Digital SAT Hard Math Practice Flashcards

If 3^(x+1) = 9^(x−2), what is x?

Answer: 5

9^(x+2) = 3^(2(x+2)) = 3^(2x+4); setting 2x−1 = 2x+4 still gives no solution. Correcting: 3^(x+1) = 9^(x−2) → x+1 = 2x−4 → x=5.

A parabola with vertex (2, −3) passes through (5, 6). What is the leading coefficient a in f(x) = a(x − 2)² − 3?

Answer: 1

Substituting (5, 6): 6 = a(5−2)² − 3 = 9a − 3, so 9a = 9 and a = 1.

The function h(t) = −4.9t² + 14.7t + 2 models the height (in meters) of a ball after t seconds. What is the total time the ball is above a height of 2 meters?

Answer: 3 seconds

Set h = 2: −4.9t² + 14.7t = 0 → t(−4.9t + 14.7) = 0 → t = 0 or t = 3; the ball is above 2 m for 3 − 0 = 3 seconds.

If the inverse of f(x) = (3x + 2)/(x − 1) is f⁻¹(x), what is f⁻¹(5)?

Answer: 7/2

Find x when f(x) = 5: (3x+2)/(x−1) = 5 → 3x + 2 = 5x − 5 → 7 = 2x → x = 7/2.

A circle in standard form is (x − 3)² + (y + 2)² = 25. A line y = 4x − 2 intersects the circle. How many intersection points are there?

Answer: 2

Substitute y = 4x − 2 into the circle: (x−3)² + (4x)² = 25 → 17x² − 6x + 9 − 25 = 0 → 17x² − 6x − 16 = 0; discriminant = 36 + 4·17·16 = 36 + 1088 = 1124 > 0 → two intersections.

Evaluate: lim_{x→2} (x³ − 8)/(x − 2).

Answer: 12

Factor: x³ − 8 = (x−2)(x² + 2x + 4); cancel (x−2) to get x² + 2x + 4; at x = 2: 4 + 4 + 4 = 12.

A rectangular box has a square base with side length s and a height of 12 − 2s, where 0 < s < 6. For what value of s is the volume maximized?

Answer: 4

V = s²(12 − 2s) = 12s² − 2s³; V′ = 24s − 6s² = 6s(4 − s) = 0 → s = 0 or s = 4; s = 4 is the maximum in (0, 6).