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ESAT Maths 2 Calculus questions

12 original ESAT Maths 2 calculus questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Tangent and normal lines from the derivativeNo stationary points: f′(x) = 0 has no real rootsSplit into powers of x, then integrateArea between a curve and a lineIntegrals of shifted functionsTrapezium rule: over- or underestimate?Integrate dy/dx and use a point to find the constantUse f′ and f″ conditions to set up simultaneous equations

Question 1ESAT Maths 2Warm-up

The tangent and the normal to the curve y=−x2+3x−5y = -x^2 + 3x - 5 are drawn at the point where x=3x = 3. At what yy-value does the tangent cross the yy-axis?
  1. A43\dfrac{4}{3}
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G77
  8. H1818
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Answer: D · Method: Tangent and normal lines from the derivative

  1. At x=3x = 3: y=−5y = -5, and dydx=−2x+3=−3\dfrac{dy}{dx} = -2x + 3 = -3.
  2. The tangent is y+5=−3(x−3)y + 5 = -3(x - 3).
  3. Put x=0x = 0: y=4y = 4.

Why the tempting wrong answers are wrong

  • A: That's where it crosses the xx-axis.
  • H: That's for the normal. The tangent has gradient mm itself.

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Question 2ESAT Maths 2Warm-up

Evaluate ∫193x−4x dx\displaystyle\int_1^{9} \frac{3x - 4}{\sqrt{x}}\,dx.
  1. A−523-\dfrac{52}{3}
  2. B3030
  3. C3636
  4. D3737
  5. E113113
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Answer: C · Method: Split into powers of x, then integrate

  1. Split the fraction: 3x−4x=3x1/2−4x−1/2\dfrac{3x - 4}{\sqrt{x}} = 3x^{1/2} - 4x^{-1/2}.
  2. Integrate: [2x3/2−8x1/2]19\left[2x^{3/2} - 8x^{1/2}\right]_1^{9}.
  3. At x=9x = 9: x3/2=27x^{3/2} = 27 and x1/2=3x^{1/2} = 3. The result is 3636.

Why the tempting wrong answers are wrong

  • B: Don't forget to subtract the value at the lower limit, x=1x = 1.
  • E: When you integrate xnx^n, divide by the new power n+1n + 1 rather than multiplying by it.

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Question 3ESAT Maths 2Warm-up

A curve has dydx=6x2+6x\dfrac{dy}{dx} = 6x^2 + 6x and passes through the point (0,5)(0, 5). What is the value of yy when x=3x = 3?
  1. A7676
  2. B8181
  3. C8585
  4. D8686
  5. E8787
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Answer: D · Method: Integrate dy/dx and use a point to find the constant

  1. Integrate: y=2x3+3x2+Ky = 2x^3 + 3x^2 + K.
  2. At (0,5)(0, 5): K=5−(0)=5K = 5 - (0) = 5.
  3. At x=3x = 3: y=81+5=86y = 81 + 5 = 86.

Why the tempting wrong answers are wrong

  • A: Check the sign when you find the constant.
  • B: Don't forget the constant of integration: use the point to find it.

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Question 4ESAT Maths 2Warm-up

Evaluate ∫193x+2x dx\displaystyle\int_1^{9} \frac{3x + 2}{\sqrt{x}}\,dx.
  1. A5959
  2. B6060
  3. C6161
  4. D6666
  5. E2603\dfrac{260}{3}
  6. F119119
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Answer: B · Method: Split into powers of x, then integrate

  1. Split the fraction: 3x+2x=3x1/2+2x−1/2\dfrac{3x + 2}{\sqrt{x}} = 3x^{1/2} + 2x^{-1/2}.
  2. Integrate: [2x3/2+4x1/2]19\left[2x^{3/2} + 4x^{1/2}\right]_1^{9}.
  3. At x=9x = 9: x3/2=27x^{3/2} = 27 and x1/2=3x^{1/2} = 3. The result is 6060.

Why the tempting wrong answers are wrong

  • D: Don't forget to subtract the value at the lower limit, x=1x = 1.
  • F: When you integrate xnx^n, divide by the new power n+1n + 1 rather than multiplying by it.

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Question 5ESAT Maths 2Warm-up

A curve has dydx=6x2+6x−3\dfrac{dy}{dx} = 6x^2 + 6x - 3 and passes through the point (−2,−8)(-2, -8). What is the value of yy when x=3x = 3?
  1. A5959
  2. B6262
  3. C6363
  4. D7272
  5. E7878
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Answer: B · Method: Integrate dy/dx and use a point to find the constant

  1. Integrate: y=2x3+3x2−3x+Ky = 2x^3 + 3x^2 - 3x + K.
  2. At (−2,−8)(-2, -8): K=−8−(2)=−10K = -8 - (2) = -10.
  3. At x=3x = 3: y=72−10=62y = 72 - 10 = 62.

Why the tempting wrong answers are wrong

  • D: Don't forget the constant of integration: use the point to find it.
  • E: Check the sign when you find the constant.

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Question 6ESAT Maths 2Standard

For which values of kk does the curve y=x3+kx2+48x−6y = x^3 + kx^2 + 48x - 6 have no stationary points?
  1. Ak<12k < 12
  2. B−12<k<12-12 < k < 12
  3. Ck>−12k > -12
  4. Dk<−12 or k>12k < -12 \text{ or } k > 12
  5. E−12≤k≤12-12 \le k \le 12
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Answer: B · Method: No stationary points: f′(x) = 0 has no real roots

  1. dydx=3x2+2kx+48\dfrac{dy}{dx} = 3x^2 + 2kx + 48. There are no stationary points when this is never zero, i.e. its discriminant is negative.
  2. (2k)2−4×3×48<0(2k)^2 - 4 \times 3 \times 48 < 0, so 4k2<5764k^2 < 576 and k2<144k^2 < 144.
  3. That's −12<k<12-12 < k < 12.

Why the tempting wrong answers are wrong

  • D: That's where the derivative has two roots, so the curve has two stationary points.
  • E: At k=±12k = \pm 12 the derivative has a repeated root, so there is one stationary point.

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Question 7ESAT Maths 2Standard

What is the area of the region enclosed by the curve y=x2−11x+22y = x^2 - 11x + 22 and the line y=−x+1y = -x + 1?
  1. A88
  2. B263\dfrac{26}{3}
  3. C293\dfrac{29}{3}
  4. D323\dfrac{32}{3}
  5. E353\dfrac{35}{3}
  6. F383\dfrac{38}{3}
  7. G643\dfrac{64}{3}
  8. H3232
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Answer: D · Method: Area between a curve and a line

  1. They meet where x2−10x+21=0x^2 - 10x + 21 = 0, i.e. (x−3)(x−7)=0(x - 3)(x - 7) = 0, so x=3x = 3 or x=7x = 7.
  2. Between these the line is above the curve, so the area is ∫37−(x−3)(x−7) dx\displaystyle\int_{3}^{7} -(x - 3)(x - 7)\,dx.
  3. That evaluates to 436=323\dfrac{4^3}{6} = \dfrac{32}{3}.

Why the tempting wrong answers are wrong

  • G: Check the integral of the difference: the x3x^3 term gives 13\tfrac{1}{3}, but the other terms reduce it.

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Question 8ESAT Maths 2Standard

The tangent and the normal to the curve y=2x2−3x+3y = 2x^2 - 3x + 3 are drawn at the point where x=2x = 2. At what xx-value does the normal cross the xx-axis?
  1. A−23-23
  2. B−4-4
  3. C11
  4. D2626
  5. E2727
  6. F2828
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Answer: E · Method: Tangent and normal lines from the derivative

  1. At x=2x = 2: y=5y = 5, and dydx=4x−3=5\dfrac{dy}{dx} = 4x - 3 = 5.
  2. The normal has gradient −15-\dfrac{1}{5}: y−5=−15(x−2)y - 5 = -\dfrac{1}{5}(x - 2).
  3. Put y=0y = 0 and solve: x=27x = 27.

Why the tempting wrong answers are wrong

  • A: Check the sign when you rearrange the line for xx.
  • C: That's where the tangent crosses. The normal is perpendicular to it, with gradient −1m-\tfrac{1}{m}.

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Question 9ESAT Maths 2Standard

What is the area of the region enclosed by the curve y=x2+2y = x^2 + 2 and the line y=−x+2y = -x + 2?
  1. A−116-\dfrac{11}{6}
  2. B−56-\dfrac{5}{6}
  3. C16\dfrac{1}{6}
  4. D13\dfrac{1}{3}
  5. E12\dfrac{1}{2}
  6. F76\dfrac{7}{6}
  7. G136\dfrac{13}{6}
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Answer: C · Method: Area between a curve and a line

  1. They meet where x2+x=0x^2 + x = 0, i.e. (x+1)x=0(x + 1)x = 0, so x=−1x = -1 or x=0x = 0.
  2. Between these the line is above the curve, so the area is ∫−10−(x+1)x dx\displaystyle\int_{-1}^{0} -(x + 1)x\,dx.
  3. That evaluates to 136=16\dfrac{1^3}{6} = \dfrac{1}{6}.

Why the tempting wrong answers are wrong

  • D: Check the integral of the difference: the x3x^3 term gives 13\tfrac{1}{3}, but the other terms reduce it.

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Question 10ESAT Maths 2Challenging

A function ff satisfies ∫25f(x) dx=−1\displaystyle\int_{2}^{5} f(x)\,dx = -1 and ∫01f(x+4) dx=5\displaystyle\int_0^{1} f(x + 4)\,dx = 5. What is ∫02(2f(x+2)−2) dx\displaystyle\int_0^{2} \big(2f(x + 2) - 2\big)\,dx?
  1. A−17-17
  2. B−16-16
  3. C−15-15
  4. D−12-12
  5. E−10-10
  6. F44
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Answer: B · Method: Integrals of shifted functions

  1. Shifting a graph left doesn't change areas: ∫01f(x+4) dx=∫45f(x) dx=5\int_0^{1} f(x + 4)\,dx = \int_{4}^{5} f(x)\,dx = 5.
  2. So ∫24f(x) dx=−1−5=−6\int_{2}^{4} f(x)\,dx = -1 - 5 = -6, and in the same way ∫02f(x+2) dx=−6\int_0^{2} f(x + 2)\,dx = -6.
  3. The integral is 2×−6−2×2=−162 \times -6 - 2 \times 2 = -16.

Why the tempting wrong answers are wrong

  • D: Don't forget the constant: ∫022 dx=4\int_0^{2} 2\,dx = 4.
  • F: ∫01f(x+4) dx\int_0^{1} f(x + 4)\,dx is the part from 44 to 55, so subtract it from the whole.

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Question 11ESAT Maths 2Challenging

The trapezium rule, with any number of equal-width strips, is used to estimate each of these integrals. Which one is always an overestimate?
  1. A∫24(6−1x)dx\displaystyle\int_{2}^{4} \left(6 - \dfrac{1}{x}\right)dx
  2. B∫25(1x+3)dx\displaystyle\int_{2}^{5} \left(\dfrac{1}{x + 3}\right)dx
  3. C∫14(6−1x)dx\displaystyle\int_{1}^{4} \left(6 - \dfrac{1}{x}\right)dx
  4. D∫24(2−x3)dx\displaystyle\int_{2}^{4} \left(2 - x^3\right)dx
  5. E∫13(x+2)dx\displaystyle\int_{1}^{3} \left(\sqrt{x + 2}\right)dx
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Answer: B · Method: Trapezium rule: over- or underestimate?

  1. If a curve bends upwards (convex), every chord lies above it, so the trapezium rule overestimates. If it bends downwards (concave), the chords lie below it and the rule underestimates.
  2. Adding a constant, stretching by a positive factor, or shifting left or right keeps the shape. Subtracting the function from a constant flips it.
  3. Only option B's curve bends upwards (convex) on its interval, so it's the overestimate.

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Question 12ESAT Maths 2Challenging

A function is defined by f(x)=ax2+bxf(x) = ax^{2} + \dfrac{b}{x}, where aa and bb are constants. It is given that f′(1)=−2f'(1) = -2 and f′′(−1)=0f''(-1) = 0. What is the value of a+ba + b?
  1. A−6-6
  2. B−5-5
  3. C−4-4
  4. D−3-3
  5. E−2-2
  6. F−1-1
  7. G00
  8. H44
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Answer: C · Method: Use f′ and f″ conditions to set up simultaneous equations

  1. f′(x)=2ax−bx−2f'(x) = 2ax - bx^{-2} and f′′(x)=2a+2bx−3f''(x) = 2a + 2bx^{-3}.
  2. f′(1)=−2f'(1) = -2 and f′′(−1)=0f''(-1) = 0 give two simultaneous equations in aa and bb.
  3. Solving: a=−2a = -2, b=−2b = -2, so a+b=−4a + b = -4.

Why the tempting wrong answers are wrong

  • G: Check which equation gives which constant.
  • H: Watch the signs when you differentiate negative powers.

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