly by x^2 + y^2 = 25. Find dy/dx at the point (3, 4).

  1. -3/4 Correct answer.
  2. 3/4
  3. -4/3
  4. 4/3

Explanation

Differentiating implicitly gives 2x + 2y(dy/dx) = 0, so dy/dx = -x/y. Substituting (3, 4) gives dy/dx = -3/4. The negative sign reflects that the circle's tangent slopes downward at that point in the first quadrant.

Question 22Differentiation and integration (calculus methods)

The derivative of a function is f'(x) = x^2 - 4x + 3. On which interval is f increasing?

  1. x < 1 or x > 3 Correct answer.
  2. x < 2
  3. 1 < x < 3
  4. x > 2

Explanation

A function increases where its derivative is positive. Factorising gives f'(x) = (x - 1)(x - 3), which is positive when x < 1 or x > 3 and negative between 1 and 3. Hence f is increasing on the outer intervals.

Question 23Differentiation and integration (calculus methods)

Using the substitution u = x^2 + 1, evaluate the integral of 2x(x^2 + 1)^3 dx.

  1. (x^2 + 1)^4/4 + C Correct answer.
  2. 2x(x^2 + 1)^4/4 + C
  3. (x^2 + 1)^3/3 + C
  4. (x^2 + 1)^4 + C

Explanation

With u = x^2 + 1, du = 2x dx, so the integral becomes the integral of u^3 du = u^4/4 + C. Substituting back gives (x^2 + 1)^4/4 + C. The factor 2x in the integrand is exactly du, which is why the substitution works cleanly.

Question 24Differentiation and integration (calculus methods)

A rectangle has perimeter 20 cm. Using calculus, find the maximum area of the rectangle.

  1. 50 cm^2
  2. 25 cm^2 Correct answer.
  3. 20 cm^2
  4. 100 cm^2

Explanation

Let the sides be x and 10 - x, so area A = x(10 - x) = 10x - x^2. Setting dA/dx = 10 - 2x = 0 gives x = 5, and the second derivative -2 confirms a maximum. The maximum area is 5 times 5 = 25 cm^2, a square.

Question 25Differentiation and integration (calculus methods)

The curve y = x^3 + ax^2 + bx has a stationary point at x = 2 with value y = 4. Find the values of a and b.

  1. a = 3, b = 0
  2. a = 0, b = -3
  3. a = -3, b = 4
  4. a = -3, b = 0 Correct answer.

Explanation

The point (2, 4) lies on the curve, so 8 + 4a + 2b = 4, giving 2a + b = -2. Stationarity at x = 2 gives 12 + 4a + b = 0, so 4a + b = -12. Subtracting yields 2a = -10, so a = -3 and then b = 0. Both conditions are needed to fix two unknowns.

Question 26Differentiation and integration (calculus methods)

The derivative of e^(3x) with respect to x is:

  1. 3e^(3x) Correct answer.
  2. e^(3x)
  3. 3xe^(3x)
  4. e^(3x)/3

Explanation

The chain rule applied to e^(u) with u = 3x gives d/dx e^(3x) = e^(3x) times du/dx = 3e^(3x). The inner derivative 3 multiplies the exponential rather than dividing it or appearing as an exponent.

Question 27Differentiation and integration (calculus methods)

The derivative of sin(5x) with respect to x is:

  1. cos(5x)
  2. 5sin(5x)
  3. -5cos(5x)
  4. 5cos(5x) Correct answer.

Explanation

Differentiating sin(u) gives cos(u) times du/dx. With u = 5x, du/dx = 5, so the derivative is 5cos(5x). The sign stays positive because the derivative of sine is positive cosine.

Question 28Differentiation and integration (calculus methods)

Evaluate the integral of (4x^3 + 2x) dx from x = 0 to x = 1.

  1. 3
  2. 5
  3. 1
  4. 2 Correct answer.

Explanation

An antiderivative is x^4 + x^2. Evaluating at x = 1 gives 1 + 1 = 2, and at x = 0 gives 0, so the definite integral equals 2.

Question 29Differentiation and integration (calculus methods)

The derivative of ln(4x) with respect to x, for x > 0, is:

  1. 1/(4x)
  2. 1/x Correct answer.
  3. 4/x
  4. 4ln(x)

Explanation

By the chain rule, d/dx ln(4x) = (1/(4x)) times 4 = 1/x. Equivalently, ln(4x) = ln 4 + ln x, and ln 4 is constant, so the derivative is 1/x.

Question 30Differentiation and integration (calculus methods)

The gradient of the curve y = x^2 e^x at x = 0 is:

  1. 0 Correct answer.
  2. 2
  3. e
  4. 1

Explanation

Using the product rule, dy/dx = 2x e^x + x^2 e^x = x e^x (2 + x). At x = 0 this factor x makes the whole expression zero, so the gradient is 0.

Question 31Differentiation and integration (calculus methods)

The derivative of tan(x) with respect to x is:

  1. -cosec^2(x)
  2. cot(x)
  3. sec(x)tan(x)
  4. sec^2(x) Correct answer.

Explanation

A standard result from the quotient sin(x)/cos(x) is that d/dx tan(x) = sec^2(x). This can also be written as 1/cos^2(x).

Question 32Differentiation and integration (calculus methods)

Find the integral of 1/x dx, for x > 0.

  1. ln|x| + C Correct answer.
  2. -1/x^2 + C
  3. 1/(2x^2) + C
  4. x ln(x) - x + C

Explanation

The antiderivative of 1/x is the natural logarithm of the absolute value of x, plus a constant of integration. Differentiating ln|x| returns 1/x, confirming the result.

Question 33Differentiation and integration (calculus methods)

A curve has gradient function dy/dx = 3x^2 - 6x. The x-coordinates of its stationary points are:

  1. x = 0 and x = 2 Correct answer.
  2. x = 0 and x = -2
  3. x = 3 and x = 6
  4. x = 1 only

Explanation

Stationary points occur where dy/dx = 0, so 3x^2 - 6x = 3x(x - 2) = 0. This gives x = 0 and x = 2.

Question 34Differentiation and integration (calculus methods)

Using the product rule, differentiate y = x^3 e^(2x) with respect to x.

  1. x^2 e^(2x)(3 + 2x) Correct answer.
  2. 3x^2 e^(2x)
  3. 6x e^(2x)
  4. x^3 e^(2x) + 2e^(2x)

Explanation

The product rule gives dy/dx = 3x^2 e^(2x) + x^3 times 2e^(2x). Factoring out x^2 e^(2x) leaves x^2 e^(2x)(3 + 2x).

Question 35Differentiation and integration (calculus methods)

Find the derivative of y = ln(x^2 + 1) with respect to x.

  1. 2x/(x^2 + 1) Correct answer.
  2. (x^2 + 1)/(2x)
  3. 1/(x^2 + 1)
  4. 2x ln(x^2 + 1)

Explanation

The chain rule gives dy/dx = (1/(x^2 + 1)) times the derivative of the inside, 2x. So the derivative is 2x/(x^2 + 1).

Question 36Differentiation and integration (calculus methods)

Evaluate the integral of sin(2x) dx from x = 0 to x = pi/4.

  1. 0
  2. 2
  3. 1/2 Correct answer.
  4. 1

Explanation

An antiderivative of sin(2x) is -(1/2)cos(2x). At x = pi/4, cos(pi/2) = 0; at x = 0, cos(0) = 1, so the value is -(1/2)(0 - 1) = 1/2.

Question 37Differentiation and integration (calculus methods)

The curve y = x^3 - 3x^2 + 4 has a stationary point at x = 2. Its nature there is:

  1. a point where the gradient is undefined
  2. a point of inflection with zero gradient
  3. a local minimum Correct answer.
  4. a local maximum

Explanation

Here dy/dx = 3x^2 - 6x = 3x(x - 2), and d^2y/dx^2 = 6x - 6. At x = 2 the second derivative equals 6, which is positive, so the stationary point is a local minimum.

Question 38Differentiation and integration (calculus methods)

Find the x-coordinate of the point on y = x^3 - 3x where the tangent is parallel to the line y = 9x + 1.

  1. x = 3 and x = -3
  2. x = 1 and x = -1
  3. x = 0 only
  4. x = 2 and x = -2 Correct answer.

Explanation

Parallel tangents have equal gradients, so set dy/dx = 3x^2 - 3 = 9, giving x^2 = 4 and x = 2 or x = -2. Both points satisfy the gradient condition.

Question 39Differentiation and integration (calculus methods)

Using the quotient rule, differentiate y = sin(x)/x, for x ≠ 0.

  1. (x cos(x) - sin(x))/x^2 Correct answer.
  2. (cos(x) - sin(x))/x^2
  3. (x sin(x) - cos(x))/x^2
  4. (sin(x) - x cos(x))/x^2

Explanation

The quotient rule gives (u'v - uv')/v^2 with u = sin(x) and v = x. This produces (x cos(x) - sin(x))/x^2.

Question 40Differentiation and integration (calculus methods)

Find the area enclosed between the curve y = x^2 - 4 and the x-axis.

  1. 8/3
  2. 16/3
  3. 32/3 Correct answer.
  4. 64/3

Explanation

The curve crosses the x-axis at x = -2 and x = 2. The enclosed area is the integral of (4 - x^2) from -2 to 2, which equals [4x - x^3/3] evaluated, giving 32/3.

Question 41Differentiation and integration (calculus methods)

A curve has gradient function dy/dx = e^(2x) and passes through the point (0, 3). Find y.

  1. y = (1/2)e^(2x) + 5/2 Correct answer.
  2. y = 2e^(2x) + 1
  3. y = (1/2)e^(2x) + 3
  4. y = e^(2x) + 2

Explanation

Integrating gives y = (1/2)e^(2x) + C. Substituting x = 0 and y = 3 gives 3 = 1/2 + C, so C = 5/2. Hence y = (1/2)e^(2x) + 5/2.

Question 42Differentiation and integration (calculus methods)

Differentiate y = (2x - 1)^5 with respect to x.

  1. 10(2x - 1)^4 Correct answer.
  2. 10(2x - 1)^5
  3. 2(2x - 1)^4
  4. 5(2x - 1)^4

Explanation

The chain rule gives 5(2x - 1)^4 times the inner derivative 2, which equals 10(2x - 1)^4. The exponent decreases by one while the inner derivative multiplies the coefficient.

Question 43Differentiation and integration (calculus methods)

The function f has derivative f'(x) = (x - 1)(x + 3). On which interval is f decreasing?

  1. x > 1
  2. -3 < x < 1 Correct answer.
  3. x < -3 or x > 1
  4. x < -3

Explanation

The derivative is negative between its roots, because the product of the two factors is negative when -3 < x < 1. Therefore f decreases on that interval.

Question 44Differentiation and integration (calculus methods)

Evaluate the integral of x cos(x) dx from x = 0 to x = pi/2.

  1. pi/2 + 1
  2. pi/2 - 1 Correct answer.
  3. pi/2
  4. 1 - pi/2

Explanation

Integrating by parts with u = x and dv = cos(x) dx gives [x sin(x) + cos(x)] from 0 to pi/2. At pi/2 this is pi/2 + 0, and at 0 it is 1, so the value is pi/2 - 1.

Question 45Differentiation and integration (calculus methods)

Using the substitution u = x^2 + 4, evaluate the integral of x/(x^2 + 4) dx from x = 0 to x = 2.

  1. 2ln(2)
  2. (1/2)ln(2) Correct answer.
  3. ln(2)
  4. (1/2)ln(8)

Explanation

With u = x^2 + 4, du = 2x dx, so the integral becomes (1/2) times the integral of 1/u du from u = 4 to u = 8. This equals (1/2)ln(8/4) = (1/2)ln(2).

Question 46Differentiation and integration (calculus methods)

A curve is defined implicitly by x^2 + xy + y^2 = 7. Find dy/dx at the point (1, 2).

  1. -4/5 Correct answer.
  2. 4/5
  3. -5/4
  4. 0

Explanation

Differentiating implicitly gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0. At (1, 2) this is 2 + 2 + (dy/dx) + 4(dy/dx) = 0, so 5(dy/dx) = -4 and dy/dx = -4/5.

Question 47Differentiation and integration (calculus methods)

A sector of a circle has radius r and perimeter 12 cm. Using calculus, find the maximum area of the sector.

  1. 9 cm^2 Correct answer.
  2. 12 cm^2
  3. 18 cm^2
  4. 6 cm^2

Explanation

Perimeter gives 2r + rθ = 12, so θ = (12 - 2r)/r. Area is (1/2)r^2θ = 6r - r^2, whose derivative 6 - 2r is zero at r = 3. The maximum area is then 18 - 9 = 9 cm^2.

Question 48Differentiation and integration (calculus methods)

Evaluate the integral of x^2 e^x dx from x = 0 to x = 1.

  1. e
  2. e - 1
  3. e - 2 Correct answer.
  4. 2e - 2

Explanation

Applying integration by parts twice gives an antiderivative of e^x(x^2 - 2x + 2). At x = 1 this is e(1) = e, and at x = 0 it is 2, so the value is e - 2.

Question 49Differentiation and integration (calculus methods)

The curve y = x^3 + ax^2 + bx has a stationary point at x = 1 with value y = -2. Find the values of a and b.

  1. a = -3, b = -9
  2. a = -3, b = 9
  3. a = 3, b = -9 Correct answer.
  4. a = 3, b = 9

Explanation

The point (1, -2) lies on the curve, so 1 + a + b = -2. Also dy/dx = 3x^2 + 2ax + b equals zero at x = 1, giving 3 + 2a + b = 0. Solving these simultaneous equations gives a = 3 and b = -9.

Question 50Differentiation and integration (calculus methods)

A curve has gradient function dy/dx = 3x^2 - 12x + 9. Find the difference in y-values between the two stationary points.

  1. 4 Correct answer.
  2. 6
  3. 8
  4. 2

Explanation

Setting 3x^2 - 12x + 9 = 0 gives x = 1 and x = 3. Integrating gives y = x^3 - 6x^2 + 9x + C, so y(1) = 4 + C and y(3) = C. The difference is 4.