▶️ Answer/Explanation
Solution
(a) For \(y = kx^{\frac{1}{2}} – 4x^2 + 2\):
First derivative: \(\frac{dy}{dx} = k \cdot \frac{1}{2}x^{-\frac{1}{2}} – 4 \cdot 2x = \frac{k}{2}x^{-\frac{1}{2}} – 8x\).
Second derivative: \(\frac{d^2y}{dx^2} = \frac{k}{2} \cdot \left(-\frac{1}{2}\right)x^{-\frac{3}{2}} – 8 = -\frac{k}{4}x^{-\frac{3}{2}} – 8\).
Answer (a): \(\frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x\), \(\frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8\)
(b) With \(k = 2\), \(y = 2x^{\frac{1}{2}} – 4x^2 + 2\).
Steps:
1. Stationary point: \(\frac{dy}{dx} = x^{-\frac{1}{2}} – 8x = 0\).
\(\frac{1}{x^{\frac{1}{2}}} = 8x \implies 1 = 8x^{\frac{3}{2}} \implies x^{\frac{3}{2}} = \frac{1}{8} \implies x = \left(\frac{1}{8}\right)^{\frac{2}{3}} = \frac{1}{4}\).
2. Find \(y\): \(y = 2 \cdot \left(\frac{1}{4}\right)^{\frac{1}{2}} – 4 \cdot \left(\frac{1}{4}\right)^2 + 2 = 1 – \frac{1}{4} + 2 = 2.75\).
3. Nature: \(\frac{d^2y}{dx^2} = -\frac{2}{4}x^{-\frac{3}{2}} – 8 = -\frac{1}{2}x^{-\frac{3}{2}} – 8\). At \(x = \frac{1}{4}\): \(x^{-\frac{3}{2}} = 8\), so \(\frac{d^2y}{dx^2} = -\frac{1}{2} \cdot 8 – 8 = -12\) (maximum).
Answer (b): \(\left(\frac{1}{4}, 2.75\right)\), maximum
(c) Tangents at A (\(x = 0.25\)) and B (\(x = 1\)) meet at \(x = 0.6\).
Steps:
1. Point A: \(y = k \cdot \frac{1}{2} – 4 \cdot \frac{1}{16} + 2 = \frac{k}{2} + 1.75\), slope \(\frac{dy}{dx} = \frac{k}{2} \cdot 2 – 8 \cdot 0.25 = k – 2\).
Tangent: \(y = (k – 2)(x – 0.25) + \frac{k}{2} + 1.75\). At \(x = 0.6\): \(y = (k – 2) \cdot 0.35 + \frac{k}{2} + 1.75 = 0.85k + 1.05\).
2. Point B: \(y = k – 4 + 2 = k – 2\), slope \(\frac{dy}{dx} = \frac{k}{2} – 8\).
Tangent: \(y = \left(\frac{k}{2} – 8\right)(x – 1) + k – 2\). At \(x = 0.6\): \(y = \left(\frac{k}{2} – 8\right)(-0.4) + k – 2 = 0.8k + 1.2\).
3. Set equal: \(0.85k + 1.05 = 0.8k + 1.2 \implies 0.05k = 0.15 \implies k = 3\).
Answer (c): \(3\)
Final Answer: \((a) \frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x, \frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8; (b) \left(\frac{1}{4}, 2.75\right), \text{maximum}; (c) 3\)
▶️ Answer/Explanation
Solution
(a) Slope of tangent at \(x = 1\): \(f'(1) = 8(2 \cdot 1 – 3)^{\frac{1}{3}} – 10 \cdot 1^{\frac{2}{3}} = 8(-1)^{\frac{1}{3}} – 10 = -8 – 10 = -18\).
Normal slope: \(\frac{1}{18}\).
Equation at \((1, 0)\): \(y = \frac{1}{18}(x – 1)\).
Answer (a): \(y = \frac{1}{18}(x – 1)\)
(b) Integrate \(f'(x) = 8(2x – 3)^{\frac{1}{3}} – 10x^{\frac{2}{3}}\):
First term: Let \(u = 2x – 3\), \(du = 2 dx\). Then \(\int 8(2x – 3)^{\frac{1}{3}} dx = 4 \int u^{\frac{1}{3}} du = 3u^{\frac{4}{3}} = 3(2x – 3)^{\frac{4}{3}}\).
Second term: \(\int -10x^{\frac{2}{3}} dx = -10 \cdot \frac{3}{5}x^{\frac{5}{3}} = -6x^{\frac{5}{3}}\).
Thus: \(f(x) = 3(2x – 3)^{\frac{4}{3}} – 6x^{\frac{5}{3}} + c\).
Using \((1, 0)\): \(0 = 3(-1)^{\frac{4}{3}} – 6 + c = 3 – 6 + c \implies c = 3\).
Answer (b): \(f(x) = 3(2x – 3)^{\frac{4}{3}} – 6x^{\frac{5}{3}} + 3\)
(c) Given \(f'(x) = 0\) as \(125x^2 – 128x + 192 = 0\).
Discriminant: \(\Delta = (-128)^2 – 4 \cdot 125 \cdot 192 = 16384 – 96000 = -79616\).
Since \(\Delta < 0\), the quadratic \(125x^2 – 128x + 192\) has no real roots and is always positive (e.g., at \(x = 1\): \(125 – 128 + 192 = 189 > 0\)).
Since \(f'(x) = 8(2x – 3)^{\frac{1}{3}} – 10x^{\frac{2}{3}}\) is given to satisfy \(125x^2 – 128x + 192 = 0\) at \(f'(x) = 0\), assume \(f'(x) = k (125x^2 – 128x + 192)\) for some constant \(k > 0\). The quadratic is always positive, so \(f'(x) > 0\) for all \(x > 0\), implying \(f\) is strictly increasing.
Answer (c): increasing
Final Answer: \((a) y = \frac{1}{18}(x – 1); (b) f(x) = 3(2x – 3)^{\frac{4}{3}} – 6x^{\frac{5}{3}} + 3; (c) \text{increasing}\)
▶️ Answer/Explanation
Solution
(a) Coordinates of A (\(x = 2\)): \(y = 2(2)^2 – 3 = 5\), so A is \((2, 5)\).
Coordinates of B (\(x = 2 + h\)): \(y = 2(2 + h)^2 – 3 = 2(4 + 4h + h^2) – 3 = 2h^2 + 8h + 5\), so B is \((2 + h, 2h^2 + 8h + 5)\).
Gradient of chord AB: \(\frac{(2h^2 + 8h + 5) – 5}{(2 + h) – 2} = \frac{2h^2 + 8h}{h} = 2(h + 4) = 2h + 8\) for \(h \neq 0\).
Answer (a): \(2h + 8\)
(b) The gradient of the curve at A is the derivative at \(x = 2\), which can be found by taking the limit of the chord’s gradient as \(h \to 0\): \(\lim_{h \to 0} (2h + 8) = 8\).
Alternatively, differentiate \(y = 2x^2 – 3\): \(\frac{dy}{dx} = 4x\). At \(x = 2\): \(4 \cdot 2 = 8\).
Answer (b): The gradient at A is 8, deduced by taking the limit of the chord’s gradient as \(h \to 0\).
Final Answer: \((a) 2h + 8; (b) 8\)
▶️ Answer/Explanation
Solution
Part (a): Find the set of values of \( x \) for which \( y \) decreases
The function decreases when \( \frac{dy}{dx} < 0 \).
Differentiate: \( y = 4 + 5x + 6x^2 – 3x^3 \)
\( \frac{dy}{dx} = 5 + 12x – 9x^2 \)
Solve: \( 5 + 12x – 9x^2 < 0 \)
Rewrite: \( 9x^2 – 12x – 5 > 0 \)
Find roots of \( 9x^2 – 12x – 5 = 0 \):
\( x = \frac{12 \pm \sqrt{144 + 180}}{18} = \frac{12 \pm \sqrt{324}}{18} = \frac{12 \pm 18}{18} \)
\( x = \frac{30}{18} = \frac{5}{3}, \quad x = \frac{-6}{18} = -\frac{1}{3} \)
The quadratic \( 9x^2 – 12x – 5 \) opens upward (coefficient of \( x^2 \) is positive). Test intervals:
– For \( x < -\frac{1}{3} \) (e.g., \( x = -1 \)): \( 9(-1)^2 – 12(-1) – 5 = 9 + 12 – 5 = 16 > 0 \)
– For \( -\frac{1}{3} < x < \frac{5}{3} \) (e.g., \( x = 0 \)): \( 9(0)^2 – 12(0) – 5 = -5 < 0 \)
– For \( x > \frac{5}{3} \) (e.g., \( x = 2 \)): \( 9(2)^2 – 12(2) – 5 = 36 – 24 – 5 = 7 > 0 \)
Thus, \( 9x^2 – 12x – 5 > 0 \) for \( x < -\frac{1}{3} \) or \( x > \frac{5}{3} \).
Answer: \( x < -\frac{1}{3} \text{ or } x > \frac{5}{3} \)
Part (b): Find the value of \( k \) such that \( y = 9x + k \) is a tangent to the curve
For \( y = 9x + k \) to be tangent, the slope of the curve must equal 9 at the point of tangency, and the line must pass through that point.
Set derivative equal to 9: \( 5 + 12x – 9x^2 = 9 \)
\( 9x^2 – 12x + 4 = 0 \)
Solve: \( x = \frac{12 \pm \sqrt{144 – 144}}{18} = \frac{12}{18} = \frac{2}{3} \)
At \( x = \frac{2}{3} \), compute \( y \):
\( y = 4 + 5\left(\frac{2}{3}\right) + 6\left(\frac{4}{9}\right) – 3\left(\frac{8}{27}\right) \)
\( = 4 + \frac{10}{3} + \frac{24}{9} – \frac{24}{27} \)
\( = 4 + \frac{10}{3} + \frac{8}{3} – \frac{8}{9} \)
\( = \frac{36}{9} + \frac{30}{9} + \frac{24}{9} – \frac{8}{9} = \frac{82}{9} \)
Tangent line at \( x = \frac{2}{3} \): \( y = 9x + k \)
\( \frac{82}{9} = 9\left(\frac{2}{3}\right) + k \)
\( \frac{82}{9} = 6 + k \)
\( k = \frac{82}{9} – \frac{54}{9} = \frac{28}{9} \)
Answer: \( k = \frac{28}{9} \)
▶️ Answer/Explanation
Solution
Part (a): Find the equation of the tangent to the curve at A
The curve is \( y = \frac{12}{\sqrt[3]{2x + 1}} = 12 (2x + 1)^{-\frac{1}{3}} \).
Differentiate using the chain rule:
\( \frac{dy}{dx} = 12 \cdot \left(-\frac{1}{3}\right) (2x + 1)^{-\frac{4}{3}} \cdot 2 \)
\( = -8 (2x + 1)^{-\frac{4}{3}} \)
At point A \( \left( \frac{7}{2}, 6 \right) \), \( x = \frac{7}{2} \):
\( 2x + 1 = 2 \cdot \frac{7}{2} + 1 = 8 \)
\( \frac{dy}{dx} = -8 (8)^{-\frac{4}{3}} \)
\( 8^{\frac{1}{3}} = 2 \Rightarrow 8^{\frac{4}{3}} = (8^{\frac{1}{3}})^4 = 2^4 = 16 \Rightarrow 8^{-\frac{4}{3}} = \frac{1}{16} \)
\( \frac{dy}{dx} = -8 \cdot \frac{1}{16} = -\frac{1}{2} \)
The slope of the tangent is \( m = -\frac{1}{2} \).
Using point-slope form at \( \left( \frac{7}{2}, 6 \right) \):
\( y – 6 = -\frac{1}{2} \left( x – \frac{7}{2} \right) \)
\( y – 6 = -\frac{1}{2} x + \frac{7}{4} \)
\( y = -\frac{1}{2} x + \frac{7}{4} + 6 = -\frac{1}{2} x + \frac{7}{4} + \frac{24}{4} = -\frac{1}{2} x + \frac{31}{4} \)
Answer: \( y = -\frac{1}{2}x + \frac{31}{4} \)
Part (b): Find the area of the region bounded by the curve and the lines \( x = 0 \), \( x = \frac{7}{2} \), and \( y = 0 \)
The area is given by:
\( \text{Area} = \int_{0}^{\frac{7}{2}} \frac{12}{\sqrt[3]{2x + 1}} \, dx \)
Substitute \( u = 2x + 1 \):
\( \frac{du}{dx} = 2 \Rightarrow dx = \frac{du}{2} \)
Limits: when \( x = 0 \), \( u = 1 \); when \( x = \frac{7}{2} \), \( u = 8 \).
The integral becomes:
\( \int_{1}^{8} \frac{12}{u^{\frac{1}{3}}} \cdot \frac{du}{2} = \int_{1}^{8} 6 u^{-\frac{1}{3}} \, du \)
Integrate: \( \int u^{-\frac{1}{3}} \, du = \frac{u^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2} u^{\frac{2}{3}} \)
\( 6 \cdot \frac{3}{2} u^{\frac{2}{3}} = 9 u^{\frac{2}{3}} \)
Evaluate: \( 9 \left[ u^{\frac{2}{3}} \right]_{1}^{8} = 9 \left( 8^{\frac{2}{3}} – 1^{\frac{2}{3}} \right) \)
\( 8^{\frac{1}{3}} = 2 \Rightarrow 8^{\frac{2}{3}} = (2)^2 = 4 \)
\( 9 (4 – 1) = 9 \cdot 3 = 27 \)
Answer: \( 27 \)
▶️ Answer/Explanation
Solution
(a) x-coordinate for gradient \( \frac{11}{2} \)
Set \( 4x – 3\sqrt{x} + 1 = \frac{11}{2} \):
\( 4x – 3\sqrt{x} – \frac{9}{2} = 0 \)
Multiply by 2: \( 8x – 6\sqrt{x} – 9 = 0 \)
Let \( u = \sqrt{x} \), so \( x = u^2 \):
\( 8u^2 – 6u – 9 = 0 \)
Solve: \( u = \frac{6 \pm \sqrt{324}}{16} = \frac{6 \pm 18}{16} \)
\( u = \frac{3}{2} \), \( u = -\frac{3}{4} \) (discard, \( u \geq 0 \))
\( x = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \)
Answer: \( x = \frac{9}{4} \)
(b) Equation through \( (4, 11) \)
Integrate \( \frac{dy}{dx} = 4x – 3x^{1/2} + 1 \):
\( y = 2x^2 – 2x^{3/2} + x + C \)
At \( (4, 11) \):
\( 11 = 2(16) – 2(8) + 4 + C \)
\( 11 = 32 – 16 + 4 + C = 20 + C \)
\( C = -9 \)
Equation: \( y = 2x^2 – 2x^{3/2} + x – 9 \)
Answer: \( y = 2x^2 – 2x^{3/2} + x – 9 \)
▶️ Answer/Explanation
Solution
Differentiate: \( y = x^2 – a x^{-1} \)
\( \frac{dy}{dx} = 2x + \frac{a}{x^2} \)
At stationary point \( x = -3 \), set \( \frac{dy}{dx} = 0 \):
\( 2(-3) + \frac{a}{9} = -6 + \frac{a}{9} = 0 \)
\( a = 54 \)
Find \( b \): \( y = (-3)^2 – \frac{54}{-3} = 9 + 18 = 27 \)
Answer: \( a = 54 \), \( b = 27 \)