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1. Answer any fifteen questions in PART- A. All questions carry
equal marks. (15 X 2 =30)
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2. Answer all questions, choosing any two sub-divisions
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each question under Part-B. All questions carry equal marks.
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(5 X 14 = 70) ( 7 + 7)
\[\underline{PART\ -\ A}\]
\[1.\ \color{green}{If\ f(x)\ =\ 3\ x\ +\ 2\ and\ A =\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix},}\ \color {green}{find\ f(A).}\ \hspace{10cm}\]
\[\color {blue}{Solution:}\ \hspace{18cm}\]
\[\hspace{3cm}\ Given\ \hspace{15cm}\]
\[\hspace{3cm}\ A =\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix}\ and\ \hspace{15cm}\]
\[\hspace{3cm}\ f(x)\ =\ 3\ x\ +\ 2\ \hspace{15cm}\]
\[f(A) = \ 3A\ +\ 2\ I, Where\ I\ is\ a\ Identity\ matrix\ of\ order\ 2\ ———- (1)\]
\[3A\ =\ 3\begin{pmatrix}
1 & 0 \\
2 & -1\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A\ =\ \begin{pmatrix}
3 & 0 \\
6 & -3\\
\end{pmatrix}\ \hspace{13cm}\]
\[I\ =\ \begin{pmatrix}
1 & 0 \\
0 & 1\\
\end{pmatrix}\ \hspace{13cm}\]
\[2I\ =\ \begin{pmatrix}
2 & 0 \\
0 & 2\\
\end{pmatrix}\ \hspace{13cm}\]
\[3A\ +\ 2I\ = \begin{pmatrix}
3 & 0 \\
6 & -3\\
\end{pmatrix}\ +\ \begin{pmatrix}
2 & 0 \\
0 & 2\\
\end{pmatrix}\ \hspace{8cm}\]
\[=\begin{pmatrix}
3\ +\ 2 & 0\ +\ 0 \\
6\ +\ 0\ & -3\ +\ 2\\
\end{pmatrix}\ \hspace{8cm}\]
\[3A\ +\ 2I = \begin{pmatrix}
5 & 0 \\
6 & -1\\
\end{pmatrix}\ \hspace{10cm}\]
\[\boxed{\color{green}{\therefore\ f(A)\ = \begin{pmatrix}
5 & 0 \\
6 & -1\\
\end{pmatrix}}}\ \hspace{10cm}\]
\[2. \ \color{green}{Find\ the\ value\ of\ x\ if\ \begin{vmatrix}
x & 1 \\
4 & x \\
\end{vmatrix}\ = 0}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ \begin{vmatrix}
x & 1 \\
4 & x \\
\end{vmatrix}\ = 0 \hspace{15cm}\]
\[x(x) – 4 (1) = 0\ \hspace{13cm}\]
\[x^2 – 4 = 0\ \hspace{13cm}\]
\[x^2 = 4\ \hspace{13cm}\]
\[x = \pm\sqrt{4}\ \hspace{13cm}\]
\[x = \pm 2\ \hspace{13cm}\]
\[\color{green}{\boxed{x\ =\ 2\ or\ x\ =\ -\ 2}}\]
\[3.\ \color{green}{Prove\ that\ \begin{pmatrix}
1 & 3 \\
2 & 6 \\
\end{pmatrix}\ is\ a\ singular\ matrix}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =\ 1(6)\ -\ 2(3) \hspace{13cm}\]
\[=\ 6\ -\ 6 \hspace{12cm}\]
\[= 0 \hspace{13cm}\]
\[A\ is\ a\ singular\ matrix\ \hspace{10cm}\]
\[4.\ \color{green}{Find\ the\ cofactor\ of\ ‘2’\ in\ \begin{pmatrix}
3 & 0 & 2 \\
5 & 1 & 7 \\
4 & 5 & -3 \\
\end{pmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[cofactor\ of\ 2\ =\ (-1)^{1\ +\ 3}\ \begin{vmatrix}
5 & 1 \\
4 & 5 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (25\ -\ 4)\ \hspace{15cm}\]
\[=\ 1(21)\ \hspace{15cm}\]
\[\color{green}{\boxed{cofactor\ of\ 2\ =\ 21}}\ \hspace{15cm}\]
\[5.\ \color{green}{Convert\ 30^0\ to\ radians}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 30^0\ \times \frac{\pi\ radians}{180^0}\]
\[=\ \hspace{2cm}\ \frac{\pi}{6}\ radians\]
\[6.\ \color{green}{Write\ any\ two\ characteristics\ of\ the\ function\ y\ =\ sin\ x}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ 1.\ \textbf{Periodicity}\ \hspace{15cm}\]
\[\hspace{2cm}\ The\ sine\ function\ is\ periodic\ with\ a\ period\ of\ 2\ \pi.\ This\ means\ that\ the\ \hspace{2cm}\\ function\ repeats\ its\ values\ every\ 2\ \pi\ nuts.\ Mathematically\ sin(x\ +\ 2\ \pi)\ =\ sin(x)\ for\ all\ values\ of\ x.\]
\[\hspace{2cm}\ 2.\ \textbf{Range and amplitude}\ \hspace{15cm}\]
\[\hspace{2cm}\ \text{The range of the sine function is the set of all possible values it can}\ \hspace{2cm}\\ \hspace{2cm}\ \text{take, which is [1, -1]. This means that sinx oscillates between -1 and}\ \hspace{2cm}\\\hspace{2cm}\ \text{1. The amplitude, which is the maximum absolute value of the function, is 1.}\]
\[7.\ \color{green}{Find\ the\ value\ of\ sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {40}^0\ sin\ {20}^0}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {40}^0\ sin\ {20}^0\ =\ Sin\ ({40}^0\ + {20}^0)\ \hspace{10cm}\]
\[=\ Sin \ {60}^0\ \hspace{10cm}\]
\[=\ \frac{\sqrt{3}}{2}\ \hspace{10cm}\]
\[\color{green}{\boxed{\therefore\ sin\ {40}^0\ cos\ {20}^0\ +\ cos\ {20}^0\ sin\ {40}^0\ =\ \frac{\sqrt{3}}{2}}}\]
\[8.\ \color{green}{Prove\ that\ \frac{sin\ 2A}{1\ +\ cos\ 2A}\ =\ tan\ A.}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ LHS\ =\ \frac{sin\ 2A}{1\ +\ cos\ 2A}\ \hspace{18cm}\]
\[=\ \frac{2\ sin\ A\ cos\ A}{2\ cos^2\ A}\ \hspace{2cm}\ W.\ K.\ T.\ sin\ 2A\ =\ 2\ sin\ A\ cos\ A\ and\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\]
\[=\ \frac{sin\ A}{cos\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[9.\ \color{green}{If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k},\ find\ 4\overrightarrow{a}\ +\ \overrightarrow{b}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Given\ \hspace{17cm}\]
\[\overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 4(5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[=\ 20\overrightarrow{i}\ +\ 8\overrightarrow{j}\ -\ 12\overrightarrow{k}\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 23\overrightarrow{i}\ +\ 6\overrightarrow{j}\ -\ 7\overrightarrow{k}\ \hspace{5cm}\]
\[10.\ \color{green}{Find\ the\ Direction\ cosines\ of\ the\ vector\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(1)^2 + (2)^2 + (-3)^2 }\]
\[ = \sqrt{(1\ +\ 4\ +\ 9) }\]
\[r =\sqrt{14}\]
\[ Direction\ cosines\ are \frac{1}{\sqrt(14)}, \frac{2}{\sqrt(14)}, \frac{-3}{\sqrt(14)} \]
\[11.\ \color{green}{Show\ that\ the\ vectors\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ + 5\overrightarrow{k} and\ -\ 2\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 4\overrightarrow{k} are\ perpendicular}\ \hspace{2cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k} \]
\[\overrightarrow{b}= – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}) .(- 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k})\]
\[=\ 1(-2)\ +\ -3(6)\ +\ 5 (4)\]
\[=\ -\ 2\ -\ 18\ +\ 20\]
\[=\ 0\]
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[12.\ \color{green}{Find}\ (i)\ \overrightarrow{k}\ . \ \overrightarrow{i}\ and\ (ii)\ \overrightarrow{k}\ \times\ \overrightarrow{i}\ \hspace{18cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ (i)\ \overrightarrow{k}\ . \ \overrightarrow{i}\ =\ 0\ and\ (ii)\ \overrightarrow{k}\ \times\ \overrightarrow{i}\ =\ \overrightarrow{j}\ \hspace{15cm}\]
\[13.\ \color{green}{Calculate\ the\ arithmetic\ mean\ of\ 10,\ 12,\ 14,\ 16\ and\ 18}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 5\ \hspace{20cm}\]
\[\Sigma x_i\ =\ 10\ +\ 12\ +\ 14\ +\ 16\ +\ 18\]
\[\Sigma x_i\ =\ 70\]
\[\bar{x}\ =\ \frac{\Sigma x_i}{n}\]
\[=\ \frac{70}{5}\]
\[\bar{x}\ =\ 14\]
\[14.\ \color{green}{The\ arithmetic\ mean\ of\ 6\ values\ is\ 45\ .\ If\ 3\ is\ added\ to\ each\ of\ the\ \hspace{5cm}\\ numbers,\ then\ find\ the\ arithmetic\ mean\ of\ the\ new\ set\ of\ values.}\ \hspace{3cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Given\ n\ =\ 6\ and\ \bar{x}\ =\ 45\ \hspace{14cm}\]
\[\bar{x} \ =\ \frac{\Sigma X}{n}\]
\[45\ =\ \frac{\Sigma X}{6}\]
\[\therefore\ \Sigma X\ =\ 270\]
\[corrected\ \Sigma X\ =\ 270\ +\ 18\]
\[corrected\ \Sigma X\ =\ 288\]
\[correct\ mean\ =\ \frac{288}{6}\ =\ 48\]
\[15.\ \color{green}{\text{If the variance of a data is 100, then find its standard deviation}}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{3cm}\ Given\ variance\ =\ 100\ \hspace{14cm}\]
\[\hspace{2cm}\ W.\ K.\ T\ variance\ =\ \sigma^2\]
\[\sigma\ =\ \sqrt{variance}\]
\[\sigma\ =\ \sqrt{100}\]
\[\hspace{3cm}\ =\ 10\]
\[\hspace{2cm}\ \boxed{standard\ deviation\ =\ 10}\]
\[16.\ \color{green}{Write\ the\ normal\ equations\ of\ the\ straight\ line\ y\ =\ a\ x\ +\ b}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
The Normal equations are
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (1)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (2)\]
\[17.\ \color{green}{A\ card\ is\ picked\ randomly\ from\ a\ pack\ of\ 52\ cards\ randomly.\ Find\ the\ probability}\ \hspace{8cm}\]\[\color{green}{of\ getting\ a\ queen\ card}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
In a standard deck of 52 playing cards, there are 4 queens(one from each suit: hearts, diamonds, clubs, and spades).
The probability PP of drawing a queen from a deck of 52 cards can be calculated as follows:
\[P(King)\ =\ \frac{Number\ of\ queens}{Total\ number\ of\ cards}\ =\ \frac{4}{52}\ =\ \frac{1}{13}\]
\[\text{So, the probability of picking a queenfrom a standard deck of 52 cards is:}\ \boxed{\frac{1}{13}}\]
\[18.\ \color{green}{A\ die\ is\ rolled\ once},\ \hspace{15cm}\]\[\color {green} {Find\ the\ probability\ of\ getting\ a\ prime\ number.}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. No. of possible outcomes = 6}\]
\[\text{The prime numbers on a die are 2, 3, and 5. No. of favourable outcomes = 3}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. The prime numbers on a die are 2, 3, and 5.}\]
\[\text{The probability P of rolling a prime number can be calculated as follows:}\]
\[P(prime\ number)\ =\ \frac{Number\ of\ favourable\ outcomes}{Number\ of\ possible\ outcomes}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting a prime number on a rolling die is:}\ \boxed{\frac{1}{2}}\]
\[19.\ \color{green}{If\ P(A)\ =\ 0.42,\ and\ P(B)\ =\ 0.48},\ \hspace{10cm}\]\[\color{green}{find\ P(\bar{A})\ and\ P(\bar{B})}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\]
\[1.\ P(\bar{A})\ =\ 1\ -\ P(A)\]
\[2.\ P(\bar{B})\ =\ 1\ -\ P(B)\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.42\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ 0.48\ \hspace{10cm}\]
\[P(\bar{A})\ =\ 1\ -\ P(A)\ =\ 1\ -\ 0.42\ =\ 0.58\]
\[P(\bar{B})\ =\ 1\ -\ P(B)\ =\ 1\ -\ 0.48\ =\ 0.52\]
\[20.\ \color{green}{Find\ P(A/B),\ If\ P(B)\ =\ 0.5,\ and\ P(A\ \cap\ B)\ =\ 0.2}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ find\ the\ conditional\ probability\ P(A/B),\ \text{We use the definitions of conditional probability.}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(B)\ =\ 0.5\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ \ 0.2\ \hspace{10cm}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\ =\ \frac{0.2} {0.5}\ =\ \frac{2}{5}\ =\ 0.4\]
\[\underline{PART\ -\ B}\]
\[21\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{If\ A =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ then\ show\ that\ A^2\ -\ 4\ A\ -\ 5I\ =0}\ \hspace{12cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ A =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{10cm}\]
\[A^2 =\begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[ = \begin{bmatrix}
1\ ร\ 1\ +\ 2 ร\ 2\ +\ 2\ ร\ 2 & 1\ ร\ 2\ +\ 2 ร\ 1\ +\ 2\ ร\ 2 & 1\ ร\ 2\ +\ 2\ ร\ 2\ +\ 2\ ร\ 1\\
2\ ร\ 1\ +\ 1 ร\ 2\ +\ 2\ ร\ 2\ & 2\ ร\ 2\ +\ 1 ร\ 1\ +\ 2\ ร\ 2\ & 2\ ร\ 2\ +\ 1 ร\ 2\ +\ 2\ ร\ 1\\
2\ ร\ 1\ +\ 2 ร\ 2\ +\ 1\ ร\ 2\ & 2\ ร\ 2\ +\ 2 ร\ 1\ +\ 1\ ร\ 2\ & 2\ ร\ 2\ +\ 2 ร\ 2\ +\ 1\ ร\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[ = \begin{bmatrix}
1\ +\ 4\ +\ 4\ & 2\ +\ 2\ +\ 4 & 2\ +\ 4 +\ 2\\
2\ +\ 2\ +\ 4\ & 4\ +\ 1\ +\ 4\ & 4\ +\ 2\ +\ 2\\
2\ +\ 4\ +\ 2\ & 4\ +\ 2\ +\ 2\ & 4\ +\ 4\ +\ 1\\
\end{bmatrix}\ \hspace{9cm}\]
\[A^2 = \begin{bmatrix}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9 \\
\end{bmatrix}\ \hspace{12cm}\]
\[4\ A\ =\ 4\ \begin{bmatrix}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[4\ A\ = \begin{bmatrix}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4 \\
\end{bmatrix}\ \hspace{10cm}\]
\[5\ I\ =\ 5\ \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}\ \hspace{12cm}\]
\[5\ I\ = \begin{bmatrix}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5 \\
\end{bmatrix}\ \hspace{10cm}\]
\[A^2\ -\ 4\ A\ -\ 5I\ =\ \begin{bmatrix}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9 \\
\end{bmatrix}\ -\ \begin{bmatrix}
4 & 8 & 8 \\
8 & 4 & 8 \\
8 & 8 & 4 \\
\end{bmatrix}\ -\ \begin{bmatrix}
5 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 5 \\
\end{bmatrix}\]
\[ = \begin{bmatrix}
9\ -\ 4\ -\ 5\ & 8\ -\ 8\ -\ 0 & 8\ -\ 8 -\ 0\\
8\ -\ 8\ -\ 0\ & 9\ -\ 4\ -\ 5\ & 8\ -\ 8\ -\ 0\\
8\ -\ 8\ -\ 0\ & 8\ -\ 8\ -\ 0\ & 9\ -\ 4\ -\ 5\\
\end{bmatrix}\ \hspace{9cm}\]
\[= \begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}\ \hspace{10cm}\]
\[\color{green}{\boxed{\therefore\ A^2\ -\ 4\ A\ -\ 5I\ =\ 0}}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Solve\ the\ following\ equations\ x + y\ +\ z\ =\ 3,\ 2x\ -\ y\ +\ z\ =\ 2\ and}\ \hspace{15cm}\\ \color{green}{3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ using\ Cramers\ Rule}\ \hspace{17cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[x + y\ +\ z\ =\ 3\ ————— (1) \hspace{7cm}\]
\[2x\ -\ y\ +\ z\ =\ 2\ \hspace{15cm}\]
\[3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 1 & 1 \\
2 & – 1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
-1 & 1 \\
2 & – 2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =\ 1(2\ -\ 2)\ – 1 (-4\ -\ 3)\ +\ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta =1(0)\ – 1\ (-\ 7)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =\ 0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\boxed{\Delta =\ 14}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
3 & 1 & 1 \\
2 & -1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =\ 3\begin{vmatrix}
-1 & 1 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 3(2\ -\ 2)\ -\ 1 (-4\ -\ 3)\ + \ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x =\ -3(0)\ – 1 (-7)\ +\ 1 (7)\
\hspace{13cm}\]
\[\Delta_x\ =\ 0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 14}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & 3 & 1 \\
2 & 2 & 1 \\
3 & 3 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & 2\\
3 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(-\ 4\ -\ 3)\ -\ 3\ (-\ 4\ -\ 3)\ +\ 1(6\ -\ 6)\
\hspace{9cm}\]
\[\Delta_y =1(-7)\ -\ 3\ (-\ 7)\ +\ 1(0)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 7\ +\ 21\ +\ 0\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ 14}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 1 & 3 \\
2 & -1 & 2 \\
3 & 2 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix}
-1 & 2 \\
2 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
3 & 3 \\
\end{vmatrix}\ +\ 3\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 1(-\ 3\ -\ 4)\ -\ 1(6\ -\ 6)\ +\ 3(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_z\ =\ 1(-\ 7)\ -\ 1 (0)\ +\ 3(7)\
\hspace{13cm}\]
\[\Delta_z =\ -\ 7\ -\ 0\ +\ 21\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 14}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{14}{14}\ =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1\ y\ =\ 1\ and\ z\ =\ 1\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 1 +\ 1\ +\ 1\]\[ =\ 3\]\[ = RHS\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ inverse\ of\ \begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Let\ A\ =\begin{bmatrix}
1 & -1 & 1 \\
2 & -3 & -3 \\
6 & -2 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =1\begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -3\\
6 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =1(3\ -\ 6)\ + 1 (-2\ +\ 18) + 1(-4\ +\ 18)\
\hspace{9cm}\]
\[ =1(-3)\ + 1 (16) + 1(14)\
\hspace{13cm}\]
\[ = -3\ + 16 + 14\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ = 27\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
-3 & -3 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (3 – 6)\ \hspace{15cm}\]
\[= (1) (-3)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
2 & -3 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (-2 + 18)\ \hspace{15cm}\]
\[= (-1) (16)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -16\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
2 & -3 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-4 + 18)\ \hspace{15cm}\]
\[= (1) (14)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (1+ 2)\ \hspace{15cm}\]
\[= (-1) (3)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = -3\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
6 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (-1- 6)\ \hspace{15cm}\]
\[= (1) (-7)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -7\ \hspace{15cm}\]
\[cofactor\ of\ -3 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
6 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-2+ 6)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ -3 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 6 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
-1 & 1 \\
-3 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3 + 3)\ \hspace{15cm}\]
\[= (1) (6)\ \hspace{15cm}\]
\[cofactor\ of\ 6 = 6\ \hspace{15cm}\]
\[cofactor\ of\ -2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
1 & 1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (-3- 2)\ \hspace{15cm}\]
\[= (-1) (-5)\ \hspace{15cm}\]
\[cofactor\ of\ -2 = 5\ \hspace{15cm}\]
\[cofactor\ of\ -1 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
1 & -1 \\
2 & -3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (-3+ 2)\ \hspace{15cm}\]
\[= (1) (-1)\ \hspace{15cm}\]
\[cofactor\ of\ -1 = -1\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
-3 & -16 & 14 \\
-3 & -7 & -4 \\
6 & 5 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{27}\ \begin{bmatrix}
-3 & -3 & 6 \\
-16 & -7 & 5 \\
14 & -4 & -1 \\
\end{bmatrix}\ \hspace{2cm}\]
\[22\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{If\ cos\ \theta\ =\ \frac{5}{13},\ then\ find\ the values of other five trigonometric}\ \hspace{10cm}\]\[\color{green}{ratios}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ cos\ \theta\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[\hspace{2cm}\ sin^2ฮธ\ +\ cos^2ฮธ\ =\ 1\ \hspace{10cm}\]
\[\hspace{2cm}\ sin^2ฮธ\ =\ 1\ -\ cos^2ฮธ\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ (\frac{5}{13})^2\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ 1\ -\ \frac{25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{169\ -\ 25}{169}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{144}{25}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sin\ \theta\ =\ \pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ tan\ ฮธ\ =\ \frac{sin\ \theta}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{\pm\ \frac{12}{5}}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{tan\ \theta\ =\ \pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cot\ ฮธ\ =\ \frac{1}{tan\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{12}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{cot\ \theta\ =\ \pm\ \frac{13}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ sec\ ฮธ\ =\ \frac{1}{cos\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\frac{5}{13}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \frac{13}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ cosec\ ฮธ\ =\ \frac{1}{sin\ \theta}\ \hspace{10cm}\]
\[\hspace{3cm}\ =\ \frac{1}{\pm\ \frac{12}{5}}\ \hspace{10cm}\]
\[\hspace{2cm}\ \boxed{sec\ \theta\ =\ \pm\ \frac{5}{12}}\ \hspace{10cm}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{If\ A\ and\ B\ are\ acute\ angles\ such\ that\ sin\ A\ =\ \frac{8}{17} \ and\ sin\ B\ =\ \frac{5}{13},}\\ \hspace{3cm}\ \color{green}{then\ prove\ that\ sin(A\ +\ B)\ =\ \frac{171}{221}}\ \hspace{8cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given\ Sin\ A\ =\ \frac{8}{17} \ and\ Sin\ B\ =\ \frac{5}{13}\ \hspace{18cm}\]
\[W.\ K.\ T\ Sin ( A + B )\ =\ Sin A\ Cos B\ +\ Cos A\ Sin B\ \hspace{15cm}\]
\[Cos\ A\ =\ ?\ ,\ Cos\ B\ = ?\ \hspace{10cm}\]
\[Cos\ A\ =\ \sqrt{1\ -\ Sin^2\ A}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{8}{17})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{64}{289}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{289\ -\ 64}{289}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{225}{289}}\ \hspace{10cm}\]
\[Cos\ A\ =\ \frac{15}{17}\ \hspace{10cm}\]
\[Cos\ B\ =\ \sqrt{1\ -\ Sin^2\ B}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ (\frac{5}{13})^2}\ \hspace{10cm}\]
\[ =\ \sqrt{1\ -\ \frac{25}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{169\ -\ 25}{169}}\ \hspace{10cm}\]
\[ =\ \sqrt{\frac{144}{169}}\ \hspace{10cm}\]
\[Cos\ B\ =\ \frac{12}{13}\ \hspace{10cm}\]
\[Sin ( A + B )\ =\ (\frac{8}{17})\ (\frac{12}{13})\ +\ (\frac{15}{17})\ (\frac{5}{13})\ \hspace{10cm}\]
\[=\ \frac{96}{221}\ +\ \frac{75}{221}\ \hspace{10cm}\]
\[=\ \frac{96\ +\ 75}{221}\ \hspace{10cm}\]
\[=\ \frac{171}{221}\ \hspace{10cm}\]
\[\boxed {\therefore\ Sin ( A + B )\ =\ \frac{171}{221}}\ \hspace{10cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Show\ that\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ cos\ A\ +\ cos\ 2A}\ =\ tan\ A}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ LHS\ =\ \frac{sin\ A\ +\ sin\ 2A}{1\ +\ Cos\ A\ +\ cos\ 2A}\ \hspace{18cm}\]
\[=\ \frac{sin\ A\ +\ 2\ sin\ A\ Cos\ A}{2\ cos^2\ A\ +\ Cos\ A}\ \hspace{2cm}\ W.\ K.\ T.\ Sin\ 2A\ =\ 2\ Sin\ A\ Cos\ A\ , and\ 1\ +\ cos\ 2A\ =\ 2\ cos^2\ A\]
\[=\ \frac{Sin\ A(1\ +\ 2\ Cos\ A)}{Cos\ A(1\ +\ 2\ Cos\ A)}\ \hspace{15cm}\]
\[=\ \frac{Sin\ A}{CosS\ A}\ \hspace{15cm}\]
\[=\ tan\ A\ =\ R.H.S\ \hspace{15cm}\]
\[23\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Prove\ that\ the\ points}\ \hspace{15cm}\]\[\color{green}{4\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}, 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 4\overrightarrow{k} and\ 3\overrightarrow{i}\ +\ 4\overrightarrow{j}\ +\ 2\overrightarrow{k}\ form\ an\ equilateral\ triangle}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}= 4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ +3\overrightarrow{j} + 4\overrightarrow{k}\]
\[\overrightarrow{OC}= 3\overrightarrow{i}\ + 4\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})\]
\[=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{AB}= -2\overrightarrow{i} +\overrightarrow{j} + \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(-2)^2 + (-1)^2 +(1)^2 }\]
\[ = \sqrt{(4 + 1 +1 }\]
\[AB = \sqrt{6}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (2\overrightarrow{i}+ 3\overrightarrow{j}+ 4\overrightarrow{k})\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 2\overrightarrow{i}- 3\overrightarrow{j}- 4\overrightarrow{k}\]
\[\overrightarrow{BC}= \overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }\]
\[ = \sqrt{(1 + 1 + 4}\]
\[BC = \sqrt{6}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k})\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{AC}= -\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(-1)^2 + (2)^2 +(-1)^2 }\]
\[ = \sqrt{(1 + 4 + 1}\]
\[AC = \sqrt{6}\]
\[AB = BC = AC = \sqrt{6}\]
\[The\ given\ triangle\ is\ an\ equilateral\ triangle\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{Show\ that\ the\ vectors\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k},\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}\ \hspace{10cm}\\ and\
2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k} are\ mutually\ perpendicular.}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}\]
\[\overrightarrow{c}\ =\ 2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[ \overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k}) .(\overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k})\]
\[=\ 2(1)\ +\ 2(- 2)\ +\ 1(2)\]
\[=\ 2\ -\ 4\ +\ 2\]
\[=\ 0\]
\[\boxed{\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors}\]
\[ \overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}) .(2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 1(2)\ +\ -\ 2(-1)\ +\ 2(-2)\]
\[=\ 2\ +\ 2 -\ 4\]
\[=\ 0\]
\[\boxed{\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors}\]
\[ \overrightarrow{c}.\overrightarrow{a}= (2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k}) .(2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k})\]
\[=\ 2(2)\ +\ -\ 1(2)\ +\ -\ 2(1)\]
\[=\ 4\ -\ 2 -\ 2\]
\[=\ 0\]
\[\boxed{\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors}\]
\[\therefore\ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Find\ the\ area\ of\ the\ triangle\ whose\ adjacent\ sides\ are\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}}\ \hspace{8cm}\\ \color{green}{and\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{a}ร\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
3 & 1 & 2\\
1 & -2 & 4\\
\end{vmatrix}\]
\[ = \overrightarrow{i}(1 . 4\ -\ 2(-2))\ -\ \overrightarrow{j}(3 . 4\ -\ 2(1))\ +\ \overrightarrow{k}(3 . (-2)\ -\ 1(1))\]
\[ = \overrightarrow{i}(4\ +\ 4)\ -\ \overrightarrow{j}(12\ -\ 2)\ +\ \overrightarrow{k}(-6\ -\ 1)\]
\[ = \overrightarrow{i}(8)\ -\ \overrightarrow{j}(10)\ +\ \overrightarrow{k}(-7)\]
\[ \overrightarrow{a}ร \overrightarrow{b}\ =\ 8\overrightarrow{i}\ -\ 10\overrightarrow{j}\ -\ 7\overrightarrow{k}\]
\[ Area\ of \ triangle\ =\ \frac{1}{2} |\overrightarrow{a}\ ร\ \overrightarrow{b}|\]
\[=\ \frac{1}{2} \sqrt{((8)^2\ +\ (-10)^2 + (-7)^2 }\]
\[=\ \frac{1}{2}\ \sqrt{(64\ +\ 100\ +\ 49)}\]
\[=\ \frac{1}{2}\ \sqrt{213}\]
\[24\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Find\ the\ arithmetic\ mean\ of\ the\ following\ data}\ \hspace{10cm}\]
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 2 | 6 | 9 | 7 | 4 | 2 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
| Marks | No. of students fi | Mid-value xi | fixi |
| 0-10 | 2 | 5 | 10 |
| 10-20 | 6 | 15 | 90 |
| 20-30 | 9 | 25 | 225 |
| 30-40 | 7 | 35 | 245 |
| 40-50 | 4 | 45 | 220 |
| 50-60 | 2 | 55 | 110 |
| N = 30 |
\[\therefore\ \bar{x}\ =\ \frac{\Sigma f_i\ x_i}{N}\ =\ \frac{860}{30}\ =\ 28.66\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{\text{Calculate the standard deviation for the following data}}\ \hspace{10cm}\]
| Items | 5 | 15 | 25 | 35 |
| Frequency | 2 | 1 | 1 | 3 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\hspace{2cm}\ Given\ n\ =\ 4\ \hspace{20cm}\]
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ d_i^2}{N}\ -\ (\frac{\Sigma f_i\ d_i}{N}})^2, \ where\ d_i\ =\ x_i\ -\ A\ and\ N\ =\ \Sigma {f_i}\]
\[A\ =\ \bar{x}\ =\ \frac{5\ +\ 15\ +\ 25\ +\ 35}{4}\ =\ \frac{80}{4}\ =\ 20\]
| xi | fi | di = xi – A | di 2 | fi di | fi di 2 |
| 5 | 2 | – 15 | 225 | – 30 | 450 |
| 15 | 1 | – 5 | 25 | – 5 | 25 |
| 25 | 1 | 5 | 25 | 5 | 25 |
| 35 | 3 | 15 | 225 | 45 | 675 |
| N=7 |
\[\sigma\ =\ \sqrt{\frac{\Sigma f_i\ d_i^2}{N}\ -\ (\frac{\Sigma f_i\ d_i}{N}})^2\]
\[=\ \sqrt{\frac{1175}{7}\ -\ (\frac{15}{7}})^2\]
\[=\ \sqrt{\frac{1175}{7}\ -\ \frac{225}{7\ \times\ 7}}\]
\[=\ \sqrt{\frac{8225\ -\ 225}{7\ \times\ 7}}\]
\[=\ \sqrt{\frac{8000}{49}}\]
\[=\ \sqrt{163.265}\]
\[\sigma\ =\ 12.79\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{Fit\ a\ straight\ line\ to\ the\ following\ data\ by\ the\ method\ of\ least\ squares}\ \hspace{8cm}\]
| x | 0 | 1 | 2 | 3 | 4 |
| y | 1 | 1 | 3 | 4 | 6 |
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Let\ the\ line\ be\ y\ =\ a\ x\ +\ b\ ——–\ (1)\]
\[The\ Normal\ equations\ are\]
\[a\ \Sigma x_i\ +\ nb\ =\ \Sigma y_i\ ——–\ (2)\]
\[a\ \Sigma x_i^2\ +\ b\ \Sigma\ x_i\ =\ \Sigma x_i\ y_i\ ——–\ (3)\]
| xi | yi | xi 2 | xi yi |
| 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 3 | 4 | 6 |
| 3 | 4 | 9 | 12 |
| 4 | 6 | 16 | 24 |
\[(2)\ becomes\ a\ (10)\ +\ 5b\ =\ 15\]
\[2\ a\ +\ b\ =\ 3\ ——–\ (4)\]
\[(3)\ becomes\ a\ (30)\ +\ b\ (10)\ =\ 43\]
\[3\ a\ +\ b\ =\ 4.3\ ——–\ (5)\]
\[Solving\ (4)\ and\ (5)\]
\[2\ a\ +\ b\ =\ 3\]
\[3\ a\ +\ b\ =\ 4.3\]
———————————-
\[a\ =\ 1.3\]
\[put\ a\ =\ 1.3\ in\ (4)\]
\[2(1.3)\ +\ b\ =\ 3\]
\[2.6\ +\ b\ =\ 3\]
\[b\ =\ 3\ -\ 2.6\]
\[b\ =\ 0.4\]
\[Eqn\ (1)\ becomes\ y\ =\ 1.3\ x\ +\ 0.4\]
\[25\ \hspace{1cm}\ (a)\ \hspace{1cm} \color{green}{Three\ coins\ are\ tossed\ simultaneously\ Find\ the\ probability\ of\ getting}\ \hspace{5cm}\]\[(i)\ at\ least\ one\ head\ \hspace{13cm}\]\[(ii)\ at\ most\ one\ head\ \hspace{13cm}\]\[(iii)\ exactly\ one\ head.\ \hspace{13cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{When three coins are tossed simultaneously, each coin has two possible outcomes: heads (H) or tails (T).}\]\[\therefore\ \text{the total number of possible outcomes when three coins are tossed is:}\]
\[2^3\ =\ 8\]
\[\text{Total number of outcomes = 8}\ \hspace{10cm}\]
\[\text{{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}\ \hspace{10cm}\]
\[(i)\ \bf{\color{green}{at\ least\ one\ head}}\ \hspace{13cm}\]
\[\text{To find the probability of getting atleast one head, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there are either one, two or three heads.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HTT, HHT, HTH, THH, HHH}}\ \hspace{10cm}\]
\[P(at\ least\ one\ head)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{5}{8}\]
\[(ii)\ \bf{\color{green}{at\ most\ one\ head}}\ \hspace{13cm}\]
\[\text{To find the probability of getting at most one head, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there is zero or one head.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{TTT, HTT}}\ \hspace{10cm}\]
\[P(Exactly\ two\ heads)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{2}{8}\ =\ \frac{1}{4}\]
\[(iii)\ \bf{\color{green}{exactly\ one\ head}}\ \hspace{13cm}\]
\[\text{To find the probability of getting exactly one head, we need to count the}\ \hspace{5cm}\]\[\text{outcomes where there is exactly one head.}\ \hspace{10cm}\]
\[\text{The favourable number of outcomes are:}\ \hspace{10cm}\]
\[\text{{HTT, THT, TTH}}\ \hspace{10cm}\]
\[P(Exactly\ one\ head)\ =\ \frac{\text{Number of favourable outcomes}}{\text{Total Number of outcomes}}\ =\ \frac{3}{8}\]
\[\hspace{2cm}\ (b)\ \hspace{1cm} \color{green}{A\ card\ is\ drawn\ at\ random\ from\ a\ pack\ of\ 52\ cards.\ Find\ the}\ \hspace{8cm}\]\[\color{green}{probability\ that\ the\ card\ is\ either\ a\ black\ card\ or\ a\ card\ with\ number\ 6.}\ \hspace{2cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{To find the probability of drawing either a black card or a card with the number 6 from a standard deck of 52 cards,}\\ \text{we need to consider both events and their possible overlap.}\ \hspace{10cm}\]
\[1.\ \text{Total number of cards in a deck: 52}\ \hspace{5cm}\]
\[\hspace{3cm}\ 2.\ \text{Number of black cards: There are 26 black cards in the deck (13 spades +}\\ \text{13 clubs}).\ \hspace{10cm}\]
\[\hspace{3cm}\ 3.\ \text{Number of cards with the number 6: There are 4 cards with the number 6}\\ \text{(one in each suit: hearts, diamonds, clubs, and spades}).\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ drawing\ a\ black\ card\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ drawing\ a\ card\ with\ on\ number\ 6\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{26}{52}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{4}{52}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ \frac{2}{52}\ (2\ such\ cards\ with \ black\ card\ and\ number\ 6\ from\ spades\ and\ clubs)\]
\[P(A\ \cup\ B)\ =\ \frac{26}{52}\ +\ \frac{4}{52}\ -\ \frac{2}{52}\ \hspace{5cm}\]
\[=\ \frac{28}{52}\ =\ \frac{7}{13}\ \hspace{3cm}\]
\[\text{So, the probability of drawing either a black card or a card with the}\\ \text{number 6 from a standard deck of 52 cards is:} \frac{7}{13}\ \hspace{5cm}\]
\[\hspace{2cm}\ (c)\ \hspace{1cm} \color{green}{A\ problem\ in\ statistics\ is\ given\ to\ two\ students\ A\ and\ B.\ The}\ \hspace{10cm}\\\ \color{green}{probability\ of\ A\ solves\ the\ problem\ is\ \frac{1}{4}\ and\ that\ of\ B\ solves\ the}\ \hspace{8cm}\\\ \color{green}{problem\ is\ \frac{2}{5}.\ If\ the\ students\ solve\ the\ problems\ independently.}\ \hspace{8cm}\\\ \color{green}{find\ the\ probability\ that\ the\ problem\ is\ solved.}\ \hspace{12cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given,\ P(A)\ =\ \frac{1}{4}\ and\ P(B)\ =\ \frac{2}{5}\ \hspace{10cm}\\\ \text{Probability that the problem is solved = Probability that A solves the problem or B solves the problem}\]
\[=\ P(A\ \cup\ B)\ \hspace{5cm}\]
\[=\ P(A)\ +\ P(B)\ -\ (P(A\ \cap\ B)\ \hspace{2cm}\]
\[=\ P(A)\ +\ P(B)\ -\ P(A)\ \cdot \ P(B)\ Since A\ and\ B\ are\ independent\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5
}\ -\ \frac{1}{4}\ \cdot \ \frac{2}{5}\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5}\ -\ \frac{1}{10}\ \hspace{2cm}\]
\[=\ \frac{5\ +\ 8\ -\ 2}{20}\ \hspace{2cm}\]
\[=\ \frac{11}{20}\ \hspace{2cm}\]
\[\text{Thus, the probability that the problem is solved by either student A, student B, or both is:}\ =\ \frac{11}{20}\]
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