The diagram shows the curves with equations ▶️ Answer/Explanation
Solution
Part (a): x-coordinates of intersection points
Equate: \( x^3 – 3x + 3 = 2x^3 – 4x^2 + 3 \)
Simplify: \( x^3 – 4x^2 + 3x = 0 \)
Factor: \( x(x – 1)(x – 3) = 0 \)
Answer: \( x = 0, 1, 3 \)
Part (b): Area of the shaded region
Functions: \( y_1 = x^3 – 3x + 3 \), \( y_2 = 2x^3 – 4x^2 + 3 \)
From \( x = 1 \) to \( x = 3 \), \( y_1 > y_2 \)
Area: \( \int_1^3 (y_1 – y_2) \, dx \)
Difference: \( y_1 – y_2 = -x^3 + 4x^2 – 3x \)
Antiderivative: \( -\frac{x^4}{4} + \frac{4x^3}{3} – \frac{3x^2}{2} \)
Evaluate: \( \left[ -\frac{x^4}{4} + \frac{4x^3}{3} – \frac{3x^2}{2} \right]_1^3 = \frac{9}{4} – \left( -\frac{5}{12} \right) = \frac{8}{3} \)
Answer: \( \frac{8}{3} \)
▶️ Answer/Explanation
Solution
Part (a): Express in vertex form
Rewrite: \( 3x^2 – 12x + 14 = 3(x^2 – 4x) + 14 \)
Complete square: \( x^2 – 4x = (x – 2)^2 – 4 \)
So: \( 3[(x – 2)^2 – 4] + 14 = 3(x – 2)^2 + 2 \)
Answer: \( 3(x – 2)^2 + 2 \)
Part (b): Least value of \( k \)
\( f(x) = 3(x – 2)^2 + 2 \), vertex at \( x = 2 \)
For \( f^{-1} \) to exist, \( f \) must be one-to-one. Increasing for \( x \geq 2 \).
Answer: \( k = 2 \)
Part (c): Inverse function
Set \( y = 3(x – 2)^2 + 2 \):
\( y – 2 = 3(x – 2)^2 \)
\( (x – 2)^2 = \frac{y – 2}{3} \)
\( x = 2 + \sqrt{\frac{y – 2}{3}} \) (positive since \( x \geq 2 \))
Answer: \( f^{-1}(x) = 2 + \sqrt{\frac{x – 2}{3}} \)
Part (d): Solve \( f(f(x)) = 29 \)
\( f(x) = 3(x – 2)^2 + 2 \)
\( f(f(x)) = 3[3(x – 2)^2]^2 + 2 = 27(x – 2)^4 + 2 \)
Set: \( 27(x – 2)^4 + 2 = 29 \)
\( (x – 2)^4 = 1 \)
\( x – 2 = \pm 1 \)
\( x = 3 \) (since \( x \geq 2 \))
Answer: \( x = 3 \)
▶️ Answer/Explanation
Solution
Part (a): Find \( k \), \( p \), and other intersection point
Line at \( \left( \frac{1}{2}, \frac{5}{2} \right) \): \( \frac{5}{2} = \frac{k}{2} + p \) (1)
Curve at \( \left( \frac{1}{2}, \frac{5}{2} \right) \): \( \frac{5}{2} = \frac{k}{8} – k + 2 \)
Simplify: \( \frac{1}{2} = \frac{k – 8k}{8} \), \( \frac{1}{2} = -\frac{7k}{8} \), \( k = \frac{4}{7} \)
From (1): \( \frac{5}{2} = \frac{4}{14} + p \), \( p = \frac{5}{2} – \frac{2}{7} = \frac{31}{14} \)
Equate: \( \frac{2}{7}x^2 – \frac{8}{7}x + 2 = \frac{4}{7}x + \frac{31}{14} \)
Multiply by 14: \( 4x^2 – 16x + 28 = 8x + 31 \)
\( 4x^2 – 24x – 3 = 0 \)
Solve: \( x = \frac{24 \pm \sqrt{576 + 48}}{8} = \frac{24 \pm 4\sqrt{39}}{8} = \frac{6 \pm \sqrt{39}}{2} \)
Other root: \( x = \frac{6 + \sqrt{39}}{2} \)
\( y = \frac{4}{7} \cdot \frac{6 + \sqrt{39}}{2} + \frac{31}{14} = \frac{12 + 2\sqrt{39} + 31}{14} = \frac{43 + 2\sqrt{39}}{14} \)
Answer: \( k = \frac{4}{7} \), \( p = \frac{31}{14} \), other point: \( \left( \frac{6 + \sqrt{39}}{2}, \frac{43 + 2\sqrt{39}}{14} \right) \)
Part (b): Possible values of \( p \) for no intersection
Equate: \( \frac{1}{2}kx^2 – 2kx + 2 = kx + p \)
\( \frac{1}{2}kx^2 – 3kx + 2 – p = 0 \)
Discriminant: \( \Delta = 9k^2 – 2k(2 – p) = 9k^2 – 4k + 2kp \)
For no roots: \( 9k^2 – 4k + 2kp < 0 \)
\( p < \frac{-9k + 4}{2} \)
Answer: \( p < \frac{4 – 9k}{2} \)