Everything you need to know about A-Level Maths Paper 1 in 131 minutes

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简要总结

本视频是针对剑桥AS数学第一卷的考前速成课程，重点讲解了十大关键考点，并通过例题详细分析了解题技巧。内容涵盖二项式展开、二次函数、三角方程、数列与级数、弧度制、圆的计算、函数、积分和微分、图像变换等多个核心主题。

  • 二项式展开：掌握展开式及系数比较法。
  • 二次函数：熟练运用判别式解决问题。
  • 三角方程：灵活运用三角恒等式。
  • 数列与级数：掌握等差数列和等比数列的公式。
  • 弧度制：能够进行弧度与角度的转换，并应用于面积和弧长计算。
  • 圆的计算：掌握圆的方程，并结合坐标几何知识解题。
  • 函数：理解定义域和值域，能够求反函数。
  • 积分和微分：掌握基本积分和微分方法，以及面积计算。
  • 图像变换：理解函数图像的平移、伸缩和反射。

二项式展开

二项式展开是AS数学第一卷的常见考点。讲解了如何展开二项式，并通过比较系数来解决问题。 第一个问题是求 $(2x + \frac{1}{4x^2})^6$ 展开式中与 x 无关的项，即常数项。通过二项式定理展开，找到 $x$ 的幂为零的项，计算出常数项为15。 第二个问题是，在 $(2x^2 + \frac{a}{x})^6$ 的展开式中，$x^6$ 和 $x^3$ 的系数相等，求非零常数 $a$ 的值。通过展开找到 $x^6$ 和 $x^3$ 的项，令其系数相等，解得 $a = \frac{3}{2}$。最后，确定展开式中 $x^6$ 的系数为0。

二次函数

二次函数是AS数学的重要组成部分，通常涉及判别式的应用。 第一个问题是，曲线 $y = kx^2 + 2x - k$ 与直线 $y = kx - 2$ 不相交，求 $k$ 的取值范围。首先，令两方程相等，得到一个关于 $x$ 的二次方程。由于曲线与直线不相交，该二次方程无实数解，即判别式 $b^2 - 4ac < 0$。解不等式得到 $\frac{2}{5} < k < 2$。 第二个问题是，将 $16x^2 - 24x + 10$ 表示为 $(4x - a)^2 + b$ 的形式，然后利用配方法解方程 $16x^2 - 24x + 10 = k$，其中 $k$ 为常数，且方程恰有一个根。通过配方，得到 $16x^2 - 24x + 10 = (4x - 3)^2 + 1$。然后，利用判别式等于零的条件，求出 $k = 1$，并最终解得 $x = \frac{3}{4}$。

三角方程

三角方程是AS数学试卷中的常见题型，需要熟练掌握三角恒等式并灵活运用。 第一个问题是，证明 $\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} = \frac{2}{\sin \theta \cos \theta}$。通过通分、化简，利用平方差公式和三角恒等式 $\sin^2 \theta + \cos^2 \theta = 1$，最终证明等式成立。然后，利用该恒等式解方程 $\frac{\tan \theta}{1 + \cos \theta} + \frac{\tan \theta}{1 - \cos \theta} = \frac{6}{\tan \theta}$，得到 $\cos \theta = \pm \frac{1}{\sqrt{3}}$，解得 $\theta = 54.7^\circ$ 和 $\theta = 125.3^\circ$。 第二个问题是，证明 $\frac{1 - 2\sin^2 \theta}{1 - \sin^2 \theta} = 1 - \tan^2 \theta$。利用三角恒等式 $1 - \sin^2 \theta = \cos^2 \theta$，将原式化简为 $\frac{1}{\cos^2 \theta} - \frac{2\sin^2 \theta}{\cos^2 \theta}$，再利用 $\sec^2 \theta = 1 + \tan^2 \theta$，最终证明等式成立。然后，利用该恒等式解方程 $\frac{1 - 2\sin^2 \theta}{1 - \sin^2 \theta} = 2\tan^4 \theta$，通过设 $y = \tan^2 \theta$，将方程转化为二次方程，解得 $\tan \theta = \pm \frac{1}{\sqrt{2}}$，最终得到 $\theta = 35.3^\circ$ 和 $\theta = 144.7^\circ$。

数列与级数

数列与级数是AS数学的常见考点，需要掌握等差数列和等比数列的公式。 第一个问题是，等差数列的首项为 $a$，公差为 -4；等比数列的首项为 $5a$，公比为 $-\frac{a}{4}$。已知等比数列的和为无穷大，且等于等差数列前8项的和，求 $a$ 的值。首先，利用等比数列求和公式 $S_\infty = \frac{a}{1 - r}$，求出等比数列的和为 $4a$。然后，利用等差数列求和公式 $S_n = \frac{n}{2}[2a + (n - 1)d]$，求出等差数列前8项的和为 $4(2a - 28)$。令两式相等，解得 $a = 28$。最后，求等差数列的第 $k$ 项为0，解得 $k = 8$。 第二个问题是，等差数列的首项为 84，公差为 -3。求第 $n$ 项为负数的最小 $n$ 值。首先，利用等差数列通项公式 $a_n = a + (n - 1)d$，令 $a_n < 0$，解得 $n > 29$。因此，最小的 $n$ 值为 30。然后，已知该等差数列前 $2k$ 项的和等于前 $k$ 项的和，求 $k$ 的值。利用等差数列求和公式，列出方程 $\frac{2k}{2}[2(84) + (2k - 1)(-3)] = \frac{k}{2}[2(84) + (k - 1)(-3)]$，解得 $k = 19$。

弧度制

弧度制是AS数学的重要内容，需要掌握弧度与角度的转换，并应用于面积和弧长计算。 题目描述：在图示中，X和Y是直线AB上的点，使得BX=9，AY=11。弧BC是以X为圆心，9厘米为半径的圆的一部分，其中CX垂直于AB。弧AC是以Y为圆心，11厘米为半径的圆的一部分。 第一个问题是，证明角XYC=0.9582弧度。利用正弦函数定义，$\sin(XYC) = \frac{9}{11}$，计算得到 $XYC = \arcsin(\frac{9}{11}) = 0.9582$ 弧度。 第二个问题是，求ABC的周长。首先，计算XY的长度，利用勾股定理得到 $XY = \sqrt{11^2 - 9^2} = \sqrt{40}$。然后，计算AB的长度，AB=11+9-$\sqrt{40}$=20-$\sqrt{40}$。接着，计算弧AC的长度，弧AC=110.9582。最后，计算弧BC的长度，弧BC=9$\frac{\pi}{2}$。将这些长度加起来，得到ABC的周长为 38.4 厘米。 第三个问题是，一个金属板ABC，AB边长为6厘米，弧AC是以B为圆心，6厘米为半径的圆的一部分，弧BC是以A为圆心，求金属板的周长和面积。由于三角形ABC是等边三角形，每个角为 $\frac{\pi}{3}$ 弧度。因此，弧AC和弧BC的长度均为 $6 \times \frac{\pi}{3} = 2\pi$。金属板的周长为 $6 + 2\pi + 2\pi = 6 + 4\pi$。金属板的面积等于两个扇形的面积减去一个三角形的面积，即 $2 \times (\frac{1}{2} \times 6^2 \times \frac{\pi}{3}) - (\frac{1}{2} \times 6 \times 6 \times \sin(\frac{\pi}{3})) = 12\pi - 9\sqrt{3}$。

圆的计算

圆的计算是AS数学的常见题型，需要掌握圆的方程，并结合坐标几何知识解题。 第一个问题是，一个圆的圆心为 (5, 2)，且经过点 (7, 5)，求圆的方程。首先，利用两点间距离公式计算半径的平方 $r^2 = (7 - 5)^2 + (5 - 2)^2 = 13$。然后，利用圆的标准方程 $(x - a)^2 + (y - b)^2 = r^2$，得到圆的方程为 $(x - 5)^2 + (y - 2)^2 = 13$。 第二个问题是，直线 $y = 5x - 10$ 与圆相交于 A 和 B 两点，求弦 AB 的长度。首先，将直线方程代入圆的方程，得到一个关于 $x$ 的二次方程。解该二次方程，得到 $x = 2$ 或 $x = 3$。然后，将 $x$ 的值代入直线方程，得到 A 点坐标为 (2, 0)，B 点坐标为 (3, 5)。最后，利用两点间距离公式计算弦 AB 的长度 $AB = \sqrt{(3 - 2)^2 + (5 - 0)^2} = \sqrt{26}$。 第三个问题是，一个圆的圆心为 (5, 1)，且经过点 (-1, -2)，点 C 使得 AC 是圆的直径，点 D 坐标为 (5, 16)，证明 DC 是圆的切线。首先，计算半径的平方 $r^2 = (5 - (-1))^2 + (1 - (-2))^2 = 45$，得到圆的方程为 $(x - 5)^2 + (y - 1)^2 = 45$。然后，求出 C 点坐标为 (11, 4)。接着，计算 BC 的斜率为 $\frac{4 - (-2)}{11 - (-1)} = \frac{1}{2}$。然后，计算 CD 的斜率为 $\frac{16 - 4}{5 - 11} = -2$。由于两斜率乘积为 -1，因此 BC 与 CD 垂直，即 DC 是圆的切线。最后，求从 D 到圆的另一条切线与圆的切点 E 的坐标，利用对称性得到 E 点坐标为 (-1, 4)。

函数

函数是AS数学的重要内容，需要理解定义域和值域，能够求反函数。 第一个问题是，已知函数 $f(x) = x^2 + 3$，$x > 0$，和 $g(x) = 2x + 1$，$x > -\frac{1}{2}$，求 $f(g(x))$。将 $g(x)$ 代入 $f(x)$，得到 $f(g(x)) = (2x + 1)^2 + 3$。然后，求 $f(g(x))$ 的反函数。首先，设 $y = (2x + 1)^2 + 3$，交换 $x$ 和 $y$，得到 $x = (2y + 1)^2 + 3$。然后，解出 $y$，得到 $y = \frac{\sqrt{x - 3} - 1}{2}$。因此，$f(g(x))$ 的反函数为 $f^{-1}(x) = \frac{\sqrt{x - 3} - 1}{2}$，定义域为 $x > 3$。最后，解方程 $f(g(x)) = g(f(x))$。首先，求出 $g(f(x)) = 2(x^2 + 3) + 1 = 2x^2 + 7$。然后，令 $f(g(x)) = g(f(x))$，得到 $(2x + 1)^2 + 3 = 2x^2 + 7$，解得 $x = 1$ 或 $x = -3$。由于 $x > 0$，因此 $x = 1$。 第二个问题是，已知函数 $f(x) = \frac{2x}{3x - 1}$，$x > \frac{1}{3}$，求 $f(x)$ 的反函数。首先，设 $y = \frac{2x}{3x - 1}$，交换 $x$ 和 $y$，得到 $x = \frac{2y}{3y - 1}$。然后，解出 $y$，得到 $y = \frac{x}{3x - 2}$。因此，$f(x)$ 的反函数为 $f^{-1}(x) = \frac{x}{3x - 2}$。然后，证明 $\frac{2}{3} + f^{-1}(x) = \frac{2x}{3x - 1}$。将 $\frac{2}{3} + f^{-1}(x)$ 通分，得到 $\frac{2(3x - 1) + 3x}{3(3x - 1)} = \frac{6x - 2 + 3x}{9x - 3} = \frac{9x - 2}{9x - 3}$。化简后得到 $\frac{2x}{3x - 1}$。最后，求 $f(x)$ 的值域。由于 $f(x) = \frac{2x}{3x - 1}$，因此 $f(x) > \frac{2}{3}$。

积分和微分

积分和微分是AS数学的核心内容，需要掌握基本积分和微分方法，以及面积计算。 第一个问题是，曲线的方程满足 $\frac{dy}{dx} = \frac{1}{(x - 3)^2} + x$，且经过点 (2, 7)，求曲线的方程。首先，将 $\frac{dy}{dx}$ 写成 $dy/dx = (x - 3)^{-2} + x$。然后，对两边积分，得到 $y = \int [(x - 3)^{-2} + x] dx = -(x - 3)^{-1} + \frac{x^2}{2} + C$。将点 (2, 7) 代入，求得 $C = 4$。因此，曲线的方程为 $y = -(x - 3)^{-1} + \frac{x^2}{2} + 4$。 第二个问题是，曲线 $y = 4\sqrt{x} - 2x$ 与直线 $y = 3 - x$ 相交于 B 和 C 两点，求阴影区域的面积。首先，求出 B 和 C 两点的横坐标。令 $4\sqrt{x} - 2x = 3 - x$，得到 $x - 4\sqrt{x} + 3 = 0$。设 $y = \sqrt{x}$，则 $y^2 - 4y + 3 = 0$，解得 $y = 1$ 或 $y = 3$。因此，$x = 1$ 或 $x = 9$。然后，求出直线与 $x$ 轴的交点为 (3, 0)。接着，计算阴影区域的面积。阴影区域的面积等于三角形的面积减去曲线与 $x$ 轴所围成的面积。三角形的面积为 $\frac{1}{2} \times 6 \times 6 = 18$。曲线与 $x$ 轴所围成的面积为 $\int_4^9 (4\sqrt{x} - 2x) dx = [\frac{8}{3}x^{\frac{3}{2}} - x^2]_4^9 = (\frac{8}{3} \times 27 - 81) - (\frac{8}{3} \times 8 - 16) = -9 - (-\frac{8}{3}) = -\frac{19}{3}$。因此，阴影区域的面积为 $18 - \frac{19}{3} = \frac{35}{3}$。

微分

微分是AS数学的重要内容，需要熟练掌握链式法则，并应用于实际问题。 第一个问题是，曲线的方程为 $y = 2 + \sqrt{25 - x^2}$，求曲线上梯度为 $\frac{4}{3}$ 的点的坐标。首先，将 $y$ 写成 $y = 2 + (25 - x^2)^{\frac{1}{2}}$。然后，求导数 $\frac{dy}{dx} = \frac{1}{2}(25 - x^2)^{-\frac{1}{2}}(-2x) = \frac{-x}{\sqrt{25 - x^2}}$。令 $\frac{dy}{dx} = \frac{4}{3}$，得到 $\frac{-x}{\sqrt{25 - x^2}} = \frac{4}{3}$。两边平方，得到 $\frac{x^2}{25 - x^2} = \frac{16}{9}$，解得 $x = \pm 4$。将 $x = -4$ 代入原方程，得到 $y = 2 + \sqrt{25 - (-4)^2} = 5$。因此，所求点的坐标为 (-4, 5)。 第二个问题是，曲线的方程为 $y = (3 - 2x)^3 + 24x$，求一阶导数和二阶导数。首先，求一阶导数 $\frac{dy}{dx} = 3(3 - 2x)^2(-2) + 24 = -6(3 - 2x)^2 + 24$。然后，求二阶导数 $\frac{d^2y}{dx^2} = -6 \times 2(3 - 2x)(-2) = 24(3 - 2x)$。接着，求驻点坐标。令 $\frac{dy}{dx} = 0$，得到 $-6(3 - 2x)^2 + 24 = 0$，解得 $x = 0.5$ 或 $x = 2.5$。将 $x = 0.5$ 代入原方程，得到 $y = (3 - 2 \times 0.5)^3 + 24 \times 0.5 = 20$。将 $x = 2.5$ 代入原方程，得到 $y = (3 - 2 \times 2.5)^3 + 24 \times 2.5 = 52$。因此，驻点坐标为 (0.5, 20) 和 (2.5, 52)。最后，判断驻点性质。将 $x = 0.5$ 代入二阶导数，得到 $\frac{d^2y}{dx^2} = 24(3 - 2 \times 0.5) = 48 > 0$，因此 (0.5, 20) 是极小值点。将 $x = 2.5$ 代入二阶导数，得到 $\frac{d^2y}{dx^2} = 24(3 - 2 \times 2.5) = -48 < 0$，因此 (2.5, 52) 是极大值点。

图像变换

图像变换是AS数学的常见题型，需要掌握函数图像的平移、伸缩和反射。 第一个问题是，已知函数 $f(x) = \frac{3}{2}\cos 2x + \frac{1}{2}$，定义域为 $[0, \pi]$，求 $f(x)$ 的值域。由于 $\cos 2x$ 的值域为 $[-1, 1]$，因此 $f(x)$ 的最大值为 $\frac{3}{2} \times 1 + \frac{1}{2} = 2$，最小值为 $\frac{3}{2} \times (-1) + \frac{1}{2} = -1$。因此，$f(x)$ 的值域为 $[-1, 2]$。然后，已知函数 $g(x) = f(x) + k$，其中 $k$ 为正数，且 $x$ 轴是曲线 $y = g(x)$ 的切线，求 $k$ 的值，并描述变换。由于 $f(x)$ 的最小值为 -1，因此要使 $x$ 轴成为 $y = g(x)$ 的切线，需要将 $f(x)$ 向上平移 1 个单位，即 $k = 1$。变换为向上平移 1 个单位。最后，求曲线 $y = f(x)$ 关于 $x$ 轴对称的曲线方程。关于 $x$ 轴对称的曲线方程为 $y = -f(x) = -\frac{3}{2}\cos 2x - \frac{1}{2}$。 第二个问题是，根据图像写出变换后的函数表达式。 第一个变换是关于 y 轴对称，表达式为 $f(-x)$。 第二个变换是在 y 轴方向上伸展 2 倍，表达式为 $2f(x)$。 第三个变换是向左平移 4 个单位，向下平移 3 个单位，表达式为 $f(x + 4) - 3$。

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0:00 okay welcome to my all of a s maths paper one for cambridge so if you're doing this particular course the as mass paper one and this is the video for you either do a crash course in the next about two two and a half hours on the key topics that you need to know we want to refresh up on the most important topics and this is exactly what i do in the video so i go through the top 10 topics that i have picked out from the paper one and to make sure that you're familiar with the exam style questions and go through two questions per topic so you're up to speed for the paper one right number one is binomial expansion so being able to expand uh binomial but also compare coefficients and use that to solve problems so i'm going to look at two questions just now going through exactly

0:46 that okay and we're going to look at our very first topic here which is binomial expansion coming up pretty often on the paper one so first thing question one to find the term independent of x i'll explain that in a moment in the expansion of two x plus one over four x squared all to the power of six so when they say the term independent of x they mean just the constant term so the term that's left that's just a number so nothing to do with x or x squared or x cubed and so on so what we're going to do here is actually expand this out and then as we go through let's see if we can find that constant term so i'm just going to use the formula that you should be familiar with from this course and i'm going to

1:32 use the following notation here so first of all we need our binomial coefficient so we're going to start with 6 c zero and all we focus on is just the first term here so two x to the power of six okay that's not going to give us a term independent of x which is unlucky let's just continue so then we have six c one and we have two x to the power of five and then multiplied by one lot of this here now just by looking at this we're gonna get x to the power of 5 term here and x squared term on the bottom x to the power of 5 divided by x squared is x cubed so this term wasn't going to work this term won't work either so let's

2:18 just keep continuing so we get 6 c2 then we get 2x to the power of 4 multiplied by 1 over 4x squared so the power of 2. remember these indices so let's get that trusty highlighter out here the four and two has to add up to six here uh six plus zero is six and then five plus well there should be a one there but i've got a highlight where there should be a one again this is how we check that we are getting the correct expansion now here let's just have a think about this if i square this then two times two is four so you get x to power four at the bottom there is next power four here that's gonna cancel and give me my term independent of x so

3:06 actually this is the term that i am going to focus on now if you don't believe me at this point then please do keep watching and you'll see dx's will disappear so we're going to have 6 c2 and now 2x to the power of 4 multiply by 1 over 4x squared to the power of 2. so let's work out this part first of all 2 to the power of 4 is equal to 16 x power 4 is x to the power of 4. if we square this then we need to square both top and bottom so 1 stays the same 1 squared is 1 4 squared is equal to 16 and x squared all to power of 2 that's equal to x to the power of four and now we need to work out six choose two now you can do this on your calculator i'm going to show you where

3:52 that is where we're going to do this i'm gonna type in six i'm gonna type in c and then we type in 2 and that will generate the binomial coefficient again on other calculators this might be a little bit easier we get 15 here so let's pop across 15 for here and notice that we get a cancellation here and here notice everything cancels leaving us just with 15 which is our answer so this is a very nice cancellation because you get the same thing top and bottom so giving you 15 as your answer right you can check the mark scheme there and they've done it maybe slightly differently to how you've seen in the mart scheme here now let's look at a slightly trickier question so question five in the expansion of two x squared plus a over x

4:40 to the power of six the coefficients so the number in front of the x squared and the x cubed are equal and we need to find the value of the non-zero constant a so a we can't assume to be zero otherwise life would be too easy so the first thing i'm going to do here is i'm going to start writing out my expansion and let's see when we get our x power of 6 term and x to the power of 3 terms so we get a over x to the power of 6 and this is approximately equal to six choose zero just using the formula here so two x squared to the power of six okay that's not going to be a term that works because we get x to power of twelve so let's keep going through so six choose 1 2x squared to the power of 5

5:28 plus a over x to the power of 1. let's think about this term here where we get a x power of 10 here an x at the bottom so x power 10 divided by x is x power of 9 so that's not going to work let's look at our next term 6 choose 2 2 x squared to the power of 4 should be a multiplied there sorry multiplied by a over x all squared let's think about this so we get 2 times 4 is 8 8 divided by x squared is 6. so that's going to give us our x power of 6 term so we'll come back to that and then six choose three two x squared to the power of three a over x to the power of three and let's

6:16 have a think about this one so two times 3 is 6. so it's going to give us an x to the power of 6 divided by x to the power of 3 that gives us an x cubed term so these are the two that we actually need to look at here now 6 choose 2 we did in the previous question that's equal to 15. that makes our life a little bit easier okay 2 to the power of 4 that's equal to 16 x squared all to power 4 is equal to x to the power of 8 and then we get a squared over x squared and in a moment we're just going to simplify this down so we get 16 times 15 that's equal to 240 so we get 240 a squared these two things cancel so x power of 8

7:02 divided by x squared gives us the x to the power of 6. so once we worked out the second term we're going to focus on that coefficient which is the number in front of the x power of 6. now let's look at this one here six choose three we're going to get to go back to that calculator and work it out okay so let's do that so we get six choose three and that's equal to twenty so we're gonna pop over 20 for that so 20 multiplied by well 2 cubed is equal to 8 so we get 8 x to the power of 6 and then we get a cubed over x cubed and let's see what we're left with so 88

7:48 times 20 is 160 so we get 160 a cubed and x power 6 divided by x power 3 is x to the power of 3 which is what we wanted so these two coefficients we're going to be looking at so then we get our coefficient here highlighted and because they are equal to each other that's what we're told in the question therefore 240 a squared is equal to 160 a cubed okay we get some cancellation on both sides here so we get 240 equal to 160 a therefore a is equal to 240 over 160 which we can simplify down

8:36 and that gives us 3 over 2 which is very nice so we get our final answer of 3 over 2. and the last part of the question is it wants to work at the coefficient of x to the power of 6 in the expansion so let's have a think about this in terms of the pattern going on so this one here we got an x to the power of 12. here even though we didn't include this we got a x to the power of 9 so here we get 12 9 6 3 and if we continue on that method on that particular pattern we'll see we keep going down in three zip and then if you think about we're going to multiply this by 3

9:23 we're actually not going to get a term that has a coefficient in x you're going to get a calculation therefore here this is why it's a one mark question so use that to help you here you're actually going to get no coefficient so it's just going to be equal to zero right you can check the mark scheme here so you can have a look through and see the calculation given and then for the last question it's just that's going to be equal to zero okay hopefully you found that useful again we're going to move on to the second topic in a moment okay and number two is working with quadratics which is quite a broad topic basically focusing on the discriminant so using the discriminant really really well also working with solving

10:09 quadratics and completing the square as well so i go through two very typical questions here uh questions from the last few years or so going through exactly this topic okay and on to our section on working with quadratics now this can come up in a variety of different forms but what's common to most of these questions is the use of the discriminant which i'll talk about a bit later on so it's testing your understanding of using discriminant and helping that with working out a quadratic so let's get started so a curve has equation y equals kx squared plus 2x minus k underline here and a line has equation y equals kx minus 2 and k is just a number it's just a constant and we need to find the set of values of k for which the curve and line do not intersect

10:55 now the way that we actually go about doing this is that we pretend they do intersect we make them equal to each other and then we work through the algebra so get a quadratic and look for no real solutions which is where our discriminant comes into play so we first will make these two things equal to each other so kx squared plus 2x minus k is equal to kx minus 2 and now to get a quadratic that we can work with we need to make the right hand side 0 the way we do this is add two and minus kx on both sides if we do this well kx squared stays the same if we minus kx from both sides we get two minus k x and then we just bring the

11:44 2 over so we get minus k plus 2 is equal to 0. this canceling here and this cancelling here now we have a quadratic that we can work with we're now going to set this up where we're looking for no real solutions i.e they do not intersect and if we want no real solutions and we need to make sure the discriminant is b squared minus 4ac which is really important is less than zero now where do a b and c come from they come from the coefficients of the x squared the x here and then the constant term like so so this is going to represent our a this is going to represent our b and this will represent part c and now we just

12:30 put it into this formula and then work through that algebra so if we do that we get 2 minus k whole squared minus of four times a which is k here lots of minus k plus two is less than zero and now it's an algebra exercise to get something that we can work with to expand the brackets we get 4 minus 4k plus k squared and then we get minus 4k multiplied by minus k plus 2 is less than zero now let's expand over here so we get four minus four k plus k squared you'd be really good at expanding these double brackets quickly minus four k times minus k that's plus

13:15 four k squared minus four k times plus two is minus 8 k almost there with our quadratic so 4k squared plus k squared is 5k squared minus 4k minus 8k minus 12k and don't forget our little plus 4 over here as well don't forget our constant term is less than zero now the way we actually go about solving a quadratic inequality is we solve it as if it were a normal inequality so we're going to factorize into two brackets a very important skill to know now because we've got five k squared and five is prime we know five k must go here and k here or the other way around and now anything about the factors of four that go in here that can generate minus twelve well they haven't got too many

14:00 choices here we've got one four or the other way around and then two two again you can play around with the numbers here to try and get to the answer um i can see here we get minus two and minus two now if you check the cross terms that's the thing you should always check here so 5k times -2 is minus 10k minus 2 times k is minus 2k and minus 10k minus 2k minus 12k so this is the correct factorization that means our critical values so we're not solving this we're just finding the critical values uh k is going to be equal to 2 or it's going to be equal to 2 over 5. now if i do a quick sketch of a quadratic again not to get a b to scale by any means

14:47 whatsoever there's our two-fifths solution there is our two solution let me put that in a slightly different color so this would be our two-fifths this would be r2 and because we're less than zero we want to consider the domain of values below the line so this part here so then to write our final solution i'll do this in black here we get two-fifths is less than k which is less than two remember it's going to be strictly less than because our inequality here is also strictly less than and we want everything below the x-axis here okay if you need to practice those skills quadratic inequalities that is a fundamental part of the course so you can check out the answer here so

15:32 this is the final answer which i got as well and the corresponding working you need to get all five marks now my second example question is a little bit different so we're going to use a bit of completing the square for the first part and we want to express 16x squared minus 24x plus 10 in this form so the way i tend to do these problems is i ignore the plus 10 for the moment i will come back to that at the end and i think right i want to generate the 16x squared minus 24x so we're told in the question this needs to be a 4x and then we need to kind of think backwards and think okay what number needs to go in here such that when i expand i get a minus 24x

16:18 okay well if i put a minus 3 here for example then 4x times -3 is minus 12x i've got 2 lots of that will give me minus 24x now if you didn't get that part that's absolutely fine let me just show you this by expanding this out in full so if i expand out 4x minus 3 all squared in 4 well 4x times 4x is 16x squared that's what we wanted 4x times -3 is -12x the same in the middle as well and -3 times minus 3 is equal to positive 9. if i collect up those middle terms i get 16x squared minus 24x plus 9. so that's how i worked out the three because i was working out okay 2 lots of

17:03 12 is 24 that's what i want now to work out how to factorize this i actually use the answer that i've got here so i get the old highlighter out here so this part here 16x squared minus 24x plus 9. that's what i get if i normally expand this but i don't want this i want a plus 10 instead of plus 9. so what do i do to 9 to get to 10 well i just add on one so all i need to do here is just not with a highlight not to be recommended but with a normal pen we get plus one so our a and b are these values here so a is equal to minus three and then b is equal to one and typical for these kind of questions we're going to need

17:48 that knowledge for the next part of the question so it is given that the equation 16x squared minus 24x plus 10 so what we had before is equal to k where k is a constant that's exactly one root which we'll come back to in a moment and we need to find the value of that root so with any quadratic i always like the right hand side to be zero so i'm going to write down 16x squared minus 24x i'm going to do a minus k on both sides here so get a plus 10 minus k is equal to zero so all i've done to both sides the original equation is just minus k on both sides so i get that right hand side 0. now using the discriminant yet again so

18:34 exactly one root so we're going to use that b squared minus 4ac from before but this time because we want exactly one root we won't make it equal to zero this time there's a really important fact to know again our a b and c like in the previous question come from the coefficient of the a the coefficient of the b and then the constant term there as well so let me just label this up for you so you can see which part is which so this is a this is b and this is c so we pop it into our formula we then get minus 24 squared minus 4ac so 4 times 16 times 10 minus k

19:22 is equal to zero this time so again we can use a calculator here so i'm just going to do minus 24 squared very quickly that's 576. again don't do that mentally minus 64 lots of 10 minus k is equal to zero so we get 576 minus 640 plus 64 k is zero 576 minus 640 that's equal to minus 64. plus 64k is equal to zero 64k is equal to 64. therefore k is equal to 1. it's very nice when you get nice numbers

20:08 you know you're probably on the right track however this is not our final answer so just be very careful with that so notice if k is equal to 1 and we put it back in our normal quadratic we get 16x squared minus 24x plus 10 is equal to 1 therefore 16x squared minus 24x plus 9 minus in 1 from both sides is equal to 0. and we've seen that before in the previous question if i go back notice that 4x minus 3 all squared was equal to that expression therefore we can say that 4x sneaky a 4x minus 3 all squared is equal to 0

20:54 using that expansion i did in the first part of the question and therefore if we want to solve this therefore four x minus three must be equal to zero okay can square root both sides and therefore x must be equal to three over four which is the actual roots that we're looking for so we're not looking for k here we're actually looking for the root itself which in this case is three over four okay you can check the solutions there as well and you also get follow through marks if you uh don't get the correct answer for the first part of the question okay and number three is trigonometric equations always a big topic on paper one so knowing your identities is really

21:39 really important but also being able to problem-solve because there could be more than one way to get to the solution that you're looking for okay and now we're going to look at trigonometric equations which is a staple of the as mass paper one so you really do need to know how to manipulate these trigonometric equations and then solve them so let's get started so question 7a and this is very typical so we need to show that this expression here is equal to 2 over sine theta cos theta and generally the way i go about doing this is i go well the left hand side is equal to tan theta and we're going to copy this out here so 1 plus cos theta plus tan theta over 1 minus cos theta

22:26 and our task is to make this look like the right hand side this is generally the approach that i take and as soon as i have two fractions here um i always think about common denominators and the first thing that comes to mind is actually multiply each by the other to actually get that a common denominator so what do i mean by that well what i'm going to do is multiply these two things together so that gives me 1 plus cosine theta multiply by 1 minus cos theta and then to work out the numerators i times this by this and this by this again very similar to how you multiply normal fractions so if we do this we get tan theta open brackets one minus cosine theta

23:11 and then plus tan theta i'll probably have to extend this defined line a little bit further one plus cos theta and now we're going to do some expanding now you may notice the bottom here of my fraction is actually a difference of two squares so imagine i write this as 1 minus x 1 plus x think of it in this way if i multiply this out i get 1 i get plus x minus x minus x squared and then these counts will give me 1 minus x squared i can use this same quadratic technique to actually simplify this in the same way which means i can go straight to one minus cos squared theta so a little shortcut

23:57 there that might save you those couple of minutes that might save you the marks in the exam here at the top however we have a more straightforward multiplying out a single bracket so tan theta times one is just tan theta tan theta times minus cos theta is just minus tan theta cos theta extend this again and then here we just get plus tan theta and then plus tan theta cos theta and something very nice has happened here well we've got plus 10 theta cos theta we have a minus tan theta cos theta so these two things cancel very helpful and we get tan theta plus tan theta which is 2 tan theta now as i'm going along i'm trying to

24:43 simplify where i can this can also be simplified so 1 minus cos squared theta remember the identity cosine cosine squared theta plus sine squared theta is always equal to 1. so i can actually rearrange this to give you sine squared theta is always equal to 1 minus cosine squared theta it's important to remember so i'm going to just replace the bottom here with side squared theta at the moment you might think we're not quite making the progress that we're hoping for we've got two but this part at the bottom is not so clear however this says something also we need to know on the course is that tan theta is equal to sine theta over cos theta so i can now replace the

25:30 tan theta with sine over cos that gives me 2 sine theta over cos theta sine squared theta and now we have cancellation of the sine thetas so how many sine theta is going sine theta one how many sine thetas go into sine squared theta well just sine theta and we actually magically generate what we needed so sine theta cos theta and this is where i just state that this is equal to the right hand side so we have achieved our objective we have made the left hand side here look like the right hand side and what generally happens in these questions is that they use this identity to then lead into some solving and this

26:15 question is no different so now we need to solve this very complicated looking equation but now we know the thing underlined here from what we literally just did is equal to 2 over sine theta cos theta so using that and this word hence gives us the clue we need to use it we can actually just make this equal to then 6 over 10 theta now at this point there's many different approaches you can take i'm going to try a cross multiplying approaching so i'm going to multiply this by this and this by this if i do that i get 2 tan theta is equal to 6 sine theta

27:01 cos theta at this point i can change tan to sine over cos like i did before so 2 sine theta over cos theta and again there's no one correct way of doing this necessarily but i'm just taking an approach and just trying to simplify as much as possible so at this point i'm going to times through by cos theta so i get 2 sine theta is equal to 6 sine theta cos squared theta i can cancel the sine theta from both sides f that gives me 2 is equal to 6 cos squared theta and now we're just working our way through it so we can divide by 6 here so cos squared theta is equal to 2 over

27:50 6 which is equal to 1 over 3. at this point i want to get cos on its own or theta on its own so i need to square root on both sides that gives me then that cosine of theta is equal to be very careful here plus minus the square root of one third and now to find our two answers all we need to do is get rid of the cos so what's the opposite of cosine inverse cosine on both sides and notice the domain here is actually quite convenient for us we only need to go between 0 and 180 so when we actually do this process i'm just going to do now so we're going to do the cosine inverse i will take the plus version first so the square root of the third

28:37 but don't forget we've also got theta is equal to inverse cosine of minus the square root of a third as well and now this is where finally we get to use our calculator and work it out okay so i'm using my gdc here but a normal calculator will be fine so what we're going to do is take inverse cosine and we're going to start with the positive version so square root and then we want the fraction button one three that gives us 54.7 and we're going to do exactly the same process now negative version so minus square root 1

29:23 over 3 and that gives us the obtuse angle so our final answers will be 54.7 and 125.3 to one decimal place so let's pop that in so we've got 54.6 degrees 1 dp 25.3 degrees also to one decimal place so you can check your answers that is exactly the same as what we've gone through and this is where these videos are very helpful because i take you directly through the method you need to solve these typical kind of trig equations okay and on to our second example here again very similar to how we approached

30:08 the question before so let's have a look at this so we've got left hand side is equal to 1 minus 2 sine squared theta over 1 minus sine squared theta and first thing i'm going to do here is use my identity so 1 minus sine square theta is equal to cosine squared theta and as i'm doing this kind of problem i'm thinking i want a tan at some point i want to try and achieve that so that's always in mind as i go through now because i've got one thing in the denominator i can actually separate this out so 1 over cos squared theta

30:55 minus 2 sine squared theta over cos squared theta and at this point we can then change this slightly so one over cos square theta is the same as sec squared theta so make sure you know this definition for second cosec sine squared over cos squared is equal to tan squared so we get minus 2 term squared theta and we're almost there so we just need to use another identity here which is important so sec squared theta is equal to one plus tan squared theta again one to know really off by heart it really does help you later on in the a

31:41 level course in general and i'm going to substitute that in and this is not the only way of solving this but seems a very convenient way because we can collect up like terms here so 1 ten square theta minus 2 tan square theta is just minus 10 squared theta which is exactly what we wanted so that is the right hand side and we have our answer like so and here's the very typical part so now we need to solve the equation using that identity from before so we worked out the left hand side was equal to one minus tan squared theta so we're going to use and exploit that now so let's work it out so you get one minus tan squared theta

32:27 is equal to 2 tan theta to the power of 4 and at this point this doesn't actually look like something we can solve however first of all i'm going to just write it in a slightly different way here and show you what i'm up to so i'm just gonna bring everything to one side which is a typical approach minus one is equal to zero and then we'll do a substitution you've seen me do this trick in different parts of the video i'm going to let y equal tan squared theta and that means then y squared if we square both sides well 2 times 2 is 4 so we get 10 theta to the power of 4. and now we can just substitute this in so we get two y squared plus y

33:13 minus one equal to zero this factorizes nicely so we get two y and y there equals zero and we just need to think about where each of the numbers go so we want a plus one here and a minus one there make sure it lines up nicely and therefore y is equal to negative one or it's equal to a half and therefore that means tan squared theta is equal to negative one or it's equal to a half at this point notice that if we tried to solve this we'd actually get a complex number so we can't actually solve this in the real sphere so this answer won't

34:00 work but we can do this one so tan square theta is equal to a half and therefore tan theta is equal to the square root plus a minus of course of one over two so to find the final answers we need to do theta is equal to ten inverse of the square root of a half and tan inverse of minus root of a half this is where we go to our calculator and work out the answer so we go to trig and we're gonna go 10 inverse and it's very similar to how we did the previous question so we're going to write it in as one over two that's our first answer in fact it's a very good thing you can

34:46 actually do on this calculator you can actually copy and paste isn't that cool again this calculator is a little bit op for this kind of question and then we get this answer here as well so notice this actually doesn't fit within the range that we wanted um we want between 0 and 180 so we just need the 35 35.3 and notice this was minus 35.3 so this way to be careful with domain this doesn't fit in the domain but remember that it does repeat every 180 degrees so if i take my minus 35.3

35:33 and add on 180 that will give me the solution within the correct domain because this is special for tan because it repeats every 180 if we do this i'm going to do it on my calculator here so i can go back to the other calculator we then get 144.7 so just to make it very clear to the examiner i'm going to put this in the box so 35.3 is correct and 144.7 like so so again you could have done this in regions as well that's absolutely fine but since this is in degrees because notice the domain here is in degrees that's why i've worked on degrees for

36:18 this question if you want to check them out scheme then have a look through here as well and hope this gives you a good fundamental understanding of how to manipulate trig equations okay and next is sequences and series which is not too dissimilar from some of the igcse topics or ad maps topics but there's a little bit of trickiness involved as well to make sure you're really really good and getting those maximum marks on this kind of question okay and let's get stuck into some secrets and series typical on the a.s mass cambridge course so the first term of an aromatic progression is equal to a and the common difference here is minus four now the first of the geometric progression is going to be five a and the common ratio required is minus a

37:04 quarter and now we're told sum to infinity so we're going to need to find that formula of the geometric progression that's equal to the sum of the first eight terms the arithmetic progression and we need to find the value of a very typical kind of question so i'm going to start with the first part here which is sum to infinity of a geometric progression this is a formula that you need to know so s infinity is equal to a over 1 minus r where a represents the first term and r represents the column ratio so if we substitute the a here for five a and the r for minus quarter and pop that in we get then five a over one minus minus a quarter if we do a little bit of

37:51 simplifying here so we get 5a over well one plus a quarter that's going to be equal to five over four and we can actually cancel five top and bottom bring the four to the top and then we're equal to 4a so the sum to infinity is equal to 4a and now we need to focus on the second part of the question which is the sum of the first eight terms of the arithmetic progression this is equal to this 4a so let's work out what s8 is first of all this is also from the formula sheet so the general formula let me write that down first before we look at s8 in general and this is a formula you need to know is equal to n over 2 open brackets 2a plus brackets n minus 1 d so

38:37 i'm just taking this from the formula sheet directly and we want to work out s 8 where the first term is equal to a so that won't change and the common difference is equal to minus four so s eight will be equal to eight over two remember n is equal to eight here open brackets two a now d is equal to minus four so we get minus four open brackets well n is also equal to 8 so 8 minus 1. and let's simplify this down slightly so 8 over 2 is equal to 4. 2a remains unchanged and now if we work this out so eight minus one is seven seven times minus four is equal to minus twenty eight and once we've got that we know that if

39:22 i get a different color pen here let's get green this here the 4a is equal to what we've just worked out here so at this point we're now exploiting the fact that these two things are equal to each other so therefore we get 4 lots of 2a minus 28. is equal to 4a and one of the reasons i didn't expand the bracket here is i can just simply cancel the 4 here and 4 here very nice and then we get 2a minus 28 is equal to a which we can rearrange here and this then gives us a is equal to 28 like so all i did here is a minus a on both sides and added 28 on both sides and from that we need to work out

40:09 whether the k term of the arithmetic progression is zero find the value of k well we now know that a is equal to 28 and if we look at the arithmetic progression we know the common difference is equal to minus four so i'm just going to write that down d equals minus four and we want to find the term such that it's equal to zero now remember the nth term of arithmetic progression here is a plus brackets m minus 1 d so i exploit this and i want to work out what it is equal to zero and then find the term in which it gets to zero so the way i'm going to do this is well a is 28 n is what i don't know so we'll call

40:54 that k here looking for the kth term and then we know d is equal to -4 and we want to work out when this is equal to zero so once we have this expression just using this formula over here we're going to do some expanding and simplifying so expand the bracket out so minus 4 times k well that's equal to minus 4k this way to be very very careful so -4 times -1 is equal to plus four and that's equal to zero like so and now a little bit of rearrangement so 28 plus four that's equal to 32 minus 4k is equal to zero and now we power through the algebra giving us then the k is equal to eight notice i'm going through these kind of processes quite quickly because you

41:41 should have that from the igcse or gcse course that you've already completed some art schemes so you can check exactly what i've done here and go through the working and that's it for this as math video that the master games are not entirely helpful so hopefully this video is helping you with this particular topic right let's look at one more question like this so the first term aromatic progression is 84 common difference is -3 so a is equal to 84 d is equal to -3 and find the smallest value of n for which the nth term is negative i'm going to take a similar approach to how i did the previous question so i'm going to use the nth term formula so a plus n

42:29 minus 1 d and i'm going to set it equal to 0 and then i can adjust to see when it needs to be negative you'll see this as we go through so we'll take 84 as our first term n is what we don't know at the moment so we leave that alone we know that d is equal to -3 and again we're going to set this to zero and see what we get so get 84 expand the bracket here to minus 3 times n is minus 3n careful here minus 1 times -3 is plus 3 which is equal to 0. 84 plus 3 is equal to 87 plus 87 is equal to zero so 3n is equal to 87 and now we have to divide here by 3. now you have a

43:16 calculator for this i'm just going to do this old school i do like a bit of old-school math so 3's into 8 go 2 remainder 2. and then 3's into 27 going 9. so n is equal to 29 this is where we have to be very careful because the 29th term is equal to 0 but we want to work out the next value such that the nth term is negative so our actual answer here and this is where we have to be very careful is that we actually want n to be 30 because that would be the smallest negative number minus three so imagine you've got your sequence you've got six three zero minus three this was the 29th term

44:01 and this is the thirtieth term the smallest negative that is possible so just be a little bit careful with those kinds of questions right on to the next part of the question and it is given that the sum of the first two k terms of this progression is equal to the sum of the first k terms and we need to find the value of k okay so we've got let's just remind ourselves here a is equal to 84 remember of this progression so what we used previously d is equal to minus three and let's work out first of all the s k terms so using the same formulas we did before here so get k over two this is k here this time

44:48 open brackets two lots of a two lots of 84 let me just write that as is plus m minus one d so k minus one multiply by minus three and let's simplify this down so it looks a little bit nicer so we get k over two about two lots of this is 168 again we've got a calculator for this uh minus three k plus three let's simplify this a little bit further so k over 2 then we get 171 minus 3k and what we're going to do now is do exactly the same process for two lots of k so we're going to substitute it in here so we get 2k over two

45:33 and then two lots of 84 doesn't change this is still 168 plus i've got 2k minus 1 minus three here two's cancel here so we'll do this as we go along this is k if we expand this out here we still get 171 here that hasn't changed and this time we get a minus 6k and what the question is saying here is that these two things are equal to each other so we underline that in green this expression here is equal to this expression here so to finish this off so therefore k lots of 171 minus 6k

46:22 is equal to k over 2 171 minus 3k now one of the reasons to keep it in bracketed form these two things cancel saving us a bit of time here so 171 minus 6k is equal to a half lots of 171 minus 3k and this point is just using a sufficient algebra as we can i'm going to times both sides by 2 here make our life a little bit easier so we get 300 342 minus 12 k is equal to 171 minus 3k subtract this from both sides add 2k on

47:08 we get 171 is equal to 9 k i'm hoping this is where everything works nicely so if we do this then so k is equal to 171 divided by nine this is where you hope you get a whole number out of it so nines into one go go nines into seventeen go one remainder eight nines in eighty one being nine so k is equal to 19. again feel free to use your calculator for these kinds of calculation as well right so you can check the marks game again not particularly helpful if you're studying this for the first time or really refreshing your knowledge of this

47:53 topic but hopefully my explanations have been clear here to help you through this particular topic and the next is radiant measure so knowing what the radian is that's obviously very useful but then applying that to finding the area finding various lengths and arcs as well and that's an important topic that also helps you later on in the a level maths course generally speaking okay and now we're going to look at some radian measure so if this video has been useful to you so far then do please think about liking and subscribing going a long way to supporting the content on the channel right question five so in the diagram here x and y are points on the line a b such a b x so this length here is equal to nine and then we've got

48:38 a y equaling 11. arc bc is part of a circle with center x so this is our center of the circle and radius 9 centimeters where cx is perpendicular to a b right angle here and arc ac is part of a circle with center y so over here and radius 11 centimeters so the way to think about this problem is kind of we have two sectors within this particular shape so i'm going to draw this in red just so you can see this nice and clearly so we have this sector first of all i'm just drawing round i'm going to do something very similar in green here so you can see the second sector as well so we also have this sector where this is the center

49:23 going down like so and then we've got going round and upwards so just keep this in mind as we go through this question because that's going to be very important now the first thing we need to do is show the angle x y c so if i mark that in x y c so this angle up here is equal to 0.5982 radians so how do we show this well we've actually got all the information that we need so if we take the green sector here well this 11 centimeters is the radius and this here yc is also going to be 11 centimeters because it's also a radius of the circle so this is going to be 11 centimeters likewise if you now look at the red sector this here 9 centimeters is a

50:11 radius therefore this one down the bottom xc must also be nine centimeters and as soon as we see this then we can simply use socatoa to actually work out what we are looking for so if i do this in blue so um we've got the angle here this side here is opposite the angle the 11 here is the hypotenuse and then we think okay which of socatoa includes both o and h well it's going to be the first one so it's going to be the saw so if we convert this now into what we need to do so the sine of the angle so sine of x y c is equal to or opposite over hypotenuse so 9 over 11. and therefore to find this

50:58 the actual angle itself what's the opposite of sine well that's going to be sine inverse on both sides so if we do this this cancels giving us x y c and then we do the inverse sine of 9 over 11. so i'm going to go to the calculator now make sure it's set in radians that's really important and let's work this out so here's my trusty calculator i'm going to use the casio for this particular question and all we need to do here is go to sign inverse so shift and sign open brackets notice this is already in radians at the top here so it says red if you want to change it you go to shift and setup and now we just put in 9 divided by 11 just to verify everything we've done here is correct and we get

51:45 0.9582 so i'm just going to write that in so zero point nine five eight two dot dot dot look at a little c for radians and therefore this is correct now on to part b which is a bit trickier so you can see it's a six mark question here is find the perimeter of abc so we need to find the perimeter of the entire shape what we're going to start on first of all is the length of a b so we're going to look at a b here and something that could be very useful to us is actually working out what x y is here so we're going to work at x y and we can simply use pythagoras because we've got the hypotenuse and a shorter side and then we want to find the other shorter side so if we use pythagoras on this

52:32 then we know that xy squared that's equal to the hypotenuse squared 11 squared minus 9 squared and if we work through this 11 squared is 121 9 squared is equal to 81 and then if we take this away we get 40. so if we actually want to work out x y itself what do we do to both sides well we need to square root on both sides so this length here is equal to the square root of 40. now i'm going to keep this as precise as possible by leaving it in third form so we can use that a bit later on and this is very important for these kind of questions not to be tempted to round too quickly so if i use a

53:17 different color lots of colors in this particular question so this yellow line representing x y and this is equal to the square root of 40. so what we want to do now is we actually want to work out the length of a b well let's think about this so if we add 11 and 9 together we do get 20 however that includes this yellow length uh twice essentially we're sort of double counting it so in order to work out a b let me get my red pen out here to work out a length of a b we need to add together the 11 and 9 but because we've double counted the yellow part because we canceled out twice we need to take that away so i'm going to minus the square root of 40 and

54:03 i'm going to keep that in this form so 20 minus root 40. so we found the first part of our perimeter remember the perimeter means the distance around the outside of the shape and that is going to be equal to 20 minus root 40. now actually working out ac and bc is the easiest part here and we need the arc formula using radians which is really important so the work out the arc all we need to do is just our theta where theta is in radians which is really really important so if we do that for the first parts let's work out ac here so let's go after here ac well that's equal to r times theta well r is equal to 11

54:48 and then the angle in terms of radians we already worked out here that's 0.9582 so i'm going to leave that like so i'll work this out a little bit later on to work out bc it's a little bit different so we need to use first we'll use nine because that's the radius and be really careful here when using the formula we do not want to put 90 degrees in that's a definitely a no-no we need that in radians so you need to know these exact values so 90 degrees is the same as pi over 2 radians the way i tend to work this out quickly is i know 180 degrees is equal to pi radians and i do all my conversions from there so if i want 90 degrees for example what do i do to 180 to get to 90 well divide by two

55:35 so i do the same to both sides and so this helps me work out these exact values very very quickly so if we put this into the formula we just get nine times pi over two and to work out the perimeter all i need to do is add all these things together so i need to do 20 minus root 40 plus 11 times 0.9582 plus and then 9 pi over 2. only at this point would i actually then reach for my calculator which i'm going to do now okay so let's type this in nice and carefully so open brackets 20 minus and we want square roots that's shifting here i'm trying to use different

56:20 calculators on the channel so you can see how each of them you are used and then we do 11 [Music] times 0.95 and then lastly we do 4.5 or 9 over 2 times pi so shift and this button here and we add all those together and then we get our final answer of 38.352 so let's go back 38.352. again rounds to one decimal place at 38.4 checking units here is equal to centimeters to one dp and we've got all six marks

57:09 right so you can check the mark scheme there and just make sure that you've picked up all those method marks although it is a six mark question notice you can pick up lots of those method marks as you go through logically through the problem right and on to our second example question six and we're not really going to use much of a calculator here in this particular question so the diagram shows a metal plate abc which the sides are the straight line a b and then you've got two arcs ac and bc light a b has length six centimeters the arc ac is part of a circle with center b and radius six centimeters now what's actually quite useful to draw in here is if i draw in a dotted line just to show exactly

57:54 where this arc is and then bc also works in the same way with the center of a so let me do that in a different color so and we've got our center a also with the arc like so and we need to first of all work out the perimeter of the plate so again very common asking you for the perimeter in these kind of questions so what we essentially need to do here i put this in black is work at the length of ac and work out the length of bc as well so if we look at this particular question notice that this six centimeters and this have to be the same so this also needs to be six centimeters and the other way around so this is six there's a radius this is also a radius

58:40 this is also six centimeters therefore we have a equilateral triangle and the angle is an equilateral triangle okay each one is 60 degrees now again we're not working in degrees here we are working in radians now 60 degrees that's equal to pi over 3 radians and using the same technique that we saw earlier over here so what do you do 180 to get 60 divided by 3 so you can do it the same way here so each of these angles is equal to pi over 3. let's mark those in and to work out the perimeter well first of all we've got the six so we've got the six at the bottom and then we need to add on using that formula for the arc it's r theta so that's

59:27 six times pi over three and then we've got the other side so that's also 6 times pi over 3. so again i'm using this formula that i mentioned earlier so arc is equal to r theta in radians and we can just simplify this down so we get 6 plus well these cancel so we get 2 pi we also get here 2 pi and so we get our final answer of 6 plus 4 pi notice we're not reaching for the calculator because it says given your answer in terms of pi so you want it left exactly like this okay so just make sure you can see the answer there like so and on to part b which is a bit trickier here for formats is find the area of the

1:00:14 plate again not reaching for our calculator giving your answer in terms of pi and root three now we're going to use the other formula it's really important as maths paper one and that is the area of a sector again it's worth remembering these off by heart actually makes your life a bit quicker and that's equal to a half r squared theta again theta also being in radians so i'm going to do a very similar technique to what we looked at before so notice if i work out the area of the sector from a and the sector from b notice we get that overlap of the triangle in the middle so one way we can actually work this out so the area of shape

1:00:59 is two lots of the area of the sectors and then minus the area of the triangle in the middle okay so let's work out the area of one of the sectors it's the same obviously go in both directions so we can just times it by two so if we put that into our formula we get a half times r squared well radius is equal to six and then times the angle which is equal to pi over three and now we need to minus the area of the triangle okay and here we need to use a formula you should be familiar with from igcse but just in case we've forgotten the area of a triangle

1:01:46 is equal to a half a b sine c so make sure you know this formula very very well and if we apply that here so we get a half times a times b so that's six times six the two sides times the sine of the angle now just being consistent here if your calculator is in radians angle in radians if it's in degrees in degrees we're using radians here so we're going to do pi over three okay so if then we just uh simplify this down a bit so two times a half cancels that's good so we get 36 pi over three which will simplify further later on and if we look at this part 6

1:02:34 times 6 is 36 36 divided by 2 is 18 so we get 18 sine pi over 3. now let's just simplify this down a bit further and just think about this so sine pi over 3 that's the same as sine 60 this is an exact value i'd really recommend you just know off by heart and i know that this is equal to root 3 over 2. if you want to know more about how i just know that then do check out the video above where i go through how this is derived again you can use your calculator on this as well so let's simplify this down so we have 36 divided by 3 that's equal to 12 so 12 pi as i just told you here this is equal to root 3 over 2. this is all good sign because we want it in terms of pi and

1:03:19 root 3. and we're almost there so then we get 12 pi 18 divided by 2 is 9 so we get 9 root 3. you could factorize this further i don't know if the mark scheme would want that they happy with exactly what i've left with so that's good but just notice here you could also factorize this and show off the examiner a bit by taking a factor of 3 out and writing it like this too that's also fine as well right you can have a look at the mark scheme so you can see the first answer six plus four pi we got before and also 6b there are lots of different alternative methods here so you may want to take one of those instead i've kind of done it this way if i've

1:04:04 taken this method but this is almost identical to the top method here as well so hope you enjoyed the radiant section of the video and then we're going to move on and next is very specific to a s mass paper one that is circle calculations so we all know exactly what the equation of a circle is but how do we actually use that in a coordinate geometry setting and this is where a lot of your knowledge from other topics also come into play okay and on to some circle calculations an important topic on the a s maths paper one so a circle with center five two which is worth underlining here passes through the point seven five and we need to find the equation of the circle now remember the general formula for the equation of

1:04:51 the circle is equal to x minus a squared plus y minus b squared is equal to r squared and this represents where the center is equal to a b and the radius is equal to r so we already have quite a bit of this information already we know that a is equal to five and b is equal to two by using the formula but now we need to work out what the radius is and a quick sketch here really helps our understanding so i draw a quick sketch here we have 5 2 which is going to be the center of our circle again none of this is to scale we have the 0.75 which is going to be say up here somewhere and imagine i draw a circle using this point here so we go around like so

1:05:37 and what we're looking for essentially is the radius so if i draw this in green the radius like so we're looking for this length now we can actually use uh pythagoras to work this out imagine i draw myself a right angle triangle you can do it either way around i'll draw it this way so let's think what do you do to two to get to five well that's going to be three and what you do to five to get to seven that's 2. so we can actually work out r squared directly by just using pythagoras so the length of the green line is equal to these two lengths squared so 3 squared plus 2 squared and if we do that well that's equal to 9 plus 4 which is equal to 30. so we get our value of r squared

1:06:24 which we need for our formula so our answer is going to be the following we're going to have x minus 5 all squared remember the 5 comes from the center of the circle plus y minus 2 all squared it comes from this 2 here and this is equal to 13. be careful make sure it's not the square root of 13 but making sure you put down just 13 because we need r squared now if we look at question b the line y equals 5x minus 10 intersects the circle at a and b so what does that actually look like well imagine i get a line in blue here having a y instep of minus 10 a very very steep line because it's got a gradient of 5 and notice it crosses the circle twice once here and once here

1:07:12 and we want to work out these two coordinates and then once you work those out then we can use a similar method using pythagoras to work out the length of the chord so that's the theory let's get back into the practice so first of all i'm going to write down the equation off the circle so x minus 5 all squared plus y minus 2 all squared is equal to 13. and we want this to intersect so the y values are going to be the same so wherever i see a y let's get the old highlighter out wherever i see a y here i'm going to replace it with 5x minus 10. keeping everything else the same now i'm not going to do i did before and use the highlighter so let's just do

1:07:58 this part now so we get x minus 5 all squared plus a y is equal to 5x minus 10 so 5x 10 minus 2 all squared is equal to 13. and let's expand some brackets so x minus five all squared is going to be equal to x squared minus 10x plus 25. if you need to practice that please do it's a really important skill now minus 10 minus 2 is equal to 5x minus 12. so we're going to need to expand that as well so we get 25 x squared uh minus 60x minus 60x is minus 120x and minus 12 times -12 is 144. again you can use your calculator to help you with these things now we're going to collect up like terms

1:08:44 so 25x squared plus x squared that's 26 x squared minus 120x minus 10x is equal to minus 130x and now we've got a little bit of calculation here to do so 144 plus 25 is equal to 169. and now we just need to take the 13 away from both sides and this is where we hope we get something quite nice coming out of this so if we do this we get 156 equals zero so we've got a quadratic to solve now the first thing i'm going to do here is look at these coefficients and see if 26 goes into them um on the surface 26 times 5 gives you

1:09:30 130 so that's good and it's an extra 26 there so 26 times 6 is 156. so if you can always check these so carrying three so 130 again you can use your calculator for this as well so six carry the three twelve 156 that's very helpful because we can take out a full factor of 26 out leaving our quadratic thing a bit easier to work with so we get minus five x plus six is equal to zero and therefore because you've got a constant term outside we can ignore that because this has to be zero in order to equal zero so essentially we just need to solve this very straightforward quadratic equals zero if we do some factorizing

1:10:16 over here we're gonna get x minus two x minus three is equal to zero so x is two and three this is why the igcse course and gcse courses are very very helpful so you can do these very standard skills quite quickly now we actually need to work out what the actual coordinates are now you can substitute into both of these either the circle or the straight line the straight line is much easier to substitute into so when x is two y is equal to 10 minus 10 is zero okay that's nice and easy and when x is equal to three y is equal to 15 minus 10 which is five so we've found our two coordinates we have two zero

1:11:04 and we have three five so those are two coordinates that we have and at this point all we need to use is the distance formula between these two points so imagine this is your two zero imagine this is your three five all we're doing now is just using pythagoras on this you can do this a couple of different ways so this would be one this would be five so the chord length will be equal to the square root of five squared plus one squared and you can use the formula if you remember that and that's equal to the square root of 26. now notice they don't want a decimal here because in the question they say exact length so as soon as they say exact length you need to make sure you

1:11:51 leave it as a cert that's really really important make sure you get all seven marks big question this one make sure you get all seven marks on this question okay so you can look through the matte scheme yourself and then just check my working and where you pick up those follow through marks if you make some errors as you go through okay and on to question nine so a circle has center at the point b five one and the point a minus one minus two lies on the circle very similar to how we did the previous question we're going to still use that formula what i'd recommend again is to draw a sketch so making sure we understand exactly conceptually what's going on because we're going to need this sketch anyway a little later on so

1:12:38 again quick sketch nothing to scale here so 5 1 is over here makes sense and then minus 1 minus 2 that's going to be down here so that's going to be our a let's just label these a and b this is minus one minus two so we get a circle that goes around this so something that looks like this again not to scale this is not right in the center but just to give us an idea of what's going on so again just like before we need to work out r squared so we need to work out this length here the way we do this because black is getting used pythagoras i tend to do this visually quite nice way of explaining it so minus one to five that's going to be six minus two to one that's gonna be three

1:13:26 so r squared will equal just six squared plus three squared that's equal to 36 plus 9 which is equal to 45. so now we use that formula remember the formula right at the back here this formula here and just substitute everything that i know so our answer for three marks that's kind of nice is x minus five all squared plus y minus one all squared and that's equal to forty like so but now we have to use that diagram in a bit more detail so point c is such that ac is a diameter of the circle and point d which is not on

1:14:12 the circle which is important to realize as coordinates 5 16 show that dc is a tangent to the circle so let's go back to my previous diagram and let's pop in this new point i'll do this in red again c so it's over this side and we don't know what it is but we'll work it out in a moment so we're going to work out this because imagine this radius goes all the way through to give you a diameter and there is a point d on the circle which reminds ourselves is 516. so 560. that's going to be up here somewhere this is d 5 16. and what they're claiming is and it kind of looks possible because if i connect up d and c continuous line all the way through that is a tangent to the circle

1:14:58 so let's go through this one stage at a time first of all we need to work out c now notice that we want six along a three up to get to the radius so we do the same thing again we go six along and three up so we do five plus six which gives us the eleven and then the one here one plus three is equal to four so we see the same pattern six and three again i can do it the other way around so you can see it so this is three and this is six because they're both radiuses or radii therefore need to be the same length so we get c is equal to eleven four and we're not sure it's a tangent to the circle and therefore this means this is a right angle here so in order to actually show use the idea it's a right angle we're going to work out the gradient of this line so we're going to

1:15:45 work out and let's just write it over here the gradient of the line p c or you can use v c as well okay so to work out the gradients again we use our formula so y two minus y one over x two minus x one so the change in y over change in x so if we do it here we can take either way around to get 11. okay that goes on the bottom right there we are so we get 4 minus minus 2. so 4 minus -2 divided by 11 minus something that's minus one yep

1:16:30 minus one so just comes from the coordinates this gives us then 6 over 12 which is the same as a half so the gradient of this diameter going across is a half therefore if this is a right angle then the gradient of this line has to be the negative reciprocal okay what do i mean by that so the gradient of cd well it needs to be this flipped and then change to a negative so if you flip this we get two over 1 and then we change it to a negative and therefore the gradient of that line is negative

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