One question at a time. Build your LC Ordinary Maths understanding.
First 50 of 199 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
Collect the x-terms on one side by subtracting 3x from both sides, giving 2x - 7 = 9. Then add 7 to both sides to get 2x = 16, and divide by 2 to obtain x = 8. Substituting 8 back into the original equation gives 33 on both sides, confirming the solution.
Multiply each term in the first bracket by each term in the second: x·x = x², x·(-3) = -3x, 4·x = 4x and 4·(-3) = -12. The two middle terms -3x and 4x combine to give +x. The constant term is negative because a positive is multiplied by a negative.
The expression is a difference of two squares, since x² - 9 = x² - 3². The standard identity a² - b² = (a - b)(a + b) then gives (x - 3)(x + 3). Expanding this product returns x² - 9, confirming the factorisation.
A negative index means the reciprocal of the corresponding positive power, so 2⁻³ = 1/2³. Since 2³ = 8, the value is 1/8. The negative sign applies to the exponent, not to the result.
Subtract 4 from both sides to get -2x < 6. Dividing both sides by -2 reverses the inequality sign, giving x > -3. Checking x = 0 satisfies the original inequality, which confirms the direction.
Substitute x = 4 into the rule to get f(4) = 3(4) - 5. That is 12 - 5 = 7. Function notation means the value inside the bracket replaces x throughout the expression.
Multiply the numerical coefficients: 2 · 4 = 8. When multiplying powers of the same base, add the indices: x³ · x⁵ = x⁸. The simplified result is therefore 8x⁸.
Rearrange into the form y = mx + c by isolating y: 4y = -3x + 12, so y = -3/4 x + 3. The coefficient of x is the slope, which is -3/4. The negative sign follows from moving the 3x term across the equals sign.
Adding the two equations eliminates y: 3x = 12, so x = 4. Substituting into x - y = 1 gives 4 - y = 1, so y = 3. Both original equations are then satisfied.
Look for two numbers that multiply to +6 and add to -5; these are -2 and -3. So the equation factorises as (x - 2)(x - 3) = 0, giving x = 2 or x = 3. Each root makes one bracket zero.
Here a = 1, b = 2 and c = -2, so the discriminant is b² - 4ac = 4 + 8 = 12. The formula gives x = (-2 ± √12)/2, and since √12 = 2√3 this simplifies to -1 ± √3.
With roots 3 and -5 the factors are (x - 3)(x + 5). Expanding gives x² + 5x - 3x - 15 = x² + 2x - 15. The sum of the roots, -2, equals the negative of the x coefficient, and the product, -15, is the constant.
Write 32 as a power of 2: 32 = 2⁵. Since the bases are equal, the indices must be equal, so x + 1 = 5 and x = 4. Checking gives 2⁵ = 32.
The numerator is a difference of squares: x² - 4 = (x - 2)(x + 2). Cancelling the common factor (x - 2) leaves x + 2. The cancelled factor cannot be zero, so x ≠ 2 must be stated.
Substitute x = 3 and y = 11 into the equation: 11 = 2(3) + k. This gives 11 = 6 + k, so k = 5. The line then has equation y = 2x + 5.
Replace every x with (x + 1): (x + 1)² - 3(x + 1). Expanding gives x² + 2x + 1 - 3x - 3, which simplifies to x² - x - 2. Care is needed to expand the squared bracket fully.
Factorise as (x - 3)(x + 2) < 0; the critical values are 3 and -2. The parabola opens upwards, so it lies below the x-axis between its roots. Hence the solution is -2 < x < 3.
Adding eliminates y: 8x = 24, so x = 3. Substituting into 3x + 2y = 16 gives 9 + 2y = 16, so 2y = 7 and y = 3.5. Checking the second equation: 15 - 7 = 8, as required.
Set 4n - 7 = 101 and solve: 4n = 108, so n = 27. Since 27 is a positive whole number, 101 is indeed a term of the sequence. Substituting n = 27 gives 4(27) - 7 = 101.
The discriminant is b² - 4ac = (-4)² - 4(2)(5) = 16 - 40 = -24. A negative discriminant means the quadratic has no real roots, since the square root of a negative number is not real.
Half of 6 is 3, and (x + 3)² = x² + 6x + 9. To keep the expression equal to x² + 6x + 1 we subtract 8, giving (x + 3)² - 8. The constant is adjusted by subtracting the square of the halved coefficient.
Equal roots require a zero discriminant: (k + 1)² - 4(1)(9) = 0, so (k + 1)² = 36. Taking square roots gives k + 1 = ±6, hence k = 5 or k = -7. Both values must be kept because the squared equation has two solutions.
Let y = 2^x, so the equation becomes y² - 5y + 4 = 0, which factorises as (y - 1)(y - 4) = 0. Thus y = 1 or y = 4, giving 2^x = 1 so x = 0, or 2^x = 4 so x = 2. Both values check in the original equation.
Factorise both parts: the numerator is (x - 3)(x + 3) and the denominator is (x + 3)(x - 2). Cancelling the common factor (x + 3) leaves (x - 3)/(x - 2). The excluded values come from where the cancelled factor and the remaining denominator are zero.
Dividing x³ - 4x² + x + 6 by (x - 2) gives the quadratic x² - 2x - 3. That factorises as (x - 3)(x + 1), since -3 and 1 multiply to -3 and add to -2. Hence the full factorisation is (x - 2)(x - 3)(x + 1).
The absolute value of a number is its distance from zero on the number line, so |−7| = 7 and |3| = 3. Adding these gives 7 + 3 = 10.
Multiply the coefficients: 3 × 4 = 12. For the variables, add the exponents of like bases: x² · x = x³ and y · y³ = y⁴. The simplified product is 12x³y⁴.
Expand the left side: 5x − 10 = 3x + 4. Subtract 3x from both sides to get 2x − 10 = 4, then add 10 to both sides: 2x = 14. Dividing by 2 gives x = 7.
Slope is the change in y divided by the change in x: (13 − 5)/(6 − 2) = 8/4 = 2. So the line rises 2 units for every 1 unit it moves to the right.
From the second equation, y = 2x − 4. Substituting into the first gives 4x + 3(2x − 4) = 18, so 10x − 12 = 18, hence x = 3. Then y = 2(3) − 4 = 2.
We need two numbers that multiply to 2 × 3 = 6 and add to 7, namely 6 and 1. Split the middle term: 2x² + 6x + x + 3, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).
Move the constant: x² + 4x = 5. Add (4/2)² = 4 to both sides: x² + 4x + 4 = 9, so (x + 2)² = 9. Thus x + 2 = ±3, giving x = 1 or x = −5.
Subtract 3x from both sides: −7 ≤ 2x + 1. Then subtract 1: −8 ≤ 2x, so x ≥ −4. Dividing by a positive number does not reverse the inequality sign.
Substitute x = −2: f(−2) = 2(−2)² − 3(−2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. Remember that (−2)² = 4 and −3(−2) = +6.
For two distinct real roots, the discriminant b² − 4ac must be positive: (−4)² − 4(k)(3) > 0, so 16 − 12k > 0, giving k < 4/3. Also, k cannot be 0 because the equation would no longer be quadratic.
Divide x³ − 5x² + 8x − 4 by (x − 2) to get x² − 3x + 2, which factorises as (x − 1)(x − 2). Hence the full factorisation is (x − 2)²(x − 1).
Factorise the numerator: 2x² − 5x − 3 = (2x + 1)(x − 3). Factorise the denominator: x² − 9 = (x − 3)(x + 3). Cancel the common factor (x − 3) to get (2x + 1)/(x + 3).
Let y = 3^x. Then y² − 10y + 9 = 0, which factorises as (y − 1)(y − 9) = 0, so y = 1 or y = 9. Thus 3^x = 1 gives x = 0, and 3^x = 9 gives x = 2.
By the factor theorem, f(1) = 0 gives 1 + a + b + 6 = 0, so a + b = −7. Also f(−2) = 0 gives −8 + 4a − 2b + 6 = 0, so 2a − b = 1. Solving these simultaneous equations gives a = 0 and b = −7.
The critical values are x = 1 (numerator zero) and x = −2 (denominator zero). Testing intervals: for x < −2 the expression is positive; for −2 < x < 1 it is negative; for x > 1 it is positive. The expression equals zero at x = 1, and x = −2 is excluded because the denominator cannot be zero, so the solution is −2 < x ≤ 1.
Differentiate term by term: the derivative of 3x² is 6x, the derivative of −5x is −5, and the constant 2 differentiates to zero. So f′(x) = 6x − 5.
By the chain rule, the derivative of sin(4x) is cos(4x) multiplied by the derivative of the inner function 4x, which is 4. This gives 4cos(4x).
An antiderivative is x³ + x. Evaluating from 0 to 2 gives (8 + 2) − 0 = 10.
Set f′(x) = 2x − 6 = 0, giving x = 3. The second derivative is 2 > 0, so this is a minimum.
The first derivative of x³ is 3x². Differentiating again gives 6x, which is the second derivative.
The derivative is dy/dx = 2x + 2. Substituting x = 1 gives 2(1) + 2 = 4, which is the slope of the tangent.
An antiderivative of 6x² is 2x³, and any constant may be added. Since 2x³ + 5 differentiates back to 6x², it is also a valid antiderivative.
Let y = 2x + 7, then solve for x: x = (y − 7)/2. Swapping the variable names gives f⁻¹(x) = (x − 7)/2.
Apply the product rule: (x²)′eˣ + x²(eˣ)′ = 2x·eˣ + x²·eˣ, which factors as eˣ(x² + 2x).
Using the quotient rule, the numerator of the derivative is 3(x − 2) − (3x + 1)(1) = 3x − 6 − 3x − 1 = −7. Dividing by (x − 2)² gives −7/(x − 2)².