Let x represent the principal in the saving amount.
R = 8% and N = 1.
∴ Interest = \(\frac { PNR }{ 100 } =\frac { x\times 1\times 8 }{ 100 } =\frac { 2x }{ 25 } \)
∴ Given \(\frac { dx }{ dt } \) = interest + Rs. 400.
∴ \(\frac { dx }{ dt } =\frac { 2x }{ 25 } +400\)

 \(\sum _{ n=1 }^{ 102 }{ { i }^{ n } } \) = (i¹+i²+i³+i⁴)+(i⁵+i⁶+i⁷+i⁸)+....+(i⁹⁷+i⁹⁸+i⁹⁹+i¹⁰⁰)+i¹⁰¹+i¹⁰²
= (i¹+i²⁺i³+i⁴)+(i¹+i²+i³+i⁴)+...+(i¹+i²+i³+i⁴)+i^-1+i^-2
= {i+(-1)+(-i)+1}+{i+(-1)+(-i)}+......+{i+(-1)+(-i)+1}+i+(-1)
= 0+0+...0+i-1
= -1+i

cosec^-1\((-\sqrt{2})\)
\(\Rightarrow -\sqrt { 2 } =cosex\theta \)
\(\Rightarrow sin\theta =\frac { -1 }{ \sqrt { 2 } } \)
\(\Rightarrow sin\theta =-sin\frac { \pi }{ 4 } \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow sin\theta =sin\left( \frac { -\pi }{ 4 } \right) \)
\(\Rightarrow \theta =-\frac { \pi }{ 4 } \)
\(\therefore\) \(\theta cosec^{ -1 }\left( -\sqrt { 2 } \right) =-\frac { \pi }{ 4 } \)

Comparing the given equation of the plane with \(\vec { r } .\vec { n } \) = p, we have \(\vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \).
We know that the perpendicular distance from the given point with position vector u to the plane \(\vec { r } .\vec { n } \)= p is given by \(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } \). Therefore, substi \(\vec { u } \)= (2, 5, -3) = \(2\hat { i } +5\hat { j } -3\hat { k } \) and \(\ \vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \) in the formula, we get
\(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } =\frac { |(2\hat { i } +5\hat { j } -3\hat { k } ).(6\hat { i } -3\hat { j } +2\hat { k } )-5| }{ |6\hat { i } -3\hat { j } +2\hat { k } | } \) = 2 unit.

Given lines are \(\vec { r } =(4\hat { i } -\hat { j } )+t(\hat { i } +2\hat { j } -2\hat { k } )\) \([\vec { r } =\vec { a } +t\vec { b } ]\)
∴ \(\vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
and \(\vec { r } =(\hat { i } -2\hat { j } +4\hat { k } )+s(-\hat { i } -2\hat { j } +2\hat { k } )\)
∴ \(\vec { d } =-\hat { i } -2\hat { j } +2\hat { k } \)
Let θ be the angle between the given lines
Then cos θ = \(\frac { \vec { b } .\vec { d } }{ |\vec { b } ||\vec { d } | } \)
= \(\frac { (\hat { i } +2\hat { j } -2\hat { k } ).(-\hat { i } -2\hat { j } +2\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2)^{ 2 } } .\sqrt { (-1)^{ 2 }+{ (-2) }^{ 2 }+{ (2) }^{ 2 } } } \)
= \(\frac { -1-4-4 }{ \sqrt { 9 } .\sqrt { 9 } } =\frac { -9 }{ 9 } \) = -1
∴ cos θ = -1
⇒ cos-1(-1)
⇒ θ = 0

\(tan^{-1}(tan\frac{5\pi}{4})\)
= \({ tan }^{ -1 }\left( tan\left( \pi +\frac { \pi }{ 4 } \right) \right) \)
= \({ tan }^{ -1 }\left( tan\frac { \pi }{ 4 } \right) \) \(\left[ \because tan\left( \pi +\theta =tan\theta \right) \right] \)
= \(\frac { \pi }{ 4 } \varepsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

\(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right| \)= 6 - 4 = 2 ≠ 0
Since A is nonsingular, A^-1 exists
A^-1 = \(\frac { 1 }{ |A| } \)
Now, adj A = \(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \)
[Interchange the entries in leading diagonal and change the sign of elements in the off diagonal]
∴ A^-1 = \(\frac{1}{2}\)\(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \).

Given z₁= 2 - i and z₂= -4+3i
z₁z₂ = (2-i)(-4+3i)
= -8 + 6i + 4i - 3i²
= -8 +10i - 3(-1)
= -8 +10i + 3 = -5 +10i
Inverse of z₁z₁ is \(\frac { 1 }{ { z }_{ 1 }{ z }_{ 2 } } \)
= \(\frac { 1 }{ -5+10i } \times \frac { -5-10i }{ -5-10i } \)
= \(\frac { -5-10i }{ (-5)^{ 2 }-(10i)^{ 2 } } \)
= \(\frac { -5-10i }{ 25-100i^{ 2 } } \)
= \(\frac { -5-10i }{ 25+100 } \)  [∵ i² = -1]
\(=\frac{\not{5}(-1-2 i)}{\not{5}\langle(2 5)}=\frac{-1-2 i}{25}\)
∴ Inverse of z₁z₂ is \(\frac { 1 }{ 25 } \) (-1-2i)
Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { 1 }{ \frac { { z }_{ 1 } }{ { z }_{ 2 } } } =\frac { { z }_{ 2 } }{ { z }_{ 2 } } \)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { { z }_{ 2 } }{ { z }_{ 1 } } =\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -4+3i }{ 2-i } \times \frac { 2+i }{ 2+i } \)
= \(\frac { -8-4i+6i+3i^{ 2 } }{ 2^{ 2 }-({ i }^{ 2 }) } \)
= \(\frac { -8+2i-3 }{ 4+1 } =\frac { -11+2i }{ 5 }\)
\( =\frac { 1 }{ 5 } \)(-11 + 2i)
∴ Inverse of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } \) is \(\frac { -11+2i }{ 5 }\)or \(\frac { 1 }{ 5 } \)(-11 + 2i)

cos^-1\([cos(-\frac{\pi}{6})]\neq \frac{\pi}{6}.\)
Since \(\frac { -\pi }{ 6 } \notin \left[ 0,\pi \right] \) which is the principal domain of cosine function. [\(\therefore \) cos -\(\theta\) = cos \(\theta\)]

\(y=\frac { 2x }{ { 2x }^{ 2-1 } } ,x\neq \pm \frac { 1 }{ \sqrt { 2 } } \)

\(y=-\frac { 1 }{ 2 } { cos }^{ 2 }x+\frac { { cos }^{ 5 }x }{ 5 } +{ xe }^{ x }-{ e }^{ x }+c\)

\(y=\frac { { x }^{ 2 } }{ 2 } +2x-13log|x+2|+c,x\varepsilon R-\{ 2\} \)

prove.

2 sq.units

100 m / s, t = 4 sec, 200 m / s, −100 m / s

Given (1 + e^2x) dy + (1 + y²)e^x dx = 0
\(\frac { dy }{ 1+{ y }^{ 2 } } =\frac { -{ e }^{ x } }{ 1+{ e }^{ 2x } } dx\)
\(\int { \frac { dy }{ 1+{ y }^{ 2 } } } =-\int { \frac { { e }^{ x }dx }{ 1+({ e }^{ x })^{ 2 } } } \)
tan^-1(y) = -tan^-1(e^x)+c
y(0) = 1 ⇒ when x = 0, y = 1;
t = e^x ⇒ dt = e^x dx
\(\int { \frac { dt }{ 1+{ t }^{ 2 } } } \) = tan^-1(t)
= tan^-1(e^x)
∴ tan^-1(1) + tan^-1(e⁰) = c;
⇒ \(\frac { \pi }{ 4 } +\frac { \pi }{ 4 } =c\Rightarrow c=\frac { \pi }{ 2 } \)
∴ (1) becomes, tan^-1(y) + tan^-1(e^x) = \(\frac { \pi }{ 2 } \) which is the required solution.

Let f(x) = cos x, x = 60° dx = 1°
f(x_o) = cos 60° = \(\frac12\) = 0.5
f'(x) = - sinx dx
f'(x_o) = - sin x_o dx
f'(60) = - sin 60° (1°)
= - (0.86603) (0.01745)
= - 0.0154
∴ f(x) = f(x_o) +f(x_o) dx 
f(61) = 0.5 - 0.0154
∴ tan 46° = f(x_o) +f(x_o) dx 
cos 61° = 0.4849

Given f(x) = x³ + bx² + ax + b
Given that Rolle's theorem holds for c = \(2+\frac { 1 }{ \sqrt { 3 } } \)
⇒ f'(c) = 0
⇒ 3c²+ 2bc+ a = 0
⇒ c = \(\frac { -2b\pm \sqrt { 4b^{ 2 }-4(3)a } }{ 6 } \)
= \(\frac { -2b\pm \sqrt { 4b^{ 2 }-12a } }{ 6 } \)
⇒ \(2+\frac { 1 }{ \sqrt { 3 } } =2\left( \frac { (-b)\pm \sqrt { { b }^{ 2 }-3a } }{ 6 } \right) \)
= \(\frac { -b\pm \sqrt { { b }^{ 2 }-3a } }{ 3 } \)
= \(\left( \frac { -b }{ 3 } \right) \pm \frac { \sqrt { { b }^{ 2 }-3a } }{ 3 } \)
⇒ \(\frac { -b }{ 3 } \) = -2 ⇒ -b = 6 ⇒ b = -6
Also, \(\frac { \sqrt { { b }^{ 2 }-3a } }{ 3 } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ \(\frac { { b }^{ 2 }-3a }{ 9 } =\frac { 1 }{ 3 } \)
⇒ b²-3a = 3
⇒ (-6)²-3a = 3
⇒ 36-3 = 3a
⇒ 33 = 3a
⇒ a = 11
∴ a = 11, b = -6

Let F(x, y) = x² – 2y² + 2xy and x(t) = cost, y(t) = sint
Then F(x, y) = cos² t - 2sin² t + 2cos t sin t  and thus F has becomes a function of one variable t. So by using chain rule, we see that
\(\frac { dF }{ dt } \) = 2 cos t(-sin t) -4 sin t cos t 2 (-sin² t + cos² t).
= -6 cos t sin t +2 ( -sin² t + cos² t)
On the other hand if we calculate
\(\frac { \partial F }{ \partial x } \frac { \partial x }{ \partial t } +\frac { \partial F }{ \partial y } \frac { \partial y }{ \partial t } \) = (2x+2y)\(\frac { d x }{ dt } \)+(2x - 4y) \(\frac { dy }{ dt } \)
= 2(cos t + sin t)(-sin t) + 2(cos t - 2sin t)(cos t)
= -6 cos t sin t +2 ( -sin² t + cos² t)
= \(\frac { dF }{ dt } \)

We need to show that u satisfies the Laplace’s equation in R². Observe that u_x(x, y) = e^-2y(-2)sin(2x) and hence u_xx (x, y) = e^-2y(-2)(2) cos(2x).
Similarly, u_y( x y) = e^-2y (-2)cos(2x) and u_yy (x, y) = (-2)(-2)e^-2ycos(2x)
Thus, u_xx + u_yy = -4e^-2y cos(2x) + 4e^-2y cos(2x) = 0.

Given f(x) = 3x² - 4x³ ; [-1, 2]
f'(x) = 12x³ - 12x²
f'(x) = 0
⇒12x³- 12x² = 0
⇒ 12x²(x-1) = 0
⇒ x = 0 or x = 1
Evaluatingf(x) at the end points x = -1, x = 2 and at the critical number x = 0, x = 1 we get
f(-1) = 3(-1)4-4 (-1)³
= 3 + 4 = 7
f(2) = 3(2)⁴ - 4(2³)
= 48 - 32 =16
f(0) = 0
f(1) = 3(1)⁴ - 4(1)³
= 3 - 4 = -1
From these values, the absolute maximum is 16 at x = 2 and the absolute minimum is -1 which occurs at x = 1.

Let the time by one man alone be x days and one woman alone be y days
∴ By the given data,
\(\frac { 4 }{ x } +\frac { 4 }{ y } =\frac { 1 }{ 3 } \)
and \(\frac { 2 }{ x } +\frac { 5 }{ y } =\frac { 1 }{ 4 } \)
put \(\frac { 1 }{ x } \) = s and \(\frac { 1 }{ y } \) = t
∴ 4s + 4t = \(\frac { 1 }{ 3 } \)
and 2s + 5t = \(\frac { 1 }{ 4 } \)
The matrix form of the system of equation is
\(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} s \\ t \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \) ⇒ AX = B where
A = \(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \)
X = A^-1B
Now |A| = \(\left| \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right| \) = 20 - 8 =12 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \left[ \frac { \begin{matrix} 1 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 4 \end{matrix} } \right] \)
=\(\frac { 1 }{ 12 } \left[ \begin{matrix} \frac { 5 }{ 3 } & -1 \\ \frac { -2 }{ 3 } & +1 \end{matrix} \right] \)
= \(\frac { 1 }{ 12 } \left[ \frac { \begin{matrix} 2 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 3 \end{matrix} } \right] =\left[ \begin{matrix} \frac { 2 }{ 3 } \times \frac { 1 }{ 12 } \\ \frac { 1 }{ 3 } \times \frac { 1 }{ 12 } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 18 } \\ \frac { 1 }{ 36 } \end{matrix} \right] \)
∴ \(\frac { 1 }{ 18 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 18 } \Rightarrow \)x = 18
t = \(\frac { 1 }{ 36 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 36 } \Rightarrow \)y = 36
one man can do 18 days
one woman can do 36 days.

Applying Gauss-Jordan method, we get
[A | I₂] = \(\left[ \begin{matrix} 0 & 5 \\ -1 & 6 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \frac { 1 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 1 \end{matrix}|\begin{matrix} 0 & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+6{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \).
So, we get A^-1 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).

Rearrange the terms as
(2x-1)(2x+ 3) (x + 3) (x - 2) + 20 = 0
⇒ (4x² + 6x - 2x- 3)(x² - 2x + 3x - 6) + 20 = 0
⇒ (4x² + 4x - 3) (x² + x - 6) + 20 = 0
put x²+ x = y
⇒ (4y - 3) (y - 6) + 20 = 0
⇒ 4y² - 24y - 3y + 18 + 20 = 0
⇒ 4y²-27y +38 = 0
⇒ (y - 2)( 4y - 19) = 0
\(y=2,\frac { 19 }{ 4 } \)

Case (i) 
When y = 2
x²+ x = 2
x² + x - 2 = 0
⇒ (x + 2)(x - 1) = 0
⇒ x = -2, 1
Case (ii) 
When \(y=\frac { 19 }{ 4 } ,{ x }^{ 2 }+x=\frac { 19 }{ 4 } \)
\(\Rightarrow { 4x }^{ 2 }+4x=19\)
\(\Rightarrow { 4x }^{ 2 }-4x-19=0\)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16-4(4)(-19) } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16+304 } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 320 } }{ 8 } \)
\(\Rightarrow x=\frac { 4\pm 8\sqrt { 5 } }{ 8 } \)
\(\Rightarrow x=\frac { -4(-1\pm 2\sqrt { 5 } ) }{ 8 } \)
\(\frac{-1 \pm 2 \sqrt{5}}{2}\)
Hence the roots are 1, -2, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

Let the encryption matrix be A =\(\left[ \begin{matrix} -1 & -1 \\ 2 & 1 \end{matrix} \right] \)
|A| = -1 + 2 = 1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 1 } \left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)
Hence the decryption matrix is \(\left[ \begin{matrix} 1 & 1 \\ -2 & -1 \end{matrix} \right] \)

So, the sequence of decoded row matrices is [8 5], [12 16]
Now the 8^th English alphabet is H.
5^th English alphabet is E.
12^th English alphabet is L.
and the 16^th English alphabet is P.
Thus the receiver reads the message as "HELP".

\(Let\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { sinx }{ 1+cosx } \right) dx } \)
\(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1+sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\left[ \because cosx=2{ cos }^{ 2 }x-1\Rightarrow 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } +\frac { sinx }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } +\frac { 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ { cos }^{ 2 }\frac { x }{ 2 } } \right) dx } \)
\(\\ [\because sin2x=2sinx\quad cosx]\)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }\left( { sec }^{ 2 }\frac { x }{ 2 } 2tan\frac { x }{ 2 } \right) dx } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { e }^{ x }(f'(x)+f(x)dx } \)
Where \(f(x)=2tan\frac { x }{ 2 } \)
\(=\frac { 1 }{ 2 } .{ e }^{ x }.f(x)={ \left[ \left( \frac { 1 }{ 2 } { e }^{ x }.2tan\frac { x }{ 2 } \right) \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ \left[ { e }^{ x }tan\frac { x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(={ e }^{ \frac { \pi }{ 2 } }tan\frac { \pi }{ 4 } -{ e }^{ 0 }tan0={ e }^{ \frac { \pi }{ 2 } }(1)\)
\(\therefore I={ e }^{ \frac { \pi }{ 2 } }\)

\({ h }({ x }_{ o })=\frac { x }{ 1+1 } =\frac { 1 }{ 2 } \)
\({ h }^{ ' }(x)=\frac { (x+1)(1)-x(1) }{ { (x+1) }^{ 2 } } \)
\(\frac { x+1-x }{ { (x+1) }^{ 2 } } =\frac { 1 }{ ({ x+1) }^{ 2 } } \)
\({ h }^{ ' }({ x }_{ o })=\frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ 4 } \)
∴ L(x) = h(x_o) + h'(x₀)(x - x_o)
\(=\frac { 1 }{ 2 } +\frac { 1 }{ 4 } (x-1)=\frac { 2+x-1 }{ 4 } =\frac { x+1 }{ 4 } \)
∴ L(x) = \(\frac { x+1 }{ 4 } \)

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

Let X be the random variable of number of tails when three coins tossed.
S = {HHH, HHT, THH, HTH, HTT, THT, TTH,TTT}
n(S) = 8
Let X denote the number of tarits occured.
X^-1 (0) {HHH} = 1
X^-1 (1) = {HHT, THT, HTH} = 3
X^-1 (2) =, {HTT, THT, TTH} = 3
X^-1 (3) = {TTT} = 1
ஃ X takes the values 0, 1, 2, 3.

This is an indeterminate form \(\frac{\infty}{\infty}\) and hence we use the l’Hôpital’s Rule to evaluate.
\(\underset{x\rightarrow 1^{-}}{lim}(\frac{log(1-x)}{cot(\pi x)})=\underset{x\rightarrow1^{-}}{lim}(\frac{-\frac{1}{1-x}}{-\pi cosec^{2}(\pi x)})\) \((\frac{\infty}{\infty})\) 
On Simplication,
\(=\underset{x\rightarrow-1}{lim}(\frac{sin^{2}(\pi x)}{\pi (1-x)})\) \((\frac{0}{0})\)
again applying the l’Hôpital Rule 
\(= \underset{x\rightarrow 1^{-}}{lim}(\frac{2\pi sin(\pi x). cos(\pi x)}{-\pi})\)
=\(\underset{x\rightarrow -1}{lim}(-2 sin(\pi x).cos (\pi x))\)
= 0.

\(f'\left( x \right) =6\times \frac { 4 }{ 3 } { x }^{ \frac { 4 }{ 3 } -1 }-3\times \frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } -1 }\) 
= \({ 8x }^{ \frac { 1 }{ 3 } }-x^{ \frac { -2 }{ 3 } }\) 
f'(x) = 0
\(\Rightarrow{ 8 }x^{ \frac { 1 }{ 3 } }-\frac { 1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow \frac { 8x-1 }{ { x }^{ \frac { 2 }{ 3 } } } =0\) 
\(\Rightarrow x=\frac { 1 }{ 8 } \)
Thus, the critical number is \(x=\frac { 1 }{ 8 } \) 
Evaluating f(x) at the end points = -1
x = 1 and at the critical number x = \(\frac { 1 }{ 8 } \) 
we get
\(f(-1)=6(-1)^{ \frac { 4 }{ 3 } }-3\left( -1 \right) ^{ \frac { 1 }{ 3 } }\) 
= 6( 1) - 3 (-1) = 6 + 3 = 9
\(f(1)=6(1)^{ \frac { 4 }{ 3 } }-3(1)^{ \frac { 1 }{ 3 } }=6-3=3\)
\(f\left( \frac { 1 }{ 8 } \right) =6\left( \frac { 1 }{ 8 } \right) ^{ \frac { 4 }{ 3 } }-3\left( \frac { 1 }{ 8 } \right) ^{ \frac { 1 }{ 3 } }\) 
=\(6\left( { 2 }^{ -3 } \right) ^{ \frac { 4 }{ 3 } }-3\left( 2^{ -3 } \right) ^{ \frac { 1 }{ 3 } }\) 
= \(\frac { 6 }{ 16 } -\frac { 3 }{ 2 } =\frac { 3 }{ 8 } -\frac { 3 }{ 2 } \) 
= \(\frac { 3-12 }{ 8 } =\frac { -9 }{ 8 } \)
From these values, the absolute maximum is 9 which occurs at x = -1 and the absolute minimum is \(-\frac { 9 }{ 8 } \) which occurs at x = \(\frac { 1 }{ 8 } \)

Let f (x) = x⁴ + 2x³-2
Then f (x) is continuous in [0, 1] and differentiable in (0, 1)  
Now, f'(x) = 4x³+6x² 
If f'(x) = 0, then
2x²(2x+3) = 0
Therefore,  \(x=0, -\frac{3}{2}\) but  \(0,-\frac{3}{2} \notin (0,1)\).
Thus, \(f'(x)>0, \forall x\in (0,1)\).
Hence by the Rolle’s theorem there do not exist \(a,b \in(0,1)\) such that, f(a) = 0 = f(b). Therefore the equation f(x ) = 0 cannot have two roots in the interval (0, 1) . But, f (0, 2) = −2 < 0 and f (1) = 1 > 0 tells us the curve y f = (x) crosses the x -axis between 0 and 1 only once by the Intermediate value theorem. Therefore the equation x⁴ + 2x³ − 2 = 0 has only one real root in the interval (0, 1) .

Observe that, f (2) = 0 = f (3) and f (x) is continuous in the interval [2, 3] and differentiable in (2, 3). Now,
\(f'(x)=\frac{x^{2}-6}{x(x^{2}+6)} \)
Therefore, \(f'{(c)}=0\) gives
\(\frac{c^{2}-6}{c(c^{2}+6)}=0\)
which implies \( c=\pm\sqrt{6}\)
Now c = \(\pm\sqrt{6} \in (2,3).\)
Observe that \(-\sqrt{6}\notin (2,3)\) and hence \(c=\pm\sqrt{6}\) satisfies the Rolle’s theorem.
Rolle’s theorem can also be used to compute the number of roots of an algebraic equation in an interval without actually solving the equation.

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Expending along R₁,
|A| = \(5\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| +1\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \)
= 5 (25 - 1)-1 (5 - 1)+ 1 (1 - 5)
= 5 (24) - 1(4) + 1(- 4)
= 120 - 4 - 4 = 120 - 8 = 112 ≠ 0
Since A is non singular, A^-1 exit
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 5 & 1 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(25-1)-(5-1)+(1-5) \\ -(5-1)+(25-1)-(5-1) \\ +(1-5)+(5-1)+(25-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -24 \\ -4 & -4 & 24 \end{matrix} \right] \)
Taking 4 common from every entry we get,
adj A = \(4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 112 } .4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
= \(\frac { 1 }{ 28 } \left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \).

Let \(\vec { \mu } \)  the position vector of an arbitrary point on the plane \(\vec { r } .(2\hat { i } -\hat { j } -2\hat { k } )\) = 6. Then, we have
\(\vec { \mu } .(2\hat { i } -\hat { j } -2\hat { k } )=6\) .........(1)
If δ is the distance between the given planes, then δ is the perpendicular distance from  \(\vec { \mu } \) to the plane
\(\vec { r } .(6\hat { i } -3\hat { j } -6\hat { k} )\) = 27
Therefore, δ = \(\frac { \left| \vec { u } .\vec { n } -p \right| }{ \left| \vec { n } \right| } =\left| \frac { \vec { u } .(6\hat { i } -3\hat { j } -6\hat { k } )-27 }{ \sqrt { 6^{ 2 }+(-{ 3) }^{ 2 }+(-{ 6 })^{ 2 } } } \right| =\left| \frac { 3(\vec { u } .(2\hat { i } -\hat { j } -2\hat { k } ))-27 }{ 9 } \right| =\left| \frac { (3(6)-27 }{ 9 } \right| =1\) unit

Let \(\vec { a } =-2\hat { i } +3\hat { j } +4\hat { k } \) and \(\vec { b } =-4\hat { i } +5\hat { j } -6\hat { k } \)
The parametric form of vector equation of a straight line passing through a point \((\vec { b } )\) and parallel to is \(\vec { b } \)is
\(\vec { r } =\vec { a } +t\vec { b } \) where \(t\in R\)
∴ \(\vec { r } =-2\hat { i } +3\hat { j } +4\hat { k } +t(-4\hat { i } +5\hat { j } -6\hat { k } ),t\in R\)
Its Cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
⇒ \(\frac { x+2 }{ -4 } =\frac { y-3 }{ 5 } =\frac { z-4 }{ -6 } \)
[∵ (x₁, y₁, z₁) is (-2, 3, 4) & (b₁, b₂, b₃) is (-4, -5, -6)]

Let −1−i = \(r(cos\ \theta +i\ sin\ \theta )\)
We have r = \(\sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
\(\alpha =tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }1=\frac { \pi }{ 4 } \)
Since the complex number −1−i lies in the third quadrant, it has the principal value,
\(\theta =\alpha -\pi =\frac { \pi }{ 4 } -\pi =-\frac { 3\pi }{ 4 } \)
Therefore,\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } \right) +isin\left( \frac { 3\pi }{ 4 } \right) \right) \)
= \(\sqrt { 2 } \left( cos\frac { 3\pi }{ 4 } -isin\frac { 3\pi }{ 4 } \right) \)
\(-1-i=\sqrt { 2 } \left( cos\left( \frac { 3\pi }{ 4 } +2k\pi \right) -isin\left( \frac { 3\pi }{ 4 } +2k\pi \right) \right) \)
Depending upon the various values of k , we get various alternative polar forms.