Correlation and Regression

The data for marks in Physics and History obtained by $10$ students is as given:

Marks in Physics	$15$	$12$	$8$	$8$	$7$	$7$	$7$	$6$	$5$	$3$
Marks in History	$10$	$25$	$17$	$11$	$13$	$17$	$20$	$13$	$9$	$15$

Using this data, compute the Karl Pearson's coefficient of correlation between the marks in Physics and History obtained by the $10$ students.

Number of students $= n = 10$

Marks in Physics $\left(x\right)$	Marks in History $\left(y\right)$
$15$	$10$
$12$	$25$
$8$	$17$
$8$	$11$
$7$	$13$
$7$	$17$
$7$	$20$
$6$	$13$
$5$	$9$
$3$	$15$
$\Sigma x = 78$	$\Sigma y = 150$

$\overline{x} = \dfrac{\Sigma x}{n} = \dfrac{78}{10} = 7.8$, $\;$ $\overline{y} = \dfrac{\Sigma y}{n} = \dfrac{150}{10} = 15$

Marks in Physics $\left(x\right)$	Marks in History $\left(y\right)$	$d_x = x - \overline{x}$	$d_y = y - \overline{y}$	$d_x \cdot d_y$	$d_x^2$	$d_y^2$
$15$	$10$	$7.2$	$- 5$	$- 36$	$51.84$	$25$
$12$	$25$	$4.2$	$10$	$42$	$17.64$	$100$
$8$	$17$	$0.2$	$2$	$0.4$	$0.04$	$4$
$8$	$11$	$0.2$	$- 4$	$- 0.8$	$0.04$	$16$
$7$	$13$	$- 0.8$	$- 2$	$1.6$	$0.64$	$4$
$7$	$17$	$- 0.8$	$2$	$- 1.6$	$0.64$	$4$
$7$	$20$	$- 0.8$	$5$	$- 4$	$0.64$	$25$
$6$	$13$	$- 1.8$	$- 2$	$3.6$	$3.24$	$4$
$5$	$9$	$- 2.8$	$- 6$	$16.8$	$7.84$	$36$
$3$	$15$	$- 4.8$	$0$	$0$	$23.04$	$0$
$\Sigma x = 78$	$\Sigma y = 150$			$\Sigma d_x d_y = 22$	$\Sigma d_x^2 = 82.56$	$\Sigma d_y^2 = 218$

Karl Pearson coefficient $= r = \dfrac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2 \cdot \Sigma d_y^2}}$

i.e. $\;$ $r = \dfrac{22}{\sqrt{82.56 \times 218}} = \dfrac{22}{\sqrt{17998.08}} = \dfrac{22}{134.16} = 0.1640$ (approx)

Matrices

Using matrices, solve the following system of equations:

$x + y + z = 6$, $\;$ $x - y + z = 2$, $\;$ $2x + y - z = 1$

The given equations are: $\;$ $x + y + z = 6$, $\;$ $x - y + z = 2$, $\;$ $2x + y - z = 1$

Writing the given equations in the matrix form $\;$ $AX = B$ $\;$ as

$\begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix}$

where $\;$ $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{bmatrix}$, $\;$ $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, $\;$ $B = \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix}$

$\begin{aligned} \left|A\right| & = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \\\\ & = 1 \left(1 - 1\right) - 1 \left(-1 - 2\right) + 1 \left(1 + 2\right) = 6 \neq 0 \end{aligned}$

$\implies$ $A$ is a non-singular matrix.

$\therefore \;$ The given system of equations has the unique solution $\;$ $X = A^{-1} B$

Cofactors of matrix $A$ are:

$A_{11} = \begin{vmatrix} -1 & 1 \\ 1 & -1 \end{vmatrix} = 1 - 1 = 0$

$A_{12} = \left(-1\right) \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = \left(-1\right) \times \left(-1 -2\right) = 3$

$A_{13} = \begin{vmatrix} 1 & -1 \\ 2 & 1 \end{vmatrix} = 1 + 2 = 3$

$A_{21} = \left(-1\right) \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = \left(-1\right) \times \left(-1 -1\right) = 2$

$A_{22} = \begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix} = -1 -2 = -3$

$A_{23} = \left(-1\right) \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} = \left(-1\right) \times \left(1 - 2\right) = 1$

$A_{31} = \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 1 + 1 = 2$

$A_{32} = \left(-1\right) \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = \left(-1\right) \times \left(1 - 1\right) = 0$

$A_{33} = \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = -1 - 1 = -2$

$\therefore \;$ $adj \left(A\right) = \begin{bmatrix} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{bmatrix}^t = \begin{bmatrix} 0 & 3 & 3 \\ 2 & -3 & 1 \\ 2 & 0 & -2 \end{bmatrix}^t = \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix}$

$\begin{aligned} \therefore \; A^{-1} = \dfrac{adj \left(A\right)}{\left|A\right|} & = \dfrac{1}{6} \begin{bmatrix} 0 & 2 & 2 \\ 3 & -3 & 0 \\ 3 & 1 & -2 \end{bmatrix} \\\\ & = \begin{bmatrix} 0 & 1/3 & 1/3 \\ 1/2 & -1/2 & 0 \\ 1/2 & 1/6 & -1/3 \end{bmatrix} \end{aligned}$

$\begin{aligned} \therefore \; X = A^{-1} B & = \begin{bmatrix} 0 & 1/3 & 1/3 \\ 1/2 & -1/2 & 0 \\ 1/2 & 1/6 & -1/3 \end{bmatrix} \begin{bmatrix} 6 \\ 2 \\ 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 0 + 2/3 + 1/3 \\ 3 - 1 + 0 \\ 3 + 1/3 - 1/3 \end{bmatrix} \\\\ i.e. \; \begin{bmatrix} x \\ y \\ z \end{bmatrix} & = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned}$

$\implies$ $x = 1$, $\;$ $y = 2$, $\;$ $z = 3$

Correlation and Regression

Find the regression coefficients $b_{yx}$ and $b_{xy}$ for the following data:
$n = 6$, $\;$ $\Sigma x = 30$, $\;$ $\Sigma y = 42$, $\;$ $\Sigma xy = 199$, $\;$ $\Sigma x^2 = 184$, $\;$ $\Sigma y^2 = 318$

$\begin{aligned} b_{yx} & = \dfrac{\Sigma xy - \dfrac{\Sigma x \Sigma y}{n}}{\Sigma x^2 - \dfrac{\left(\Sigma x\right)^2}{n}} \\\\ & = \dfrac{199 - \dfrac{30 \times 42}{6}}{184 - \dfrac{\left(30\right)^2}{6}} \\\\ & = \dfrac{199 - 210}{184 - 150} = \dfrac{-11}{34} = - 0.3235 \end{aligned}$

$\begin{aligned} b_{xy} & = \dfrac{\Sigma xy - \dfrac{\Sigma x \Sigma y}{n}}{\Sigma y^2 - \dfrac{\left(\Sigma y\right)^2}{n}} \\\\ & = \dfrac{199 - \dfrac{30 \times 42}{6}}{318 - \dfrac{\left(42\right)^2}{6}} \\\\ & = \dfrac{199 - 210}{318 - 294} = \dfrac{-11}{24} = - 0.4583 \end{aligned}$

Analytical Geometry - Conics - Ellipse

Find the equation of ellipse whose latus-rectum is $5$ and eccentricity is $\dfrac{2}{3}$.

Latus-rectum of ellipse $= \dfrac{2b^2}{a} = 5$ (given)

$\implies$ $b^2 = \dfrac{5 a}{2}$ $\;\;\; \cdots \; (1)$

Eccentricity $= e = \dfrac{2}{3}$ (given)

For an ellipse, $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (2)$

Substituting the values of $b^2$ and $e$ in equation $(2)$ gives,

$\dfrac{5a}{2} = a^2 \left[1 - \left(\dfrac{2}{3}\right)^2\right]$

i.e. $\;$ $\dfrac{5}{2} = \dfrac{5a}{9}$

i.e. $\;$ $a = \dfrac{9}{2}$ $\implies$ $a^2 = \dfrac{81}{4}$

Substituting the value of $a$ in equation $(1)$ gives,

$b^2 = 5 \times \dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{45}{4}$

The equation of ellipse is $\;\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$

i.e. $\;$ $\dfrac{4 x^2}{81} + \dfrac{4 y^2}{45} = 1$

Definite Integration

Evaluate: $\;$ $\displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$

Let $\;$ $I = \displaystyle \int_{2}^{3} \dfrac{x^3 + 1}{x \left(x - 1\right)} \; dx$

$\begin{aligned} i.e. \; I & = \int_{2}^{3} \dfrac{\left(x - 1\right) \left(x^2 + x + 1\right)}{x \left(x - 1\right)} \; dx \;\;\; \left[\text{Note: } a^3 - b^3 = \left(a - b\right) \left(a^2 + ab + b^2\right)\right] \\\\ & = \int_{2}^{3} \dfrac{x^2 + x + 1}{x} \; dx \\\\ & = \int_{2}^{3} x \; dx + \int_{2}^{3} dx + \int_{2}^{3} \dfrac{1}{x} \; dx \\\\ & = \left[\dfrac{x^2}{2}\right]_{2}^{3} + \left[x\right]_{2}^{3} + \left[\log x\right]_{2}^{3} \\\\ & = \dfrac{1}{2} \left(3^2 - 2^2\right) + \left(3 - 2\right) + \log 3 - \log 2 \\\\ & = \dfrac{7}{2} + \log \left(\dfrac{3}{2}\right) \end{aligned}$

Indefinite Integration

Evaluate: $\;$ $\displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $I = \displaystyle \int \sin^{-1} \left(\dfrac{2x}{1 + x^2}\right) \; dx$

Let $\;$ $x = \tan \theta$

Then, $\;$ $dx = \sec^2 \theta \; d\theta$

$\begin{aligned} \therefore \; I & = \int \sin^{-1} \left(\dfrac{2 \tan \theta}{1 + \tan^2 \theta}\right) \sec^2 \theta \; d\theta \\\\ & = \int \sin^{-1} \left[\sin \left(2 \theta\right)\right] \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin \left(2A\right) = \dfrac{2 \tan A}{1 + \tan^2 A}\right] \\\\ & = \int 2 \theta \; \sec^2 \theta \; d\theta \;\;\; \left[\text{Note: } \sin^{-1} \left[\sin \left(A\right)\right] = A\right] \\\\ & = 2 \left\{\theta \int \sec^2 \theta \; d\theta - \int \left[\dfrac{d}{d \theta} \left(\theta\right) \times \int \sec^2 d \theta\right] d \theta \right\} \\\\ & \left[\text{Note: Integration by parts: } \int u \cdot v \; dx = u \int v \; dx - \int \left\{\dfrac{du}{dx} \times \int v \; dx \right\} dx\right] \\\\ & = 2 \left\{\theta \; \tan \theta - \int \tan \theta \; d\theta \right\} + c \;\;\; \left[\text{Note: } \int \sec^2 x \; dx = \tan x + c\right] \\\\ & = 2 \; \theta \; \tan \theta - 2\log \left|\sec \theta\right| + c \;\;\; \left[\text{Note: } \int \tan x \; dx = \log \left|\sec x\right| + c\right] \\\\ & = 2 \; \theta \; \tan \theta - \log \left|\sec \theta\right|^2 + c \\\\ & = 2 \; \theta \; \tan \theta - \log \left|1 + \tan^2 \theta\right| + c \;\;\; \left[\text{Note: } 1 + \tan^2 x = \sec^2 x\right] \end{aligned}$

$\because \;$ $x = \tan \theta$ $\implies$ $\theta = \tan^{-1}x$

$\therefore \;$ $I = 2 \; x \; \tan^{-1}x - \log \left|1 + x^2\right| + c$

Inverse Trigonometric Functions

Solve: $\;$ $\cos^{-1} \left[\sin \left(\cos^{-1}x\right)\right] = \dfrac{\pi}{3}$

Given: $\;$ $\cos^{-1} \left[\sin \left(\cos^{-1}x\right)\right] = \dfrac{\pi}{3}$

i.e. $\;$ $\sin \left(\cos^{-1}x\right) = \cos \left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ $\;\;\; \cdots \; (1)$

Let $\;$ $\cos^{-1}x = p$ $\;\;\; \cdots \; (2)$

$\therefore \;$ In view of equation $(2)$ equation $(1)$ becomes,

$\sin \left(p\right) = \dfrac{1}{2}$

i.e. $\;$ $p = \sin^{-1} \left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$

Substituting the value of $p$ in equation $(2)$, we get,

$\cos^{-1}x = \dfrac{\pi}{6}$

$\therefore \;$ $x = \cos \left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$

Matrices

If $A = \begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}$, show that $\;$ $A^t A^{-1} = \begin{bmatrix} \cos 2x & - \sin 2x \\ \sin 2x & \cos 2x \end{bmatrix}$

$A = \begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}$

Transpose of $A = A^t = \begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}^t$

i.e. $\;$ $A^t = \begin{bmatrix} 1 & - \tan x \\ \tan x & 1 \end{bmatrix}$

i.e. $\;$ $A^t = \begin{bmatrix} 1 & \dfrac{- \sin x}{\cos x} \\ \dfrac{\sin x}{\cos x} & 1 \end{bmatrix}$

$A^{-1} = \dfrac{1}{\left|A\right|} \times adj \left(A\right)$

$\left|A\right| = \begin{vmatrix} 1 & \tan x \\ - \tan x & 1 \end{vmatrix} = 1 + \tan^2 x = \sec^2 x$

Cofactors of matrix $A$ are:

$A_{11} = 1$, $\;$ $A_{12} = \tan x$, $\;$ $A_{21} = - \tan x$, $\;$ $A_{22} = 1$

$\therefore \;$ $adj \left(A\right) = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}^t = \begin{bmatrix} 1 & \tan x \\ - \tan x & 1 \end{bmatrix}^t = \begin{bmatrix} 1 & - \tan x \\ \tan x & 1 \end{bmatrix}$

$\begin{aligned} \therefore \; A^{-1} & = \dfrac{1}{\sec^2 x} \begin{bmatrix} 1 & - \tan x \\ \tan x & 1 \end{bmatrix} \\\\ & = \cos^2 x \begin{bmatrix} 1 & \dfrac{- \sin x}{\cos x} \\ \dfrac{\sin x}{\cos x} & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} \cos^2 x & - \sin x \cos x \\ \sin x \cos x & \cos^2 x \end{bmatrix} \end{aligned}$

$\begin{aligned} \therefore \; A^t A^{-1} & = \begin{bmatrix} 1 & \dfrac{- \sin x}{\cos x} \\ \dfrac{\sin x}{\cos x} & 1 \end{bmatrix} \begin{bmatrix} \cos^2 x & - \sin x \cos x \\ \sin x \cos x & \cos^2 x \end{bmatrix} \\\\ & = \begin{bmatrix} \cos^2 x - \dfrac{\sin x}{\cos x} \times \sin x \cos x & - \sin x \cos x - \dfrac{\sin x}{\cos x} \times \cos^2 x \\ \dfrac{\sin x}{\cos x} \times \cos^2 x + \sin x \cos x & \dfrac{\sin x}{\cos x} \times \left(- \sin x \cos x\right) + \cos^2 x \end{bmatrix} \\\\ & = \begin{bmatrix} \cos^2 x - \sin^2 x & - 2 \sin x \cos x \\ 2 \sin x \cos x & \cos^2 x - \sin^2 x \end{bmatrix} \\\\ & = \begin{bmatrix} \cos 2x & - \sin 2x \\ \sin 2x & \cos 2x \end{bmatrix} \end{aligned}$

Equation of a Line; Section Formula

A line $AB$ meets the X-axis at $A$ and the Y-axis at $B$. $P \left(4, -1\right)$ divides $AB$ in the ratio $1 : 2$.

1. Find the coordinates of $A$ and $B$.
2. Find the equation of the line through $P$ and perpendicular to $AB$.

1. Given: $\;$ Line $AB$ meets the X-axis at $A$ and the Y-axis at $B$.
   
   $\therefore \;$ Let the coordinates of point $A$ be $\left(a, 0\right)$
   
   and the coordinates of point $B$ be $\left(0, b\right)$.
   
   Given: $\;$ $P \left(4, -1\right)$ divides $AB$ in the ratio $1 : 2$
   
   $\therefore \;$ We have by section formula,
   
   $4 = \dfrac{1 \times 0 + 2 \times a}{1 + 2}$
   
   i.e. $\;$ $12 = 2a$ $\implies$ $a = 6$
   
   and $\;$ $-1 = \dfrac{1 \times b + 2 \times 0}{1 + 2}$ $\implies$ $b = -3$
   
   $\therefore \;$ $A = \left(6, 0\right)$; $\;$ $B = \left(0, -3\right)$



2. Slope of $AB = m_1 = \dfrac{-3 - 0}{0 - 6} = \dfrac{1}{2}$
   
   Slope of line perpendicular to $AB = m_2 = \dfrac{-1}{m_1} = -2$
   
   The required line passes through the point $P \left(4, -1\right)$
   
   $\therefore \;$ Equation of the required line is
   
   $y + 1 = -2 \left(x - 4\right)$
   
   i.e. $\;$ $y + 1 = -2x + 8$
   
   i.e. $\;$ $2x + y = 7$

Matrices

If $A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, then find $k$ such that $A^2 = kA - 2 I$

$A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}$, $\;$ $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 2 \times 2$ $\;$ identity matrix

$\begin{aligned} A^2 = A \times A & = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \times \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \\\\ & = \begin{bmatrix} 3 \times 3 - 2 \times 4 & 3 \times \left(-2\right) - 2 \times \left(-2\right) \\ 4 \times 3 - 2 \times 4 & 4 \times \left(-2\right) - 2 \times \left(-2\right) \end{bmatrix} \\\\ & = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \end{aligned}$

$\begin{aligned} kA & = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \\\\ & = \begin{bmatrix} 3k & - 2k \\ 4k & -2k \end{bmatrix} \end{aligned}$

$\begin{aligned} 2I & = 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \end{aligned}$

Given: $\;$ $A^2 = kA - 2I$

i.e. $\;$ $\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$

When two matrices are equal, their corresponding elements are equal.

$\;$ We have, $\;$ $1 = 3k - 2$; $\;$ $-2 = -2 k$; $\;$ $4 = 4k$; $\;$ $-4 = -2k - 2$

These equations give $\;$ $k = 1$

Similarity - Maps and Models

A map is drawn to a scale of $1 : 20000$

1. The distance between the two towns on the map is $9 \; cm $. Calculate the actual distance between the towns in km.
2. Calculate the actual area in $m^2$ which represents $8 \; cm^2 $ on the map.

1. Scale of map is $\;$ $1 : 20000$
   
   i.e. $\;$ $1 \; cm$ on the map corresponds to an actual distance of $\;$ $20000 \; cm$
   
   $\therefore \;$ $9 \; cm$ on the map corresponds to an actual distance of $\;$ $\dfrac{9 \times 20000}{1} = 180000 = 1.8 \times 10^5 \; cm$
   
   Now, $\;$ $10^5 \; cm = 1 \; km$
   
   $\therefore \;$ $1.8 \times 10^5 \; cm = 1.8 \; km$
   
   $\therefore \;$ Actual distance between the towns $= 1.8 \; km$



2. Scale of map is $\;$ $1 : 20000$
   
   i.e. $\;$ $1 \; cm$ on the map corresponds to an actual distance of $\;$ $20000 = 2 \times 10^4 \; cm$
   
   i.e. $\;$ $1 \; cm^2$ on the map corresponds to an actual area of $\;$ $\left(2 \times 10^4\right)^2 = 4 \times 10^8 \; cm^2$
   
   $\therefore \;$ $8 \; cm^2$ on the map corresponds to an actual area of $\;$ $\dfrac{8 \times 4 \times 10^8}{1} = 3.2 \times 10^9 \; cm^2$
   
   Now, $\;$ $10^4 \; cm^2 = 1 \; m^2$
   
   $\therefore \;$ $3.2 \times 10^9 \; cm^2 = \dfrac{3.2 \times 10^9 \times 1}{10^4} = 3.2 \times 10^5 \; m^2$
   
   $\therefore \; $ Actual area $= 3.2 \times 10^5 \; m^2$

Factor and Remainder Theorems

The polynomials $2x^3 - 7x^2 + ax - 6$ and $x^3 - 8x^2 + \left(2a + 1 \right)x - 16$ leave the same remainder when divided by $\left(x - 2\right)$. Find the value of $a$.

Let $\;$ $f \left(x\right) = 2x^3 - 7x^2 + ax - 6$, $\;$ $g \left(x\right) = x^3 - 8x^2 + \left(2a + 1\right)x - 16$

By remainder theorem,

remainder obtained when $f \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = f \left(2\right)$

remainder obtained when $g \left(x\right)$ is divided by $\left(x - 2\right)$ is $ = g \left(2\right)$

Given: $\;$ $f \left(2\right) = g \left(2\right)$ $\;\;\; \cdots \; (1)$

Now, $\;$ $f \left(2\right) = 2 \times 2^3 - 7 \times 2^2 + 2a - 6$

i.e. $\;$ $f \left(2\right) = 2a - 18$ $\;\;\; \cdots \; (2a)$

and $\;$ $g \left(2\right) = 2^3 - 8 \times 2^2 + \left(2a + 1\right) \times 2 - 16$

i.e. $\;$ $g \left(x\right) = 4a - 38$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,

$2a - 18 = 4a - 38$

i.e. $\;$ $2a = 20$ $\implies$ $a = 10$

Quadratic Equations

Find $k$ for which $x = 3$ is a solution of the quadratic equation $\left(k + 2\right)x^2 - kx + 6 = 0$. Thus find the other root of the equation.

$\because \;$ $x = 3$ is a solution of the quadratic equation $\left(k + 2\right)x^2 - kx + 6 = 0$, we have

$\left(k + 2\right) \times 3^2 - k \times 3 + 6 = 0$

i.e. $\;$ $9k + 18 - 3k + 6 = 0$

i.e. $\;$ $6k = - 24$ $\implies$ $k = -4$

Substituting the value of $k$, the given quadratic equation becomes,

$\left(-4 + 2\right)x^2 - \left(-4\right)x + 6 = 0$

i.e. $\;$ $2 x^2 - 4 x - 6 = 0$

i.e. $\;$ $x^2 - 2x - 3 = 0$

i.e. $\;$ $x^2 - 3x + x - 3 = 0$

i.e. $\;$ $x \left(x - 3\right) + 1 \left(x - 3\right) = 0$

i.e. $\;$ $\left(x + 1\right) \left(x - 3\right) = 0$

i.e. $\;$ $x + 1 = 0 \;$ or $\;$ $x - 3 = 0$

i.e. $\;$ $x = -1 \;$ or $\;$ $x = 3$

$\therefore \;$ The other root of the equation is $x = -1$.

ALTERNATIVELY

Comparing $\;$ $x^2 - 2x - 3 = 0$ $\;$ with standard quadratic equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 1, \; b = -2, \; c = -3$

Given: $\;$ One root of the quadratic equation $= \alpha = 3$

Let the other root of the quadratic equation $= \beta$

Sum of roots $= \alpha + \beta = \dfrac{-b}{a}$

i.e. $\;$ $3 + \beta = \dfrac{2}{1} = 2$

$\implies$ $\beta = -1$

$\therefore \;$ The other root of the equation is $\beta = -1$.

Mensuration

A solid cylinder of radius $7 \; cm$ and height $14 \; cm$ is melted and recast into solid spheres of radius $3.5 \; cm$. Find the number of spheres formed.

Radius of cylinder $= r = 7 \; cm$

Height of cylinder $= h = 14 \; cm$

Volume of cylinder $= V_c = \pi r^2 h = \pi \times 7^2 \times 14 \; cm^3$

Radius of sphere $= R = 3.5 \; cm$

Volume of sphere $= V_s = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3.5^3 \; cm^3$

Let the number of spheres formed $= n$

$\because \;$ the cylinder is melted and recast into a number of spheres,

$\text{Volume of cylinder} = n \times \text{Volume of sphere}$

$\begin{aligned} \therefore \; n & = \dfrac{\text{Volume of cylinder}}{\text{Volume of sphere}} \\\\ & = \dfrac{\pi \times 7^2 \times 14}{\dfrac{4}{3} \times \pi \times 3.5^3} \\\\ & = \dfrac{3 \times 7^2 \times 14}{4 \times 3.5^3} = 12 \end{aligned}$

$\therefore \;$ Number of spheres formed $= 12$

Arithmetic Progression

The first and the last term of an A.P are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?

First term of A.P $= a = 17$

Common difference of A.P $= d = 9$

Let the $n^{th}$ term be the last term.

Last term of A.P $= t_n = a + \left(n - 1\right)d = 350$ $\;\;\; \cdots \; (1)$

Substituting the values of $a$ and $d$ in equation $(1)$ we have,

$17 + \left(n - 1\right) \times 9 = 350$

i.e. $\;$ $n - 1 = \dfrac{333}{9}$

i.e. $\;$ $n = \dfrac{333}{9} + 1 = \dfrac{342}{9} = 38$

$\therefore \;$ Number of terms in A.P $= n = 38$

Sum of $n$ terms of A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

$\therefore \;$ $S_{38} = \dfrac{38}{2} \left[2 \times 17 + \left(38 - 1\right) \times 9\right]$

i.e. $\;$ $S_{38} = 19 \left(34 + 333\right) = 6973$

Factor Theorem

Find the value of $a$ and $b$ if $\left(x - 1\right)$ and $\left(x - 2\right)$ are factors of $x^3 - ax + b$.

Let $f \left(x\right) = x^3 - ax + b$

$\because \;$ $\left(x - 1\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(1\right) = 0$

i.e. $\;$ $f \left(1\right) = 1^3 - a \times 1 + b = 0$

i.e. $\;$ $a - b = 1$ $\;\;\; \cdots \; (1)$

$\because \;$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\implies$ by factor theorem $\;$ $f \left(2\right) = 0$

i.e. $\;$ $f \left(2\right) = 2^3 - a \times 2 + b = 0$

i.e. $\;$ $2a - b = 8$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously we have,

$a = 7$ $\;$ and $\;$ $b = 6$

Matrices

If $\;$ $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, then show that $\;$ $A^2 - 5 A + 7 I = 0$

$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, $\;$ $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 2 \times 2$ $\;$ identity matrix

$\begin{aligned} A^2 = A \times A & = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 3 \times 3 + 1 \times \left(-1\right) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times \left(-1\right) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \end{aligned}$

$\begin{aligned} 5A & = 5 \times \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \\\\ & = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \end{aligned}$

$\begin{aligned} 7 I & = 7 \times \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \end{aligned}$

$\begin{aligned} \therefore \; A^2 - 5 A + 7 I & = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \\\\ & = \begin{bmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 7 \end{bmatrix} \\\\ & = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \\\\ & = 0 \;\;\; \left[\text{null matrix}\right] \end{aligned}$

$\therefore \;$ $A^2 - 5A + 7 I = 0$ $\;\;\;$ Hence proved.

Quadratic Equations

A train covers a distance of $90 \; km$ at an uniform speed. Had the speed been $15 \; kmph$ more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.

Let the original speed of the train be $x \; kmph$

Original time taken to cover $90 \; km$ $= t_1 = \dfrac{\text{distance}}{\text{speed}} = \dfrac{90}{x}$ hour

New speed of train $= x + 15 \; kmph$

New time taken to cover $90 \; km$ $= t_2 = \dfrac{90}{x + 15}$ hour

As per question, $\;$ $t_2 = t_1 - 30 \; \text{minutes} = t_1 - \dfrac{1}{2} \; \text{hour}$

i.e. $\;$ $\dfrac{90}{x + 15} = \dfrac{90}{x} - \dfrac{1}{2}$

i.e. $\;$ $\dfrac{90}{x + 15} = \dfrac{180 - x}{2x}$

i.e. $\;$ $180x = 180x - x^2 + 2700 - 15x$

i.e. $\;$ $x^2 + 15x - 2700 = 0$

i.e. $\;$ $x^2 + 60x - 45x - 2700 = 0$

i.e. $\;$ $x \left(x + 60\right) - 45 \left(x + 60\right) = 0$

i.e. $\;$ $\left(x + 60\right) \left(x - 45\right) = 0$

i.e. $\;$ $x + 60 = 0$ $\;$ or $\;$ $x - 45 = 0$

i.e. $\;$ $x = -60$ $\;$ or $\;$ $x = 45$

Since the speed of the train cannot be negative,

$\therefore \;$ original speed of train $= 45 \; kmph$

Coordinate Geometry

A straight line passes through the points $P \left(-1, 4\right)$ and $Q \left(5, -2\right)$. It intersects the coordinate axes at points $A$ and $B$. $M$ is the midpoint of segment $AB$. Find:

1. the equation of the line;
2. the coordinates of points $A$ and $B$;
3. the coordinates of $M$.

1. Given: $\;$ $P \left(x_1, y_1\right) \equiv \left(-1, 4\right)$, $\;$ $Q \left(x_2, y_2\right) \equiv \left(5, -2\right)$
   
   Slope of line $PQ = m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-2 - 4}{5 - \left(-1\right)} = \dfrac{-6}{6} = -1$
   
   $\therefore \;$ Equation of line $PQ$ is: $\;$ $y - y_1 = m \left(x - x_1\right)$
   
   i.e. $\;$ $y - 4 = -1 \left(x + 1\right)$
   
   i.e. $\;$ $y - 4 = -x - 1$
   
   i.e. $\;$ $x + y = 3$



2. Let the line $PQ$ intersect the X-axis at the point $A$ and the Y-axis at the point $B$.
   
   Then, the coordinates of point $A$ are $\left(a, 0\right)$ and
   
   the coordinates of point $B$ are $\left(0, b\right)$.
   
   When the line $PQ$ cuts the X-axis, putting $y = 0$ in the equation of $PQ$ gives $\;$ $x = 3$
   
   When the line $PQ$ cuts the Y-axis, putting $x = 0$ in the equation of $PQ$ gives $\;$ $y = 3$
   
   $\therefore \;$ $A = \left(3, 0\right)$; $\;$ $B = \left(0, 3\right)$



3. $M$ is the midpoint of $AB$.
   
   $\therefore \;$ Coordinates of $M = \left(\dfrac{3 + 0}{2}, \dfrac{0 + 3}{2}\right)$
   
   i.e. $\;$ $M = \left(\dfrac{3}{2}, \dfrac{3}{2}\right)$

Geometric Progression

The sum of three numbers in a geometric progression is $\dfrac{39}{10}$ and their product is $1$. Find the numbers.

Let the three numbers in GP be $\;$ $\dfrac{a}{r}, \; a, \; ar$

where $\;$ $a = $ first term of GP and $\;$ $r = $ common ratio of GP

As per question,

$\dfrac{a}{r} + a + ar = \dfrac{39}{10}$ $\;\;\; \cdots \; (1)$

and $\;$ $\dfrac{a}{r} \times a \times ar = 1$ $\implies$ $a^3 = 1$ $\implies$ $a = 1$

Substituting the value of $\;$ $a$ $\;$ in equation $(1)$ we have,

$\dfrac{1}{r} + 1 + r = \dfrac{39}{10}$

i.e. $\;$ $\dfrac{1 + r^2}{r} = \dfrac{39}{10} - 1 = \dfrac{29}{10}$

i.e. $\;$ $10 + 10 r^2 = 29 r$

i.e. $\;$ $10 r^2 - 29 r + 10 = 0$

i.e. $\;$ $10 r^2 - 25 r - 4 r + 10 = 0$

i.e. $\;$ $5r \left(2r - 5\right) - 2 \left(2r - 5\right) = 0$

i.e. $\;$ $\left(5r - 2\right) \left(2r - 5\right) = 0$

i.e. $\;$ $5r - 2 = 0$ $\;$ or $\;$ $2r - 5 = 0$

i.e. $\;$ $r = \dfrac{2}{5}$ $\;$ or $\;$ $r = \dfrac{5}{2}$

$\therefore \;$ The three numbers in GP are:

$\dfrac{1}{2/5}, \; 1, \; 1 \times \dfrac{2}{5}$ $\;$ or $\;$ $\dfrac{1}{5/2}, \; 1, \; 1 \times \dfrac{5}{2}$

i.e. $\;$ $\dfrac{5}{2}, \; 1, \; \dfrac{2}{5}$ $\;$ or $\;$ $\dfrac{2}{5}, \; 1, \; \dfrac{5}{2}$

Trigonometry

Prove that: $\;$ $\dfrac{\cos A}{1 + \sin A} + \tan A = \sec A$

$\begin{aligned} LHS & = \dfrac{\cos A}{1 + \sin A} + \tan A \\\\ & = \dfrac{\cos A}{1 + \sin A} + \dfrac{\sin A}{\cos A} \\\\ & = \dfrac{\cos^2 A + \sin^2 A + \sin A}{\cos A \left(1 + \sin A\right)} \\\\ & = \dfrac{1 + \sin A}{\cos A \left(1 + \sin A\right)} \;\;\; \left[\text{Note: } \sin^2 A + \cos^2 A = 1\right] \\\\ & = \dfrac{1}{\cos A} \\\\ & = \sec A = RHS \end{aligned}$

Hence proved.

Statistics

The median of the observations $11, \; 12, \; 14, \; \left(x - 2\right), \; \left(x + 4\right), \; \left(x + 9\right), \; 32, \; 38, \; 47$, arranged in ascending order, is $24$. Find the value of $x$ and hence find the mean.

Given observations: $\;$ $11, \; 12, \; 14, \; \left(x - 2\right), \; \left(x + 4\right), \; \left(x + 9\right), \; 32, \; 38, \; 47$

Number of observations $= n = 9 $ (odd number of observations)

Median value $= \left(\dfrac{n + 1}{2}\right)^{th}$ term

$\therefore \;$ Median $= \left(\dfrac{9 + 1}{2}\right)^{th}$ term $= 5^{th}$ term $= x + 4$

Given: $\;$ Median $= 24$

$\implies$ $x + 4 = 24$ $\implies$ $x = 20$

$\therefore \;$ The given observations are: $\;$ $11, \; 12, \; 14, \; 18, \; 24, \; 29, \; 32, \; 38, \; 47$

Mean $= \dfrac{11 + 12 + 14 + 18 + 24 + 29 + 32 + 38 + 47}{9} = \dfrac{225}{9} = 25$

Probability

One card is drawn at random from a well shuffled deck of $52$ cards. Find the probability of drawing

1. a jack of spades;
2. a black queen.

Sample space $= S = 52$ cards

$\therefore \;$ Number of elements in sample space $= n \left(S\right) = 52$

1. Let $A = $ event of selecting a jack of spades.
   
   Number of elements in $A = n \left(A\right) = 1$
   
   $\therefore \;$ Probability of event $A = P \left(A\right) = \dfrac{n \left(A\right)}{n \left(S\right)} = \dfrac{1}{52}$



2. Let $B = $ event of selecting a black queen.
   
   Number of elements in $B = n \left(B\right) = 2$
   
   $\therefore \;$ Probability of event $B = P \left(B\right) = \dfrac{n \left(B\right)}{n \left(S\right)} = \dfrac{2}{52} = \dfrac{1}{26}$

Circle

In the figure, lines $AB$ and $CD$ pass through the center $O$ of a circle.

If $\angle AOD = 75^\circ$ and $\angle OCE = 40^\circ$, find $\angle CDE$ and $\angle OBE$.

Given: $\;$ $O$ is the center of the circle and line $CD$ passes through $O$.

$\implies$ $CD$ is a diameter of the given circle.

Then, $\;$ $\angle CED = 90^\circ$ $\;\;\;$ [angle in a semicircle]

Given: $\;$ $\angle OCE = 40^\circ$

Points $D$, $O$ and $C$ are collinear points.

$\therefore \;$ $\angle DCE = \angle OCE = 40^\circ$

Now, in right triangle $CED$,

$\angle DCE + \angle CED + \angle CDE = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $40^\circ + 90^\circ + \angle CDE = 180^\circ$

$\implies$ $\angle CDE = 50^\circ$

Given: $\;$ $\angle AOD = 75^\circ$

Now, $\;$ $\angle COM = \angle AOD = 75^\circ$ $\;\;\;$ [vertically opposite angles are equal]

In $\triangle OCM$,

$\angle OCM + \angle CMO + \angle COM = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $40^\circ + \angle CMO + 75^\circ = 180^\circ$

i.e. $\;$ $\angle CMO = 65^\circ$

Now, $\;$ $\angle EMB = \angle CMO = 65^\circ$ $\;\;\;$ [vertically opposite angles are equal]

In $\triangle MBE$,

$\angle CED = \angle EMB + \angle MBE$ $\;\;\;$ [exterior angle is equal to the sum of the two interior opposite angles]

$\therefore \;$ $\angle MBE = \angle CED - \angle EMB = 90^\circ - 65^\circ = 25^\circ$

But $\;$ $\angle MBE = \angle OBE$ $\;\;\;$ [Points $O$, $M$ and $B$ are collinear points]

$\implies$ $\angle OBE = 25^\circ$

Constructions

Construct an equilateral triangle $ABC$ with side $4 \; cm$. Draw a circle touching the sides of the triangle.

Construct an equilateral triangle with each side equal to $4 \; cm$.

Draw perpendicular bisectors $PQ$ and $RS$ of sides $AB$ and $BC$ respectively.

The perpendicular bisectors intersect at point $O$.

With $O$ as center and $OA$ as radius, draw a circle which passes through all the vertices of the equilateral triangle $ABC$.

Circle

In the figure, $O$ is the center of the circle and $SP$ is the tangent.

If $\angle SRT = 65^\circ$, find the values of $x$, $y$ and $z$.

$\angle TSP = \angle TSR = 90^\circ$ $\;\;\;$ [angle between the radius of a circle and the tangent at the point of contact]

Given: $\;$ $\angle SRT = 65^\circ$

In right $\triangle TSR$,

$\angle TSR + \angle STR + \angle SRT = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $90^\circ + x + 65^\circ = 180^\circ$ $\implies$ $x = 25^\circ$

Now, $\;$ $OT = OQ$ $\;\;\;$ [radii of the same circle]

$\therefore \;$ $\angle OQT = \angle OTQ = 25^\circ$ $\;\;\;$ [angles opposite equal sides are equal]

In $\triangle OTQ$,

$\angle OTQ + \angle OQT + \angle TOQ = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $25^\circ + 25^\circ + \angle TOQ = 180^\circ$ $\implies$ $\angle TOQ = 130^\circ$

Now, $\;$ $\angle TOQ + \angle OSR = 180^\circ$ $\;\;\;$ [linear pair]

i.e. $\;$ $130^\circ + y = 180^\circ$ $\implies$ $y = 50^\circ$

In right $\triangle OSP$,

$\angle OSP + \angle SPO + \angle POS = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]

i.e. $\;$ $90^\circ + z + 50^\circ = 180^\circ$ $\implies$ $z = 40^\circ$

$\therefore \;$ $x = 25^\circ, \; y = 50^\circ, \; z = 40^\circ$

Trigonometry

A tower subtends an angle $\alpha$ on the same level as the foot of the tower and at a second point $h$ meters above the first, the depression of the foot of the tower is $\beta$. Show that the height of the tower is $h \tan \alpha \cot \beta$.

In the figure,

$AT = $ tower of height $H$

$B = $ a point on the same level as the foot of the tower at a distance $x$ from $A$

$C = $ a point $h$ meters above $B$

From the figure,

in $\triangle ATB$, $\;$ $\tan \alpha = \dfrac{AT}{AB} = \dfrac{H}{x}$

in $\triangle ACB$, $\;$ $\tan \beta = \dfrac{BC}{AB} = \dfrac{h}{x}$

$\therefore \;$ $\dfrac{\tan \alpha}{\tan \beta} = \dfrac{H / x}{h / x} = \dfrac{H}{h}$

i.e. $\;$ $H = \dfrac{\tan \alpha}{\tan \beta} \times h$

i.e. $\;$ $H = h \tan \alpha \cot \beta$

Mensuration

A conical vessel of radius $6 \; cm$ and height $8 \; cm$ is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. Find the fraction of water that overflows.

Figure shows the cross-sectional view of the conical vessel (triangle ABC) and the sphere (circle at center $C_1$).

Since the sphere touches the sides of the conical vessel when it is just immersed, the sides of the cone are tangents to the circle.

In the figure,

radius of the conical vessel $= CM = MB = r = 6 \; cm$

height of the conical vessel $= AM = h = 8 \; cm$

slant height of the conical vessel $= AC = \ell = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \; cm$

radius of the sphere $= C_1 K = R \; cm$

In triangles $AKC_1$ and $AMC$,

$\angle AKC_1 = \angle AMC = 90^\circ$

[angle between the tangent to a circle and the radius of the circle at the point of contact]

$\angle A = \angle A$ $\;\;\;$ [common angle]

$\therefore \;$ By AA postulate, $\;$ $\triangle AKC_1 \sim \triangle AMC$

$\therefore \;$ $\dfrac{MC}{KC_1} = \dfrac{AC}{AC_1}$ $\;\;\;$ [corresponding parts of similar triangles are in proportion]

i.e. $\;$ $\dfrac{r}{R} = \dfrac{\ell}{8 - R}$

i.e. $\;$ $\dfrac{6}{R} = \dfrac{10}{8 - R}$

i.e. $\;$ $48 - 6R = 10R$

i.e. $\;$ $16 R = 48$ $\implies$ $R = 3 \; cm$

Now,

Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$

Volume of conical vessel $= V_C = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \pi \times 6^2 \times 8 = 96 \pi \; cm^3$

$\therefore \;$ Fraction of water that overflows

$= \dfrac{V_C - V_S}{V_C} = \dfrac{96 \pi - 36 \pi}{96 \pi} = \dfrac{60}{96} = \dfrac{5}{8}$ of the total volume of water

Statistics

The following table shows the distribution of marks obtained by a group of $400$ students in an examination.

Marks less than	$10$	$20$	$30$	$40$	$50$	$60$	$70$	$80$	$90$	$100$
Number of students	$5$	$10$	$30$	$60$	$105$	$180$	$270$	$355$	$390$	$400$

Using a scale of $1 cm = 10 $ marks and $1 cm = 50$ students, plot these values and draw a smooth curve through these points.

Estimate from the graph the median marks, the lower quartile marks, the upper quartile marks and the inter-quartile range.

Marks less than	Class Interval	Number of students (Cumulative frequency)
$10$	$0 - 10$	$5$
$20$	$10 - 20$	$10$
$30$	$20 - 30$	$30$
$40$	$30 - 40$	$60$
$50$	$40 - 50$	$105$
$60$	$50 - 60$	$180$
$70$	$60 - 70$	$270$
$80$	$70 - 80$	$355$
$90$	$80 - 90$	$390$
$100$	$90 - 100$	$400$

Number of students $= N = 400$

Taking marks (class intervals) along X-axis and number of students (cumulative frequency) along Y-axis, draw an ogive.

Median $= \left(\dfrac{N}{2}\right)^{th} \text{value} = \left(\dfrac{400}{2}\right)^{th} \text{value} = 200^{th} \; \text{value} = 62.5$

$\therefore \;$ Median marks $= 62.5$

Lower quartile $= Q_1 = \left(\dfrac{N}{4}\right)^{th} \text{value} = \left(\dfrac{400}{4}\right)^{th} \text{value} = 100^{th} \; \text{value} = 49$

$\therefore \;$ Lower quartile $= 49$

Upper quartile $= Q_3 = \left(\dfrac{3N}{4}\right)^{th} \text{value} = \left(\dfrac{3 \times 400}{4}\right)^{th} \text{value} = 300^{th} \; \text{value} = 73.5$

$\therefore \;$ Upper quartile $= 73.5$

Inter-quartile range $= Q_3 - Q_1 = 73.5 - 49 = 24.5$

Commercial Mathematics - Shares and Dividends

A person invests ₹ $16500$ partly in $10\% \;$ ₹ $100$ shares at ₹ $130$ and partly in $8\% \;$ ₹ $100$ shares at ₹ $120$. If the total annual income from these shares is ₹$1180$, find the investment in each kind of shares.

Let the money invested in $10\% \;$ ₹ $100$ shares at ₹ $130$ $= $ ₹ $x$

Let the money invested in $8\% \;$ ₹ $100$ shares at ₹ $120$ $= $ ₹ $\left(16500 - x\right)$

For ₹ $100$ shares at ₹ $130$:

Nominal Value (N.V) of each share $= $ ₹ $100$

Market Value (M.V) of each share $= $ ₹ $130$

$\therefore \;$ Number of shares bought $= \dfrac{\text{money invested}}{M.V} = \dfrac{x}{130}$

Dividend on each share $= 10\% \text{ of N.V } = \dfrac{10}{100} \times 100 = $ ₹ $10$

$\therefore \;$ Dividend (income) from $\dfrac{x}{130}$ shares $= \dfrac{x}{130} \times 10 = $ ₹ $\dfrac{x}{13}$

For ₹ $100$ shares at ₹ $120$:

Nominal Value (N.V) of each share $= $ ₹ $100$

Market Value (M.V) of each share $= $ ₹ $120$

$\therefore \;$ Number of shares bought $= \dfrac{\text{money invested}}{M.V} = \dfrac{16500 - x}{120}$

Dividend on each share $= 8\% \text{ of N.V } = \dfrac{8}{100} \times 100 = $ ₹ $8$

$\therefore \;$ Dividend (income) from $\dfrac{16500 - x}{120}$ shares

$=$ ₹ $ \dfrac{16500 - x}{120} \times 8 = $ ₹ $\left(\dfrac{16500 - x}{15}\right)$

$\therefore \;$ Total dividend $= $ ₹ $\left(\dfrac{x}{13} + \dfrac{16500 - x}{15}\right)$

Given: $\;$ Total dividend $= $ ₹ $1180$

i.e. $\;$ $\dfrac{x}{13} + \dfrac{16500 - x}{15} = 1180$

i.e. $\;$ $15 x + 214500 - 13x = 230100$

i.e. $\;$ $2x = 15600$ $\implies$ $x = 7800$

$\therefore \;$ Amount invested in $10\% \;$ ₹ $100$ shares at ₹ $130 = $ ₹ $7800$

Amount invested in $8\% \;$ ₹ $100$ shares at ₹ $120 = $ ₹ $\left(16500 - 7800\right) = $ ₹ $8700$

Factor and Remainder Theorems

If $\;$ $2x^3 + ax^2 + bx - 2$ $\;$ when divided by $\;$ $2x - 3$ $\;$ and $\;$ $x + 3$ $\;$ leaves remainder $7$ and $-20$ respectively, find the values of $a$ and $b$.

Let $f \left(x\right) = 2x^3 + ax^2 + bx - 2$

Given: $\;$ $f \left(x\right)$ when divided by $2x - 3$ leaves remainder $7$

$\therefore \;$ By remainder theorem, $\;$ $f \left(\dfrac{3}{2}\right) = 7$

i.e. $\;$ $2 \times \left(\dfrac{3}{2}\right)^3 + a \times \left(\dfrac{3}{2}\right)^2 + b \times \dfrac{3}{2} - 2 = 7$

i.e. $\;$ $\dfrac{27}{4} + \dfrac{9a}{4} + \dfrac{3b}{2} = 9$

i.e. $\;$ $9a + 6b = 36 - 27 = 9$

i.e. $\;$ $3a + 2b = 3$ $\;\;\; \cdots \; (1)$

Given: $\;$ $f \left(x\right)$ when divided by $x + 3$ leaves remainder $-20$

$\therefore \;$ By remainder theorem, $\;$ $f \left(-3\right) = -20$

i.e. $\;$ $2 \times \left(-3\right)^3 + a \times \left(-3\right)^2 + b \times \left(-3\right) - 2 = -20$

i.e. $\;$ $9a - 3b = -18 + 54 = 36$

i.e. $\;$ $3a - b = 12$ $\;\;\; \cdots \; (2)$

Solving equations $(1)$ and $(2)$ simultaneously, we get,

$3b = -9$ $\implies$ $b = -3$

Substituting the value of $b$ in equation $(2)$, we get,

$3a + 3 = 12$

i.e. $\;$ $3a = 9$ $\implies$ $a = 3$

$\therefore \;$ $a = 3$, $\;$ $b = -3$

Matrices

Evaluate: $\;$ $\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ - \tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \text{cosec } 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix}$

$\begin{bmatrix} 2 \cos 60^\circ & -2 \sin 30^\circ \\ - \tan 45^\circ & \cos 0^\circ \end{bmatrix} \times \begin{bmatrix} \cot 45^\circ & \text{cosec } 30^\circ \\ \sec 60^\circ & \sin 90^\circ \end{bmatrix} $

$= \begin{bmatrix} 2 \times \dfrac{1}{2} & - 2 \times \dfrac{1}{2} \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$

$ = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$

$ = \begin{bmatrix} 1 \times 1 -1 \times 2 & 1 \times 2 - 1 \times 1 \\ -1 \times 1 + 1 \times 2 & -1 \times 2 + 1 \times 1 \end{bmatrix}$

$ = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} $

Statistics

For the following distribution draw a histogram.

Weight (in kg)	$44 - 47$	$48 - 51$	$52 - 55$	$56 - 59$	$60 - 63$	$64 - 67$
Frequency	$23$	$25$	$37$	$18$	$7$	$2$

Estimate the mode from the histogram.

Here, the class intervals given are in inclusive form and need to be converted into exclusive form.

$\begin{aligned} \text{Adjustment factor} & = \dfrac{1}{2} \left(\text{upper limit of a class} - \text{lower limit of previous class}\right) \\\\ & = \dfrac{1}{2} \left(48 - 47\right) = 0.5 \end{aligned}$

$\therefore \;$ The adjusted class intervals are

Weight in kg (Inclusive form)	Weight in kg (Exclusive form)	Frequency
$44 - 47$	$43.5 - 47.5$	$23$
$48 - 51$	$47.5 - 51.5$	$25$
$52 - 55$	$51.5 - 55.5$	$37$
$56 - 59$	$55.5 - 59.5$	$18$
$60 - 63$	$59.5 - 63.5$	$7$
$64 - 67$	$63.5 - 67.5$	$2$

In the histogram, the highest rectangle represents the maximum frequency (or modal class).

Inside this rectangle, draw lines $AC$ and $BD$ diagonally from the upper corners $A$ and $D$ of adjacent rectangles.

Let the point of intersection of $AC$ and $BD$ be $K$.

Draw $KL$ perpendicular to the horizontal axis.

The value of point $L$ on the horizontal axis represents the mode and the class interval in which point $L$ lies is the modal class.

From the figure, mode $= 53.1 \; kg$; $\;$ modal class $= 51.5 - 55.5$

Mensuration

A cylindrical vessel of radius $4 \; cm$ contains water. A solid sphere of radius $3 \; cm$ is lowered into water until it is completely immersed. Find the rise in the water level in the vessel.

Let the initial level of water in the cylindrical vessel $= h \; cm$

Radius of the cylindrical vessel $= r = 4 \; cm$

Initial volume of water $= \pi r^2 h = \pi \times 4^2 \times h = 16 \pi h \; cm^3$

Let the radius of the sphere $= R = 3 \; cm$

Volume of sphere $= \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$

Let the new height of water $= H \; cm$

New volume of water $= \pi r^2 H = 16 \pi H \; cm^3$

$\therefore \;$ We have,

$16 \pi h + 36 \pi = 16 \pi H$

i.e. $\;$ $16 \pi \left(H - h\right) = 36 \pi$

i.e. $\;$ $H - h = \dfrac{36}{16} = 2.25 \; cm$

i.e. $\;$ Rise in water level $= 2.25 \; cm$

Equation of a Line

$ABCD$ is a rhombus. The coordinates of $A$ and $C$ are $\left(3,6\right)$ and $\left(-1,2\right)$ respectively. Write down the equation of $BD$.

Given: $\;$ $ABCD$ is a rhombus with coordinates $A \left(x_1, y_1\right) = \left(3,6\right)$ and $C \left(x_2, y_2\right) = \left(-1,2\right)$

$AC$ is the diagonal of the rhombus $ABCD$

Slope of diagonal $AC = m_1 = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{2 - 6}{-1 - 3} = 1$

The diagonals of a rhombus are perpendicular to each other and bisect each other.

$\therefore \;$ Slope of diagonal $BD = \dfrac{-1}{m_1} = -1$

Midpoint of $AC = M = \left(\dfrac{3 - 1}{2}, \dfrac{6 + 2}{2}\right) = \left(1, 4\right)$

Now, $BD$ also passes through $M \left(1,4\right)$ and has slope $m_2 = -1$

$\therefore \;$ Equation of $BD$ is

$y - 4 = -1 \left(x - 1\right)$

i.e. $\;$ $y - 4 = -x + 1$

i.e. $\;$ $x + y = 5$

Ratio and Proportion

If $a : b = c : d$, prove that $\left(abcd\right) \left(a^{-2} + b^{-2} + c^{-2} + d^{-2}\right) = a^2 + b^2 + c^2 + d^2$

Given: $\;$ $a : b = c : d$

$\implies$ $ad = bc$ $\;\;\; \cdots \; (1)$

Now,

$\begin{aligned} LHS & = \left(abcd\right) \left(a^{-2} + b^{-2} + c^{-2} + d^{-2}\right) \\\\ & = \left(abcd\right) \left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} + \dfrac{1}{d^2}\right) \\\\ & = \left(abcd\right) \left(\dfrac{b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2}{a^2 b^2 c^2 d^2}\right) \\\\ & = \dfrac{b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2}{abcd} \\\\ & = \dfrac{bcd}{a} + \dfrac{acd}{b} + \dfrac{abd}{c} + \dfrac{abc}{d} \\\\ & = bc \left(\dfrac{a}{d} + \dfrac{d}{a}\right) + ad \left(\dfrac{c}{b} + \dfrac{b}{c}\right) \\\\ & = ad \left(\dfrac{a}{d} + \dfrac{d}{a}\right) + bc \left(\dfrac{c}{b} + \dfrac{b}{c}\right) \;\;\; \left[\text{in view of equation (1)}\right] \\\\ & = a^2 + d^2 + c^2 + b^2 \\\\ & = RHS \end{aligned}$

Hence proved.

Arithmetic Progression

Which term of the A.P $\;$ $5, \; 12, \; 19, \; 26, \; 33, \; \cdots$ $\;$ will be $35$ more than its $12^{th}$ term?

Given A.P : $\;$ $5, \; 12, \; 19, \; 26, \; 33, \; \cdots$

First term of A.P $= a = 5$

Common difference of A.P $= d = 7$

Let $n^{th}$ term of A.P be $35$ more than its $12^{th}$ term.

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d = 5 + \left(n - 1\right) \times 7$

$12^{th}$ term of A.P $= t_{12} = 5 + \left(12 - 1\right) \times 7 = 82$

As per question,

$5 + \left(n - 1\right) \times 7 = 82 + 35$

i.e. $\;$ $5 + 7n - 7 = 117$

i.e. $\;$ $7n = 105$ $\implies$ $n = 15$

$\therefore \;$ The $15^{th}$ term of the given A.P will be $35$ more than its $12^{th}$ term.

Arithmetic Progression

If the sum of first $n$, $2n$ and $3n$ terms of an A.P be $S_1$, $S_2$ and $S_3$ respectively, then prove that $\;$ $S_3 = 3 \left(S_2 - S_1\right)$

Let the first term of A.P $= a$

Let the common difference of A.P $= d$

Sum of first $n$ terms of A.P $= S_1 = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$ $\;\;\; \cdots \; (1)$

Sum of first $2n$ terms of the A.P $= S_2 = \dfrac{2n}{2} \left[2a + \left(2n - 1\right)d\right]$

i.e. $\;$ $S_2 = n \left[2a + \left(2n - 1\right)d\right]$ $\;\;\; \cdots \; (2)$

Sum of first $3n$ terms of the A.P $= S_3 = \dfrac{3n}{2} \left[2a + \left(3n - 1\right)d\right]$ $\;\;\; \cdots \; (3)$

From equations $(1)$ and $(2)$,

$\begin{aligned} S_2 - S_1 & = n \left[2a + \left(2n - 1\right)d\right] - \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right] \\\\ & = n \left\{2a + 2nd - d - \dfrac{1}{2} \left(2a + nd - d\right) \right\} \\\\ & = n \left\{2a + 2nd - d - a - \dfrac{nd}{2} + \dfrac{d}{2} \right\} \\\\ & = n \left\{a + \dfrac{3nd}{2} - \dfrac{d}{2} \right\} \\\\ & = \dfrac{n}{2} \left[2a + \left(3n - 1\right)d\right] \end{aligned}$

$\therefore \;$ $3 \left(S_2 - S_1\right) = \dfrac{3n}{2} \left[2a + \left(3n - 1\right)d\right] = S_3$ $\;\;\;$ [by equation $(3)$]

i.e. $\;$ $S_3 = 3 \left(S_2 - S_1\right)$

Hence proved.

Coordinate Geometry

On a graph, plot $A \left(4,6\right)$ and $B \left(2,3\right)$. Find the image of $A$ when reflected in the line $y = 0$. Name it $A'$. Find the coordinates of $B'$, the image of $B$ when reflected in the line $AA'$. Give a geometrical name for the figure $AB'A'B$. Calculate the area of the figure $AB'A'B$.

The points $A \left(4,6\right)$ and $B \left(2,3\right)$ are plotted.

Image of $A$ reflected in the line $y = 0$ (i.e. X axis) is $A' \left(4, -6\right)$

Image of $B$ reflected in the line $AA'$ is $B' \left(6, 3\right)$

$AB'A'B$ is a kite.

Area of the figure $AB'A'B = 2 \times \text{Area} \left(\triangle AB'A'\right)$

$\text{Area} \left(AB'A'\right) = \dfrac{1}{2} \times 12 \times 2 = 12 $ sq units

$\therefore \;$ Area of figure $AB'A'B = 2 \times 12 = 24$ sq units

Commercial Mathematics - Banking

A person deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is $8\%$ per annum and the person gets ₹ $8088$ from the bank after $3$ years, find the amount of money deposited each month.

Let money deposited per month $= P = $ ₹ $x$

Rate of interest $= r = 8 \%$ per annum

Number of months $= n = 36$ $\;\;\;$ [$3$ years $= 36$ months]

Interest $= I = P \times \dfrac{n \left(n + 1\right)}{2 \times 12} \times \dfrac{r}{100}$

i.e. $\;$ $I = $ ₹ $x \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} = $ ₹ $4.44 \; x$

Money deposited in $36$ months $= $ ₹ $36 \; x$

$\therefore \;$ Maturity value (M.V) $= $ ₹ $36 \; x + $ ₹ $4.44 \; x = $ ₹ $40.44 \; x$

Given: $\;$ M.V $= $ ₹ $8088$

$\implies$ $40.44 \; x = 8088$ $\implies$ $x = \dfrac{8088}{40.44} = 200$

$\therefore \;$ Money deposited per month $= $ ₹ $200$

Section Formula

Find the ratio in which the line joining $A \left(6,5\right)$ and $B \left(4,-3\right)$ is divided by the line $y = 2$. Also, find the coordinates of the point of intersection.

Let any point on the line $y = 2$ be $\;$ $P \left(p, 2\right)$

Let point $P$ divide the line joining $A \left(6, 5\right)$ and $B \left(4, -3\right)$ in the ratio $k : 1$

Then by section formula,

$p = \dfrac{4k + 6}{k + 1}$ $\;\;\; \cdots \; (1a)$

$2 = \dfrac{-3k + 5}{k + 1}$ $\;\;\; \cdots \; (1b)$

Solving equation $(1b)$ for $k$ gives,

$2k +2 = -3k + 5$

i.e. $\;$ $5k = 3$ $\implies$ $k = \dfrac{3}{5}$

i.e. $\;$ The line $y = 2$ divides the line joining the points $A$ and $B$ in the ratio $3:5$

Substituting the value of $k$ in equation $(1a)$ gives

$p = \dfrac{4 \times \dfrac{3}{5} + 6}{\dfrac{3}{5} + 1} = \dfrac{42}{8} = \dfrac{21}{4}$

$\therefore \;$ The point of intersection is $\left(\dfrac{21}{4}, 2\right)$

Matrices

If $A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} -2 & 4 \\ 3 & 1 \end{bmatrix}$, find a matrix $X$ such that $3 A + 4 X = 5 B$

Given: $\;$ $A = \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix}$, $\;$ $B = \begin{bmatrix} -2 & 4 \\ 3 & 1 \end{bmatrix}$

The matrix equation is $\;$ $3 A + 4 X = 5 B$

i.e. $\;$ $4 X = 5 B - 3 A$

$\implies$ $X = \dfrac{5}{4} B - \dfrac{3}{4}A$ $\;\;\; \cdots \; (1)$

Now,

$\begin{aligned} \dfrac{5}{4} B & = \dfrac{5}{4} \begin{bmatrix} -2 & 4 \\ 3 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} -2 \times \dfrac{5}{4} & 4 \times \dfrac{5}{4} \\ 3 \times \dfrac{5}{4} & 1 \times \dfrac{5}{4} \end{bmatrix} \\\\ & = \begin{bmatrix} -10/4 & 20 / 4 \\ 15 / 4 & 5 / 4 \end{bmatrix} \;\;\; \cdots \; (2a) \end{aligned}$

$\begin{aligned} \dfrac{3}{4} A & = \dfrac{3}{4} \begin{bmatrix} 2 & 0 \\ -3 & 1 \end{bmatrix} \\\\ & = \begin{bmatrix} 2 \times \dfrac{3}{4} & 0 \times \dfrac{3}{4} \\ -3 \times \dfrac{3}{4} & 1 \times \dfrac{3}{4} \end{bmatrix} \\\\ & = \begin{bmatrix} 6/4 & 0 \\ -9 / 4 & 3 / 4 \end{bmatrix} \;\;\; \cdots \; (2b) \end{aligned}$

$\therefore \;$ In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,

$\begin{aligned} X & = \begin{bmatrix} -10/4 & 20/4 \\ 15/4 & 5/4 \end{bmatrix} - \begin{bmatrix} 6/4 & 0 \\ -9/4 & 3/4 \end{bmatrix} \\\\ & = \begin{bmatrix} -16/4 & 20/4 \\ 24/4 & 2/4 \end{bmatrix} \\\\ & = \begin{bmatrix} -4 & 5 \\ 6 & 1/2 \end{bmatrix} \end{aligned}$

Geometric Progression

The $2^{nd}$ and $5^{th}$ terms of a G.P are $\dfrac{1}{2}$ and $\dfrac{1}{16}$ respectively. Find the sum of the first eight terms of the G.P. Give the answer as a fraction.

Let the first term of G.P be $= a$

Let the common ratio of G.P be $= r$

$n^{th}$ term of G.P $= t_n = ar^{n-1}$

Given: $\;$ $2^{nd}$ term of G.P $= t_2 = ar = \dfrac{1}{2}$ $\;\;\; \cdots \; (1)$

$5^{th}$ term of G.P $= t_5 = ar^4 = \dfrac{1}{16}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ we have,

$\dfrac{ar^4}{ar} = \dfrac{1/16}{1/2}$

i.e. $\;$ $r^3 = \dfrac{1}{8} = \left(\dfrac{1}{2}\right)^3$

$\implies$ $r = \dfrac{1}{2}$

Substituting the value of $r$ in equation $(1)$ gives,

$a \times \dfrac{1}{2} = \dfrac{1}{2}$ $\implies$ $a = 1$

Sum of $n$ terms of G.P $= S_n = \dfrac{a \left(1 - r^n\right)}{1 - r}$ $\;$ when $\;$ $r < 1$

$\therefore \;$ Sum of first $8$ terms of G.P $= S_8 = \dfrac{1 \left[1 - \left(\dfrac{1}{2}\right)^8\right]}{1 - \dfrac{1}{2}}$

i.e. $\;$ $S_8 = \dfrac{1 - \dfrac{1}{256}}{\dfrac{1}{2}} = \dfrac{255 / 256}{1 / 2} = \dfrac{255}{128}$

Factor Theorem

Use the factor theorem to factorize completely: $\;$ $x^3 + x^2 -4x - 4$

Let $\;$ $f \left(x\right) = x^3 + x^2 - 4 x - 4$

$f \left(2\right) = 2^3 + 2^2 -4 \times 2 - 4 = 0$

$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$ $\;\;\;$ [By factor theorem]

Now,

$\begin{array}{lllll} x - 2 & ) & x^3 + x^2 - 4x - 4 & ( & x^2 + 3x + 2 \\ & & x^3 - 2x^2 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ 3x^2 - 4x & & \\ & & \ \ \ \ \ \ \ \ 3x^2 - 6x & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x - 4 & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x - 4 & & \\ & & - - - - - - - - - - - - & & \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \end{array}$

$\begin{aligned} \therefore \; x^3 + x^2 - 4x - 4 & = \left(x - 2\right) \left(x^2 + 3x + 2\right) \\\\ & = \left(x - 2\right) \left(x^2 + 2 x + x + 2\right) \\\\ & = \left(x - 2\right) \left[x \left(x + 2\right) + 1 \left(x + 2\right)\right] \\\\ & = \left(x - 2\right) \left(x + 1\right) \left(x + 2\right) \end{aligned}$

Statistics

Using step deviation method, calculate the mean for the following data:

Height (in cm)	Number of boys
$135 - 140$	$4$
$140 - 145$	$9$
$145 - 150$	$18$
$150 - 155$	$28$
$155 - 160$	$24$
$160 - 165$	$10$
$165 - 170$	$5$
$170 - 175$	$2$

Class size $= $ Upper class limit $-$ Lower class limit $= h = 5$

Let assumed mean $= A = 152.5$

Height (in cm)	Number of boys (frequency) $\left(f_i\right)$	Mid-value $\left(x_i\right)$	deviation $= x_i - A $	$ t_i = \dfrac{x_i - A}{h} $	$ f_i \times t_i $
$135 - 140$	$4$	$137.5$	$-15$	$-3$	$-12$
$140 - 145$	$9$	$142.5$	$-10$	$-2$	$-18$
$145 - 150$	$18$	$147.5$	$-5$	$-1$	$-18$
$150 - 155$	$28$	$152.5$	$0$	$0$	$0$
$155 - 160$	$24$	$157.5$	$5$	$1$	$24$
$160 - 165$	$10$	$162.5$	$10$	$2$	$20$
$165 - 170$	$5$	$167.5$	$15$	$3$	$15$
$170 - 175$	$2$	$172.5$	$20$	$4$	$8$

$\Sigma f_i = 100$, $\;$ $\Sigma f_i \times t_i = 19$

Mean $= A + \dfrac{\Sigma f_i \times t_i}{\Sigma f_i} \times h$

i.e. $\;$ Mean $= 152.5 + \dfrac{19}{100} \times 5 = 153.45$

Commercial Mathematics - Shares and Dividends

A person buys $500$, ₹ $20$ shares at a discount of $20\%$ and receives a return of $10\%$ on her money. Calculate the amount invested and the rate of dividend paid by the company.

Number of shares $= 500$

Nominal value of each share $= N.V = $ ₹ $20$

Market value of each share $= M.V = $ ₹ $20 - 20 \%$ of ₹ $20 = $ ₹ $20 \; - $ ₹ $4 = $ ₹ $16$

Rate of return $= 10\%$

Amount invested $=$ Number of shares $\times$ M.V of 1 share

i.e. $\;$ Amount invested $= 500 \times $ ₹ $16 = $ ₹ $8000$

Now, $\;$ $\text{Rate of return} \times M.V = \text{Rate of dividend} \times N.V$

$\therefore \;$ $\text{Rate of dividend} = \dfrac{\text{Rate of return} \times M.V}{N.V}$

i.e. $\;$ $\text{Rate of dividend} = \dfrac{10}{100} \times 16 \times \dfrac{1}{20} = \dfrac{8}{100} = 8\%$

Quadratic Equations

Find the set of values of $k$ for which the equation $kx^2 + 2x + 1 = 0$ has distinct real roots.

The standard equation $\;$ $ax^2 + bx + c = 0$ $\;$ has distinct real roots when $\;$ $b^2 - 4ac > 0$

Given equation: $\;$ $kx^2 + 2x + 1 = 0$

Comparing the given quadratic equation with the standard quadratic equation, we have,

$a = k$, $\;$ $b = 2$, $\;$ $c = 1$

$\therefore \;$ For real distinct roots, $\;$ $2^2 - 4 \times k \times 1 > 0$

i.e. $\;$ $4 - 4k > 0$

i.e. $\;$ $1 - k > 0$

i.e. $\;$ $1 > k$

$\implies$ $k < 1$

$\therefore \;$ The given quadratic equation has distinct real roots for $\;$ $\left\{k \mid k < 1, \; k \in R \right\}$

Linear Inequations

Solve the inequation $\;$ $-3 + x \leq \dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x; \; x \in I$ $\;$ and represent the solution set on a number line.

Consider $\;$ $-3 + x \leq \dfrac{8x}{3} + 2$

i.e. $\;$ $-3 - 2\leq \dfrac{8x}{3} - x$

i.e. $\;$ $-5 \leq \dfrac{5x}{3}$

i.e. $\;$ $-1 \leq \dfrac{x}{3}$

i.e. $\;$ $-3 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x$

i.e. $\;$ $\dfrac{8x}{3} - 2x \leq \dfrac{14}{3} - 2$

i.e. $\;$ $\dfrac{2x}{3} \leq \dfrac{8}{3}$

i.e. $\;$ $2x \leq 8$

i.e. $\;$ $x \leq 4$ $\;\;\; \cdots (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- 3 \leq x \leq 4$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -3 \leq x \leq 4, \; x \in I \right\}$

Quadratic Equations

$50$ is divided into two parts such that the sum of their reciprocals is $\dfrac{1}{12}$. Find the two parts.

Let $50$ be divided into $x$ and $50 - x$

Reciprocal of the parts: $\;$ $\dfrac{1}{x}$ $\;$ and $\;$ $\dfrac{1}{50 - x}$

Given: Sum of reciprocal of parts $= \dfrac{1}{12}$

i.e. $\;$ $\dfrac{1}{x} + \dfrac{1}{50 - x} = \dfrac{1}{12}$

i.e. $\;$ $\dfrac{50 - x + x}{x \left(50 - x\right)} = \dfrac{1}{12}$

i.e. $\;$ $600 = 50 x - x^2$

i.e. $\;$ $x^2 - 50 x + 600 = 0$

i.e. $\;$ $x^2 - 30x - 20x + 600 = 0$

i.e. $\;$ $x \left(x - 30\right) - 20 \left(x - 30\right) = 0$

i.e. $\;$ $\left(x - 20\right) \left(x - 30\right) = 0$

$\implies$ $x = 20$ $\;$ or $\;$ $x = 30$

$\therefore \;$ The two parts are $\;$ $20$ $\;$ and $\;$ $30$

Ratio and Proportion

Solve for $x$ when $\;\;$ $\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x}- \sqrt{a - x}} = b$

Given: $\;$ $\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x}- \sqrt{a - x}} = b = \dfrac{b}{1}$

By componendo-dividendo we have,

$\dfrac{\left(\sqrt{a + x} + \sqrt{a - x}\right) + \left(\sqrt{a + x} - \sqrt{a - x}\right)}{\left(\sqrt{a + x} + \sqrt{a - x}\right) - \left(\sqrt{a + x} - \sqrt{a - x}\right)} = \dfrac{b + 1}{b - 1}$

i.e. $\;$ $\dfrac{2 \sqrt{a + x}}{2 \sqrt{a - x}} = \dfrac{b + 1}{b - 1}$

i.e. $\;$ $\dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{b + 1}{b - 1}$

Squaring both sides we get,

$\dfrac{a + x}{a - x} = \dfrac{\left(b + 1\right)^2}{\left(b - 1\right)^2}$

i.e. $\;$ $\dfrac{a + x}{a - x} = \dfrac{b^2 + 2b + 1}{b^2 - 2b + 1}$

By componendo-dividendo we have,

$\dfrac{\left(a + x\right) + \left(a - x\right)}{\left(a + x\right) - \left(a - x\right)} = \dfrac{\left(b^2 + 2b + 1\right) + \left(b^2 - 2b + 1\right)}{\left(b^2 + 2b + 1\right)- \left(b^2 - 2b + 1\right)}$

i.e. $\;$ $\dfrac{2a}{2x} = \dfrac{2 \left(b^2 + 1\right)}{4b}$

i.e. $\;$ $\dfrac{a}{x} = \dfrac{b^2 + 1}{2b}$

$\implies$ $x = \dfrac{2ab}{b^2 + 1}$