Full Grade 11 test

The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.

On a symmetric domain, a function is even if f(−x) = f(x) and odd if f(−x) = −f(x). Since (−x)² + 4 = x² + 4, x² + 4 is even. Since (−x)³ = −x³, x³ is odd.

Expand (k + 1)³ and compare it with k³. Write each summand as a difference of consecutive cubes. Identify which cubes cancel in the sum and retain both boundary terms.

Let AB = u and AD = v. Express AM and AN using midpoint relations, then expand their dot product. Use the given angle's cosine for the mixed term.

Apply change of base. Put the natural logarithm of the argument in the numerator and that of the base in the denominator. Reduce the fraction.

Every logarithm argument must be > 0, with base a > 0 and a ≠ 1. logₐ u = r means u = aʳ. Equal logarithms of the same base have equal positive arguments. Check all arguments and the original equation after finding x.

For a > 0 and a ≠ 1, equal powers of a have equal exponents. From a^(x + b) = aᶜ obtain x + b = c, hence x = c − b. For aˣ = v, x = logₐ v when v > 0.

The vertex is B, so use vectors BA and BC. The sign of their dot product determines whether the angle is less than, equal to or greater than 90°.

Change the base to a or to e. Bring the exponents outside the logarithms, keeping track of the coefficients in numerator and denominator.

Check whether one divisor is a multiple of the other. Multiples of that divisor are then also multiples of the other divisor. Check equality separately; if inclusion holds in neither direction, none of the first statements applies.

The left side of √(x + a) = b is nonnegative, so require b ≥ 0 and x + a ≥ 0. Squaring gives x = b² − a. Check the candidate in the original equation because squaring can introduce extraneous roots.

The fractional exponent ^m⁄_n means that you may first take the nth root and then raise the result to the power m.

Rule: a^m/n = (ⁿ√a)^m = ⁿ√(a^m).

1. Look at the denominator n of the exponent: it gives the root index.
2. Take the nth root of the base.
3. Raise the resulting number to the power m shown by the numerator.

In this question, the base is generated as an exact nth power, so the root is an integer.

Rearrange the dot-product formula for the cosine. Divide the dot product by the product of both lengths and reduce the fraction.

Express the ratio of the given terms as a power of the common ratio, using the index difference. Take the positive ratio, recover the first term, and apply the sum formula.

Each year retains a fraction 1 − p/100 of that year starting amount. After n years the value is S × (1 − p/100)ⁿ. Apply each decrease to the already reduced amount.

Calculate differences between several consecutive terms and, when the preceding term is nonzero, their ratios. A constant difference indicates an arithmetic progression; a constant ratio indicates a geometric progression. Check every pair.

logₐ b = c means aᶜ = b. A logarithm is the exponent required on the base. Require a > 0, a ≠ 1 and b > 0. Thus a^(logₐ b) = b; the common logarithm has base 10.

Factor out x and factor the difference of squares. Find the valid part of the interval, then count its integers, excluding any root.

Find elements belonging only to A and only to B separately. Subtract the intersection from each set size, then add the remaining counts.

Split |kx + b| = r into two equations: kx + b = r and kx + b = −r. Solve both equations and give both solutions.