Home / CIE A level -Pure Mathematics 1 : Topic :  1.7 Differentiation: limit of the gradients: Exam Questions Paper 1

Question

The equation of a curve is \(y = kx^{\frac{1}{2}} – 4x^2 + 2\), where \(k\) is a constant.
(a) Find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\) in terms of \(k\).
(b) Given \(k = 2\), find the coordinates of the stationary point and determine its nature.
(c) Points A and B on the curve have x-coordinates 0.25 and 1 respectively. For a different value of \(k\), the tangents to the curve at points A and B meet at a point with x-coordinate 0.6. Find this value of \(k\).

▶️ 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: \((a) \frac{dy}{dx} = \frac{k}{2}x^{-\frac{1}{2}} – 8x, \frac{d^2y}{dx^2} = -\frac{k}{4}x^{-\frac{3}{2}} – 8; (b) \left(\frac{1}{4}, 2.75\right), \text{maximum}; (c) 3\)

Question

A function \(f\) with domain \(x > 0\) is such that \(f'(x) = 8(2x – 3)^{\frac{1}{3}} – 10x^{\frac{2}{3}}\). The curve with equation \(y = f(x)\) passes through the point \((1, 0)\).
(a) Find the equation of the normal to the curve at the point \((1, 0)\).
(b) Find \(f(x)\).
(c) Given that \(f'(x) = 0\) can be expressed as \(125x^2 – 128x + 192 = 0\), determine whether \(f\) is an increasing function, a decreasing function, or neither.

▶️ 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: \((a) y = \frac{1}{18}(x – 1); (b) f(x) = 3(2x – 3)^{\frac{4}{3}} – 6x^{\frac{5}{3}} + 3; (c) \text{increasing}\)

Question

The equation of a curve is \(y = 2x^2 – 3\). Two points A and B with x-coordinates 2 and \(2 + h\) respectively lie on the curve.
(a) Find and simplify an expression for the gradient of the chord AB in terms of \(h\).
(b) Explain how the gradient of the curve at the point A can be deduced from the answer to part (a), and state the value of this gradient.

▶️ 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: \((a) 2h + 8; (b) 8\)

Question

The equation of a curve is \( y = 4 + 5x + 6x^2 – 3x^3 \).
(a) Find the set of values of \( x \) for which \( y \) decreases as \( x \) increases.
(b) It is given that \( y = 9x + k \) is a tangent to the curve. Find the value of the constant \( k \).

▶️ 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: \( k = \frac{28}{9} \)

Question

The diagram shows part of the curve with equation \( y = \frac{12}{\sqrt[3]{2x + 1}} \). The point A on the curve has coordinates \( \left( \frac{7}{2}, 6 \right) \).
Curve Diagram
(a) Find the equation of the tangent to the curve at A. Give your answer in the form \( y = mx + c \).
(b) Find the area of the region bounded by the curve and the lines \( x = 0 \), \( x = \frac{7}{2} \), and \( y = 0 \).

▶️ 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: \( 27 \)

Question

Curve has \( \frac{dy}{dx} = 4x – 3\sqrt{x} + 1 \).
(a) Find x-coordinate where gradient is \( \frac{11}{2} \).
(b) Curve passes through \( (4, 11) \). Find equation.

▶️ 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: \( y = 2x^2 – 2x^{3/2} + x – 9 \)

Question

Curve \( y = x^2 – \frac{a}{x} \) has stationary point at \( (-3, b) \). Find \( a \) and \( b \).

▶️ 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: \( a = 54 \), \( b = 27 \)

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