P(0.2≤X < 0.6)
= \(\int _{ 0.2 }^{ 0.6 }{ f(x)dx } =\int _{ 0.2 }^{ 0.6 }{ x.dx } =\left[ \frac { { x }^{ 2 } }{ 2 } \right] _{ 0.2 }^{ 0.6 }\)  
= 0.18 - 0.02
= 0.16

\(\left| \left( \overline { 1+i } \right) \left( 2+3i \right) \left( 4i-3 \right) \right| =\left| \left( \overline { 1+i } \right) \right| \left| 2+3i \right| \left| 4i-3 \right| \) (\(\because \) |z₁z₂z₃|=|z₁|z₂||z₃|)
= |1+i| |2+3i| |-3+4i|  \(\left( \because |z|=\left| \overline { z } \right| \right) \)
= \(\left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 } } \right) \left( \sqrt { \left( 3 \right) ^{ 2 }+{ 4 }^{ 2 } } \right) \).
\(=(\sqrt{2})(\sqrt{13})(\sqrt{25})=5 \sqrt{26}\)

2z+3w
= 2(5-2i)+3(-1+3i)
= 10-4i-3+9i
= (10-3)+ i(-4+9)
= 7+5i

\(\frac { 2i }{ 3+4i } \)
Let z = \(\frac { 2i }{ 3+4i } \)
|z| = \(\left| \frac { 2i }{ 3+4i } \right| =\frac { |2i| }{ |3+4i| } =\frac { \sqrt { 2^{ 2 } } }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =\frac { 2 }{ \sqrt { 9+16 } } \)
= \(\frac { 2 }{ \sqrt { 25 } } =\frac { 2 }{ 5 } \).

To find the real and imaginary parts of \(\frac { 3+4i }{ 5-12i } \) first it should be expressed in the rectangular form x + iy .
To simplify the quotient of two complex numbers, multiply the numerator and denominator by the conjugate of the denominator to eliminate i in the denominator
\(\frac { 3+4i }{ 5-12i } =\frac { (3-4i)(5+12i) }{ (5-12i)(5+12i) } \)
= \(\frac { (15-48)+(20+36)i }{ { 5 }^{ 2 }+{ 12 }^{ 2 } } \)
= \(\frac { -33+56i }{ 169 } =-\frac { 33 }{ 169 } +i\frac { 56 }{ 169 } \)
Therefore,\(\frac { 3+4i }{ 5-12i } =-\frac { 33 }{ 169 } +i\frac { 56 }{ 169 } \) This is in the x + iy form.
Hence real part is \(-\frac { 33 }{ 169 } \) and imaginary part is \(\frac { 56 }{ 169 } \)

\(Let\quad t=2x\Rightarrow dt=2dx\Rightarrow \frac { dt }{ 2 } =dx\)
\(\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { sin }^{ 6 }t\quad dt=\frac { 1 }{ 2 } { I }_{ 6 } } \)

\([\int _{ 0 }^{ \pi /2 }{ { sin }^{ n }xdx=\frac { n-1 }{ n } { I }_{ n-2 }, } n\ge 2]\)
\(\\ =\frac { 1 }{ 2 } \times \frac { 5 }{ 6 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 2 } \times \frac { \pi }{ 2 } =\frac { 5\pi }{ 64 } \)

Let A = \(\left[ \begin{matrix} 4 & 3 \\ -3 & -1 \\ 6 & 7 \end{matrix}\begin{matrix} 1 & -2 \\ -2 & 4 \\ -1 & 2 \end{matrix} \right] \). Then A is a matrix of order 3 \(\times\) 4. So ρ(A)  ≤ min {3, 4} = 3.
The highest order of minors of A is 3. We search for a non-zero third-order minor of A. But we find that all of them vanish. In fact, we have
\(\left| \begin{matrix} 4 & 3 & 1 \\ -3 & -1 & -2 \\ 6 & 7 & -1 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 3 & -2 \\ -3 & -1 & 4 \\ 6 & 7 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 4 & 1 & -2 \\ -3 & -2 & 4 \\ 6 & -1 & 2 \end{matrix} \right| \) = 0; \(\left| \begin{matrix} 3 & 1 & -2 \\ -1 & -2 & 4 \\ 7 & -1 & 2 \end{matrix} \right| \) = 0.
So, ρ(A) < 3. Next, we search for a non-zero second-order minor of A.
We find that \(\left| \begin{matrix} 4 & 3 \\ -3 & -1 \end{matrix} \right| \) = -4 + 9 = 5 ≠ 0. So, ρ(A) = 2.

LHS =  \(\cfrac { a+b\omega +c{ \omega }^{ 2 } }{ b+c\omega +{ a\omega }^{ 2 } } +\cfrac { a+b\omega +{ c\omega }^{ 2 } }{ c+a\omega +b{ \omega }^{ 2 } } \)
\(\cfrac { a\omega ^{ 3 }+b\omega +c{ \omega }^{ 2 } }{ b+c\omega +{ a\omega }^{ 2 } } +\cfrac { a\omega ^{ 3 }+b\omega ({ \omega }^{ 3 })+{ c\omega }^{ 2 } }{ c+a\omega +b{ \omega }^{ 2 } } \) [∵ ω³ = 1]

= ω + ω² = -1              [ ∵ 1 + ω + ω² = 0]

Let p(-x) = x⁹- 5x⁵ + 4x⁴ + 2x² + 1 = 0
P(x) has only one sign change. It has atmost one positive roots.
Also p(-x) = (-x)⁹ - 5(-x)⁵ + 4(-x)4 +2(-x)² +1 = 0
p(-x) = -x⁹+ 5x⁵+4x⁴+ 2x²+1 = 0
It has only one sign change.
It has atmost one negative roots.
Clearly 0 is not a root.
So maximum number of real roots is 3 and hence there are atleast six imaginary solutions.

\(y=\frac { 1 }{ 2 } log(x+1)+(tan^{ -1 }+x)^{ 2 }+c\)

prove.

(i) m + n∈Z, ∀m, n∈Z. Hence + is a binary operation on Z.
(ii) Also m + n = n + m,∀m, n∈Z. So the commutative property is satisfied
(iii) ∀m, n, p∈Z, m+ (n + p) = (m+ n) + p. Hence the associative property is satisfied.
(iv) m + e = e + m = m ⇒ e = 0. Thus ヨ 0∈Z⋺(m+ 0) = (0 + m) = m. Hence the existence of identity is assured.
(v) m + m' = m'+ m = 0 ⇒ m' = −m. Thus ∀∈Z,ョ−m∈Z ⋺ m+ (−m) = (−m) + m = 0. Hence, the existence of inverse property is also assured. Thus we see that the usual addition + on Z satisfies all the above five properties.

Given z(x, y) = x³ - 3x²y³ ; x = se^t; y = se^-t
\(\frac { \partial z }{ \partial s } \) = 3x² - 6xy³; \(\frac { \partial z }{ \partial y } \) = -9x²y²
∴ \(\frac { \partial z }{ \partial s } \) 3s²e^2t - 6 se^t s³ e^-3t
\(\frac { \partial z }{ \partial y } \) = 9s²e^2t.s²e^-2t
\(\frac { \partial z }{ \partial y } \) = -9s⁴
\(\frac { \partial z }{ \partial s} \) = e^t; \(\frac { \partial y }{ \partial s} \) = e^-t
\(\frac { \partial x }{ \partial t} \) = se^t; \(\frac { \partial y }{ \partial s} \) = -se^-t 
By chain rule;
\(\therefore \frac { \partial z }{ \partial s } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial u }{ \partial y } .\frac { dy }{ dx } \)
= (3s²e^2t - 6 s⁴ e^-2t) (e^t) + (-9s⁴) (e^-t)
= 3s²e^3t - 6 s⁴ e^-t -9s⁴ e^-t
\(\frac { \partial z }{ \partial s } \) = 3s²e^3t - 15 s⁴ e^-t
By chain rule;
\(\frac { \partial z }{ \partial t } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
= (3s²e^2t - 6 s⁴e-^2t) (se^t) + (-9 s⁴) (-se^-t)
= 3s³e^3t - 6 s⁵e^-t + 9 s⁵ e^-t
= 3s³ e^3t + 3 s⁵e^-t
= 3s³(e^3t + s²e^-t)

By definition the cumulative distribution function for discrete random variable is
\(F(x)P\left( X\le x \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ 1 })\) 
\(P(X<1)=0\) for -∞
\(F(1)=P\left( X\le 1 \right) =\underset { { x }_{ 1 }\le x }{ \Sigma } P(X={ x }_{ i })=\sum _{ -\infty }^{ 1 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) =0+\frac { 1 }{ 12 } =\frac { 1 }{ 12 } \)
\(F(2)=P\left( X\le 2 \right) =\sum _{ -\infty }^{ 2 }{ P\left( X=x \right) } =P\left( X\le 1 \right) +P\left( X=1 \right) +P\left( X=2 \right) \) 
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } =\frac { 1 }{ 2 } \) 
\(F(3)=P\left( X\le 3 \right) =\sum _{ -\infty }^{ 3 }{ P(X=x) } =P\left( X<1 \right) +P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) \)
= \(0+\frac { 1 }{ 2 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } =\frac { 11 }{ 12 } \) 
\(F(4)=P\left( X\le 4 \right) =\sum _{ -\infty }^{ 4 }{ P\left( X=x \right) } =P\left( X=1 \right) +P\left( X=2 \right) +P\left( X=3 \right) +P(X=4)\)
= \(0+\frac { 1 }{ 12 } +\frac { 5 }{ 12 } +\frac { 5 }{ 12 } +\frac { 1 }{ 12 } =1\)
\(F(x)= \begin{cases}0, & -\infty

(ii) \(P(X\le 3)=F(3)\frac { 11 }{ 12 } \)
(iii) \(P(X\ge 2)=1-P\left( X<2 \right) =1-P(X\le 1)=1-F(1)=1-\frac { 1 }{ 12 } =\frac { 11 }{ 12 } \)

Given w (x, y, z) = x² +y² +y², 
x = e^t, y = e^t sin t, z = e^t cos t
\(\frac { \partial u }{ \partial x } \) = 2x; \(\frac { \partial u }{ \partial y} \) = 2y; \(\frac { \partial u }{ \partial z} \) = 2z
\(\frac { \partial u }{ \partial x } \) = 2e^t
\(\frac { \partial u }{ \partial y} \) = 2e^t sin t
\(\frac { \partial u }{ \partial z} \) = 2e^t cos t
\(\frac{dx}{dt}\) = e^t
\(\frac{dy}{dt}\) = e^t cas t + sin t e^t
⇒ \(\frac{dz}{dt}\) = et (- sin t ) + cos t et
By chain rule
\(\frac { dw }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { dw }{ \partial y } .\frac { dy }{ dt } +\frac { \partial w }{ \partial z } .\frac { dz }{ dt } \)
∴ \(\frac { dw }{ dt } \) = 2e^t(e^t) + 2e^t sin t (e^t cos t +sin t e^t) - e^t sin t + 2e^t cas t (e^t cos t - e^t sin t )

= e^2t [2 + 2] = 4e^2t

x cos y dy = e^x(x log x + 1)dx
\(\Rightarrow cos\ y\ dy={ e }^{ x }\frac { (x\ log+1) }{ x } dx\)
\(=\left[ { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) \right] dx\)
\(\Rightarrow \int { cos\quad y\quad dy } =\int { { e }^{ x }\left( log\quad x+\frac { 1 }{ x } \right) dx } \)
Taking integration on both sides, we get 
 \( \int \cos y d y=\int e^x\left[\log x+\frac{1}{x}\right] d x  \)
 RHS 
\( \int e^x\left[\log x+\frac{1}{x}\right] d x\)
 Take \(\mathrm{f}(x)=\log x \Rightarrow f^{\prime}(x)=\frac{1}{x} \)
This of the form  \(\int e^x\left[f(x)+f^{\prime}(x)\right] d x=e^x f(x)+C \)
 \(\therefore \int e^x\left[\log x+\frac{1}{x}\right] d x=\mathrm{e}^x \log x+\mathrm{C}\)
Substituting in (1), we get 
\(\sin y=e^x \log x+C  \)