Additional Mathematics 4037/11 — October/November 2010

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Trigonometry · Calculus · Straight-line graphs · Quadratic functions · Factors of polynomials · Series · +3 more

Q1Quadratic functionsFree sample

The equation of a curve is given by $y = 2x^2 + ax + 14$ , where $a$ is a constant.

Given that this equation can also be written as $y = 2(x - 3)^2 + b$ , where $b$ is a constant, find

(i)

the value of $a$ and of $b$ ,

DifficultyMedium-Easy

Worked solution

Approach

Expand $y = 2(x - 3)^2 + b$ and compare coefficients with $y = 2x^2 + ax + 14$ .

Working

2(x - 3)^2 + b = 2\left(x^2 - 6x + 9\right) + b = 2x^2 - 12x + (18 + b)

Comparing coefficients with $y = 2x^2 + ax + 14$ :

a = -12

18 + b = 14 \quad \Rightarrow \quad b = -4

Answer

a = -12, \qquad b = -4

Final answer

a = -12, b = -4

Detailed explanation

Walkthrough

The two forms describe the same curve, so expanding the completed-square form must reproduce the original expression. Expanding $(x-3)^2 = x^2 - 6x + 9$ , multiplying by 2 gives $2x^2 - 12x + 18$ , and adding $b$ shifts only the constant term. Matching the coefficient of $x$ immediately gives $a = -12$ , and matching the constant terms, $18 + b = 14$ , gives $b = -4$ .

Key Takeaways

Completed-square form makes the coefficients readable by direct comparison; the squared bracket determines the linear coefficient and the outside constant adjusts the constant term.

Common Mistakes

Sign errors when expanding $(x-3)^2$ (writing $+6x$ instead of $-6x$ ); forgetting to multiply the constant 9 by 2 before comparing; equating $b$ to 14 instead of solving $18 + b = 14$ .

Things to Be Careful About

Both values are required for full marks — one B1 each. Keep exact integer values; no rounding is involved.

Techniques used

expand the completed-square formcompare coefficients of like powers of x

(ii)

the minimum value of $y$ .

DifficultyEasy

Worked solution

Approach

In the form $y = 2(x - 3)^2 + b$ , the squared term is never negative, so the least value of $y$ occurs when it is zero.

Working

Since $2(x - 3)^2 \geq 0$ for all $x$ , with equality at $x = 3$ :

y_{\min} = b = -4

Answer

-4

Final answer

-4

Detailed explanation

Walkthrough

The completed-square form shows that $y$ equals a non-negative quantity plus $b$ . The smallest possible value therefore occurs when the bracket vanishes, at $x = 3$ , giving $y = b = -4$ . This uses the value of $b$ found in part (i).

Key Takeaways

For $y = A(x - h)^2 + k$ with $A > 0$ , the minimum value is simply $k$ , attained at $x = h$ .

Common Mistakes

Giving the $x$ -coordinate of the minimum instead of the minimum value of $y$ ; quoting $+4$ through a sign slip.

Things to Be Careful About

The mark scheme allows follow-through on the candidate's own $b$ value, but the correct answer here is exactly $-4$ .

Techniques used

read the minimum value from the completed-square form

The rest of this paper

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Q8Logarithmic and exponential functions · Straight-line graphs7M
Q9Calculus8M
Q10Circular measure · Trigonometry10M
Q11Calculus10M
Q12Calculus · Straight-line graphs22M

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