Core 1 A Level Maths
Core 1 A Level Maths – Introduction
The old Core 1 A Level Maths curriculum was a part of the A Level Mathematics syllabus in the United Kingdom. It covered various topics in mathematics, primarily focusing on algebra, calculus, and numerical methods. This curriculum aimed to provide students with a strong foundation in mathematical concepts and problem-solving skills.
The old Core 1 A Level Maths curriculum was replaced by the new A Level Mathematics curriculum, which was introduced in September 2017. The new curriculum introduced changes in the content and assessment structure, aiming to align the A Level Mathematics syllabus with the needs of higher education and the demands of modern mathematics
Core 1 A Level Maths - Questions 1 - 3
Q1.
Solutions
Part a) 16^{\frac{1}{2}}=\sqrt{ } 16=4
Part b) 16^{\frac{3}{2}}=\left(16^{\frac{1}{2}}\right)^3=4^3=64
Q2
Solutions
\begin{aligned} & \int\left(6 x^2+x^{\frac{1}{2}}\right) \mathrm{d} x \\ & =6 \frac{x^3}{3}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c \\ & =2 x^3+\frac{2}{3} x^{\frac{3}{2}}+c \end{aligned}Q3
Solutions
Part a)
a_2=2 a_1-1=4-1=3a_3=2 a_2-1=6-1=5
Part b)
a_4=2 a_3-1=10-1=9\begin{aligned} & a_5=2 a_4-1=18-1=17 \\ & 5 \\ & \sum \quad a_r=a_1+a_2+a_3+a_4+a_5=2+3+5+9+17=36 \\ & r=1 \end{aligned}
Core 1 A Level Maths - Questions 4 - 6
Solutions
Part a)
(5+\sqrt{ } 2)^2=(5+\sqrt{ } 2)(5+\sqrt{ } 2)=25+10 \sqrt{ } 2+2=27+10 \sqrt{ } 2Part b)
(5-\sqrt{ } 2)^2=(5-\sqrt{ } 2)(5-\sqrt{ } 2)=25-10 \sqrt{ } 2+2=27-10 \sqrt{ } 2\begin{aligned} & (5+\sqrt{ } 2)^2-(5-\sqrt{ } 2)^2 \\ & =(27+10 \sqrt{ } 2)-(27-10 \sqrt{ } 2) \\ & =27+10 \sqrt{ } 2-27+10 \sqrt{ } 2 \\ & =20 \sqrt{ } 2 \end{aligned}
Q5
Solutions
\begin{aligned} & x-3 y=6 \\ & x=6+3 y \end{aligned}Substituting 3 x y+x=24:
\begin{aligned} & 3 y(6+3 y)+(6+3 y)=24 \\ & 18 y+9 y^2+6+3 y=24 \\ & 9 y^2+21 y-18=0 \end{aligned}
\begin{aligned} & 3 y^2+7 y-6=0 \\ & (3 y-2)(y+3)=0 \\ & y=\frac{2}{3}, y=-3 \end{aligned}
Sub into x=6+3 y
\begin{aligned} & y=\frac{2}{3} \Rightarrow x=6+2=8 \\ & y=-3 \Rightarrow x=6-9=-3 \\ & x=-3, y=-3 \text { or } x=8, y=\frac{2}{3} \end{aligned}
Q6.
Solutions
Part a)
\text { Gradient of } l_1=\frac{y_2-y_1}{x_2-x_1}=\frac{4-8}{5-(-3)}=-\frac{4}{8}=-\frac{1}{2}\begin{aligned} & y-y_1=m\left(x-x_1\right) \\ & y-4=-\frac{1}{2}(x-5) \\ & y-4=-\frac{1}{2} x+\frac{5}{2} \\ & \frac{1}{2} x+y-\frac{13}{2}=0 \\ & x+2 y-13=0 \end{aligned}Part b)
For perpendicular lines m_1 m_2=-1
m_1=-\frac{1}{2} \text {, so } m_2=2
Equation for l_2 is y=2 x
Part c)
\text { Substitute } y=2 x \text { into } x+2 y-13=0 \text { : }\begin{aligned} & x+4 x-13=0 \\ & 5 x=13 \\ & x=2 \frac{3}{5} \\ & y=2 x=5 \frac{1}{5} \end{aligned}
Coordinates of P are \left(2 \frac{3}{5}, 5 \frac{1}{5}\right)
Core 1 A Level Maths – Questions 7 – 8
Solutions
Part ai) x^2+4 x+c=(x+2)^2-4+c=(x+2)^2+(c-4)
So a=2
aii) b=c-4
aiii) \begin{aligned} & (x+2)^2-4+c=0 \\ & (x+2)^2=4-c \\ & 4-c>0 \\ & c<4 \end{aligned}
bi) 3 x<20-x
\begin{aligned} & 3 x+x<20 \\ & 4 x<20 \\ & x<5 \end{aligned}bii) Solve x^2+4 x-21=0
\begin{aligned} & (x+7)(x-3)=0 \\ & x=-7, x=3 \end{aligned}x^2+4 x-21>0 \text { when } x<-7 \text { or } x>3Part c)
Conclusion:
The old Core 1 A Level Maths curriculum was a component of the A Level Mathematics syllabus in the UK. However, it was replaced by the new A Level Mathematics curriculum in September 2017, which brought updates to the content and assessment structure.