Raju's Resource Hub
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1. Answer any fifteen questions in PART- A. All questions carry
equal marks. (15 X 2 =30)
2. Answer all questions, choosing any two sub-divisions
each question under Part-B. All questions carry equal marks.
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(5 X 14 = 70) ( 7 + 7)
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{If\ A =\begin{pmatrix}
1 & 2 \\
3 & 5\\
\end{pmatrix}\ and\ B =\begin{pmatrix}
– 5 & 7 \\
0 & 4\\
\end{pmatrix}},\ \hspace{15cm}\]
\[\color {green}{\ find}\ 2A\ +\ B\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[ 2A = 2\begin{pmatrix}
1 & 2 \\
3 & 5\\
\end{pmatrix}\ \hspace{13cm}\]
\[2A = \begin{pmatrix}
2 & 4 \\
6 & 10\\
\end{pmatrix}\ ———- (1)\ \hspace{8cm}\]
\[B = \begin{pmatrix}
-5 & 7 \\
0 & 4\\
\end{pmatrix}\ ———- (2)\ \hspace{8cm}\]
\[2A\ +\ B = \begin{pmatrix}
2 & 4 \\
6 & 10\\
\end{pmatrix}\ +\ \begin{pmatrix}
-5 & 7 \\
0 & 4\\
\end{pmatrix}\ \hspace{8cm}\]
\[=\begin{pmatrix}
2\ -\ 5 & 4\ +\ 7 \\
6\ +\ 0 & 10\ +\ 4\\
\end{pmatrix}\ \hspace{8cm}\]
\[2A\ +\ B = \begin{pmatrix}
-3 & 11 \\
6 & 14\\
\end{pmatrix}\ \hspace{10cm}\]
\[2. \ \color{green}{Find\ the\ cofactor\ of\ ‘2’\ in}\ \begin{pmatrix}
1 & 0 & – 1 \\
2 & 3 & 4 \\
7 & 8 & – 2 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{2\ +\ 1}\ \begin{vmatrix}
0 & – 1 \\
8 & – 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (0\ -\ (- 8)\ \hspace{15cm}\]
\[=\ – 1(8)\ \hspace{15cm}\]
\[\color{green}{\boxed{cofactor\ of\ 2\ =\ – 8}}\ \hspace{15cm}\]
\[3.\ \color{green}{Prove\ that\ the\ matrix \begin{pmatrix}
1 & – 2 \\
– 2 & 4\\
\end{pmatrix}\ is\ a\ singular\ matrix}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ A =\begin{pmatrix}
1 & – 2 \\
– 2 & 4\\
\end{pmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 1(4) – (-2)(- 2) \hspace{13cm}\]
\[=\ 4\ -\ 4 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[4.\ \color{green}{Find\ the\ adjoint\ of\ \begin{pmatrix}
1 & 5 \\
3 & 6 \\
\end{pmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ A\ =\begin{pmatrix}
1 & 5 \\
3 & 6 \\
\end{pmatrix}\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ (6)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 1\ = \ 6\ \hspace{15cm}\]
\[cofactor\ of\ 5 = (-1)^{1\ +\ 2}\ (3)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ -1 =\ -\ 3\ \hspace{15cm}\]
\[cofactor\ of\ 3\ = (-1)^{2\ +\ 1}\ (5)\ \hspace{15cm}\]
\[= (-1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 3\ =\ -\ 5\ \hspace{15cm}\]
\[cofactor\ of\ 6\ = (-1)^{2\ +\ 2}\ (1)\ \hspace{15cm}\]
\[= (1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 6\ = 1\ \hspace{15cm}\]
\[\therefore\ cofactor\ matrix\ = \begin{pmatrix}
6 & -\ 3 \\
-\ 5 & 1 \\
\end{pmatrix}\ \hspace{15cm}\]
\[Adj.\ A = \begin{pmatrix}
6 & -\ 5 \\
-\ 3 & 1 \\
\end{pmatrix}\ \hspace{15cm}\]
\[5.\ \color{green}{Convert\ \frac{\pi}{4}\ in\ to\ degrees}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ \frac{\pi}{4}\ \times \frac{180^0}{\pi}\]
\[=\ \hspace{2cm}\ 45^0\]
\[6.\ \color{green}{Write\ any\ two\ characteristics\ of\ the\ function\ y\ =\ e^x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Exponential Growth}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ exhibits\ exponential\ growth.\ As\ x\ increases,\ y\ increases\ very\ rapidly\]
\[\hspace{3cm}\ Conversely,\ as x\ decreases,\ y\ approaches\ zero\ but\ never\ reaches\ it\]
\[\hspace{2cm}\ 2.\ \textbf{Derivative and Integral properties}\ \hspace{15cm}\]
\[\hspace{3cm}\ The\ function\ y\ =\ e^x\ \text{is unique in that it is its own derivative and integral}\]
\[\hspace{3cm}\ \frac{d}{dx}\ e^x\ =\ e^x\ and\ \int{e^x}\ dx\ =\ e^x\ +\ c\]
\[7.\ \color{green}{Find\ the\ value\ of\ \frac{Tan\ {25}^0\ +\ Tan\ {20}^0}{1\ -\ Tan\ {25}^0\ Tan\ {20}^0}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ W.\ K.\ T\ \frac{Tan A\ +\ Tan B}{1\ -\ Tan A\ Tan B}\ =\ Tan(A + B)\ \hspace{18cm}\]
\[\frac{Tan\ {25}^0\ +\ Tan\ {20}^0}{1\ -\ Tan\ {25}^0\ Tan\ {20}^0}\ =\ Tan({25}^0\ +\ {20}^0)\ =\ Tan{45}^0\ =\ 1\ \hspace{10cm}\]
\[8.\ \color{green}{Find\ the\ value\ of\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ 2\ Sin\ 15^{0}\ Cos\ 15^{0}\ =\ Sin\ 2(15^{0})\ \hspace{15cm}\]
\[=\ Sin\ 30^{0}\ \hspace{15cm}\]
\[=\ \frac{1}{2}\ \hspace{15cm}\]
\[\boxed {2\ Sin\ 30^{0}\ Cos\ 30^{0}\ =\ \frac{1}{2}}\]
\[9.\ \color{green}{If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k},}\ \color {green} {find\ the\ value\ of\ 4\overrightarrow{a}\ +\ \overrightarrow{b}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 4(5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[=\ 20\overrightarrow{i}\ +\ 8\overrightarrow{j}\ -\ 12\overrightarrow{k}\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 23\overrightarrow{i}\ +\ 6\overrightarrow{j}\ -\ 7\overrightarrow{k}\ \hspace{5cm}\]
\[10.\ \color{green}{Find\ the\ Direction\ cosines\ of\ 3\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 3\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(3)^2 + (2)^2 + (4)^2 }\]
\[ = \sqrt{(9\ +\ 4\ +\ 16) }\]
\[r =\sqrt{29}\]
\[ Direction\ cosines\ are \frac{3}{\sqrt(29)}, \frac{2}{\sqrt(29)}, \frac{4}{\sqrt(29)} \]
\[11.\ \color{green}{Find\ the\ projection\ of\ the\ vector\ 2\overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k} on\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 2\overrightarrow{k}} \ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k} \]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2 \overrightarrow{j}\ -\ 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\ \frac{(2\overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k}).(\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 2\overrightarrow{k})}{\sqrt{(1)^2 + (-2)^2 + (-2)^2 }}\]
\[=\ \frac{2(1)\ +\ 1(-2)\ +\ (-2) (-2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{2\ -\ 2\ +\ 4}{\sqrt{9}}\]
\[\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{4}{3}}\]
\[12.\ \color{green}{Show\ that\ the\ vectors\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k}\ and\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ are\ parallel}.\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ show\ \overrightarrow{a}×\overrightarrow{b} =\ 0\ \hspace{10cm}\]
\[\overrightarrow{a}\ =\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k} \]
\[\overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
4 & -2 & -6\\
2 & -1 & -3\\
\end{vmatrix}\]
\[ = \overrightarrow{I}(6\ -\ 6)\ -\overrightarrow{j}(-12\ +\ 12)\ +\ \overrightarrow{k}(-4\ +\ 4)\]
\[ = \overrightarrow{i}(0) -\overrightarrow{j}(0)+\overrightarrow{k}(0)\]
\[\boxed{\overrightarrow{a}×\overrightarrow{b} =\ 0}\ \hspace{7cm}\]
\[\therefore\ The\ given\ vectors\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ parallel\]
\[13.\ \color{green}{Calculate\ the\ arithmetic\ mean\ of\ 10,\ 12,\ 14,\ 16\ and\ 18}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ Given\ n\ =\ 5\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 10\ +\ 12\ +\ 14\ +\ 16\ +\ 18\]
\[\Sigma x_i\ =\ 70\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{70}{5}\]
\[\bar{x}\ =\ 14\]
\[14.\ \color{green}{The\ arithmetic\ mean\ of\ 10\ numbers\ is\ 20\ ,\ find\ the\ sum\ of\ the}\ \hspace{10cm}\]\[\color{green}{values}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 10\ and\ \bar{x}\ =\ 20\ \hspace{17cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[20\ =\ \frac{\Sigma X}{10}\]
\[\therefore\ \Sigma X\ =\ 200\]
\[15.\ \color{green}{\text{If the standard deviation of a data is 6.8,}}\ \color{green}{find\ its\ variance}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ \sigma\ =\ 6.8\ \hspace{20cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\hspace{3cm}\ =\ (6.8)^2\]
\[\hspace{2cm}\ \boxed{variance\ =\ 46.24}\]
\[16.\ \color{green}{Write\ down\ the\ normal\ equations\ to\ fit\ a\ straight\ line\ y\ =\ a\ x\ +\ b.}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
The Normal equations are
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (1)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (2)\]
\[17.\ \color{green}{\text{ An integer is chosen at random from the integers 1 to 10. }}\ \hspace{10cm}\]\[\color {green} {Find\ the\ probability\ that\ it\ is\ an\ even\ number.}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given an integer is chosen at random from 1 to 10. No. of possible outcomes = 10}\]
\[\text{The even numbers from 1 to 10 are 2, 4, 6, 8, 10. No. of favourable outcomes = 5}\]
\[\text{The probability P of rolling a prime number can be calculated as follows:}\]
\[P(even\ number)\ =\ \frac{Number\ of\ favourable\ outcomes}{Number\ of\ possible\ outcomes}\ =\ \frac{5}{10}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting an even number from the integers 1 to 10 is:}\ \boxed{\frac{1}{2}}\]
\[18.\ \color{green}{While\ a\ die\ is\ rolled\ once,}\ \hspace{15cm}\]\[\color {green} {Find\ the\ probability\ of\ getting\ an\ even\ number}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. No. of possible outcomes = 6}\]
\[\text{The even numbers on a die are 2, 4, and 6. No. of favourable outcomes = 3}\]
\[\text{The probability P of rolling an odd number can be calculated as follows:}\]
\[P(even\ number)\ =\ \frac{No.\ of\ favourable\ outcomes}{No.\ of\ possible\ outcomes}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting an odd number on a rolling die is:}\ \boxed{\frac{1}{2}}\]
\[19.\ \color{green}{If\ A\ and\ B\ are\ two\ events\ such\ that\ P(A)\ =\ 0.42,\ and\ P(B)\ =\ 0.48,}\ \hspace{10cm}\]\[\color{green}{find\ P(\bar{A})\ and\ P(\bar{B})}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\]
\[1.\ P(\bar{A})\ =\ 1\ -\ P(A)\]
\[2.\ P(\bar{B})\ =\ 1\ -\ P(B)\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.42\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ 0.48\ \hspace{10cm}\]
\[P(\bar{A})\ =\ 1\ -\ P(A)\ =\ 1\ -\ 0.42\ =\ 0.58\]
\[P(\bar{B})\ =\ 1\ -\ P(B)\ =\ 1\ -\ 0.48\ =\ 0.52\]
\[20.\ \color{green}{If\ A\ and\ B\ are\ two\ independent\ events\ such\ that\ P(A)\ =\ 0.4\ and\ P(A\ \cup\ B)\ =\ 0.9.}\ \hspace{10cm}\]\[\color {green} {Find\ P(B)}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.4\ \hspace{10cm}\]
\[\bullet\ P(A\ \cup\ B)\ =\ 0.9\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ 0\ (A\ and\ B\ are\ independent\ events)\ \hspace{5cm}\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[0.9\ =\ 0.4\ +\ P(B)\ -\ 0\ \hspace{5cm}\]
\[P(B)\ =\ 0.9\ -\ 0.4\ \hspace{5cm}\]
\[\boxed{P(B)\ =\ 0.5}\]
\[\underline{PART\ -\ B}\]
\[21\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{If\ A =\ \begin{pmatrix}
4 & 6 & 2 \\
0 & 1 & 5 \\
0 & 3 & 2 \\
\end{pmatrix}\ ,\ B =\begin{pmatrix}
0 & 1 & – 1 \\
3 & – 1 & 4 \\
– 1 & 2 & 1 \\
\end{pmatrix}}\ \hspace{7cm}\]\[\color {green}{Prove\ that\ (A\ +\ B)^T\ =\ A^T\ +\ B^T}\ \hspace{5cm}\]
\[\color {blue}{Solution:}\ Given\ A =\ \begin{pmatrix}
4 & 6 & 2 \\
0 & 1 & 5 \\
0 & 3 & 2 \\
\end{pmatrix}\ ,\ B =\begin{pmatrix}
0 & 1 & – 1 \\
3 & – 1 & 4 \\
– 1 & 2 & 1 \\
\end{pmatrix}\ \hspace{5cm}\]
\[A + B=\begin{pmatrix}
4 + 0 & 6 + 1 & 2 – 1 \\
0 + 3 & 1 – 1 & 5 + 4 \\
0 – 1 & 3 + 2 & 2 + 1 \\
\end{pmatrix}\ \hspace{6cm}\]
\[A + B=\ \begin{pmatrix}
4 & 7 & 1 \\
3 & 0 & 9\\
-1 & 5 & 3\\
\end{pmatrix}\ \hspace{6cm}\]
\[(A + B)^T\ =\ \begin{pmatrix}
4 & 3 & – 1 \\
7 & 0 & 5 \\
1 & 9 & 3\\
\end{pmatrix}\ \hspace{2cm}\ ——– (1)\]
\[A^T =\begin{pmatrix}
4 & 0 & 0 \\
6 & 1 & 3 \\
2 & 5 & 2 \\
\end{pmatrix}\ \hspace{10cm}\]
\[B^T =\begin{pmatrix}
0 & 3 & – 1 \\
1 & – 1 & 2 \\
– 1 & 4 & 1 \\
\end{pmatrix}\ \hspace{10cm}\]
\[A^T\ +\ B^T\ =\begin{pmatrix}
4 + 0 & 0\ +\ 3 & 0\ -\ 1\\
6\ +\ 1 & 1 – 1 & 3 + 2 \\
2 – 1 & 5 + 4 & 2 + 1 \\
\end{pmatrix}\ \hspace{6cm}\]
\[A^T\ +\ B^T\ =\begin{pmatrix}
4 & 3 & – 1\\
7 & 0 & 5 \\
1 & 9 & 3 \\
\end{pmatrix}\ \hspace{2cm}\ ———- (2)\]
\[From\ (1)\ and\ (2), It\ is\ concluded\ that (A\ +\ B)^T\ =\ A^T\ +\ B^T\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[\hspace{4cm} \color{green}{4\ x\ +\ y\ +\ z\ =\ 6,\ 2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ and\ x\ +\ y\ +\ z\ =\ 3}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ \hspace{15cm}\]
\[x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
4 & 1 & 1 \\
2 & -1 & -2\\
1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =4\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =4(-1\ +\ 2)\ – 1 (2\ +\ 2)\ +\ 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta\ =\ 4(1)\ – 1 (4)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =4\ -\ 4\ +\ 3\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 3}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
6 & 1 & 1 \\
-6 & -1 & -2 \\
3 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =6\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
-6 & -2 \\
3 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
-6 & -1\\
3 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 6(-1\ +\ 2) – 1 (\ -6\ +\ 6)\ +\ 1(-6\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x\ =\ 6(1)\ – 1 (0)\ +\ 1(-3)\
\hspace{13cm}\]
\[\Delta_x = 6\ +\ 0\ -3\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 3}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
4 & 6 & 1 \\
2 & -6 & -2 \\
1 & 3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 4\begin{vmatrix}
-6 & -2 \\
3 & 1\\
\end{vmatrix}\ -\ 6\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -6\\
1 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 4(-\ 6\ +\ 6)\ -\ 6 (2\ +\ 2)\ +\ 1(6\ +\ 6)\
\hspace{9cm}\]
\[\Delta_y\ =\ 4(0)\ -\ 6 (4)\ +\ 1(12)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 0\ -\ 24\ +\ 12\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ -12}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
4 & 1 & 6 \\
2 & -1 & -6 \\
1 & 1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 4\begin{vmatrix}
-1 & -6 \\
1 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -6 \\
1 & 3 \\
\end{vmatrix}\ +\ 6\begin{vmatrix}
2 & – 1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 4(-\ 3\ +\ 6)\ -\ 1 (6\ +\ 6)\ +\ 6(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_z\ =\ 4(3)\ -\ 1 (12)\ +\ 6(3)\
\hspace{13cm}\]
\[\Delta_z\ =\ 12\ -\ 12\ +\ 18\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 18}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-12}{3} =\ -4\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{18}{3} =\ 6\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ -4\ and\ z = 6\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 4(1) – 4 + 6\]\[ = 4 – 4 + 6 = 6\]\[ = RHS\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ inverse\ of\ \begin{pmatrix}
2 & 1 & 1 \\
1 & 0 & 2 \\
4 & 2 & 2 \\
\end{pmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\color {blue}{Solution:}\ Let\ A\ =\begin{pmatrix}
2 & 1 & 1 \\
1 & 0 & 2 \\
4 & 2 & 2 \\
\end{pmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 2\begin{vmatrix}
0 & 2 \\
2 & 2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
1 & 2 \\
4 & 2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
1 & 0\\
4 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =\ 2(0\ -\ 4)\ -\ 1 (2\ -\ 8)\ +\ 1(2\ -\ 0)\
\hspace{9cm}\]
\[ =\ 2(-4)\ -\ 1 (-6)\ +\ 1(2)\
\hspace{13cm}\]
\[ =\ -8\ +\ 6\ + 2\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ does\ not\ exist\ \hspace{10cm}\]
\[22\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{If\ sin\ \theta\ =\ \frac{3}{5},\ then\ find\ \text{the values of other five trigonometric}}\ \hspace{10cm}\]\[\color{green}{ratios}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ sin\ \theta\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2θ\ +\ cos^2θ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ cos^2θ\ =\ 1\ -\ cos^2θ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{5}{13})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{169\ -\ 25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{144}{13}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cos\ \theta\ =\ \pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ θ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{5}{13}}{\frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ θ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ θ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \pm\ \frac{1}{\frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{13}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ θ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{13}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{If\ A\ and\ B\ are\ acute\ angles\ such\ that\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}},\hspace{15cm}\\ \color {green} {then\ prove\ that\ Cos(A\ +\ B)\ =\ \frac{33}{65}}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ Given\ Sin\ A\ =\ \frac{3}{5} \ and\ Cos\ B\ =\ \frac{12}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Cos( A + B )\ =\ Cos A\ Cos B\ -\ Sin A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Sin\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{3}{5})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25\ -\ 9}{25}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{16}{25}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{4}{5}\ \hspace{10cm}\]
\[Sin\ B\ =\ \sqrt{1\ -\ Cos^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{12}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 144}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{25}{169}}\ \hspace{10cm}\]
\[Sin\ B\ =\ \frac{5}{13}\ \hspace{10cm}\]
\[Cos( A + B )\ =\ (\frac{4}{5})\ (\frac{12}{13})\ -\ (\frac{3}{5})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{48}{65}\ -\ \frac{15}{65}\ \hspace{10cm}\]
\[=\ \frac{48\ -\ 15}{65}\ \hspace{10cm}\]
\[=\ \frac{33}{65}\ \hspace{10cm}\]
\[\boxed {Cos ( A + B )\ =\ \frac{33}{65}}\ \hspace{10cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Show\ that\ \frac{1\ +\ cos\ 2A\ +\ sin\ 2A}{1\ -\ cos\ 2A\ +\ sin\ 2A}\ =\ cot\ A}\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ LHS\ =\ \frac{1\ +\ cos\ 2A\ +\ sin\ 2A}{1\ -\ cos\ 2A\ +\ sin\ 2A}\ \hspace{18cm}\]
\[=\ \frac{2\ cos^2\ A\ +\ 2\ sin\ A\ cos\ A}{2\ sin^2\ A\ +\ 2\ sin\ A\ cos\ A}\ \hspace{1cm}\ W.\ K.\ T.\ sin\ 2A\ =\ 2\ sin\ A\ cos\ A\ ,\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\ and\ 1\ -\ cos\ 2A\ =\ 2\ sin^2\ A\]
\[=\ \frac{2\ cos\ A(cos\ A\ +\ sin\ A)}{2\ sin\ A(sin\ A\ +\ cos\ A)}\ \hspace{15cm}\]
\[=\ \frac{con\ A}{sin\ A}\ \hspace{15cm}\]
\[=\ cot\ A\ =\ R.H.S\ \hspace{15cm}\]
\[23\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Prove\ that\ the\ points}\ \hspace{15cm}\]\[\color{green}{3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\ ,\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k} and\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ form\ an\ isosceles\ triangle}\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(2)^2\ +\ (2)^2\ +\ (-1)^2 }\]
\[ = \sqrt{4\ +\ 4\ +\ 1}\]
\[ =\ \sqrt{9}\]
\[AB =\ 3 \]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 5\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{BC}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2\ +\ (-2)^2\ +\ (2)^2 }\]
\[ = \sqrt{1\ +\ 4\ +\ 4}\]
\[ =\ \sqrt{9}\]
\[BC =\ 3 \]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ 4\overrightarrow{i}\ +\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(4)^2\ +\ (0)^2\ +\ (1)^2 }\]
\[ = \sqrt{16\ +\ 0\ +\ 1}\]
\[AC = \sqrt{17}\]
\[\boxed{AB = BC\ \neq\ AC}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Prove\ that\ \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}\ and\ 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} are\ mutually\ orthogonal.}\ \hspace{13cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}\]
\[ \overrightarrow{b}= \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{c}= 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}) .(\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k})\]
= 1(1) + 2(1) + 1(-3)
= 0
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}) .( 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})\]
= 1(7) + 1(-4) – 3(1)
= 7 – 4 – 3
= 0
\[\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.\]
= 7(1) -4 (2) + 1 (1)
\[ \overrightarrow{c}.\overrightarrow{a}= (7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}) .(\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k})\]
= 7 – 8 + 1
= 0
\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ are}\ \hspace{8cm}\]\[\color{green}{\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\ , 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ – (\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k})\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ -\ \overrightarrow{i}\ – 3\overrightarrow{j}\ – 2\overrightarrow{k}\]
\[\boxed{\overrightarrow{AB}\ =\ \overrightarrow{i}\ – 4\overrightarrow{j}\ -\ \overrightarrow{k}}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – (2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\overrightarrow{k}\]
\[\boxed{\overrightarrow{BC}\ =\ -\ 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\ \begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
1 & -4 & -1\\
-3 & 3 & 2\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -8 + 3)\ -\ \overrightarrow{j}(2 – 3)\ +\ \overrightarrow{k}(3 – 12)\]
\[ = \overrightarrow{i}(-5) -\overrightarrow{j}(-1)+\overrightarrow{k}(-9)\]
\[\boxed{\overrightarrow{AB}× \overrightarrow{BC}\ =\ -5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 9\overrightarrow{k}}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-5 )^2\ +\ (1)^2\ + (-9)^2 }=\sqrt{25 + 1 + 81 }=\sqrt{107}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[\boxed{Area\ of \ triangle\ =\frac{\sqrt{107}}{2}\ sq. units}\]
\[24\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Find\ the\ arithmetic\ mean\ of\ the\ following\ data}\ \hspace{10cm}\]
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 |
| Frequency | 2 | 5 | 1 | 3 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 2 | 5 | 10 |
| 10-20 | 5 | 15 | 75 |
| 20-30 | 1 | 25 | 25 |
| 30-40 | 3 | 35 | 105 |
| N = 11 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{215}{11}\ =\ 19.55\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{\text{Find the standard deviation of the following data:}}\ \hspace{10cm}\]
| Class interval | 4 – 8 | 8 – 12 | 12 – 16 | 16 – 20 |
| frequency | 3 | 6 | 4 | 7 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N})^2}, \ where\ N\ =\ \Sigma {f_i}\]
| Marks | fi | xi | fixi | xi 2 | fixi2 |
| 4 -8 | 3 | 6 | 18 | 36 | 108 |
| 8 -12 | 6 | 10 | 60 | 100 | 600 |
| 12 – 16 | 4 | 14 | 56 | 196 | 784 |
| 16 – 20 | 7 | 18 | 126 | 324 | 2268 |
| N = 20 |
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ x_i^2}{N}\ -\ (\frac{\Sigma f_i\ x_i}{N})^2}\]
\[=\ \sqrt{\frac{3760}{20}\ -\ (\frac{-260}{20})^2}\]
\[=\ \sqrt{188\ -\ (13)^2}\]
\[=\ \sqrt{188\ -\ 169}\]
\[=\ \sqrt{19}\]
\[\sigma\ =\ 4.36\ (approximately)\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Fit\ a\ straight\ line\ to\ the\ following\ data:}\ \hspace{15cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 10 | 14 | 19 | 26 | 31 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 10 | 0 | 0 |
| 1 | 14 | 1 | 14 |
| 2 | 19 | 4 | 38 |
| 3 | 26 | 9 | 78 |
| 4 | 31 | 16 | 124 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 99\]
\[2\ a\ +\ b\ =\ 19.8\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 254\]
\[3\ a\ +\ b\ =\ 25.4\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 19.8\]
\[3\ a\ +\ b\ =\ 25.4\]
———————————-
\[a\ =\ 5.6\]
\[put\ a\ =\ 5.6\ in\ (4)\]
\[2(5.6)\ +\ b\ =\ 19.8\]
\[11.2\ +\ b\ =\ 19.8\]
\[b\ =\ 19.8\ -\ 11.2\]
\[b\ =\ 8.6\]
\[Eqn\ (1)\ becomes\ y\ =\ 5.6\ x\ +\ 8.6\]
\[25\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Two\ dice\ are\ rolled\ together.\ Find\ the\ probability}\ \hspace{10cm}\\\ \color{green}{for\ getting\ the\ sum\ of\ the\ numbers\ on\ on\ the\ faces\ are\ 10\ and\ 7}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[When\ two\ dice\ are\ thrown\ simultaneously,\ each\ die\ has\ 6\ faces,\ so\ there\ are\ a\ \hspace{2cm}\\\ total\ of\ 6\ \times\ 6\ =\ 36\ possible\ outcomes\ \hspace{12cm}\]
\[1.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 10}}\ \hspace{10cm}\]
\[\text{The possible outcomes where the sum of the two dice is 6 are:}\ \hspace{5cm}\]
\[\bullet\ (4,\ 6)\ \hspace{15cm}\]
\[\bullet\ (5,\ 5)\ \hspace{15cm}\]
\[\bullet\ (6,\ 4)\ \hspace{15cm}\]
\[\text{There are 3 such outcomes.}\ \hspace{10cm}\]
\[A\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 10\ \hspace{10cm}\]
\[P(A)\ =\ \frac{3}{36}\ \hspace{10cm}\]
\[2.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 7}\ \hspace{7cm}\]
\[\text{The possible outcomes where the numbers on both dice are the same are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 6)\ \hspace{15cm}\]
\[\bullet\ (2,\ 5)\ \hspace{15cm}\]
\[\bullet\ (3,\ 4)\ \hspace{15cm}\]
\[\bullet\ (4,\ 3)\ \hspace{15cm}\]
\[\bullet\ (5,\ 2)\ \hspace{15cm}\]
\[\bullet\ (6,\ 1)\ \hspace{15cm}\]
\[\text{There are 6 such outcomes.}\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 10\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[3.\ \bf{\color{green}{probability\ of\ sum\ of\ the\ numbers\ on\ on\ the\ faces\ are\ 10\ and\ 7}}\ \hspace{2cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{3}{36}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ 0\ \hspace{10cm}\]
\[P(A\ \cup\ B)\ =\ \frac{3}{36}\ +\ \frac{6}{36}\ -\ 0\ \hspace{5cm}\]
\[=\ \frac{9}{36}\ =\ \frac{1}{4}\ \hspace{3cm}\]
\[\text{So, the probability of getting a sum of 10 and 7 on both dice is:}\ =\ \frac{1}{4}\ \hspace{1cm}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{A\ card\ is\ selected\ at\ random\ from\ a\ pack\ of\ 52\ cards.\ Find\ the}\ \hspace{8cm}\]\[\color{green}{probability\ that\ the\ card\ is\ either\ a\ black\ card\ or\ a\ card\ with\ number\ 6.}\ \hspace{2cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{To find the probability of drawing either a black card or a card with the number 6 from a standard deck of 52 cards,}\\ \text{we need to consider both events and their possible overlap.}\ \hspace{10cm}\]
\[1.\ \text{Total number of cards in a deck: 52}\ \hspace{5cm}\]
\[\hspace{3cm}\ 2.\ \text{Number of black cards: There are 26 black cards in the deck (13 spades +}\\ \text{13 clubs}).\ \hspace{10cm}\]
\[\hspace{3cm}\ 3.\ \text{Number of cards with the number 6: There are 4 cards with the number 6}\\ \text{(one in each suit: hearts, diamonds, clubs, and spades}).\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ drawing\ a\ black\ card\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ drawing\ a\ card\ with\ on\ number\ 6\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{26}{52}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{4}{52}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ \frac{2}{52}\ (2\ such\ cards\ with \ black\ card\ and\ number\ 6\ from\ spades\ and\ clubs)\]
\[P(A\ \cup\ B)\ =\ \frac{26}{52}\ +\ \frac{4}{52}\ -\ \frac{2}{52}\ \hspace{5cm}\]
\[=\ \frac{28}{52}\ =\ \frac{7}{13}\ \hspace{3cm}\]
\[\text{So, the probability of drawing either a black card or a card with the}\\ \text{number 6 from a standard deck of 52 cards is:} \frac{7}{13}\ \hspace{5cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{A\ problem\ in\ statistics\ is\ given\ to\ two\ students\ A\ and\ B.\ The}\ \hspace{10cm}\\\ \color{green}{probability\ of\ A\ solves\ the\ problem\ is\ \frac{1}{4}\ and\ that\ of\ B\ solves\ the}\ \hspace{8cm}\\\ \color{green}{problem\ is\ \frac{2}{5}.\ If\ they\ solve\ the\ problem\ independently,}\ \hspace{8cm}\\\ \color{green}{find\ the\ probability\ that\ the\ problem\ is\ solved.}\ \hspace{12cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given,\ P(A)\ =\ \frac{1}{4}\ and\ P(B)\ =\ \frac{2}{5}\ \hspace{10cm}\\\ \text{Probability that the problem is solved = Probability that A solves the problem or B solves the problem}\]
\[=\ P(A\ \cup\ B)\ \hspace{5cm}\]
\[=\ P(A)\ +\ P(B)\ -\ (P(A\ \cap\ B)\ \hspace{2cm}\]
\[=\ P(A)\ +\ P(B)\ -\ P(A)\ \cdot \ P(B)\ Since A\ and\ B\ are\ independent\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5
}\ -\ \frac{1}{4}\ \cdot \ \frac{2}{5}\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5}\ -\ \frac{1}{10}\ \hspace{2cm}\]
\[=\ \frac{5\ +\ 8\ -\ 2}{20}\ \hspace{2cm}\]
\[=\ \frac{11}{20}\ \hspace{2cm}\]
\[\text{Thus, the probability that the problem is solved by either student A, student B, or both is:}\ =\ \frac{11}{20}\]
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