TMUA Calculus questions

Answer: B · Method: Tangent and normal lines from the derivative

1. At $x = 3$: $y = 6$, and $\dfrac{dy}{dx} = 4x - 4 = 8$.
2. The tangent is $y - 6 = 8(x - 3)$.
3. Put $x = 0$: $y = -18$.

Why the tempting wrong answers are wrong

• D: That's where it crosses the $x$-axis.
• E: That's for the normal. The tangent has gradient $m$ itself.

Answer: C · Method: Split into powers of x, then integrate

1. Split the fraction: $\dfrac{6x - 6}{\sqrt{x}} = 6x^{1/2} - 6x^{-1/2}$.
2. Integrate: $\left[4x^{3/2} - 12x^{1/2}\right]_1^{4}$.
3. At $x = 4$: $x^{3/2} = 8$ and $x^{1/2} = 2$. The result is $16$.

Why the tempting wrong answers are wrong

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

Answer: A · Method: Split into powers of x, then integrate

1. Split the fraction: $\dfrac{5x + 6}{\sqrt{x}} = 5x^{1/2} + 6x^{-1/2}$.
2. Integrate: $\left[\dfrac{10}{3}x^{3/2} + 12x^{1/2}\right]_1^{9}$.
3. At $x = 9$: $x^{3/2} = 27$ and $x^{1/2} = 3$. The result is $\dfrac{332}{3}$.

Why the tempting wrong answers are wrong

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

Answer: B · Method: Integrate dy/dx and use a point to find the constant

1. Integrate: $y = -x^3 - 3x^2 + x + K$.
2. At $(1, -8)$: $K = -8 - (-3) = -5$.
3. At $x = 2$: $y = -18 - 5 = -23$.

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.

Answer: C · Method: Simplify an integral with a substitution

1. Substitute $u = 1 - x$, so $x = 1 - u$ and the limits swap, which cancels the minus sign from $dx = -du$.
2. The integral becomes $\int_0^1 (1-u)u^4\,du = \tfrac{1}{5} - \tfrac{1}{6} = \tfrac{1}{30}$.

Answer: C · Method: Optimise with one variable

1. By symmetry the corners are at $x = \pm t$, so the area is $A = 2t(12 - t^2) = 24t - 2t^3$ for $0 < t < \sqrt{12}$.
2. $A'(t) = 24 - 6t^2 = 0$ gives $t = 2$, and $A(2) = 4 \times 8 = 32$. Since $A'' < 0$ there, this is the maximum.

Answer: C · Method: Area between a curve and a line

1. They meet where $x^2 = 2x + 3$, so $x = -1$ or $x = 3$.
2. Area $= \int_{-1}^{3} (2x + 3 - x^2)\,dx = \left[x^2 + 3x - \tfrac{x^3}{3}\right]_{-1}^{3} = 9 - \left(-\tfrac{5}{3}\right) = \tfrac{32}{3}$.

Answer: D · Method: Integrate a modulus by splitting the interval

1. $x^2 - 2x = x(x - 2)$ is negative for $0 < x < 2$ and positive for $2 < x < 3$, so split the integral there.
2. $\int_0^2 (2x - x^2)\,dx = 4 - \tfrac{8}{3} = \tfrac{4}{3}$ and $\int_2^3 (x^2 - 2x)\,dx = (9 - 9) - \left(\tfrac{8}{3} - 4\right) = \tfrac{4}{3}$.
3. The total is $\tfrac{8}{3}$.

Why the tempting wrong answers are wrong

• A: $0$ is the integral without the modulus: the part below the axis cancels the part above. The modulus makes both parts count positively.

Answer: A · Method: Tangent and normal lines from the derivative

1. At $x = -1$: $y = 6$, and $\dfrac{dy}{dx} = -2x - 4 = -2$.
2. The normal has gradient $\dfrac{1}{2}$: $y - 6 = \dfrac{1}{2}(x + 1)$.
3. Put $y = 0$ and solve: $x = -13$.

Why the tempting wrong answers are wrong

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

Answer: D · Method: No stationary points: f′(x) = 0 has no real roots

1. $\dfrac{dy}{dx} = 3x^2 + 2kx + 27$. There are no stationary points when this is never zero, i.e. its discriminant is negative.
2. $(2k)^2 - 4 \times 3 \times 27 < 0$, so $4k^2 < 324$ and $k^2 < 81$.
3. That's $-9 < k < 9$.

Why the tempting wrong answers are wrong

• A: That's where the derivative has two roots, so the curve has two stationary points.
• E: At $k = \pm 9$ the derivative has a repeated root, so there is one stationary point.

Answer: D · Method: Trapezium rule: over- or underestimate?

1. The graph of $y = 2^x$ bends upwards (it is convex), so each straight chord of a trapezium lies above the curve: the trapezium rule gives an overestimate.
2. Translating the graph, stretching it vertically by a positive factor, or reflecting it in the $y$-axis keeps it bending upwards, so A, B, C and E are all overestimates.
3. $y = 5 - 2^x$ is a reflection in the $x$-axis (then a translation), so it bends downwards. Its chords lie below the curve, so the trapezium rule underestimates: D.

Why the tempting wrong answers are wrong

• A: Adding $3$ raises every trapezium by the same amount, so the error doesn't change sign.
• E: Reflecting in the $y$-axis keeps the curve bending upwards, so this is still an overestimate.

Answer: C · Method: Integrals of shifted functions

1. The graph of $y = f(x + 2)$ is $y = f(x)$ shifted left by $2$, so $\int_0^1 f(x + 2)\,dx = \int_2^3 f(x)\,dx = 3$.
2. Then $\int_1^2 f(x)\,dx = 4 - 3 = 1$, and in the same way $\int_0^1 f(x + 1)\,dx = \int_1^2 f(x)\,dx = 1$.
3. So the integral is $3 \times 1 - \int_0^1 2x\,dx = 3 - 1 = 2$.

Why the tempting wrong answers are wrong

• D: You also need to subtract $\int_0^1 2x\,dx = 1$.