X B(n, p)
Given 4P(X = 4) = P(X = 2) and n = 6.
4. [⁶C₄p⁴ (1 - p)²] = ⁶C₂p² q⁴
⇒ 4p² = q² 
⇒ 4(1-q²) = q²  
4(1-2q+q²) = q²  
⇒ 3q² - 8q +4 = 0
⇒ (q - 2)(3q - 2) = 0
\( -q=\frac { 2 }{ 3 } \ \ \ (q\neq2)\) 
\(p = 1- q=\cfrac { 1 }{ 3 } \)  
Distribution
P(X = x) = nC_xp^x (1- p)^n-x, x = 0, 1,2, ... n
(i) \(P(X=x)= ^6C_{ x }\left( \frac { 1 }{ 3 } \right) ^{ x }\left( \frac { 2 }{ 3 } \right) ^{ 6-x }\) , x = 0,1,2..n
(ii) \(mean=np=6\times \frac { 1 }{ 3 } =2\) 
(iii) standard deviation = \(\sqrt { npq } =\sqrt { 2 \times \frac {2}{ 3 } } \) 
= \( { \frac { 2 }{ \sqrt 3 } } \)

First let us find w_x, w_y, and w_z
Now w_x = 2xy + z², w_y = 2yz +x² and w_z = 2zx + y².
Thus,by (15), the differential is
dw = (2xy + z² )dx + (2yz + x² )dy+ (2zx + y² )dz.

Let \(f(x)={ sin }^{ 2 }x\)
\(f(-x)={ (sin(-x)) }^{ 2 }={ sin }^{ 2 }x=f(x)\)
\(\therefore \int _{ -\frac { \pi }{ 4 } }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } =2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { sin }^{ 2 }xdx } \)
\(cos\ 2x=1-2{ sin }^{ 2 }x\)
\(2{ sin }^{ 2 }x=1-cos2\)
\({ sin }^{ 2 }x=\frac { 1-cos\quad 2x }{ 2 } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( \frac { 1-cos2x }{ 2 } \right) dx } \)
\(=\frac { 2 }{ 2 } { \left[ x-\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\)
\(=\frac { \pi }{ 4 } -\frac { sin\frac { \pi }{ 4 } }{ 2 } -0+\frac { sin0 }{ 2 } \)
\(\\ =\frac { \pi }{ 4 } -\frac { 1 }{ 2 } =\frac { \pi -2 }{ 4 } \)

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

Equation of the given curve is y² - 4xy = x² + 5..(1)
Differentiating with respect to 'x' we get,
\(2y\frac { dy }{ dx } -4\left[ x\frac { dy }{ dx } +y(1) \right] =2x\)
⇒ \(2y\frac { dy }{ dx } -4x\frac { dy }{ dx } -4y=2x\)
⇒ \(\frac { dy }{ dx } \) (2y -4x) = 2x + 4y
⇒ \(\frac { dy }{ dx } \) = \(\frac { x+2y }{ y-2x } \)
Since the tangent to the curve is horizontal, \(\frac { dy }{ dx } \) = 0 
ஃ \(\frac { x+2y }{ y-2x } \) = 0
⇒ x+ 2y = 0
⇒ x = -2y .....(2)
Substituting (2) in (1) we get,
y² - 4(-2y)y = (-1y)² + 5
⇒ y² + 8y² = 4y² + 5
⇒ y² = 4y² + 5
⇒ 5y² = 5
⇒ y² = 1
⇒ y = 土 1
From (2), When y = 1, x = - 2
When y = -1, x = 2
∴ The required points are (2, -1) and (-2,1)

Given 2cos α = x+\(\frac { 1 }{ x } \)
⇒ 2cos α = \(\frac { { x }^{ 2 }+1 }{ x } \)
⇒ x²+1 = 2xcos α
⇒ x²-2x cos α+1 = 0
⇒ \(\frac { 2cos\alpha \pm \sqrt { (-2cos\alpha )^{ 2 }-4(1)(1) } }{ 2 } \)
= \(\frac { 2cos\alpha \pm \sqrt { 4cos^{ 2 }\alpha -4 } }{ 2 } \)  \(\left[ \because \frac { b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
= \(\frac { 2cos\alpha \pm \sqrt { -sin^{ 2 }\alpha } }{ 2 } \)
= \(\frac { 2cos\alpha \pm isin\alpha }{ 2 } \) [∵ sin²α+cos²α = 1]
⇒ x² = cos α ± sin α
Also, 2cos β = y+\(\frac { 1 }{ y } \)
⇒ 2cos β = \(\frac { { y }^{ 2 }+1 }{ y } \)
⇒ y²-2y cos β+1 = 0
⇒ \(\frac { 2cos\beta \pm \sqrt { (-2cos^{ 2 }\beta ^{ 2 }-4(1)(1) } }{ 2 } \)
=\(\frac { 2cos\beta \pm \sqrt { 4cos^{ 2 }\beta } -4 }{ 2 } =\frac { 2cos\beta \pm 2isin\beta }{ 2 } \)
⇒ y = cosβ ± i sinβ
\({ x }^{ m }{ y }^{ n }+\cfrac { 1 }{ { x }^{ m }{ y }^{ n } } =2cos(m\alpha +n\beta )\)
x^myⁿ = (cos α + i sin mα) (cos nβ + i sin nβ)
cos(mα+nβ)+i sin(mα+nβ)
\(\frac { 1 }{ { x }^{ m }{ y }^{ n } } \) = cos(mα+nβ)-i sin(mα+nβ)

= 2cos(mα+nβ)

  
Vector area of quadrilateral ABCD
= vector area of ΔABC + vector area of ΔACD
\(\frac { 1 }{ 2 } (\vec { AB } \times \vec { AC } )+\frac { 1 }{ 2 } (\vec { AC } \times \vec { AD } )\)
= \(-\frac { 1 }{ 2 } (\vec { AC } \times \vec { AB } )+\frac { 1 }{ 2 } (\vec { AC } \times \vec { AD } )\)
\(\left[ \because \ \vec { b } \times \vec { a } =-(\vec { a } \times \vec { b } ) \right] \)
= \(\frac { 1 }{ 2 } \vec { AC } \times (-\vec { AB } +\vec { AD } )\)
= \(\frac { 1 }{ 2 } \vec { AC } \times (\vec { BA } +\vec { AD } )\)  \([\because \vec { AB } =-\vec { BA } ]\)
= \(\frac { 1 }{ 2 } \vec { AC } \times \vec { BD } \) [By Δ law of addition]
∴ Area of the quadrilateral ABCD = \(\frac { 1 }{ 2 } \vec { AC } \times \vec { BD } \)

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.

The roots of x³+2x2+3x+4 = 0 are ∝, β, ૪
∴ ∝+β+૪ = -co-efficient of x² = -2 ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3 ....(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4 ...(3)
Form a cubic equation whose roots are 2∝, 2β, 2૪
2∝+2β+2૪ = 2(∝+β+૪) = 2(-2) = -4 [from (1)]
4∝β+4β૪+4૪∝ = 4(∝β+β૪+୪∝) = 4(3) = 12 [from (2)]
(2∝)(2β)(2૪) = 8(∝β૪) = 8(-4) = -32 [from (3)]
∴ The required cubic equation is
x³-(2∝+2β+2૪)x² + (2∝β+2β૪+2୪∝)x - (2∝)(2β)(2૪) = 0
⇒ x³+(-4)x²+12x+32 = 0
⇒ x³+4x²+12x+32 = 0

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

a *b = min (a, b) on A = {1,2,3,4, 5}  Let a,b ∈A
A = {1,2, 3, 4, 5}
a*b = min {(a, b)}
Now, 1,2 ∈ A \(\Rightarrow\)1 * 2 = 1 ∈ A
3, 5 ∈ A \(\Rightarrow\) 3 * 5 = 3 ∈ A
Hence * is a binary operation on A

Given that \(\frac{dy}{dx}+2y\) = e^-x
This is a linear differential equation
Here P = 2 ; Q = e^−x.
\(\int { pdx } =\int { 2dx } =2x\)
Thus, I.F.\(={ e }^{ \int { pdx } }={ e }^{ 2x }\)
Hence the solution of (1) is \({ ye }^{ \int { pdx } }=\int { { Qe }^{ \int { Pdx } }dx+C } \)
That is, \({ ye }^{ 2x }=\int { { e }^{ -x }{ e }^{ 2x }dx+C } or\quad { ye }^{ 2x }={ e }^{ x }+C\quad or\quad y={ e }^{ -x }+{ Xe }^{ -2x }\) required solution

Given y = x⁴ + 2x² - x
\(\frac { dy }{ dx } \) = 4x³ + 4x - 1
Slope of the tangent at x = 1 is
m = \(\left( \frac { dy }{ dx } \right) \)(x = 1)
= 4(1)³+ 4 (1) - 1
= 4+4-1 = 7
∴ m = 7

\(y\left( \frac { dy }{ dx } \right) =\frac { x }{ \left( \frac { dy }{ dx } \right) +{ \left( \frac { dy }{ dx } \right) }^{ 3 } } \)
is the given differential equation.
\(\Rightarrow y{ \left( \frac { dy }{ dx } \right) }^{ 2 }+{ \left( \frac { dy }{ dx } \right) }^{ 4 }=x\)
The highest derivative is 1 and its maximum power is 4.
∴ Order 1, degree 4.

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Using the given value for z₁ and z₂ the value of \(\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { 3-2i }{ 6+4 } =\frac { 3-2i }{ 6+4i } \times \frac { 6-4i }{ 6-4i } \)
= \(\frac { \left( 18-8 \right) +i\left( 12-12 \right) }{ { 6 }^{ 2 }+{ 4 }^{ 2 } } =\frac { 10-24i }{ 52 } =\frac { 10 }{ 52 } =\frac { 24i }{ 52 } \)
= \(\frac { 5 }{ 26 } -\frac { 6 }{ 13 } i\)

Given M = \(\left\{ \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) :x\in R-\{ 0\} \right\} \) and * be the matrix multipilication.
Let A \(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \) and B = \(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \in M\)
Where x, y ∈R-{0}.
\(A*B=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \)
\(\\ =\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \in M\)
[∵ 2xy ∈R-{0}]
∴ M is closed under *.
Commutative property:
we know A*B =\(\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) ..(1)\)
Let x,y∈R-{0}
Now B + A \(=\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
\(=\left( \begin{matrix} xy+xy & xy+xy \\ xy+xy & xy+xy \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) \\ \)
From (1) &(2), A*B = B*A
∴ *has commutative property on M
Associative property:
Let A = \(\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) \)
B =\(\left( \begin{matrix} y & y \\ y & y \end{matrix} \right) \) and
C = \(\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
for x, y, z ∈R-{0}
\((A*B)*C=\left( \begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix} \right) *\left( \begin{matrix} z & z \\ z & z \end{matrix} \right) \)
\(=\left( \begin{matrix} 2xyz+2xyz & 2xyz+2xyz \\ 2xyz+2xyz & 2xyz+2xyz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(1)\)
Now\(A*(B*C)=A*\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} x & x \\ x & x \end{matrix} \right) *\left( \begin{matrix} 2yz & 2yz \\ 2yz & 2yz \end{matrix} \right) \)
\(=\left( \begin{matrix} 4xyz & 4xyz \\ 4xyz & 4xyz \end{matrix} \right) ...(2)\\ \)
From (1)&(2), (a*B)*C = A*B*C)
Since matrix multiplication is associative, this axiom holds good for M.

Also the above result can be proved without using truth tables. This proof will be provided after studying the logical equivalence

Given f(x, y) = \(\frac { 3x }{ y+sinx \ } \)
\({ f }_{ x }=\frac { \partial f }{ \partial x } \)
= \(\frac { (y+sin \ x)(3)-3x(cos \ x) }{ { (y+sin \ x) }^{ 2 } } \)
\(=\frac { 3y+3sin \ x-3x \ cos \ x }{ (y+sin{ x })^{ 2 } } \)
\({ f }_{ yx }=\frac { \partial ^{ 2 }f }{ \partial y\partial x } \)
\(=\frac { (y+{ sinx) }^{ 2 }[3]-(3y+3sinx-3xcosx)(2)(y+sinx)(1) }{ { (y+sinx) }^{ 3 } } \)
= \(\frac { (y+{ sinx) }-[3y+3sinx-6y-6sinx+6xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ yx }=\frac { -3y-3sinx+6xcosx }{ { (y+sinx) }^{ 3 } } \) ......(1)
\({ f }_{ y }={ [3x[-1][y+sinx] }^{ -2 }\)
\(=\frac { -3x }{ (y+sin{ x) }^{ 2 } } \)
\(\therefore { f }_{ xy }=-3\left[ \frac { ({ y+sinx) }^{ 2 }(1)-x(2)(y+sinx)(cosx) }{ { (y+sinx) }^{ 4 } } \right] \)
\({ f }_{ xy }=\frac { -3(y+sinx)[y+sinx-2xcosx] }{ { (y+sinx) }^{ 4 } } \)
\({ f }_{ xy }=\frac { -3(y+sinx-2xcosx) }{ ({ y+sinx })^{ 3 } } \) ...(2)
From (1) and (3)
f_xy = f^yx

First we shall calculate \(\frac { \partial F }{ \partial x } \)(x, y) then we evaluate it at (−1, 3) As we have already observed we find the derivative with respect to x holding y as a constant. That is,
\(\frac { \partial f }{ \partial x } (x,y)=\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial x } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial x } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial x } +\frac { \partial (7) }{ \partial x } \)
= 3x² y + y² +0
= 3x² y + y² .
so, \(\frac { \partial F }{ \partial x } \) ( -1, 3) = 3( -1)² 3 + 3² = 18.
Next similarly we find partial derivative with respect to y.
\(\frac { \partial F }{ \partial y } \) (x, y) = \(\frac { \partial \left( { x }^{ 3 }y+{ y }^{ 2 }x+7 \right) }{ \partial y } =\frac { \partial \left( { x }^{ 3 }y \right) }{ \partial y } +\frac { \partial \left( { y }^{ 2 }x \right) }{ \partial y } +\frac { \partial (7) }{ \partial y } \)
= x³ + 2yx + 0
= x³ + 2yx.
Hence we have \(\frac { \partial F }{ \partial y } \) (-2, 1) = (-2)³ + 2(1)( -2) = -12.
Note that in the above example \(\frac { \partial F }{ \partial x } \) (x, y) = 3x² y + y² which is again a function of two variables. 
So, we can take the partial derivative of this function with respect to x or y.
For instance, if we take G(x, y) = 3x² y+y² then we find  \(\frac { \partial F }{ \partial x } \) = 6xy. Since G(x, y) = \(\frac { \partial F }{ \partial x } \), we have \(\frac { \partial G }{ \partial x } \)=\(\frac { \partial }{ \partial x } \)\(\left( \frac { \partial f }{ \partial x } \right) \) = 6xy.
We denote this as \(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } \) which is called the second order partial derivative of F with respect to x. 
Also, \(\frac { \partial F }{ \partial y } \) = 3x² + 2y. Since G (x, y) = \(\frac { \partial F }{ \partial x } \) we have  \(\frac { \partial G }{ \partial y } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \) = 3x² + 2y. 
We denote this as \(\frac { { \partial }^{ 2 }F }{ \partial y\partial x } \) which is called the mixed partial derivative of F with respect to x, y. 
Similarly we can also calculate \(\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \right) \) = 3x²+2y.
Also, if we differentiate \(\frac { \partial F }{ \partial y } \) partially with respect to y we obtain \(\frac { { \partial }^{ 2 }F }{ { \partial y }^{ 2 } } \) which is called the second order partial derivatives of F with respect to y.
So for any function F defined on any subset {(x, y) | a < x < b, c < y < d} ⊂ R² we have the following notation
\(\frac { { \partial }^{ 2 }F }{ { \partial x }^{ 2 } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ xx' }\frac { { \partial }^{ 2 }F }{ { \partial x\partial y } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ xy }\)
\(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) ={ F }_{ yx' }\frac { { \partial }^{ 2 }F }{ { { \partial y }^{ 2 } } } =\frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial y } \right) ={ F }_{ yy }\)
All the above are called second order partial derivatives of F. 
Similarly we can define higher order partial derivatives.
For example, \(\frac { { \partial }^{ 2 }F }{ { \partial y\partial x } } =\frac { \partial }{ \partial y } \left( \frac { \partial }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) and \(\frac { { \partial }^{ 2 }F }{ { \partial x\partial y\partial x } } =\frac { \partial }{ \partial x } \left( \frac { \partial F }{ \partial y } \left( \frac { \partial F }{ \partial x } \right) \right) \) 
Next we shall see more examples on partial differentiation.

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\sqrt { tan\ x } +\sqrt { cot\ x } )dx } \) Then we get
\(I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( \sqrt { \frac { sinx }{ cosx } } +\sqrt { \frac { cosx }{ sinx } } \right) dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { sinxcosx } } dx } =\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx+cosx }{ \sqrt { 2sinxcosx } } dx } \)
\(=\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{2 } }{ \frac { (sinx+cosx)dx }{ \sqrt { 1-sinx-cosx{ ) }^{ 2 } } } } \)
Put u = sin x − cos x 
Then, du = (cos x + sin x)dx .
When x = 0, u = −1
When x =\(\frac{\pi}{2}\), u = 1
\(\therefore I=\sqrt { 2 } \int _{ -1 }^{ 1 }{ \frac { du }{ \sqrt { 1-{ u }^{ 2 } } } =\sqrt { 2 } { [{ sin }^{ -1 }u] }_{ -1 }^{ 1 }=\sqrt { 2 } \left[ { sin }^{ -1 }(1)-{ sin }^{ -1 }(-1)) \right] } =\pi \sqrt { 2 } \)

\(\frac { dy }{ dx } +\frac { 3y }{ x } =\frac { 1 }{ { x }^{ 2 } } \), given that y = 2 when x = 1 
This is a linear differential equation.
\(\therefore P=\frac { 3 }{ x } ;Q=\frac { 1 }{ { x }^{ 2 } } \)
\(\int { pdx } =3\int { \frac { 1 }{ x } dx=3logx=log{ x }^{ 3 } } \)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ x }^{ 3 } }={ x }^{ 3 }\)
\(\therefore\) The solution is \({ ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow { yx }^{ 3 }=\int { \frac { 1 }{ { x }^{ 2 } } .{ x }^{ 3 } } dx+c\)
\(\Rightarrow { yx }^{ 3 }=\int { xdx+c } \)
\(\Rightarrow { yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +c...(1)\)
When x = 1, y = 2
\(\Rightarrow 2{ (1) }^{ 3 }=\frac { 1 }{ 2 } +c\Rightarrow 2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } \)
\({ yx }^{ 3 }=\frac { { x }^{ 2 } }{ 2 } +\frac { 3 }{ 2 } \)
\({ 2x }^{ 3 }y={ x }^{ 2 }+3\)

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 \((\vec { a } \times \vec { b } )\times \vec { c } \)\(​​\neq \)\((\vec { a } \times \vec { b } )\times \vec { c } \)
 

Let P(x) be the polynomial under consideration.
The number of sign changes for P(x) and P(−x) are zero and hence it has no positive roots and no negative roots. Clearly zero is not a root. Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots.

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 \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

Equation of the circle is x² + y² − 6x + 6y − 8 = 0
∴ Equation of the tangent at (x₁, y₁) is
xx₁ + yy₁ - \(\frac 62\) (x + x₁) + \(\frac 62\) (y + y₁) - 8 = 0
Given (x₁, y₁) is (2, 2)
Equation of the tangent at (2, 2) is
x(2) + y(2) - 3(x + 2) + 3(y + 2) - 8 = 0
\(\Rightarrow 2 x-2 y-3 x+\not 6+3 y+\not 6-8=0\)
⇒ -x + 5y - 8 = 0
⇒ x − 5y + 8 = 0
Equation of the normal is
yx₁ - xy₁ + g(y - y₁) - f(x - x₁) = 0
⇒ y(2) -x(2) - 3(y - 2) -3(x - 2) = 0
∵ 2g = -6
⇒ g = -3
2f = 6
⇒ f = 3
⇒ 2y - 2x - 3y + 6 - 3x + 6 = 0
⇒ -5x - y + 12 = 0
Another method for Normal:
Equation of tangent is perpendicular to normal
x - 5y + 8 = 0
Perpendicular equation be 5x + y + k = 0
At (2, 2)
10 + 2 + k = 0
k = -12
Therefore 5x + y - 12 = 0 be equation of normal.

Given F (\(\alpha\)) = \(\left[ \begin{matrix} \cos { \alpha } & 0 & \sin { \alpha } \\ 0 & 1 & 0 \\ -\sin { \alpha } & 0 & \cos { \alpha } \end{matrix} \right] \)
Expanding along R₁ we get,
|F(\(\alpha\))| = cos \(\alpha\) \(\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| -0+sin\alpha \left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \)
= cos \(\alpha\) (cos - 0) + sin \(\alpha\) (0 + sin \(\alpha\))
= cos² + sin² \(\alpha\) = 1 ≠ 0
Since F (\(\alpha\)) is a non-singular matrix, [F(\(\alpha\))]^-1 exists
Now, adj (F(\(\alpha\))) = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ sin\alpha & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & sin\alpha \\ 0 & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & 0 \\ sin\alpha & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & sin\alpha \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & sin\alpha \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & 0 \\ 0 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(cos\alpha -0) & -(0) & +(0+sin\alpha ) \\ -(0) & +(cos^{ 2 }\alpha +sin^{ 2 }\alpha & -(0) \\ +(0-sin\alpha ) & -(0) & +(cos-0) \end{matrix} \right] ^{ T }\)
\(\left[ \begin{matrix} cos\alpha & 0 & +sin\alpha \\ 0 & 1 & 0 \\ -sin\alpha & 0 & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
∴ F(\(\alpha\))^-1 = \(\frac { 1 }{ |F(\alpha )| } \) adj (F(\(\alpha\)))
[F(\(\alpha\))]^-1 = \(\frac { 1 }{ 1 } \left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(1)
Now, F(-\(\alpha\))=\(\left[ \begin{matrix} cos(-\alpha ) & 0 & sin(-\alpha ) \\ 0 & 1 & 0 \\ -s9n(-\alpha ) & 0 & cos(-\alpha ) \end{matrix} \right] \)
=\(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(2)
[∵ cos \(\alpha\) is an even function, cos (-\(\alpha\)) = cos \(\alpha\) and sin \(\alpha\) is an odd function, sin (-\(\alpha\)) = -sin\(\alpha\)]
From (1) and (2)
[F(\(\alpha\))]^-1 = F (-\(\alpha\))

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)