This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Question 1 Given K = sqrt(1-3x)-x. 1.1 Undefined. For K to be undefined, either the denominator is zero or the expression under the square root is negative. Step 1: Set the denominator to zero. -x = 0 x = 0 Step 2: Set the expression under the square root to be negative. 1-3x < 0 1 < 3x (1)/(3) < x Step 3: Combine the conditions. K is undefined if x=0 or x > (1)/(3). x=0 or x > (1)/(3) 1.2 Real. For K to be real, the expression under the square root must be non-negative, and the denominator must be non-zero. Step 1: Ensure the expression under the square root is non-negative. 1-3x 0 1 3x x (1)/(3) Step 2: Ensure the denominator is non-zero. -x 0 x 0 Step 3: Combine the conditions. K is real if x (1)/(3) and x 0. x (1)/(3), x 0 1.3 Positive. For K to be positive, K > 0. This means sqrt(1-3x)-x > 0. The numerator sqrt(1-3x) must be real and positive, and the denominator -x must also be positive. Step 1: Ensure the numerator is positive. For sqrt(1-3x) to be positive, 1-3x > 0. 1 > 3x x < (1)/(3) Step 2: Ensure the denominator is positive. -x > 0 x < 0 Step 3: Combine the conditions. Both conditions x < (1)/(3) and x < 0 must be met. The stricter condition is x < 0. x < 0 Question 2 2.1 Simplify (3x-2)(9x^2+6x+4) Step 1: Recognize the pattern as the difference of cubes formula, (a-b)(a^2+ab+b^2) = a^3-b^3. Here, a=3x and b=2. (3x-2)((3x)^2 + (3x)(2) + 2^2) Step 2: Apply the formula. (3x)^3 - 2^3 Step 3: Calculate the powers. 27x^3 - 8 27x^3 - 8 2.2 Simplify 2x^-2 ÷ x^-5 + (x^-1)^-3 Step 1: Simplify the division term using the exponent rule a^m ÷ a^n = a^m-n. 2x^-2 ÷ x^-5 = 2x^-2 - (-5) = 2x^-2+5 = 2x^3 Step 2: Simplify the second term using the exponent rule (a^m)^n = a^mn. (x^-1)^-3 = x^(-1) × (-3) = x^3 Step 3: Add the simplified terms. 2x^3 + x^3 = 3x^3 3x^3 2.3 Simplify (a^-2-b^-2) ÷ (a^-1+b^-1) Step 1: Rewrite terms with negative exponents as fractions. a^-2-b^-2 = (1)/(a^2) - (1)/(b^2) = (b^2-a^2)/(a^2b^2) a^-1+b^-1 = (1)/(a) + (1)/(b) = (b+a)/(ab) Step 2: Perform the division by multiplying by the reciprocal of the divisor. (b^2-a^2)/(a^2b^2) ÷ (b+a)/(ab) = (b^2-a^2)/(a^2b^2) × (ab)/(b+a) Step 3: Factor the difference of squares in the numerator, b^2-a^2 = (b-a)(b+a). ((b-a)(b+a))/(a^2b^2) × (ab)/(b+a) Step 4: Cancel common factors (b+a) and ab. ((b-a))/(ab) (b-a)/(ab) Question 3 3.1 Factorise the following expression fully: 5x^3 + (40)/(x^3) Step 1: Factor out the common factor of 5. 5(x^3 + (8)/(x^3)) Step 2: Recognize the expression inside the parenthesis as a sum of cubes, A^3+B^3 = (A+B)(A^2-AB+B^2), where A=x and B=(2)/(x). 5(x^3 + ((2)/(x))^3) Step 3: Apply the sum of cubes formula. 5(x + (2)/(x))(x^2 - x((2)/(x)) + ((2)/(x))^2) Step 4: Simplify the terms inside the second parenthesis. 5(x + (2)/(x))(x^2 - 2 + (4)/(x^2)) 5(x + (2)/(x))(x^2 - 2 + (4)/(x^2)) 3.2 Solve simultaneously for x and y if: x^2(y-3)+36(3-y)=0 Step 1: Rewrite the term 36(3-y) to have a common factor with x^2(y-3). Note that (3-y) = -(y-3). x^2(y-3) - 36(y-3) = 0 Step 2: Factor out the common term (y-3). (y-3)(x^2-36) = 0 Step 3: Set each factor equal to zero to find the possible values for x and y. y-3 = 0 or x^2-36 = 0 Step 4: Solve for y. y = 3 Step 5: Solve for x. x^2 = 36 x = ±sqrt(36) x = ± 6 The solutions are y=3 (for any real x) or x=6 (for any real y) or x=-6 (for any real y). More precisely, the equation holds if y=3 (and x can be any real number) OR if x=6 or x=-6 (and y can be any real number). y=3 or x=6 or x=-6 Send me the next one 📸