Exam-Style Problems

9709 P12 - Jun 2021 - Q3 - 4 marks

87

The equation of a curve is \(y=(x-3)\sqrt{x+1}+3\). The following points lie on the curve. Non-exact values are rounded to 4 decimal places.

Point	Coordinates
A	\((2,k)\)
B	\((2.9,2.8025)\)
C	\((2.99,2.9800)\)
D	\((2.999,2.9980)\)
E	\((3,3)\)

(a) Find k, giving your answer correct to 4 decimal places. [1]

(b) Find the gradient of AE, giving your answer correct to 4 decimal places. [1]

The gradients of BE, CE and DE, rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively.

(c) State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at E. [2]

9709 P13 - Nov 2021 - Q3 - 5 marks

1154

(a) Express \(5y^2-30y+50\) in the form \(5(y+a)^2+b\), where a and b are constants. [2]

The function f is defined by \(f(x)=x^5-10x^3+50x\) for \(x\in\mathbb R\).

(b) Determine whether f is an increasing function, a decreasing function or neither. [3]

9709 P13 - Jun 2021 - Q2 - 4 marks

1155

The function f is defined by \[f(x)=\frac13(2x-1)^{3/2}-2x\quad\text{for }1/2<x<a.\] It is given that f is a decreasing function.

Find the maximum possible value of the constant a. [4]

9709 P13 - Jun 2011 - Q4

1183

Differentiate \(\frac{2x^3 + 5}{x}\) with respect to \(x\).

9709 P1 - Jun 2003 - Q3

1184

Differentiate \(4x + \frac{6}{x^2}\) with respect to \(x\).

9709 P13 - Nov 2025 - Q3 - 3 marks

9667

A curve is \(y=f(x)=\frac12x^{2/3}(x-2)^2\). Points on it are \[A(8,72),\quad B(8.001,k),\quad C(8.01,72.300388),\quad D(8.1,75.038882).\] Non-exact ordinates are given to 6 decimal places.

(a) Find \(k\), to 6 decimal places. [1]

The chord gradients below are given to 4 decimal places.

Chord	AB	AC	AD
Gradient	30.0039	30.0388

(b) Find the gradient of \(AD\), to 4 decimal places. [1]

(c) State what the table suggests about \(f^{\prime}(8)\). [1]

9709 P12 - Nov 2024 - Q3 - 5 marks

10059

On \(y=2x^2-3\), A and B have x-coordinates 2 and 2+h.

(a) Find and simplify the gradient of chord AB in terms of h. [3]

(b) Explain how to deduce the gradient at A and state it. [2]

9709 P11 - Jun 2024 - Q4 - 3 marks

10091

The curve is \(y=f(x)=(2x-1)\sqrt{3x-2}-2\). Its points are listed below; the y-coordinates are rounded to 5 decimal places.

Point	x	y
A	2	4
B	2.0001	k
C	2.001	4.00625
D	2.01	4.06261
E	2.1	4.63566
F	3	11.22876

(a) Find k to 5 decimal places. [1]

The gradients of AB, AC, AD and AF are 6.2501, 6.2511, 6.2608 and 7.2288 respectively.

(b) Calculate the gradient of AE to 4 decimal places. [1]

(c) Estimate f′(2). [1]

A polynomial with a repeated stationary factor · Maths4U Original · 8 marks

16843

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-DIFF-001

For real x, let \(f(x)=3x^4-8x^3+6x^2-5\).

Find f′(x) and express it as a product of factors. [3]
Find f″(x). [2]
Find every point on the curve where the tangent is horizontal. [3]

Solution

Hint

Differentiate term by term, then factor the first derivative before solving for horizontal tangents.

Worked solution

(a)

Differentiate each nonconstant term: \(\displaystyle f'=12x^{3}-24x^{2}+12x\). Taking out 12x leaves \(\displaystyle x^{2}-2x+1=(x-1)^{2}\), so \(f^{\prime}=12x(x-1)^2\).

Answer: \(f^{\prime}(x)=12x^3-24x^2+12x=12x(x-1)^2\).

(b)

Differentiate the expanded first derivative to get \(f^{\prime\prime}=36x^2-48x+12\).

Answer: \(36x^2-48x+12\).

(c)

Horizontal tangents require \(\displaystyle 12x(x-1)^{2}=0\), so \(\displaystyle x=0\) or 1. Substituting into the original function gives \(\displaystyle f(0)=-5\) and \(\displaystyle f(1)=3-8+6-5=-4\). The points are (0,−5) and (1,−4). A repeated root is counted once.

Answer: \((0,-5)\) and \((1,-4)\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Differentiate the quartic and cubic terms correctly.
1 mark: Differentiate the quadratic and constant terms correctly.
1 mark: Factor the complete derivative as \(\displaystyle 12x(x-1)^{2}\).

Alternative methods: Expand and then factor, or use an equivalent correct differentiation method.

Common error: Differentiating the constant as a nonzero term or losing the repeated factor.

(b) · 2 marks

1 mark: Differentiate the cubic and quadratic terms of f′ correctly.
1 mark: Give the complete second derivative \(\displaystyle 36x^{2}-48x+12\).

Alternative methods: Differentiating an equivalent factored expression correctly is accepted.

Common error: Reusing the original exponents instead of differentiating f′.

(c) · 3 marks

1 mark: Find both and only stationary inputs 0 and 1.
1 mark: Obtain the corresponding ordinates −5 and −4.
1 mark: Give the two points, with no duplicated root.

Alternative methods: Solving the unfactored derivative and checking each point is accepted.

Common error: Giving only x-values or assuming the repeated root is not a stationary point.

Total: 8 marks.

A radical and a reciprocal · Maths4U Original · 8 marks

16844

Notes & worked examples

Info

Maths4U Original · M4U-ORIG-DIFF-002

For x>0, let \(\displaystyle g(x)= \sqrt{3x+ 1}+ \frac{2}{x}\).

Find g′(x). [3]
Find g″(x). [3]
Find g′(1) and interpret its sign for the curve near \(\displaystyle x=1\). [2]

Solution

Hint

Rewrite the radical and reciprocal as powers. Keep the factor 3 from the linear inner function at each differentiation.

Worked solution

(a)

Write \(\displaystyle g= (3x+ 1)^{\frac{1}{2}}+ 2x^{- 1}\). The radical contributes \(\displaystyle \tfrac{1}{2}(3x+ 1)^{- \frac{1}{2}}(3)\), while the reciprocal contributes \(\displaystyle -2x^{−2}\). Hence \(\displaystyle g^{\prime }= \frac{3}{2\sqrt{3x+ 1}}- \frac{2}{x^{2}}\), valid for x>0.

Answer: \(\dfrac3{2\sqrt{3x+1}}-\dfrac2{x^2}\).

(b)

Differentiate \(\displaystyle (\frac{3}{2})(3x+ 1)^{- \frac{1}{2}}\): the coefficient becomes \(\displaystyle (\frac{3}{2})(- \frac{1}{2})(3)= - \frac{9}{4}\) and the exponent is \(\displaystyle - \frac{3}{2}\). The second term \(\displaystyle -2x^{−2}\) differentiates to \(\displaystyle 4x^{−3}\). Thus \(\displaystyle g^{\prime \prime }= - \frac{9}{[4(3x+ 1)^{\frac{3}{2}}]}+ \frac{4}{x^{3}}\).

Answer: \(\displaystyle - \dfrac{9}{4(3x+ 1)^{\frac{3}{2}}}+ \dfrac{4}{x^{3}}\).

(c)

\(\displaystyle g^{\prime }(1)= \frac{3}{2\sqrt{4}}- 2= \frac{3}{4}- 2= - \frac{5}{4}\). The derivative is continuous and negative at this interior point of the positive domain, so it stays negative in a sufficiently small neighbourhood. The curve is decreasing there; this statement does not assert that it decreases for every positive x.

Answer: \(\displaystyle g^{\prime }(1)= - \frac{5}{4}\); the curve is decreasing near \(\displaystyle x=1\).

Maths4U indicative marking guidance

Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.

(a) · 3 marks

1 mark: Apply the outer half-power derivative correctly.
1 mark: Include the inner multiplier 3.
1 mark: Differentiate \(\displaystyle \frac{2}{x}\) as \(\displaystyle - \frac{2}{x^{2}}\) and give the complete result.

Alternative methods: Any algebraically equivalent exact expression on the stated domain is accepted.

Common error: Omitting the inner derivative or giving a positive derivative for \(\displaystyle \frac{2}{x}\).

(b) · 3 marks

1 mark: Obtain the radical-part exponent \(\displaystyle - \frac{3}{2}\).
1 mark: Obtain its full coefficient \(\displaystyle - \frac{9}{4}\), including the second inner multiplier.
1 mark: Give the reciprocal-part derivative \(\displaystyle \frac{4}{x^{3}}\) and the complete result.

Alternative methods: Equivalent denominator or negative-index notation is accepted.

Common error: Applying the inner derivative only during the first differentiation.

(c) · 2 marks

1 mark: Evaluate the exact gradient \(\displaystyle - \frac{5}{4}\).
1 mark: Interpret it as a negative slope/decreasing behaviour near the stated input.

Alternative methods: A correct local tangent-slope interpretation is accepted.

Common error: Treating the negative gradient as a negative ordinate or claiming global decrease.

Total: 8 marks.