Past Exam Questions

(iii) To express \(6x - x^2\) in the form \(a - (x-b)^2\), we complete the square:

\(6x - x^2 = -(x^2 - 6x)\)

\(= -(x^2 - 6x + 9 - 9)\)

\(= -(x-3)^2 + 9\)

Thus, \(6x - x^2 = 9 - (x-3)^2\).

(iv) To find \(h^{-1}(x)\), start with \(y = 6x - x^2\).

Rearrange to express \(x\) in terms of \(y\):

\(y = 9 - (x-3)^2\)

\((x-3)^2 = 9 - y\)

\(x-3 = \pm \sqrt{9-y}\)

Since \(x \geq 3\), we take the positive root:

\(x = 3 + \sqrt{9-y}\)

Thus, \(h^{-1}(x) = 3 + \sqrt{9-x}\).

(ii) To express \(f(x) = x^2 - 3x\) in the form \((x-a)^2 - b\), complete the square:

\(x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4}\).

Thus, \(a = \frac{3}{2}\) and \(b = \frac{9}{4}\).

(iii) The minimum value of \(f(x) = (x - \frac{3}{2})^2 - \frac{9}{4}\) is \(-\frac{9}{4}\), so the range is \(f(x) \geq -\frac{9}{4}\).

(iv) The function \(f(x) = x^2 - 3x\) is not one-to-one because it is a quadratic function, which is symmetric about its vertex. Therefore, \(f\) does not have an inverse.

(v) Solve \(g(x) = x - 3\sqrt{x} = 10\).

Let \(y = \sqrt{x}\), then \(x = y^2\).

Substitute into the equation: \(y^2 - 3y = 10\).

Rearrange to form a quadratic: \(y^2 - 3y - 10 = 0\).

Factorize: \((y - 5)(y + 2) = 0\).

Thus, \(y = 5\) or \(y = -2\).

Since \(y = \sqrt{x} \geq 0\), \(y = 5\).

Therefore, \(x = y^2 = 25\).

(i) To find the set of values of \(x\) for which \(f(x) > 15\), solve the inequality:

\(x^2 - 2x > 15\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 15 = 0\)

Factorize the quadratic:

\((x - 5)(x + 3) = 0\)

The solutions are \(x = 5\) and \(x = -3\).

Test intervals around the roots to find where \(f(x) > 15\):

For \(x < -3\), choose \(x = -4\):

\(f(-4) = 16 + 8 = 24 > 15\)

For \(x > 5\), choose \(x = 6\):

\(f(6) = 36 - 12 = 24 > 15\)

Thus, \(x < -3\) and \(x > 5\).

(ii) To find the range of \(f\), complete the square:

\(f(x) = x^2 - 2x = (x-1)^2 - 1\)

The minimum value of \((x-1)^2\) is 0, occurring at \(x = 1\).

Thus, the minimum value of \(f(x)\) is \(-1\).

Therefore, the range of \(f\) is \(y \geq -1\).

Since \(f\) is a quadratic function and not one-to-one, it does not have an inverse.

(a) The function \(f(x) = -3x^2 + 2\) is a downward-opening parabola. The vertex occurs at \(x = 0\), but since \(x \leq -1\), the maximum value of \(f(x)\) is at \(x = -1\). Calculating \(f(-1) = -3(-1)^2 + 2 = -1\). Therefore, the range of \(f\) is \(y \leq -1\).

(b) To find the inverse, start with \(y = -3x^2 + 2\). Rearrange to solve for \(x\):

\(3x^2 = 2 - y\)

\(x^2 = \frac{2-y}{3}\)

\(x = \pm \sqrt{\frac{2-y}{3}}\)

Since \(x \leq -1\), we take the negative root:

\(x = -\sqrt{\frac{2-y}{3}}\)

Thus, \(f^{-1}(x) = \left\{ -\sqrt{\frac{2-x}{3}} \right\}\).

(i) To express \(2x^2 + 8x - 10\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 + 8x - 10 = 2(x^2 + 4x) - 10\)

\(= 2((x+2)^2 - 4) - 10\)

\(= 2(x+2)^2 - 8 - 10\)

\(= 2(x+2)^2 - 18\)

Thus, \(a = 2\), \(b = 2\), \(c = -18\).

(ii) The least value of \(y\) occurs at the vertex of the parabola, which is at \(x = -2\). Substituting \(x = -2\) into the equation:

\(y = 2(-2)^2 + 8(-2) - 10 = 8 - 16 - 10 = -18\)

Thus, the least value of \(y\) is \(-18\) when \(x = -2\).

(iii) To find the set of values of \(x\) for which \(y \geq 14\), solve:

\(2x^2 + 8x - 10 \geq 14\)

\(2x^2 + 8x - 24 \geq 0\)

\(x^2 + 4x - 12 \geq 0\)

\((x+6)(x-2) \geq 0\)

The solution is \(x \geq 2\) or \(x \leq -6\).

(iv) For \(f\) to be one-one, the domain must be restricted to where the function is either increasing or decreasing. The vertex is at \(x = -2\), so the smallest \(k\) is \(-2\).

(v) To find \(f^{-1}(x)\), solve \(y = 2x^2 + 8x - 10\) for \(x\):

\(y = 2(x+2)^2 - 18\)

\(y + 18 = 2(x+2)^2\)

\(\frac{y+18}{2} = (x+2)^2\)

\(x+2 = \pm \sqrt{\frac{y+18}{2}}\)

Since \(x \geq -2\), choose the positive root:

\(x = \sqrt{\frac{y+18}{2}} - 2\)

Thus, \(f^{-1}(x) = \sqrt{\frac{x+18}{2}} - 2\).

(a) To express \(f(x) = -2x^2 - 8x - 13\) in the form \(-2(x + a)^2 + b\), complete the square:

\(f(x) = -2(x^2 + 4x) - 13\)

\(= -2((x + 2)^2 - 4) - 13\)

\(= -2(x + 2)^2 + 8 - 13\)

\(= -2(x + 2)^2 - 5\)

Thus, \(a = 2\) and \(b = -5\).

(b) The vertex form \(-2(x + 2)^2 - 5\) shows the maximum value is \(-5\) when \(x = -2\). Since \(x < -3\), \(f(x) < -7\).

(c) To find \(f^{-1}(x)\), start with \(y = -2(x + 2)^2 - 5\).

Rearrange to solve for \(x\):

\(y + 5 = -2(x + 2)^2\)

\((x + 2)^2 = -\frac{y + 5}{2}\)

\(x + 2 = \pm \sqrt{-\frac{y + 5}{2}}\)

\(x = -2 \pm \sqrt{-\frac{y + 5}{2}}\)

Thus, \(f^{-1}(x) = -2 - \sqrt{\frac{x+5}{2}}\) or \(f^{-1}(x) = -2 + \sqrt{\frac{x+5}{2}}\).

To find \(f^{-1}(x)\), start by setting \(y = f(x) = 2 - \frac{3}{4x - p}\).

Rearrange to solve for \(x\):

\(y - 2 = -\frac{3}{4x - p}\)

\((y - 2)(4x - p) = -3\)

\(4xy - 8x = py - 2p - 3\)

\(4x(y - 2) = p(y - 2) - 3\)

\(4x = \frac{p(y - 2) - 3}{y - 2}\)

\(x = \frac{p(y - 2) - 3}{4(y - 2)}\)

\(x = \frac{p}{4} - \frac{3}{4(y - 2)}\)

Thus, \(f^{-1}(x) = \frac{p}{4} - \frac{3}{4x - 8}\).

For \(f^{-1}(x) \equiv f(x)\), equate the expressions:

\(2 - \frac{3}{4x - p} = \frac{p}{4} - \frac{3}{4x - 8}\)

Equating the coefficients gives \(p = 8\).

(a) To express \(f(x) = 2x^2 - 16x + 23\) in the form \(2(x + a)^2 + b\), complete the square:

Start with \(2(x^2 - 8x) + 23\).

Complete the square for \(x^2 - 8x\):

\(x^2 - 8x = (x-4)^2 - 16\).

Thus, \(2(x^2 - 8x) = 2((x-4)^2 - 16) = 2(x-4)^2 - 32\).

So, \(f(x) = 2(x-4)^2 - 32 + 23 = 2(x-4)^2 - 9\).

(b) The vertex form \(2(x-4)^2 - 9\) shows the minimum value is \(-9\) when \(x = 4\). Since \(x < 3\), the range is \(y > -7\).

(c) To find \(f^{-1}(x)\), start from \(y = 2(x-4)^2 - 9\).

\((x-4)^2 = \frac{y+9}{2}\).

\(x-4 = \pm \sqrt{\frac{y+9}{2}}\).

Since \(x < 3\), choose the negative root: \(x = 4 - \sqrt{\frac{y+9}{2}}\).

Thus, \(f^{-1}(x) = 4 - \sqrt{\frac{x+9}{2}}\).

(a) Start with \(y = \frac{x^2 - 4}{x^2 + 4}\). Clear the denominator: \((x^2 + 4)y = x^2 - 4\). Expand and rearrange: \(x^2 y + 4y = x^2 - 4\). Isolate \(x^2\): \(x^2 (1 - y) = 4y + 4\). Solve for \(x^2\): \(x^2 = \frac{4y + 4}{1 - y}\). Solve for \(x\): \(x = \frac{\sqrt{4y + 4}}{\sqrt{1 - y}}\). Thus, \(f^{-1}(x) = \frac{4x + 4}{1 - x}\).

(b) Start with \(1 - \frac{8}{x^2 + 4}\). Use a common denominator: \(\frac{x^2 + 4 - 8}{x^2 + 4} = \frac{x^2 - 4}{x^2 + 4}\). Therefore, the range of \(f\) is \(0 < f(x) < 1\).

(c) The composite function \(ff\) cannot be formed because the range of \(f\) does not include the whole of the domain of \(f\) (or any of it).

(a) To express \(-3x^2 + 12x + 2\) in the form \(-3(x-a)^2 + b\), complete the square:

\(-3(x^2 - 4x) + 2\)

\(-3((x-2)^2 - 4) + 2\)

\(-3(x-2)^2 + 12 + 2\)

\(-3(x-2)^2 + 14\)

(b) The vertex form \(-3(x-2)^2 + 14\) shows the maximum value is 14 when \(x = 2\). Thus, the largest possible value of \(k\) is 2.

(c) Given \(k = -1\), substitute into \(-3(x-2)^2 + 14\) to find the range:

\(-3((-1)-2)^2 + 14 = -3(9) + 14 = -27 + 14 = -13\)

Thus, the range is \(y \leq -13\).

(d) To find \(f^{-1}(x)\), start from \(y = -3(x-2)^2 + 14\):

\((x-2)^2 = \frac{14-y}{3}\)

\(x-2 = \pm \sqrt{\frac{14-y}{3}}\)

Since \(x \leq 2\), take the negative root:

\(x = 2 - \sqrt{\frac{14-y}{3}}\)

Thus, \(f^{-1}(x) = 2 - \sqrt{\frac{14-x}{3}}\).

(a) Start with \(y = \frac{2x}{3x-1}\). Rearrange to find \(x\) in terms of \(y\):

\(y(3x-1) = 2x\)

\(3xy - y = 2x\)

\(3xy - 2x = y\)

\(x(3y - 2) = y\)

\(x = \frac{y}{3y-2}\)

Thus, \(f^{-1}(x) = \frac{-x}{3x-2}\).

(b) Start with \(\frac{2}{3} + \frac{2}{3(3x-1)}\):

\(\frac{2}{3} + \frac{2}{3(3x-1)} = \frac{2(3x-1) + 2}{3(3x-1)}\)

\(= \frac{6x}{3(3x-1)}\)

\(= \frac{2x}{3x-1}\)

(c) The range of \(f\) is \(f(x) > \frac{2}{3}\).

(i) To express \(2x^2 - 12x + 7\) in the form \(2(x + a)^2 + b\), complete the square:

\(2x^2 - 12x + 7 = 2(x^2 - 6x) + 7\)

\(= 2((x - 3)^2 - 9) + 7\)

\(= 2(x - 3)^2 - 18 + 7\)

\(= 2(x - 3)^2 - 11\)

Thus, \(a = -3\) and \(b = -11\).

(ii) The range of \(f\) is \(y > -11\) because the vertex of the parabola is at its minimum point.

(iii) For \(g\) to have an inverse, it must be one-to-one. The function is one-to-one for \(x \leq 3\), so the largest value of \(k\) is \(k < 3\).

(iv) To find \(g^{-1}(x)\), start with:

\(y = 2(x - 3)^2 - 11\)

\(y + 11 = 2(x - 3)^2\)

\(\frac{y + 11}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 11}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 11}{2}}\)

Thus, \(g^{-1}(x) = 3 - \sqrt{\frac{x + 11}{2}}\).

(a) The function \(f(x) = 1 + \frac{3}{x-2}\) has a vertical asymptote at \(x = 2\) and a horizontal asymptote at \(y = 1\). As \(x \to \infty\), \(f(x) \to 1\). Therefore, the range of \(f\) is \(y > 1\).

(b) To find \(f^{-1}(x)\), start with \(y = 1 + \frac{3}{x-2}\).

Rearrange to find \(x\):

\(y - 1 = \frac{3}{x-2}\)

\((x-2)(y-1) = 3\)

\(x-2 = \frac{3}{y-1}\)

\(x = \frac{3}{y-1} + 2\)

Thus, \(f^{-1}(x) = \frac{3}{x-1} + 2\).

The domain of \(f^{-1}\) is \(x > 1\) because the original range of \(f\) was \(y > 1\).

To find the inverse function \(f^{-1}(x)\), we start with the equation:

\(y = \sqrt{\frac{x+3}{2}} + 1\)

Rearrange to make \(x\) the subject:

\(y - 1 = \sqrt{\frac{x+3}{2}}\)

Square both sides:

\((y - 1)^2 = \frac{x+3}{2}\)

Multiply both sides by 2:

\(2(y - 1)^2 = x + 3\)

Expand \((y - 1)^2\):

\(2(y^2 - 2y + 1) = x + 3\)

\(2y^2 - 4y + 2 = x + 3\)

Rearrange to solve for \(x\):

\(x = 2y^2 - 4y + 2 - 3\)

\(x = 2y^2 - 4y - 1\)

Thus, \(f^{-1}(x) = 2x^2 - 4x - 1\).

For the domain of \(f^{-1}\), since \(f(x) = \sqrt{\frac{x+3}{2}} + 1\) is defined for \(x \geq -3\), the range of \(f(x)\) is \(y \geq 1\). Therefore, the domain of \(f^{-1}\) is \(x \geq 1\).

(i) To express \(f(x) = 4x^2 - 24x + 11\) in the form \(a(x-b)^2 + c\), complete the square:

\(f(x) = 4(x^2 - 6x) + 11\)

\(= 4((x-3)^2 - 9) + 11\)

\(= 4(x-3)^2 - 36 + 11\)

\(= 4(x-3)^2 - 25\)

The vertex form is \(4(x-3)^2 - 25\), so the vertex is \((3, -25)\).

(ii) The range of \(g(x) = 4x^2 - 24x + 11\) for \(x \leq 1\) is determined by the vertex, which is the minimum point. Thus, the range is \(g(x) \geq -9\).

(iii) To find \(g^{-1}(x)\), start with \(y = 4(x-3)^2 - 25\).

\((x-3)^2 = \frac{1}{4}(y + 25)\)

\(x-3 = \pm \frac{1}{2}\sqrt{y + 25}\)

Since \(x \leq 1\), choose the negative root:

\(x = 3 - \frac{1}{2}\sqrt{y + 25}\)

Thus, \(g^{-1}(x) = 3 - \frac{1}{2}\sqrt{x + 25}\).

The domain of \(g^{-1}\) is \(x \geq -9\) because the range of \(g\) is \(g(x) \geq -9\).

Let \(y = \frac{3}{2x+5}\).

Rearrange to make \(x\) the subject: \(2x + 5 = \frac{3}{y}\).

Multiply both sides by \(y\): \(y(2x + 5) = 3\).

Expand: \(2xy + 5y = 3\).

Rearrange to isolate \(x\): \(2xy = 3 - 5y\).

Divide by \(2y\): \(x = \frac{3 - 5y}{2y}\).

Thus, \(f^{-1}(x) = \frac{3 - 5x}{2x}\).

Let \(y = f(x) = (3x + 2)^3 - 5\).

To find \(f^{-1}(x)\), solve for \(x\) in terms of \(y\):

\(y = (3x + 2)^3 - 5\)

Add 5 to both sides: \(y + 5 = (3x + 2)^3\)

Take the cube root: \(\sqrt[3]{y + 5} = 3x + 2\)

Subtract 2: \(\sqrt[3]{y + 5} - 2 = 3x\)

Divide by 3: \(x = \frac{\sqrt[3]{y + 5} - 2}{3}\)

Thus, \(f^{-1}(x) = \frac{\sqrt[3]{x + 5} - 2}{3}\).

The domain of \(f^{-1}\) is the range of \(f\), which is \(x \geq 3\).

(i) To express \(f(x) = 2x^2 - 8x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 4x) + 11\)

\(= 2((x - 2)^2 - 4) + 11\)

\(= 2(x - 2)^2 - 8 + 11\)

\(= 2(x - 2)^2 + 3\)

(ii) The range of \(f\) is \(\geq 3\) because the vertex of the parabola is at its minimum point \((2, 3)\).

(iii) \(f\) does not have an inverse because it is not one-to-one. It is a quadratic function, which means it is symmetric and has two x-values for each y-value.

(iv) The largest value of \(A\) for which \(g\) has an inverse is \(A = 2\), as this makes the function one-to-one on the interval \(x \leq 2\).

(v) To find \(g^{-1}(x)\), start with \(y = 2(x - 2)^2 + 3\) and solve for \(x\):

\(y - 3 = 2(x - 2)^2\)

\(\frac{y - 3}{2} = (x - 2)^2\)

\(x - 2 = \pm \sqrt{\frac{y - 3}{2}}\)

Since \(x \leq 2\), choose the negative root:

\(x = 2 - \sqrt{\frac{y - 3}{2}}\)

Thus, \(g^{-1}(x) = 2 - \sqrt{\frac{x - 3}{2}}\).

The range of \(g^{-1}\) is \(\leq 2\) because the inverse function reflects the domain of \(g\).

\(Let y = (2x − 3)³ − 8.\)

To find the inverse, solve for x in terms of y:

\((2x − 3)³ = y + 8\)

\(2x − 3 = \sqrt[3]{y + 8}\)

\(2x = \sqrt[3]{y + 8} + 3\)

\(x = \frac{\sqrt[3]{y + 8} + 3}{2}\)

\(Thus, f⁻¹(x) = \frac{\sqrt[3]{x + 8} + 3}{2}.\)

The range of f is the domain of f⁻¹. Calculate the range of f for 2 ≤ x ≤ 4:

\(For x = 2, f(2) = (2(2) − 3)³ − 8 = −7.\)

\(For x = 4, f(4) = (2(4) − 3)³ − 8 = 117.\)

Therefore, the domain of f⁻¹ is −7 ≤ x ≤ 117.

(iii) To express \(x^2 - 6x\) in the form \((x-p)^2 - q\), we complete the square:

\(x^2 - 6x = (x-3)^2 - 9\)

Thus, \(p = 3\) and \(q = 9\).

(iv) To find \(h^{-1}(x)\), start with \(y = (x-3)^2 - 9\).

Solve for \(x\):

\(y + 9 = (x-3)^2\)

\(x - 3 = \pm \sqrt{y + 9}\)

Since \(x \geq 3\), we take the positive root:

\(x = \sqrt{y + 9} + 3\)

Thus, \(y = h^{-1}(x) = \sqrt{x + 9} + 3\).

The domain of \(h^{-1}\) is \(x \geq -9\) because the expression under the square root must be non-negative.