Mathematics · Oct / Nov 2021 · Paper 21 · Question 9

x

−

y

…

0

⇒

x

−

...

xy

+

y

…4

0

⇒

x

−

2

xy

+

y

…4

0

x

+

y

⇒

…

xy

2

x

y

+

States for example when

x

= −

8,

y

= −

2,

= −,

5

xy =

4

so

2

x

+

y

<

xy

2

(a)

M1: Sets up a correct inequality and attempts to expand (

)

2

x

−

y

leading to three terms.

A1:

Correct expanded equation.

A1:

Rearranges to the required equation with no errors seen.

If working in reverse allow the first M and A (if steps correct) but require also a minimal

conclusion for the final A.

(b)

B1:

Gives a suitable example with both sides evaluated correctly and a minimal conclusion.

There is no need to refer to x and y in the conclusion, so long as it has been shown the

−−

8

2

=

−

5

…

4

=

16

=

−

8

×−

2

required inequality does not hold. E.g.

QED is

2

fine.

A common response most likely to score 2 out of 3 marks

(

)

2

2

2

9.(a)

x

+

y

x

+

y

x

+

...

xy

+

y

…

xy

⇒

…

xy

⇒

…

xy

Alt 1

2

4

4

x

2

2

xy

y

2

…

0

(

x

y

)

2

…

0

⇒

−

+

⇒

−

States both of the following o.e

(

)

2

x

−

y

… as it is a square number

0

x

+

y

so

…

xy

is true

2

x

+

y

…

xy

M1: Assumes

true and attempts to square obtaining at least three terms.

2

A1:

Correct expansion and rearranges the inequality correctly to factorise to a perfect square.

A1:

A complete conclusion given.

9.(a)

States (

)

2

2

2

x

−

y

…

0

⇒

x

−

2

xy

+

y

…

0

Alt 2

2

2

(

)

2

Rearranges

⇒

x

+

2

xy

+

y

…

4

xy

⇒

x

+

y

…4

4

xy

States that as x, y positive, so x+ y > 0 (and xy > 0)

x

+

y

(

)

⇒

x

+

y

…4

4

xy

⇒

…4

xy

2

M1: Sets up an inequality using an appropriate perfect square and expands to at least three

terms.

A1:

Makes a correct rearrangement and factors the left hand side to produce the equation

shown.

A1:

Makes a full conclusion justifying why the square root gives x + y.

π

π

π

2

π

tan

2

2

x

+

=

3

⇒

2

x

+

= ±

,

±