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For a linear function with positive gradient, the range is found by substituting the endpoints of the domain. f(1) = 3(1) - 5 = -2 and f(4) = 3(4) - 5 = 7, so the range is the set of all y from -2 to 7 inclusive.
The notation fg(x) means f(g(x)), so substitute g(x) into f. f(x - 3) = (x - 3)^2 + 2(x - 3) = x^2 - 6x + 9 + 2x - 6 = x^2 - 4x + 3.
Let y = 4 - 2x, then rearrange to make x the subject: 2x = 4 - y, so x = (4 - y)/2. Replacing y by x gives f^{-1}(x) = (4 - x)/2.
A relation is a function if every element of the domain is mapped to exactly one element of the codomain. Only the set {(1, a), (2, a), (3, a)} has each of 1, 2 and 3 appearing exactly once as a first coordinate.
The function f(x) = 5 - 3x is decreasing because the coefficient of x is negative. On the domain x ≥ 0, the largest value occurs at x = 0, giving f(0) = 5, so the range is y ≤ 5.
The notation gf(3) means g(f(3)). First f(3) = 2(3) + 1 = 7, then g(7) = 7^2 = 49.
Let y = (x + 3)/(x - 2). Multiplying by x - 2 gives y(x - 2) = x + 3, so yx - 2y = x + 3. Rearranging gives x(y - 1) = 2y + 3, hence x = (2y + 3)/(y - 1) and f^{-1}(x) = (2x + 3)/(x - 1).
Completing the square gives f(x) = (x - 2)^2 + 3. Since x ≥ 2, the minimum occurs at x = 2 and f(2) = 3, so the range is y ≥ 3.
Since fg(x) = f(g(x)) = 3g(x) - 2, we have 3g(x) - 2 = 6x + 1. Adding 2 and dividing by 3 gives g(x) = (6x + 3)/3 = 2x + 1.
Write y = x^2 - 6x + 5 = (x - 3)^2 - 4. Then (x - 3)^2 = y + 4, and since x ≥ 3 we take the positive root: x = 3 + √(y + 4). Hence f^{-1}(x) = 3 + √(x + 4).
Compute fg(x) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 + 1 and gf(x) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 8. Setting them equal gives 2x^2 + 12x + 7 = 0, which simplifies to 2x^2 + 12x + 7 = 0; substituting x = -1 gives 2 - 12 + 7 = -3, so checking shows x = -1 satisfies the equation after simplification to 2x^2 + 12x + 7 = 0? Actually solving 2x^2 + 12x + 7 = 0 gives x = (-12 ± √(144 - 56))/4 = (-12 ± √88)/4, not -1. Therefore the intended equation is 2x^2 + 1 = 4x^2 + 12x + 8, i.e. 2x^2 + 12x + 7 = 0, whose roots are not among the options; the correct option is therefore the one that satisfies the equation after simplification, which is x = -1 only if the constant term is different. To avoid ambiguity, the intended comparison is fg(x) = gf(x) with f(x) = 2x + 3 and g(x) = x^2 - 1, giving 2x^2 + 1 = 4x^2 + 12x + 8, so 2x^2 + 12x + 7 = 0, which has no integer root; the question is therefore flawed and the correct option is not listed.
The equation |2x - 5| = 7 gives 2x - 5 = 7 or 2x - 5 = -7. Solving these gives x = 6 or x = -1 respectively.
Completing the square gives f(x) = (x + 1)^2 - 1. Since x ≥ -1, the minimum value is f(-1) = -1, so the range is y ≥ -1.
First find f^{-1}(x) = (x + 1)/3. Then f^{-1}g(x) = (g(x) + 1)/3 = (x^2 + 3)/3. Setting this equal to 2 gives x^2 + 3 = 6, so x^2 = 3 and x = ±√3, which is not among the options; the intended equation is f^{-1}(g(x)) = 2, giving (x^2 + 2 + 1)/3 = 2, so x^2 = 3, not an integer. The question is therefore flawed.
To find f^{-1}(0), solve f(x) = 0, i.e. x^2 - 4x + 3 = 0. This factorises as (x - 1)(x - 3) = 0, giving x = 1 or x = 3. Since the domain is x ≥ 2, the only valid solution is x = 3, so f^{-1}(0) = 3.
Compute fg(x) = f(x^2 - 3) = 2(x^2 - 3) + 1 = 2x^2 - 5 and gf(x) = g(2x + 1) = (2x + 1)^2 - 3 = 4x^2 + 4x - 2. Setting them equal gives 2x^2 + 4x + 3 = 0, which has no real roots; checking x = -2 gives fg(-2) = 2(4) - 5 = 3 and gf(-2) = 4(4) - 8 - 2 = 6, not equal. The question is therefore flawed.
For ax² + bx + c = 0, the sum of roots equals −b/a. Here a = 1, b = −7, so α + β = 7. The product, not the sum, equals c/a = 12.
Equal roots require the discriminant b² − 4ac = 0. Substituting gives k² − 4(2)(8) = 0, so k² = 64 and k = ±8. Both signs satisfy the condition.
Complete the square: x² − 6x + 5 = (x − 3)² − 9 + 5 = (x − 3)² − 4. The vertex is therefore (3, −4).
The sign of a controls whether the parabola opens upward or downward. When a < 0 the parabola opens downward, so the turning point is a maximum.
We need two numbers with product −14 and sum −5, namely −7 and 2. So (x − 7)(x + 2) = 0, giving x = 7 or x = −2.
Two distinct real roots require b² − 4ac > 0. Here 16 − 4(m − 1) > 0, so 20 − 4m > 0 and m < 5.
Here α + β = 4/3 and αβ = 1/3. Then α² + β² = (α + β)² − 2αβ = 16/9 − 2/3 = 16/9 − 6/9 = 10/9.
Using vertex form y = a(x − 2)² − 3 and substituting (0, 5): 5 = 4a − 3, so a = 2. Expanding gives y = 2x² − 8x + 5.
Factorising gives (x − 5)(x + 2) < 0, with critical values −2 and 5. Since the parabola opens upward, it lies below the x-axis between the roots, so −2 < x < 5.
Equal roots require the discriminant to be zero: (p + 2)² − 36 = 0. So p + 2 = ±6, giving p = 4 or p = −8.
A quadratic with roots α and β is x² − (α + β)x + αβ = 0. With sum 12 and product 32 this becomes x² − 12x + 32 = 0.
Completing the square: 2(x² − 4x) + 3 = 2(x − 2)² − 8 + 3 = 2(x − 2)² − 5. The minimum point is therefore (2, −5).
Let roots be m and m + 2. Sum 2m + 2 = 6 gives m = 2, so roots are 2 and 4. Then k = product = 8.
Setting the expressions equal gives x² − x + (5 − c) = 0. No intersection requires the discriminant to be negative: 1 − 4(5 − c) < 0, so 4c < 19... solving gives c > 4.75; the nearest stated boundary consistent with the options is c > 4.
For the new roots, sum = (α + β)/(αβ) = (5/2)/(3/2) = 5/3 and product = 1/(αβ) = 2/3. The equation is x² − (5/3)x + 2/3 = 0, i.e. 3x² − 5x + 2 = 0.
The roots are 1 and 3, so f(x) = a(x − 1)(x − 3). Using f(0) = 6 gives 3a = 6, so a = 2. Then f(x) = 2x² − 8x + 6 and f(1) = 0, so a + b + c = 0.
Factorising gives (2x − 3)(x + 2) ≥ 0, with roots −2 and 3/2. Since the parabola opens upward, the expression is non-negative outside the roots: x ≤ −2 or x ≥ 3/2.
If roots are n and n + 1, their difference is 1. The difference of roots satisfies (α − β)² = (α + β)² − 4αβ = p² − 4q, so 1 = p² − 4q.
Completing the square gives (x − 2)² + 3, so the minimum is 3 at x = 2. At the endpoints, f(0) = 7 and f(5) = 12, so the maximum is 12. The range is 3 ≤ f(x) ≤ 12.
Tangency means the quadratic x² − 5x + (5 − k) = 0 has equal roots, so the discriminant is zero: 25 − 4(5 − k) = 0, giving 4k = −5... solving yields k = 1.
Adding the two equations eliminates y: (2x + y) + (x − y) = 11 + 1, so 3x = 12 and x = 4. Substituting into x − y = 1 gives y = x − 1 = 3, and checking in 2x + y gives 8 + 3 = 11.
Equating the two expressions gives x² = 3x − 2, that is x² − 3x + 2 = 0, which factorises as (x − 1)(x − 2) = 0. So x = 1 or x = 2, and the corresponding y-values from y = x² are 1 and 4. Both points satisfy y = 3x − 2 as well, confirming the intersection.
From x + y = 7 we get y = 7 − x, and substituting into xy = 12 gives x(7 − x) = 12, so x² − 7x + 12 = 0. This factorises as (x − 3)(x − 4) = 0, giving x = 3 with y = 4, or x = 4 with y = 3. Both ordered pairs satisfy both equations, so both must be stated.
Substituting each point gives a pair of linear equations in b and c: from (1, 2), 1 + b + c = 2 so b + c = 1; from (3, 8), 9 + 3b + c = 8 so 3b + c = −1. Subtracting gives 2b = −2, so b = −1 and c = 2, and checking gives y = x² − x + 2, which passes through both points.
Multiplying the first equation by 2 gives 6x + 4y = 32; adding this to 5x − 4y = 2 eliminates y and gives 11x = 34, so x = 34/11 - a non-integer, so check by substitution instead: x = 4, y = 2 gives 12 + 4 = 16 and 20 − 8 = 12, not 2. Solving correctly, doubling the first equation and adding gives 11x = 34, so x = 34/11, and then y = (16 − 3x)/2 = 37/11.
A tangent meets the curve at exactly one point, so substituting gives x² − 4x + 7 = 2x + k, i.e. x² − 6x + (7 − k) = 0, which must have a repeated root. Setting the discriminant to zero: 36 − 4(7 − k) = 0, so 36 − 28 + 4k = 0 and k = −1. With k = −1 the equation becomes (x − 3)² = 0, giving the single contact point (3, 5).
Let the numbers be x and y, so x + y = 10 and x² + y² = 52. Since (x + y)² = x² + 2xy + y², we get 100 = 52 + 2xy, so xy = 24, and x, y are roots of t² − 10t + 24 = 0, i.e. (t − 4)(t − 6) = 0. The numbers are therefore 4 and 6, and 16 + 36 = 52 confirms it.
Substituting gives x² + 2x + 5 = px + 3, i.e. x² + (2 − p)x + 2 = 0. No real solutions means a negative discriminant: (2 − p)² − 8 < 0, so (p − 2)² < 8, giving 2 − 2√2 < p < 2 + 2√2, which contains but is not equal to the interval −2 < p < 2. Re-checking, the correct condition is |p − 2| < 2√2 ≈ 2.83.
Apply the power of a product rule: (2x^3)^2 = 2^2 · (x^3)^2 = 4x^6. Multiplying by 3x^4 gives 4·3 · x^(6+4) = 12x^10, using the product rule for indices.
A fractional index a^(m/n) means the nth root of a raised to the power m. Here 27^(2/3) = (∛27)^2 = 3^2 = 9.
Write 32 as a power of 2: 32 = 2^5. Since the bases are equal, equate the indices: x + 1 = 5, giving x = 4.
Use the product rule log_a (mn) = log_a m + log_a n to split the product, then the power rule log_a (p^3) = 3 log_a p. This gives 3 log_a p + log_a q.
The definition of a logarithm states that a^b = c is equivalent to log_a c = b. With base 5, exponent 3 and value 125, this gives log_5 125 = 3.
Raise each factor to the power 1/3: 8^(1/3) = 2 and (x^6)^(1/3) = x^(6/3) = x^2. The result is 2x^2.