Standard +0.3 This is a standard discriminant problem requiring students to apply b²-4ac < 0 for no real roots, then solve a quadratic inequality. While it involves multiple steps (discriminant condition, algebraic manipulation, inequality solving including the k=0 case), these are well-practiced AS-level techniques with no novel insight required, making it slightly easier than average.
Realises that k = 0 will give no real roots as equation becomes
3 = 0 (proof by contradiction)
B1
3.1a
(For k (cid:122)0)quadratic has norealroots provided
Answer
Marks
Guidance
b2 (cid:31)4ac so16k2 (cid:31)12k
M1
2.4
4k(4k(cid:16)3)(cid:31)0with attempt at solution
M1
1.1b
So 0(cid:31)k (cid:31) 3, which together with k = 0 gives 0 k (cid:31) 3*
Answer
Marks
Guidance
4 4
A1*
2.1
(4 marks)
Notes:
B1: Explains why k = 0 gives no real roots
M1: Considers discriminant to give quadratic inequality – does not need the k (cid:122)0 for this
mark
M1: Attempts solution of quadratic inequality
A1*: Draws conclusion, which is a printed answer, with no errors (dependent on all three
previous marks)
Answer
Marks
Guidance
Question
Scheme
Marks
11 (a)
Answer
Marks
Guidance
Way 1
Since x and y are positive, their square roots are real and so
( x (cid:16) y)2(cid:116)0 giving x(cid:16)2 x y (cid:14) y(cid:116)0
M1
2.1
(cid:63) 2 xy (cid:100) x(cid:14) y provided x and y are positive and so
x(cid:14) y
xy (cid:100) *
Answer
Marks
Guidance
2
A1*
2.2a
(2)
Way 2
Longer
Answer
Marks
Guidance
method
Since (x(cid:16) y)2(cid:116)0 for real values of x and y, x2 (cid:16)2xy(cid:14) y2(cid:116)0 and
so 4xy(cid:100) x2 (cid:14)2xy(cid:14) y2 i.e. 4xy(cid:100)(x(cid:14) y)2
M1
2.1
(cid:63) 2 xy (cid:100) x(cid:14) y provided x and y are positive and so
x(cid:14) y
xy (cid:100) *
Answer
Marks
Guidance
2
A1*
2.2a
(2)
Answer
Marks
Guidance
(b)
Let x = (cid:16)3 and y = (cid:16)5 then LHS = 15 and RHS= (cid:16)4 so as
15 (cid:33)(cid:16)4 result does not apply
B1
2.4
(1)
(3 marks)
Notes:
(a)
M1: Need two stages of the three stage argument involving the three stages, squaring, square
rooting terms and rearranging
A1*: Need all three stages making the correct deduction to achieve the printed result
(b)
B1: Chooses two negative values and substitutes, then states conclusion
Answer
Marks
Guidance
Question
Scheme
Marks
Question 10:
10 | Realises that k = 0 will give no real roots as equation becomes
3 = 0 (proof by contradiction) | B1 | 3.1a
(For k (cid:122)0)quadratic has norealroots provided
b2 (cid:31)4ac so16k2 (cid:31)12k | M1 | 2.4
4k(4k(cid:16)3)(cid:31)0with attempt at solution | M1 | 1.1b
So 0(cid:31)k (cid:31) 3, which together with k = 0 gives 0 k (cid:31) 3*
4 4 | A1* | 2.1
(4 marks)
Notes:
B1: Explains why k = 0 gives no real roots
M1: Considers discriminant to give quadratic inequality – does not need the k (cid:122)0 for this
mark
M1: Attempts solution of quadratic inequality
A1*: Draws conclusion, which is a printed answer, with no errors (dependent on all three
previous marks)
Question | Scheme | Marks | AOs
11 (a)
Way 1 | Since x and y are positive, their square roots are real and so
( x (cid:16) y)2(cid:116)0 giving x(cid:16)2 x y (cid:14) y(cid:116)0 | M1 | 2.1
(cid:63) 2 xy (cid:100) x(cid:14) y provided x and y are positive and so
x(cid:14) y
xy (cid:100) *
2 | A1* | 2.2a
(2)
Way 2
Longer
method | Since (x(cid:16) y)2(cid:116)0 for real values of x and y, x2 (cid:16)2xy(cid:14) y2(cid:116)0 and
so 4xy(cid:100) x2 (cid:14)2xy(cid:14) y2 i.e. 4xy(cid:100)(x(cid:14) y)2 | M1 | 2.1
(cid:63) 2 xy (cid:100) x(cid:14) y provided x and y are positive and so
x(cid:14) y
xy (cid:100) *
2 | A1* | 2.2a
(2)
(b) | Let x = (cid:16)3 and y = (cid:16)5 then LHS = 15 and RHS= (cid:16)4 so as
15 (cid:33)(cid:16)4 result does not apply | B1 | 2.4
(1)
(3 marks)
Notes:
(a)
M1: Need two stages of the three stage argument involving the three stages, squaring, square
rooting terms and rearranging
A1*: Need all three stages making the correct deduction to achieve the printed result
(b)
B1: Chooses two negative values and substitutes, then states conclusion
Question | Scheme | Marks | AOs
The equation $kx^2 + 4kx + 3 = 0$, where $k$ is a constant, has no real roots.
Prove that
$$0 \leqslant k < \frac{3}{4}$$
[4]
\hfill \mbox{\textit{Edexcel AS Paper 1 Q10 [4]}}