Mathematics practice questions

Answer: A) 4, at x = 2

• ACorrect. f′(x) = −2x + 4 = 0 at x = 2, and f(2) = 4.
• BIncorrect. Set the derivative to zero.
• CIncorrect. That's a root, not the maximum.
• DIncorrect. Recheck: −4 + 8 = 4.

Answer: A) 5 and 2

• ACorrect. A diagonal matrix's eigenvalues are its diagonal entries.
• BIncorrect. Those are the trace and an off-diagonal entry.
• CIncorrect. 10 is the determinant.
• DIncorrect. Those are the off-diagonal entries.

Answer: A) 18

Use cofactor expansion along the first row with alternating signs: 2(11) − 1(4) + 0 = 18.

• ACorrect. Expanding along row 1: 2(3·4 − 1·1) − 1(1·4 − 1·0) + 0 = 22 − 4 = 18.
• BIncorrect. This adds the second cofactor term instead of subtracting it, ignoring the alternating (+, −, +) sign pattern.
• CIncorrect. This is just the product of the main-diagonal entries (2·3·4). That shortcut works only for triangular matrices.
• DIncorrect. This keeps only the first cofactor term, 2(12 − 1), and drops the contribution of the second entry in row 1.

Answer: B) 6

Take the base-2 logarithm: n = log2(64) = 6, since 2^6 = 64.

• AIncorrect. 32 is 64 divided by 2. Solving for an exponent needs a logarithm, n = log2(64), not division. Also, 2^32 is about 4.3 billion.
• BCorrect. n = log2(64) = 6, because 2^6 = 2·2·2·2·2·2 = 64.
• CIncorrect. 8 is the square root of 64 (8^2 = 64), but 2^8 = 256.
• DIncorrect. 2^4 = 16, not 64. Note that 4^3 = 64, which can make 4 look right by mistake.

Answer: B) 1

• AIncorrect. That's the integrand at x = 1.
• BCorrect. \([x^3]_0^1 = 1\).
• CIncorrect. That's the derivative at x = 1.
• DIncorrect. Recheck the antiderivative.

Answer: C) 10

• AIncorrect. That's only two terms.
• BIncorrect. A ratio below 1 converges.
• CCorrect. 1/(1 − 0.9).
• DIncorrect. That's the ratio.

Answer: B) 25%

• AIncorrect. Divide by the original 80.
• BCorrect. (100 − 80)/80.
• CIncorrect. That's 80/100.
• DIncorrect. That's the ratio, not the increase.

Answer: A) 12.5

Linear interpolation: y = y1 + (y2 − y1)(x − x1)/(x2 − x1) = 10 + 10(1/4) = 12.5.

• ACorrect. The slope is (20 − 10)/(6 − 2) = 2.5 per unit of x, so the value at x = 3 is 10 + 2.5(3 − 2) = 12.5.
• BIncorrect. 15 is the average of the two tabulated values, which only applies at x = 4, the midpoint of the interval. x = 3 is one quarter of the way from 2 to 6, not halfway.
• CIncorrect. This applies the one-unit step from the wrong end of the interval (20 − 2.5), which is the value at x = 5, not x = 3.
• DIncorrect. This subtracts the slope instead of adding it (10 − 2.5). The quantity increases with x, so the value at x = 3 must be greater than 10.

Answer: B) 6

• AIncorrect. That counts jobs, not orderings.
• BCorrect. 3! = 3 × 2 × 1.
• CIncorrect. That's 3².
• DIncorrect. That's 3³.

Answer: C) 0

The dot product is the sum of the products of matching components: (3)(4) + (4)(−3) = 0. A zero dot product means the vectors are orthogonal.

• AIncorrect. 25 is the squared magnitude of either vector (3² + 4² = 25), not the dot product of the two.
• BIncorrect. This drops the negative sign: 3·4 + 4·3 = 24. The second product is 4·(−3) = −12.
• CCorrect. A·B = (3)(4) + (4)(−3) = 12 − 12 = 0. A zero dot product means the vectors are perpendicular.
• DIncorrect. This adds all the components (3 + 4 + 4 − 3 = 8). The dot product multiplies matching components and then adds the products.