\(-\sqrt { 3 } -i\)

r = 2 and \(\alpha =\frac { \pi }{ 6 } \)
Since the complex number lies in the fourth quadrant, has the principal value
\(\theta =\alpha -\pi =\frac { \pi }{ 6 } -\pi =-\frac { 5\pi }{ 6 } \)
Therefore, the modulus and principal argument of \(-\sqrt { 3 } -i\)  are 2 and -\(\frac { 5\pi }{ 6 } \) respectively.

\(\theta =\frac { 2\pi }{ 3 } \)
When θ = \(\frac { \pi }{ 3 } \)
Roration of z is \(ze^{ i\theta }=ze^{ i2\frac { \pi }{ 3 } }\) [using (1)]
= \(2\sqrt { 2 } e^{ -i\frac { \pi }{ 4 } }.{ e }^{ i2\frac { \pi }{ 3 } }\)   [using (1)]
= \(2\sqrt { 2 } e^{ i\left( i\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \right) }=2\sqrt { 2 } e^{ i5\frac { \pi }{ 12 } }\)

We have = \(\frac { z+3 }{ z-5i } =\frac { 1+4i }{ 2 } \)
\(\Rightarrow\) 2(z + 3) = (1 + 4i) (z− 5i)
\(\Rightarrow\) 2z + 6 = (1 + 4i)z + 20−5i
\(\Rightarrow\) (2−1−4i)z = 20− 5i− 6 
\(\Rightarrow\) \(z=\frac { 14-5i }{ 1-4i } =\frac { \left( 14-5i \right) \left( 1+4i \right) }{ \left( 1-4i \right) \left( 1+4i \right) } =\frac { 34+51i }{ 17 } =2+3i\)

 We know that sin^-1: [-1, 1] \(\rightarrow \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)is given by
sin⁻¹x = y if and only if x = sin y for −1\(\le x\le \) and -\(\frac { \pi }{ 2 } \le y \le \frac { \pi }{ 2 } \). Thus,
 \({ Sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) \)= \(\frac{\pi}{4}\), Since \(\frac{\pi}{4}\)\(\in \left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)and sin \(\frac { \pi }{ 4 } =\frac { 1 }{ \sqrt { 2 } } \)

(z+w)
= (5-2i) + (-1+3i)
= (5-1) + i(-2+3)
= 4+i(1)
= 4+i

Since the given function is a probability mass function, the total probability is one. That is \(\underset { x }{ \Sigma } f(x)=1\)
 From the given data k + 2k + 6k + 5k + 6k +10k+1
\(30k=1\Rightarrow k=\frac { 1 }{ 30 } \) 
Therefore the probability mass function is

x	1	2	3	4	5	6
f(x)	\(\cfrac { 1 }{ 30 } \)	\(\cfrac { 2 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 5 }{ 30 } \)	\(\cfrac { 6 }{ 30 } \)	\(\cfrac { 10 }{ 30 } \)

(i) P(2 < X < 6) = f(3)+ f(4)+ f(5) = \(\frac { 6 }{ 30 } +\frac { 5 }{ 30 } +\frac { 6 }{ 30 } =\frac { 17 }{ 30 } \) 
(ii) P(2≤X≤5) = f(2)+f(3)+f(4) = \(\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 13 }{ 30 } \) 
(iii) P(2≤4) = f(1)+f(2)+f(3)+f(4) = \(\frac { 1 }{ 30 } +\frac { 2 }{ 30 } +\frac { 6 }{ 30 } +\frac { 5 }{ 30 } =\frac { 14 }{ 30 } \) 
(iv) P(3>X) = f(4)+f(5)+f(6) = \(\frac { 5 }{ 30 } +\frac { 6 }{ 30 } +\frac { 10 }{ 30 } =\frac { 21 }{ 30 } \)   

Let x and y be the length, breadth of the printed rectangular page

Given xy = 24
\(\Rightarrow y=\frac { 24 }{ x } \)    ..(1)
Length of the page with margin
= x+1+1 = x+2
breadth of the page with margin
= y + 1.5 + 1.5 = y + 3
Area of the page = (x + 2) (y + 3)
Let f(x) = (x + 2) (y + 3)
= \((x+2)\left( \frac { 24 }{ x } +3 \right) \) 
= \(24+3x+\frac { 48 }{ x } +30\) 
= \(3x+\frac { 48 }{ x } +30\) 
\(f'\left( x \right) =3-\frac { 48 }{ { x }^{ 2 } } \)
\(\Rightarrow { x }^{ 2 }=16\Rightarrow x=\pm 4\)
ஃ The critical number are 4,-4
\(f''\left( x \right) =-48\left( \frac { -2 }{ { x }^{ 3 } } \right) =\frac { 96 }{ { x }^{ 3 } } \) 
\(f''\left( 4 \right) =\frac { 96 }{ 64 } >0\) 
ஃ f(x) is minimum when x = 4
When \(x=4,y=\frac { 24 }{ 4 } =6\)  [From (1)]
ஃ Length of the page = x + 2 = 4 + 2 = 9 cm
Breadth of the page = y + 3 = 6 + 3 = 6 cm

x⁴+ 3x³- 3x - 1 = 0
Sum of the co-efficients = 1 + 3 - 3.- 1 = 0 
⇒x = 1 is a root ⇒(x - 1) is a factor

[Using synthetic division]
∴ x = 1, 1 are the roots and the remaining factor
\(\Rightarrow x=\frac { -3\pm \sqrt { 9-4(1)(1) } }{ 2 } \)
\(\Rightarrow x=\frac { -3\pm \sqrt { 5 } }{ 2 } \)
∴ The roots are 1, -1, \(\frac { -3+\sqrt { 5 } }{ 2 } ,\frac { -3-\sqrt { 5 } }{ 2 } \).

2x + 3y - z = 9, x + y + z = 9, 3x - y - z = -1
The matrix form of the system is
\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ X = A^-1N
|A| = \(\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \)
= 2(-1+1)-3(-1-3)-1(-1-3)
= 0-3(-4) = 12+4 = 16
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & -1 \\ -1 & -1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ 3 & -1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & -1 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(-1+1) & -(-1-3) & +(-1-3) \\ -(-3-1) & +(2+3) & -(-2-9) \\ +(3+1) & -(2+1) & +(2-3) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 0 & 4 & -4 \\ 4 & 1 & 11 \\ 4 & -3 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
= \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0+36-4 \\ 36+9+3 \\ -36+99+1 \end{matrix} \right] =\frac { 1 }{ 16 } \left[ \begin{matrix} 32 \\ 48 \\ 64 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
∴ x = 2, y = 3, z = 4
∴ Solution set is {2, 3, 4}

By definition, \(\vec { a } \times \vec { b } \) \(=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 3 & -1 & 3 \end{matrix} \right| =7\hat { i } -7\hat { k } \)
Then, \((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 7 & 0 & -7 \\ 2 & -5 & 1 \end{matrix} \right| =-35\hat { i } -21\hat { j } -35\hat { k } \)......(1)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & -1 & 3 \\ 2 & -5 & 1 \end{matrix} \right| =14\hat { i } +3\hat { j } -13\hat { k } \)
\(\vec { a } (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 14 & 3 & -13 \end{matrix} \right| =-33\hat { i } -54\hat { j } -48\hat { k } \)....(2)
Therefore, equations (1) and (2) show that \((\vec { a } \times \vec { b } )\times \vec { c } \)\(​​\neq \)\((\vec { a } \times \vec { b } )\times \vec { c } \)
 

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

If A is orthogonal, then AA^T = A^TA = I₃. So, we have
AA^T = I₃ ⇒ \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] \frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 45+{ a }^{ 2 } & 6b+6+6a & 12-3c+3a \\ 6b+6+6a & { b }^{ 2 }+40 & 2b-2c+18 \\ 12-3c+3a & 2b-2c+18 & { c }^{ 2 }+13 \end{matrix} \right] =49\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left\{ \begin{matrix} 45+{ a }^{ 2 }=49 \\ \begin{matrix} { b }^{ 2 }+40=49 \\ \begin{matrix} { c }^{ 2 }+13=49 \\ \begin{matrix} 6b+6+6a=0 \\ \begin{matrix} 12-3c+3a=0 \\ 2b-2c+18=0 \end{matrix} \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right\} \) ⇒ \(\left\{ \begin{matrix} { a }^{ 2 }=4,{ b }^{ 2 }=9,{ c }^{ 2 }=36 \\ a+b=-1,a-c=-4,b-c=-9 \end{matrix} \right\} \) ⇒ a = 2, b = −3, c = 6
So we get A = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \) and hence, we get A^-1 = A^T = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \).

Let the general equation of the circle be
x² +y² +2gx + 2fy + c = 0 ........ (1)
It passes through points (1, 1), (2, -1) and(3, 2) .
Therefore,
2g+2f+c = -2 ........ (2)
4g−2f+c = -5 ........(3)
6g+4f+c = -13 ...... (4)
(2) – (3) gives −2g + 4f = 3 ... (5)
(4) – (3) gives 2g + 6f = -8  ....(6)
(5) + (6) gives f = \(-\frac { 1 }{ 2 } \) 
Substituting f = \(-\frac { 1 }{ 2 } \) in (6), g = -\(\frac { 5 }{ 2 } \)
Substituting f = \(-\frac { 1 }{ 2 } \) and g = - \(\frac { 5 }{ 2 } \) in (2), c = 4 .
Therefore the required equation of the circle is
x²+ y²+ 2\(\left( -\frac { 5 }{ 2 } \right) x+2\left( -\frac { 1 }{ 2 } \right) \)y+4 = 0 and x² + y² − 5x − y + 4 = 0.

Taking u = 1−x² and v = e^{−\(\lambda\)x}, and applying the Bernoulli’s formula, we get
\(I=\int _{ -1 }^{ 1 }{ { e }^{ -\lambda x } } (1-{ x }^{ 2 })dx={ \left[ (1-{ x }^{ 2 })\left( \frac { { e }^{ -\lambda x } }{ -\lambda } \right) -(-2x)\left( \frac { { e }^{ -\lambda x } }{ { \lambda }^{ 2 } } \right) +(-2)\left( \frac { { e }^{ -\lambda x } }{ -{ \lambda }^{ 3 } } \right) \right] }_{ -1 }^{ 1 }\)
\(=2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 2 } } \right) +2\left( \frac { { e }^{ -\lambda } }{ { \lambda }^{ 3 } } \right) +2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 2 } } \right) -2\left( \frac { { e }^{ \lambda } }{ { \lambda }^{ 3 } } \right) \)
\(=\frac { 2 }{ { \lambda }^{ 2 } } ({ e }^{ \lambda }+{ e }^{ -\lambda })-\frac { 2 }{ { \lambda }^{ 3 } } ({ e }^{ \lambda }-{ e }^{ -\lambda })\)

We have,
\(\frac{d}{dt}W(t)=-20\times(1-0.1t)\)
Therefore at t = 2, \(\frac{d}{dt}W(t)=-16\)
That is, the person forgets at the rate of 16 words after 2 days of studying.

Treating \((\vec { a } \times \vec { b } )\) as the first vector on the right hand side of the given equation and using the vector triple product expansion, we get
\((\vec { a } \times \vec { b } )\times (\vec { a } \times \vec { c } )=((\vec { a } \times \vec { b } ).\vec { c } )\vec { a } -((\vec { a } \times \vec { b } ).\vec { a } )\vec { c } =(\vec { a } .\vec { b } \times \vec { c } ))\vec { a } \)

Let ABCD be a parallelogram such that its diagonals AC and BD are equal.
Taking A as the origin, let the p.v. of B and D be \(\vec { b } \)and \(\vec { d } \) respectively.
Then \(\vec { AB } =\vec { b } \) and \(\vec { AD } =\vec { d } \)
Using triangle law of addition of vectors is ΔABC, we get
\(\vec { AB } +\vec { BC } =\vec { AC } \)
⇒ \(\vec { AB } +\vec { AD } =\vec { AC } \)
⇒ \(\vec { b } +\vec { d } =\vec { AC } \)
Using triangle law of addition of vectors in ΔABD, we get \(\vec { AB } +\vec { BD } =\vec { AD } \)
⇒ \(\vec { b } +\vec { BD } =\vec { d } \)
⇒ \(\vec { BD } =\vec { d } -\vec { b } \)
In parallelogram ABCD we have AC = BD
⇒ \(|\vec { AC } |=|\vec { BD } |\)
⇒ \(|\vec { AC } |^{ 2 }=|\vec { BD } |^{ 2 }\)
⇒ \(|\vec { b } +\vec { d } |^{ 2 }=|\vec { d } -\vec { b } |^{ 2 }\)

⇒ \(4(\vec { b } .\vec { d } )=0\Rightarrow \vec { b } .\vec { d } =0\)=0
⇒ \(\vec { b } \bot \vec { d } \)
⇒ \(\vec { AB } \bot \vec { AD } \)
Hence, ABCD is a rectangle.

Area of the circle = 9π sq.units

πr² = 9π ⇒ r² ⇒ 9 ⇒ r ⇒ 3
Diameters are x + y = 5 ................(1)
and x - y = 1 .................(2)
We know that centre is the point of intersection of diameters.
∴ To find the centre, solve (1) and (2).

⇒ 2x = 6
⇒ x = 3
∴ (1) ⇒ 3 + y = 5
⇒ y = 5 - 3 = 2
∴ Centre is (3, 2)
Hence, equation of the circle is
(x - h)² + (y - k)² = r2
⇒ (x - 3)² + (y - 2)² = 32
\(\Rightarrow x^{2}-6 x+9+y^{2}-4 y+4=9\)
⇒ x² + y² − 6x − 4y + 4 = 0

The condition for the line y = mx + c to be a tangent to the circle x² + y² = a² is c² = a²(1 + m²) from  
Then \(c=\pm \sqrt { 9\left( 1+16 \right) } \)
\(c=\pm 3\sqrt { 17 } \)

For the given A, We get |A| = (2)(7) - (9)(1) = 14 - 9 = 5. So, A^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -9 \\ -1 & 2 \end{matrix} \right] =\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 9 }{ 5 } \\ -\frac { 1 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \).
Then, (A^-1)^T = \(\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 1 }{ 5 } \\ -\frac { 9 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).           ....(1)
For the given A, We get A^T = \(\left[ \begin{matrix} 2 & 1 \\ 9 & 7 \end{matrix} \right] \). So |A^T| = (2)(7) - (1)(9) = 5
Then, (A^T)^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).             ...(2)
From (1) and (2), we get (A^-1)^T = (A^T)^-1. Thus, we have verified the given property.

We compute |adj A| = \(\left| \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right| \) = 9.
So, we get
A^-1 = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(A) = \(\frac { 1 }{ \sqrt { 9 } } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] =\pm \frac { 1 }{ 3 } \left[ \begin{matrix} -1 & 2 & 2 \\ 1 & 1 & 2 \\ 2 & 2 & 1 \end{matrix} \right] \).