ACE THE TMUA IN 30 DAYS

Line (IV) is incorrect as we are dividing by 0.00…01, which was defined to be 0. This causes a division by 0 error.

Here is a detailed algebraic explanation for this question:

To maximise a, we can let b = 1, so log₁₀ b = 0.

Therefore, 32log₁₀ a = 2

So log₁₀ a = 1/16

So a = 10^1/16

sin²(3x – 60) varies between 0 and 1, so if sin²(3x – 60) = 1,

f(x) = 3² = 9.

If sin²(3x – 60) = 0, f(x) = (-4)² = 16

Hence the maximum value of f(x) is 16.

9^x-1 = 3^2(x-1)

If we let y = 3^x-1,

Then we get the quadratic:

y² – 30y + 81 = 0

So, y = 27 or y = 3.

So, 3^x-1 = 27 and x = 4

Or 3^x-1 = 3 and x = 2

So, the sum of the roots is 6.

We can equate the common ratios, so 1 / (√x + 2) = (√x – 2) / 1

So x = 5

So r = √5 – 2

As the sum to infinity is a / (1-r), the sum is (√5 + 2) / (1 – (√5 – 2))

So the answer is B.

“A only if B” means “if A, then B”. So we want a function with integral of 0 from -a to a, for all a, but is not symmetrical at x = 0.

x² does not have an integral of 0 for any a, so we can eliminate I.

x³ does have an integral of 0 for any a, and is not symmetric about x = 0 so II is a counterexample.

tan(x) initially appears to also have an integral of 0, but if a = π/2, then tan(x) is undefined so this is not valid for every value of a, so we can eliminate III.

The student correctly differentiates in step I, and finds the discriminant in step II. The student correctly identifies this means p(x) has no turning points and due to it being a positive cubic, this results in it only having one real root.

We need to figure out how many different ways we can combine the variables raised to different powers, such that the sum of their exponents equals 7. To do this, we consider different cases based on how many a terms are chosen.

If no a terms are chosen (i.e. the power of  a is 0), then there are 8 possible combinations: b⁷, b⁶c, b⁵c², b⁴c³, b³c⁴, b²c⁵, bc⁶, c⁷

If one a term is chosen, the powers of b and c must add to 6, leaving 7 combinations. If we continue this pattern, adding one more a term each time, we are left with
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 terms

Using a method called the multinomial expansion will prove to be useful in binomial-like questions with 3 or more terms in the bracket. This video by R2Drew2 is recommended to understand the method here:

The x² term can be made by either: picking one 2x² term and three 1 terms; or two 3x terms and two 1 terms. The first case is is 4! / (3! 1!) × 2 × 1³ = 8. The second case is is 4! / (2! 2!) × 3² = 54. Adding these up gives 62.

This question can be approached in 2 ways. Expanding the brackets is possible in this case as the power is only 3, though this will become extremely long for any higher power. Instead, using a method called the multinomial expansion will prove to be useful in binomial-like questions with 3 or more terms in the bracket. This video by R2Drew2 is recommended to understand the method here:

Using this method, we first spot the  ∑_i=0³ xⁱ just means (1 + x + x^{2 +} x³), so to find any x³ terms, we could find the constant term from (1 + 2x + 3x²)³ and multiply it by x³, take the x² multiplied by x, etc.

Looking at (1 + 2x + 3x²)³:

The x³ term can be made by either: picking one 3x² term, one 2x term, and one 1 term; or three 2x terms. This is 3! / (1! 1! 1!) × 3 × 2 = 36

and  3! / (3!) × 2 × 2 × 2 = 8, meaning the coefficient of x³ is 44.

The x² term can be made by either: picking one 3x² term, two 1 terms; or two 2x terms, and one 1 term. This is 3! / (2! 1!) × 3 = 9 or  3! / (2! 1!) × 2 × 2 = 12, so the coefficient of x² is 21.

The x term can only be made by picking 2x term, and two 1 term. This is 3! / (2! 1!) × 2 = 6, so the coefficient of x is 6.

The constant term can only be made by picking three 1 terms. This is 3! / (3!)  = 1, so the coefficient of x is 1. Adding all these terms up gives 72.

Remove the log by taking 4 to the power of both sides so:

4^x = 2^x+1 – 3

2^2x = 2^x+1 – 3

2^2x = 2(2^x) – 3

2^x = 3 or -1, reject -1 as 2^x > 0

So log (2^x) = log (3)

x = log (3) / log (2)

For (I): Not sufficient: (x/2 – d)(2x – e)(x – f) could lead to a repeated root if any of these are equal: 2f = d; 4e = d; 2e = f.

For (II): Not sufficient: The cubic can have 2 distinct turning points but if both are above or below the x axis, we will only have 1 root. We would additionally need one of these to be strictly positive and the other one strictly negative for this condition to be sufficient.

For (III): Sufficient: x³ + ax² + bx + c where c = o can be factorised into

x(x² + ax + b)

Solving for the quadratic, we get x = (-a ± √(a² – 4b) )/2.

If b < 0, √(a² – 4b) will be larger than a. This means one root will be strictly positive and the other will be strictly negative (and a third at 0), guaranteeing 3 distinct solutions.

For (I): Knowing points A and B initially doesn’t seem to guarantee that we can find the centre. However, we know the points are labelled anticlockwise, we know they are not opposite and do form a side. However, it seems like C could be in 2 places, though in every case, one C will have the points labelled clockwise and in the other case, the points are labelled anticlockwise. Knowing A and C is sufficient to find the centre P, simply by taking the midpoint of both points.

For (II): A rhombus has 4 equal sides, so knowing a single side allows us to find all 4. However, just knowing the length of these sides is insufficient to pinpoint the coordinates of any of the other points, so this is insufficient.

For (III): Similar to (II), just knowing the length of these sides and even area is insufficient to pinpoint the coordinates of any of the other points, so this is also insufficient.

First, we expand (2 + x²)²

(2 + x²)² = 4 + 4x² + x⁴

To make x², we can either take a constant multiplied by x², or x⁴ multiplied by x^-2.

Expanding (x^-2 – 3x²)³, the important terms are:

• -9x^-2 (We pick two x^-2 terms and one – 3x² term, multiplied by 3 choose 1.)
• 27x² (We pick one x^-2 term and two – 3x² terms, multiplied by 3 choose 2.)

Then, multiply out:

• • From the x⁴ term in (2 + x²)² and the -9x^-2 term in (x^-2 – 3x²)³:

x⁴ × (-9x^-2) = -9x²

• • From the constant 4 in (2 + x²)² and the 27x² term in (x^-2 – 3x²)³:

4 × 27x² = 108x²

So the coefficient of x² is:

108x² – 9x² = 99x²

We can consider ranges for the LHS and RHS.

As 0 ≤ sin¹²(x) ≤ 1, the LHS ranges from -7 to 9.

As -1 ≤ cos(x) ≤ 1, the RHS ranges from 9 to 25.

Therefore, the only overlapping value is both sides are 9.

Solving for RHS = 9, we get cos(x) = -1, so x = 180° is the only value in the range.

Subbing in x = 180° to the LHS, we get LHS = 9, so x = 180° is the only solution.

The sequence is a, a + d, a + 2d if we let d < 0. Adding these first 3 terms results in 3a + 3d = 27, so a + d = 9. Squaring these 3 terms and adding them results in the second equation a² + (a + d)² + (a + 2d)². Expanding the brackets results in 3a² + 6ad + 5d² = 293. As a = 9 – d, we get d² = 25 so d = ±5. However, we said d < 0 so d = -5, so a = 14. This means the nth term is 19 – 5n so the 100th term is 19 – 5(100) which is -481.

We can use an interesting method where we take the base of a log and the input of the log to the same power, so log₂(√(2x)) turns into log₁₆(4x²) if we raise both to the power of 4. We take both sides to the power of 16, removing the logs and arrive at the quadratic 4x² + 12x – 16, which gives solutions 1 and -4. However, when we sub x = -4 into the question, we get log₂(√-8), which is not a real number so only x = 1 is valid, so there is 1 solution.

We write out the first few terms

• k = 21: cos(50° + 45 × 21°) = cos(995°) = cos(275°)
• k = 22: cos(50° + 45 × 22°) = cos(1040°) = cos(320°)
• k = 23: cos(50° + 45 × 23°) = cos(1085°) = cos(5°)
• k = 24: cos(50° + 45 × 24°) = cos(1130°) = cos(50°)
• k = 25: cos(50° + 45 × 25°) = cos(1175°) = cos(95°)

From here, we notice that the terms repeat every 8 steps because cosine has a period of 360°. This means the same sequence of angles repeats every 8 values of k.

So the full pattern of 8 terms is:

• cos(50°)
• cos(95°)
• cos(140°)
• cos(185°) = -cos(5°)
• cos(230°) = -cos(50°)
• cos(275°) = -cos(95°)
• cos(320°) = cos(40°)
• cos(5°)

Summing these 8 terms results in 0 because each positive cosine term cancels with a corresponding negative cosine term.

Calculate the total number of terms:

We are summing from k = 21 to k = 201, so there are:

201 – 21 + 1 = 181 terms

This means there are 22 full cycles of 8 terms (which sum to 0) and 5 remaining terms.

We can group all the numbers into groups of 8, apart from the terms 21 to 24 and the 201st term, so the sum is cos(275°) + cos(320°) + cos(5°) + cos(50°) + cos(95°)

As cos(275°) = – cos(95°), these terms cancel. Finally, we can simplify cos(320°) = cos(40°), to arrive at the answer:

cos(5°) + cos(40°) + cos(50°)