1

\(\int \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}} d x =\frac{1}{2} \int \frac{1}{t} d t \)
\(=\frac{1}{2} \log |t|=\frac{1}{2} \log \left|e^{2 x}+e^{-2 x}\right|+C\)

Matrix is skew symmetric, \(A=\left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right] A'=\left[ \begin{matrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right] =-A \)
As A' = -A, so matrix A is skew-symmetric.

\( P(B / A)=\frac{P(A \cap B)}{P(A)} \)
\(\Rightarrow 0.6 \times 0.4=P(A \cap B) \)
\(\Rightarrow P(A \cap B)=0.24 \)
\( P(A \cup B)=P(A)+P(B)-P(A \cap B) \)
\(=0.4+0.7-0.24=0.86 \)

Given, \(f(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\) 
To prove, f(x).f(y) = f(x + y).
For f(y), replace x with y, so thatf(x) becomes f(y).
LHS = f(x). f(y)
\( =\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{cc} \cos x \cos y-\sin x \sin y+0 \\ \sin x \cos y+\cos x \sin y+0 \\ 0+0+0 \end{array}\right. \)
\(\left.\begin{array}{cc} -\cos x \sin y-\sin x \cos y+0 & 0+0+0 \\ -\sin x \sin y+\cos x \cos y+0 & 0+0+0 \\ 0+0+0 & 0+0+1 \end{array}\right]\) 
[multiplying rows by columns]
\(=\left[\begin{array}{ccc} \cos (x+y) -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right] \)
\(\left[\begin{array}{rc} \because \cos (A+B)=\cos A \cos B-\sin A \sin B \\ \text { and } \sin (A+B)=\sin A \cos B+\cos A \sin B \end{array}\right]\)
\(=f(x+y)=\mathrm{RHS} \)

The given system of equation is
x + 3y = 5 and
2x + 6y = 8.
The given system of equations can be written in the form of  A X = B, where 
\(A=\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right], X=\left[\begin{array}{l} x \\ y \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 8 \end{array}\right] \).
Now, |A| = 1(6) - 3(2) = 6 - 6 = 0
∴ A  is a singular matrix.
\(\text { Now, }(\operatorname{adj} A)=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\)
\((a d j A) B=\left[\begin{array}{cc} 6 & -3 \\ -2 & 1 \end{array}\right]\left[\begin{array}{l} 5 \\ 8 \end{array}\right]=\left[\begin{array}{c} 30-24 \\ -10+8 \end{array}\right]=\left[\begin{array}{l} 6 \\ -2 \end{array}\right] \neq O\)
Thus, the solution of the given system of equations does not exist. Hence, the system of equation is inconsistent 

\(64 \% , as \mathrm{P} (same fact )=P(A B or \bar{A} \bar{B}) \)
\(=\frac{70}{100} \times \frac{85}{100}+\frac{30}{100} \times \frac{15}{100} \)
\(=\frac{5950+450}{10000}=\frac{6400}{10000}=64 \%\)

Let x, y and z be the values of resourcefulness, competence and determination respectively. Then the system of equation is
4x + 3y + 2z = 0.37
5x + 3y + 4z = 0.47
x + y + z = 0.12
Matrix equation is
\(\left[ \begin{matrix} 4 & 3 & 2 \\ 5 & 3 & 4 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0.37 \\ 0.47 \\ 0.12 \end{matrix} \right] \)
i.e., AX = B
|A| =\(\left[ \begin{matrix} 4 & 3 & 2 \\ 5 & 3 & 4 \\ 1 & 1 & 1 \end{matrix} \right] \)
= 4(3-4)-3(5-4)+2(5-3)
= -3 \(\neq \)0
\(\therefore\) A^-1  exists
adj A = \(\left[ \begin{matrix} -1 & -1 & 2 \\ -1 & 2 & -1 \\ 6 & -6 & -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} -1 & -1 & 6 \\ -1 & 2 & -6 \\ 2 & -1 & -3 \end{matrix} \right] \)
X = A^-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ -3 } \left[ \begin{matrix} -1 & -1 & 6 \\ -1 & 2 & -6 \\ 2 & -1 & -3 \end{matrix} \right] \left[ \begin{matrix} 0.37 \\ 0.47 \\ 0..12 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0.04 \\ 0.05 \\ 0.03 \end{matrix} \right] \)
x = 4,000, y = 5,000, z = 3,000
i.e., Rs. 4,000 for resourcefulness, Rs. 5,000 for competence and 3,000 for determination.
Value: One more value loke sincerity, kindness etc.

(i)Let 'x', 'y' and 'z' be the amount invested in three investments.
Then
x + y + z = 65000  ......(1)
\({6x\over100}+{8y\over100}+{9z\over100}=4800\)
\(\Rightarrow\) 6x + 8y + 9z = 480000  ...(2)
\({9z\over100}=600+{8y\over100}\)
\(\Rightarrow\) 0x - 8y + 9z = 60000
These can be written as AX = B where:
\(A=\begin{bmatrix} 1&1&1\\6&8&9\\0&-8&9\end{bmatrix},X=\begin{bmatrix} x\\y\\z\end{bmatrix}\)
and \(B=\begin{bmatrix} 650000\\480000\\60000\end{bmatrix}\)
(ii) Now \(|A|=\begin{bmatrix}1&1&1\\6&8&9\\0&-8&9 \end{bmatrix}\)
= 1.(72 + 72) - 6(9+8)
= 144 - 102 = 42 ≠ 0
Hence, the equations have a unique solution.
(iii) No.We are not fools because most of such companies are frauds.

Let E₁ = Event that 5 or 6 is shown on die
and E₂ = Event that 1, 2, 3 or 4 is shown on die
Here, n(E₁) = 2andn(E₂) = 4
Also, n(S) = 6
\(\therefore P\left(E_{1}\right)=\frac{2}{6}=\frac{1}{3}\) 
and \(P\left(E_{2}\right)=\frac{4}{6}=\frac{2}{3}\) 
Let E = The event that exactly one head show up.
\(\therefore P\left(\frac{E}{E_{1}}\right)=P\) (exactly one head show up when coin is tossed thrice)
\(=P\{H T T, T H T, T T H\}=\frac{3}{8}\) 
\(\left[\because \text { total number of outcomes }=2^{3}=8\right]\)
\(P\left(\frac{E}{E_{2}}\right)=P\) (head shows up when coin is tossed once) \(=\frac{1}{2}\) 
The probability that the girl threw 1, 2, 3 or 4 with the die, if she obtained exactly one head, is given by
\(P\left(\frac{E_{2}}{E}\right)=\frac{P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)}\) 
[by Baye's theorem]
\(=\frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8}+\frac{2}{3} \times \frac{1}{2}}=\frac{\frac{1}{3}}{\frac{1}{8}+\frac{1}{3}}=\frac{8}{8+3}=\frac{8}{11}\)

Maximise Z = 70x + 40y
Subject to the constraints
         \(\begin{array}{r} 3 x+2 y \leq 9 \end{array}\),
         \(\begin{array}{r} 3 x+y \leq 9 \end{array}\)
  and   \(\begin{array}{r} x \geq 0, y \geq 0 \end{array}\)
Now, considering the inequations as equations, we get
3x + 2y = 9         .....(i)
3x + y  =9           ....(ii)
Table for line 3x + 2y = 9 is

So, it passes through the points (3, 0) and (0, 9/2) on putting (0, 0) in the inequality 3x + 2y \(\leq\) 9.
0 \(\leq\) 9 (which is true)
So, the half plane is towards the origin.
Table for line 3x + y = 9 is

On putting (0, 0) in the inequality 3x + y \(\leq\) 9
0\(\leq\)9 (which is true)
So, the half plane is towards the origin.
Also, x \(\geq\) 0,y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (3, 0).
The graphical representation of the above system of inequations is given below

Clearly, feasible region is OABO, whose corner points are O(0, 0), A(3, 0) and B (0, 9/2).

In the table, we find that maximum value of Z is 210 at the point A(3, 0).

We have \(f(\alpha)=\left[\begin{array}{ccc}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{array}\right]\) ..(i)
For f(-ß), replace α in f(α) by -ß
\( \therefore f(-\beta) =\left[\begin{array}{ccc} \cos (-\beta) & -\sin (-\beta) & 0 \\ \sin (-\beta) & \cos (-\beta) & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{ccc} \cos \beta & \sin \beta & 0 \\ -\sin \beta & \cos \beta & 0 \\ 0 & 0 & 1 \end{array}\right]\)
By LHS \(=f(\alpha) f(-\beta)\)
\( =\left[\begin{array}{ccc} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos \beta & \sin \beta & 0 \\ -\sin \beta & \cos \beta & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{ccc} \cos \alpha \cos \beta+\sin \alpha \sin \beta & \cos \alpha \sin \beta-\sin \alpha \cos \beta & 0 \\ \sin \alpha \cos \beta-\cos \alpha \sin \beta & \sin \alpha \sin \beta+\cos \alpha \cos \beta & 0 \\ 0 & 0 & 1 \end{array}\right]\)
\(=\left[\begin{array}{ccc} \cos (\alpha-\beta) & -\sin (\alpha-\beta) & 0 \\ \sin (\alpha-\beta) & \cos (\alpha-\beta) & 0 \\ 0 & 0 & 1 \end{array}\right]\)
Now, for \(f(\alpha-\beta)\), replace \(\alpha\) in f(\(\alpha\)) by \((\alpha-\beta)\)
RHS \(=f(\alpha-\beta)\)
\(=\left[\begin{array}{ccc} \cos (\alpha-\beta) & -\sin (\alpha-\beta) & 0 \\ \sin (\alpha-\beta) & \cos (\alpha-\beta) & 0 \\ 0 & 0 & 1 \end{array}\right]\)
From Eqs. (ii) and (iii), we get
LHS = RHS
Hence proved. 

\(\frac{4}{17}\)

Let the events E₁ and E₂ be the events when lost card is a diamond and not a diamond respectively.
\(P({ E }_{ 1 })=\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \)and
\(P({ E }_{ 2 })=\frac { 39 }{ 52 } =\frac { 3 }{ 4 } \)
Let A be the event:
"two cards drawn from the remaining pack are diamonds"
\(P(A/{ E }_{ 1 })=\frac { 12\times 11 }{ 51\times 50 } \)
\(P(A/{ E }_{ 2 })=\frac { 13\times 12 }{ 51\times 50 } \)
By Bayes' Theorem
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) }{ \left( \frac { 1 }{ 4 } \right) \left( \frac { 12\times 11 }{ 51\times 50 } \right) +\left( \frac { 3 }{ 4 } \right) \left( \frac { 13\times 12 }{ 51\times 50 } \right) } \)
\(=\frac { 12\times 11 }{ 12\times 11+3\times 13\times 12 } \)
\(=\frac { 132 }{ 132+468 } =\frac { 132 }{ 600 } =\frac { 11 }{ 50 } \)

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

\(We\quad have:\quad 2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
\(\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\( \Rightarrow \begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} y+2 & 6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
Equating corresponding elements:
2x + 2 = 8 and y + 2 = 5
\(\Rightarrow \) 2x = 6 and y = 3.
Hence, x = 3 and y = 3.

We have: \(\left[ \begin{matrix} 2x & 3 \end{matrix} \right] \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix}\left[ \begin{matrix} x \\ 8 \end{matrix} \right] =0\)
\(\Rightarrow \left[ \begin{matrix} 2x-9 & 4x \end{matrix} \right] \left[ \begin{matrix} x \\ 8 \end{matrix} \right] =0\)
\(\Rightarrow \left[ 2{ x }^{ 2 }-9x+32x \right] =[0]\)
\( \Rightarrow [2{ x }^{ 2 }+23x]=[0]\)
\(\Rightarrow 2{ x }^{ 2 }+23x=0\)
\(\Rightarrow x(2x+23)=0.\)
Hence,  \(x=0,\quad -\frac { 23 }{ 2 } .\)

The sample space has 216 outcomes.
\(\text { Now } \quad \mathrm{A}=\left\{\begin{array}{lllll} (1,1,4) & (1,2,4) & \ldots & (1,6,4) & (2,1,4) & (2,2,4) & \ldots (2,6,4) \\ (3,1,4) & (3,2,4) & \ldots &(3,6,4) & (4,1,4) & (4,2,4) & \ldots(4,6,4) \\ (5,1,4) & (5,2,4) & \ldots & (5,6,4) & (6,1,4) & (6,2,4) & \ldots(6,6,4) \end{array}\right\}\)
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)} and A ∩ B = {(6,5,4)}.
\(\text { Now }P(B)=\frac{6}{216} \text { and } P(A \cap B)=\frac{1}{216} \)
Then  \(P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { \frac { 1 }{ 216 } }{ \frac { 6 }{ 216 } } =\frac { 1 }{ 6 }\)

The shaded region is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner points

A, B and C are (0,5), (4,3) and (0,6) respectively. Now we evaluate Z = 200x + 500y at these points.
Hence, minimum value of Z is 2300 attained at the point (4, 3)

Given \(P(A)=\frac { 1 }{ 2 } ,P(A\cup B)=\frac { 3 }{ 5 } \) and P(B) = p
\(\therefore \) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
\(\Rightarrow \)  \(\frac { 3 }{ 5 } =\frac { 1 }{ 2 } +p-P(A\cap B)\)
\(\Rightarrow \) \(P(A\cap B)=p+\frac { 1 }{ 2 } -\frac { 3 }{ 5 } =p-\frac { 1 }{ 10 } \)
(i) When A and B are mutually exclusive,
then \(P(A\cap B)=0\Rightarrow p-\frac { 1 }{ 10 } =0\)
\(\Rightarrow \)  \(p=\frac { 1 }{ 10 } \)
(ii) When A and B are independent,
then \(P(A\cap B)=P(A)P(B)\)
\(\Rightarrow \) \(p-10=\frac { 1 }{ 2 } p\Rightarrow p-\frac { p }{ 2 } =\frac { 1 }{ 10 } \)
\(\Rightarrow \) \(\frac { p }{ 2 } =\frac { 1 }{ 10 } \)
Hence, \(p=\frac { 1 }{ 5 } \)

\(\int { \sec { x } } (\sec { x } +\tan { x } )dx\)
\( =\int { \sec ^{ 2 }{ x } } dx+\int { \sec { x } } \tan { x } dx\)
\(=\tan { x } +\sec { x } +C\)

-12 at (4,0).

As if A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) then \(\left| A \right| =\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
\(KA=\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\) and \(\left| KA \right| =\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\)
\(={ K }^{ 2 }\begin{vmatrix} a & b \\ c & d \end{vmatrix}={ K }^{ 2 }|A|\)

As \(\frac { d }{ dx } \left( \frac { { 2 }^{ x } }{ { log }_{ e }2 } \right) \)
\(=\frac { 1 }{ { log }_{ e }2 } .{ 2 }^{ x }.{ log }_{ e }2 ={ 2 }^{ x }\)

Hint The order of SA is 3 x m and 2B is 3 x n, where m = 7L
\(\therefore\) The order of SA - 2B is 3 x m or 3 x 7L

\(\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=\pm 35\)