Let f(a) = b , then \(5^{44-6}=b\)
\(\Rightarrow a^{2}-4 a=\log _{1} b \rightarrow a^{t}-4 a-\ln g_{0} b=0\)
\(\Rightarrow a=\frac{4 \pm \sqrt{16+4 \log _{0} b}}{2}\)
Hence \(f^{-1}(\alpha)=2+\sqrt{4+\log _{t} a}\)

We have, x = \(\frac { at }{ 1+{ t }^{ 2 } } ,y=\frac { { at }^{ 2 } }{ 1+{ t }^{ 2 } } \)
\(\frac { dx }{ dt } =\frac { { (1+t) }^{ 2 }a-at(2t) }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\Rightarrow\)  \(\frac { dx }{ dt } =\)\(\frac { a+a{ t }^{ 2 }-2a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
=\(\frac { a-a{ t }^{ 2 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
and \(\frac { dy }{ dt } =\frac { { (1+{ t }^{ 2 } })^{ 2 }2at-a{ t }^{ 2 }(2t) }{ { (1+{ t }^{ 2 } })^{ 2 } } \)
\(\Rightarrow\)\(\frac { dy }{ dt } =\frac { 2at+2a{ t }^{ 3 }-2a{ t }^{ 3 } }{ { (1+{ t }^{ 2 }) }^{ 2 } } =\frac { 2at }{ { (1+{ t }^{ 2 }) }^{ 2 } } \)
\(\therefore\) \(\frac { dy }{ dx } =\frac { 2at }{ { (1+{ t }^{ 2 } })^{ 2 } } \times \frac { { (1+{ t }^{ 2 } })^{ 2 } }{ a-a{ t }^{ 2 } } \)
\(=\frac { 2at }{ a-a{ t }^{ 2 } } =\frac { 2t }{ 1-{ t }^{ 2 } } \)
\({ (\frac { dy }{ dx } ) }_{ att=2 }\) = \(\frac { 2(2) }{ 1-4 } \)
= -4/3

Given, equations of line is \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{8-z}{-6}\) or \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z-8}{6}\) and the point is (-2, 4, -5)
Now, cartesian equations of the required line is \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)   ...(i)
[\(\because\) in parallel lines, direction ratios are proportional] and vector equation of the required line is
\(\vec{r}=(-\hat{2} i+\hat{4} j-\hat{5} k)+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

\(\frac { dy }{ dx } +y={ e }^{ -x }\)
Let \(\frac { dy }{ dx } +Py=Q\)
Where P = 1, Q = e^-x
\(y\left( I.F. \right) =\int { I.F.\times Qdx } \)
\(\Rightarrow I.F.={ e }^{ \int { Pdx } }={ e }^{ \int { Pdx } }={ e }^{ x }\)
\(\Rightarrow y{ e }^{ x }=\int { { e }^{ x }.{ e }^{ -x } } dx=\int { dx } \)
\(\Rightarrow y{ e }^{ x }=x+C\)

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=\overset\wedge i+2\overset\wedge j+\overset\wedge k+3\overset\wedge i+\overset\wedge j-\overset\wedge k\)
\(\Rightarrow \overset\rightarrow c=4\overset\wedge i+3\overset\wedge j\)
\(\Rightarrow \overset\rightarrow d=\overset\rightarrow a-\overset\rightarrow b\)
=(i+2j+k)-(3i+j-k)
\(\Rightarrow\overset\rightarrow d =-2\overset\wedge i+\overset\wedge j+2\overset\wedge k\)
Projection \(\overset\rightarrow c \) on \(\overset\rightarrow d\)\(=\frac{\overset\rightarrow c.\overset\rightarrow d}{\left| \overset\rightarrow d \right| }\)
\(=\frac{(4\overset\wedge i+3\overset\wedge j).(-2\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{\left|-2\overset\wedge i+\overset\wedge j+2\overset\wedge k \right| }\)
\(=\frac{-8+3}{\sqrt{4+1+4}}=-\frac{5}{3}\)

Since E and F are independent, we have
P(E ∩ F) = P(E) . P(F) ....(1)
From the venn diagram in Fig 13.3, it is clear that E ∩ F and E ∩ F′ are mutually exclusive events and also E =(E ∩ F) ∪ (E ∩ F′). Therefore P(E) = P(E ∩ F) + P(E ∩ F′)
or P(E ∩ F′) = P(E) − P(E ∩ F)
= P(E) − P(E) . P(F) (by (1))
= P(E) (1−P(F))
= P(E). P(F′)
Hence, E and F′ are independent

\(R.H.S.=2{ sin }^{ -1 }\sqrt { \frac { 1-x }{ 2 } } \quad \quad \begin{cases} Let\quad x=cos\theta \\ \Rightarrow \theta ={ cos }^{ -1 }x \end{cases}\)
\(=2{ sin }^{ -1 }\left( \sqrt { \frac { 1-cos\theta }{ 2 } } \right) \)
\( \begin{cases} as\quad cos\theta =1=2{ sin }^{ 2 }\frac { \theta }{ 2 } \\ \Rightarrow 1-cos\theta =2sin^{ 2 }\frac { \theta }{ 2 } \end{cases}\)
\(=2{ sin }^{ -1 }\left( sin\frac { \theta }{ 2 } \right) \)
\(=2\left( \frac { \theta }{ 2 } \right) =\theta ={ cos }^{ -1 }x\)
L.H.S = R.H.S

\(\int { { cos }^{ 3 } } xdx=\frac { 1 }{ 4 } \int { (cos } 3x+3cosx)dx\)
\([\because cos3x=4cos^{ 3 }x-3cosx]\)
\(=\frac { 1 }{ 4 } \int { cos3x } dx+\int { \frac { 3 }{ 4 } } cosxdx\)
\(=\frac { 1 }{ 4 } \left( \frac { sin3x }{ 3 } \right) +\frac { 3 }{ 4 } sinx+C\)
 \(=\frac { 1 }{ 12 } sin3x+\frac { 3 }{ 4 } sinx+C\)

A function \(f:A\rightarrow B\) is said to be one-one if
\(a\neq b\)
\(\Rightarrow f(a)\neq f(b)for\quad all\quad a,b,\in A\)
or f(a) = f(b)
a = b for all a,b,A
In other words, is one-one, if no two elements of A have same image, i.e., no two elements of A are mapped to same elements of B.

Let A be a skew-symmetric matrix. Then by definition \({ A }^{ \prime }=-A\)
\(\Rightarrow\) the (i, j)^th element of \({ A }^{ \prime }\) = the (i, j)^th element of (- A)
\(\Rightarrow\) the (j, i)^th element of A = - the (i, j)^th element of A
For the diagonal elements i = j \(\Rightarrow\) the (i, j)the element of A = - the (i, j)^th element of A.
\(\Rightarrow\)  the (i, j)^th element of A = 0
Hence the diagonal elements are all zero.

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

\({ A }^{ 2 }=\left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=l;{ A }^{ 6 }=({ A }^{ 2 })^{ 3 }={ l }^{ 3 }=l \)

\( \int _{ 2 }^{ 5 }{ (\left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| ) } dx,....(1)\)
Now \(\int _{ 2 }^{ 5 }{ \left| x-2 \right| } dx=\int _{ 2 }^{ 5 }{ (x-2) } dx\)
\(={ \left[ \frac { { x }^{ 2 } }{ 2 } -2x \right] }_{ 2 }^{ 5 }=\left( \frac { 25 }{ 2 } -10 \right) -\left( \frac { 4 }{ 2 } -4 \right) =\frac { 5 }{ 2 } +2=\frac { 9 }{ 2 } .\)
And \(\int _{ 2 }^{ 5 }{ \left| x-3 \right| } dx=\int _{ 2 }^{ 3 }{ \left| x-3 \right| } dx+\int _{ 3 }^{ 5 }{ \left| x-3 \right| } dx\)
\(=\int _{ 2 }^{ 3 }{ -(x-3) } dx+\int _{ 3 }^{ 5 }{ (x-3) } dx\)
\(={ \left[ \frac { { x }^{ 2 } }{ 2 } -3x \right] }_{ 2 }^{ 3 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -3x \right] }_{ 3 }^{ 5 }\)
\(=-\left[ \left( \frac { 9 }{ 2 } -9 \right) -\left( \frac { 4 }{ 2 } -6 \right) \right] +\left[ \left( \frac { 25 }{ 2 } -15 \right) -\left( \frac { 9 }{ 2 } -9 \right) \right] \)
\(=-\left[ -\frac { 9 }{ 2 } +4 \right] +\left[ -\frac { 5 }{ 2 } +\frac { 9 }{ 2 } \right] \)
\(=\frac { 1 }{ 2 } +2=\frac { 5 }{ 2 } \)
Finally \(\int _{ 2 }^{ 5 }{ \left| x-5 \right| } dx=\int _{ 2 }^{ 5 }{ -(x-5) } dx\)
\(=-{ \left[ \frac { { x }^{ 2 } }{ 2 } -5x \right] }_{ 2 }^{ 3 }\)
\(=-\left[ \left( \frac { 9 }{ 2 } -15 \right) -\left( \frac { 4 }{ 2 } -10 \right) \right] \)
\(=-\left[ -\left( \frac { 21 }{ 2 } +8 \right) \right] =\frac { 5 }{ 2 } .\)
ஃ From (1), \(\int _{ 2 }^{ 5 }{ [\left| x-2 \right| +\left| x-3 \right| +\left| x-5 \right| ] } dx\)
\(=\frac { 9 }{ 2 } +\frac { 5 }{ 2 } +\frac { 5 }{ 2 } \frac { 19 }{ 2 } .\)

The region determined by the constraints, x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0 is as follows.

There is no feasible region and thus, Z has no maximum value.

Let \(x_{ 1 },x_{ 2 }\in R\)
Now \(f(x_{ 1 })f(x_{ 2 })\Rightarrow ax_{ 1 }+b=ax_{ 2 }+b\)
\(\Rightarrow \) \(ax_{ 1 }=ax_{ 2 }\Rightarrow x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) 'f' is one-one.
Let \(y\in R.\) Let \(y=f(x_{ 0 })\)
Then \(ax_{ 0 }+b=y\Rightarrow x_{ 0 }=\frac { y-b }{ a } \Rightarrow x_{ 0 }\in R\).
\(f(x_{ 0 })=ax_{ 0 }+b=a\left( \frac { y-b }{ a } \right) +b=y.\)
\(\therefore \) For each \(y\in R\) , there exists \(x_{ 0 }\in R\) such that \(f(x_{ 0 })=y\).
\(\therefore \) 'f' is onto.
Hence, 'f' is a bijection.

Let 'P' be the population at any time.
By the question, \(\frac { dP }{ dt } =kP\)
\(\Rightarrow\) \(\frac { dP }{ P } =k\quad dt\) 
|Variables Separable
Integrating,  \(log\quad P=kt+C\)...(1) [\(\because\)P>0]
Let  \(t=0\) referred to \(1999\)
When  \(t=0,\quad P=20000,\)
\(\therefore\) \(log\quad 20000=0+C\)                         
\(\Rightarrow\) \(C=log\quad 20000\)..(2)
When  \(t=5,P=25000,\)
\(\therefore \)  \(log\quad 25000=5k\quad +C\)
\(\Rightarrow\)  \(log\quad 25000=5k+log20000\)
\(\Rightarrow\)  \(5k=log\frac { 25 }{ 20 } \Rightarrow k=\frac { 1 }{ 5 } log\frac { 5 }{ 4 } \)
Using (2) and (3) in (1), we get:
\(logP=\frac { 1 }{ 5 } \left( log\frac { 5 }{ 4 } \right) t+log20000\)
When \(t=10,\)
then \(log\quad P\quad =\frac { 1 }{ 5 } \left( log\frac { 5 }{ 4 } \right) \left( 10 \right) +log20000\)
\(=2log\frac { 5 }{ 4 } +log 20000\)
\(=log\frac { 25 }{ 16 } +log20000=log\frac { 25 }{ 16 } \times 20000\)
\(=log\quad 31250.\)
hence, the population was 31,250 in 2009

We have: x sin(a + y) + sin a cos(a+y) = 0 ......(1)
\(x=-\frac { sin\quad a\quad cos(a+y) }{ sin\quad (a+y) } ...(2)\)
Diff.w.r.t.x,
\(x\ cos(a+y)\left( 0+\frac { dy }{ dx } \right) +sin(a+y).1+sin\quad a(-sin(a+y))\left( 0+\frac { dy }{ dx } \right) =0\)
\(\left[ x\ cos(a+y)-sin\ a\ sin(a+y) \right] \frac { dy }{ dx } =-sin(a+y)\)
\(=\left[ -\frac { sin\quad a\quad cos(a+y) }{ sin\quad (a+y) } cos(a+y)-sinasin(a+y) \right] \frac { dy }{ dx } \)
\(=-sin(a+y)\)
\(=-sina\left[ { cos }^{ 2 }(a+y)+{ sin }^{ 2 }(a+y) \right] \frac { dy }{ dx } \)
\(=-{ sin }^{ 2 }(a+y)\)
\(=-sin\quad a(1)\frac { dy }{ dx } =-{ sin }^{ 2 }(a+y)\)
Hence, \(\frac { dy }{ dx } =\frac { { sin }^{ 2 }(a+y) }{ sin\quad a } \) which is true.

The given equation are:
5x-y+4z=5
2x+3y+5z=2
5x-2y+6z=-1
Here A= \(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
Now |A|=\(\begin{bmatrix}5&-1&4\\2&3&5\\5&-2&6 \end{bmatrix}\)
=5(18+10)+1.(12-25)+4(-4-15)
=140-13-76=\(51\neq0\)
Hence, the given system of equations is consistent.

The correct answer is C.
It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.
Therefore, \(A={ \left[ { a }_{ ij } \right] }_{ m\times n }\) is a square matrix, if m = n.

As three given lines are mutually perpendicular to each other, therefore,
\(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0 \)
\(l_{1} l_{3}+m_{1} m_{3}+n_{1} n_{3}=0 \)
\(l_{2} l_{3}+m_{2} m_{3}+n_{2} n_{3}=0 \)
Let a be angle between lines with direction cosines
\(\frac{l_{1}+l_{2}+l_{3}}{\sqrt{3}}, \frac{m_{1}+m_{2}+m_{3}}{\sqrt{3}}, \frac{n_{1}+n_{2}+n_{3}}{\sqrt{3}} \text { and } l_{1}, m_{1}, n_{1},\)
\(\text { then, } \cos \alpha=l_{1}\left(\frac{l_{1}+l_{2}+l_{3}}{\sqrt{3}}\right)+m_{1}\left(\frac{m_{1}+m_{2}+m_{3}}{\sqrt{3}}\right) +n_{1}\left(\frac{n_{1}+n_{2}+n_{3}}{\sqrt{3}}\right) \)
\(\Rightarrow \cos \alpha=\frac{l_{1}^{2}+m_{1}^{2}+n_{1}^{2}}{\sqrt{3}} \)
\(cos\alpha =\frac { 1 }{ \sqrt { 3 } } \) Similarly, find angle with other lines.

\(log\left| cosec\frac { y }{ x } -cot\frac { y }{ x } \right| =-log\left| x \right| =logc\Rightarrow x\left[ cosec\frac { y }{ x } -cot\frac { y }{ x } \right] =c\) is the required solution.

Let the events be:
E₁ : Item from machine A
E₂ : Item from machine B
And A : Item is defective.
\(P({ E }_{ 1 })=\frac { 60 }{ 100 } =\frac { 3 }{ 5 } \)
\(P({ E }_{ 2 })=\frac { 40 }{ 100 } =\frac { 2 }{ 5 } \)
Also   \(P(A/{ E }_{ 1 })=\frac { 2 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 1 }{ 100 } \)
By Bayes' Theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { \left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) }{ \left( \frac { 3 }{ 5 } \right) \left( \frac { 2 }{ 100 } \right) +\left( \frac { 2 }{ 5 } \right) \left( \frac { 1 }{ 100 } \right) } \)
\(=\frac { 2 }{ 6+2 } =\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \)

The required area  of the region BOB′RFSB is enclosed by the ellipse and the lines x = 0 and x = ae.
Note that the area of the region BOB′RFSB
\(=2 \int_{0}^{a e} y d x=2 \frac{b}{a} \int_{0}^{a e} \sqrt{a^{2}-x^{2}} d x \)
\(=\frac{2 b}{a}\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{0}^{a e} \)
\(=\frac{2 b}{2 a}\left[a e \sqrt{a^{2}-a^{2} e^{2}}+a^{2} \sin ^{-1} e\right] \)
\(=a b\left[e \sqrt{1-e^{2}}+\sin ^{-1} e\right] \)

Let numbers be x, y, z then
x + y + z = -1
2y + 3z = 5
x + y – z = -1
\(x=-\frac { 7 }{ 2 } \), y = \(\frac { 5 }{ 2 } \), z = 0

Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as:
\(a * b=\left\{\begin{array}{ll} a+b, & \text { if } a+b<6 \\ a+b-6 & \text { if } a+b \geq 6 \end{array}\right.\)
An element e ∈ X is the identity element for the operation *, if a * e = a = e * a ∀ a ∈ X
For a ∈ X we observed that
a * 0 = a + 0 =a   [a ∈ X ⇒ a + 0 < 6]
0 * a = 0 + a = a        [a ∈ X ⇒ 0 + a < 6]
a * 0 = a = 0 * a ∀ a ∈ X
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b∈ X such that a * b = 0 = b * a.
ie \(\left\{\begin{array}{ll} a+b=0=b+a & \text { if } a+b<6 \\ a+6-6=0=b+a-6 & \text { if } a+b \geq 6 \end{array}\right.\)
i.e.,
a = −b or b = 6 − a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ −b.

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved

\({ A }^{ -1 }=\left[ \begin{matrix} 1/a & 0 & 0 \\ 0 & 1/b & 0 \\ 0 & 0 & 1/c \end{matrix} \right] \)
The inverse is impossible. It is possible when a, b, c are all real and\(\ne\)0, otherwise |A| becomes singular
i.e., |A| = 0
and \({ A }^{ -1 }=\frac { adj\quad A }{ |A| } \)
will become meaningless.
Value: Anybody can progress till he continues working, otherwise his progress will be halted.

Total number of oranges = 25
number of good oranges = 20
number of bad oranges = 5
Probability of getting a bad orange,
\(P(B)=\frac { 5 }{ 25 } =\frac { 1 }{ 5 } \)
Now  \(p=\frac { 1 }{ 5 } \)
\(\\ q=1-p=\frac { 4 }{ 5 } \)
Let X be the random variable of "Number of bad oranges".
\(\Rightarrow X=0,1,2,3,4\)
\(P(X=0)={ n }_{ { C }_{ r } }{ \left( p \right) }^{ n-r }{ \left( q \right) }^{ r }\)
\(P(X=0)={ 4 }_{ { C }_{ 0 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 4 }=\frac { 256 }{ 625 } \)
\(P(X=1)=4\times \frac { 64 }{ 125 } \times \frac { 1 }{ 5 } =\frac { 256 }{ 625 } \)
\(P(X=2)={ 4 }_{ { C }_{ 2 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 }{ \left( \frac { 1 }{ 5 } \right) }^{ 2 }\)
\(=6\times \frac { 16 }{ 625 } =\frac { 96 }{ 625 } \)
\(P(X=3)={ 4 }_{ { C }_{ 3 } }\left( \frac { 4 }{ 5 } \right) { \left( \frac { 1 }{ 5 } \right) }^{ 3 }\)
\(=4\times \frac { 4 }{ 625 } =\frac { 16 }{ 625 } \)
\(P(X=4)={ 4 }_{ { C }_{ 4 } }{ \left( \frac { 4 }{ 5 } \right) }^{ 0 }{ \left( \frac { 1 }{ 5 } \right) }^{ 4 }\)
\(=\frac { 1 }{ 625 } \)
\(\therefore\)   Probability distribution is

\(\therefore Mean=\sum { XP(X) } \)
\(=\frac { 500 }{ 625 } =\frac { 20 }{ 25 } =\frac { 4 }{ 5 } \)
\(Var.\quad (X)=E({ X }^{ 2 })-[E(X){ ] }^{ 2 }\)
\(=\frac { 800 }{ 625 } -\frac { 16 }{ 25 } \)
\(Var.(X)=\frac { 32 }{ 25 } -\frac { 16 }{ 25 } =\frac { 16 }{ 25 } \)

Let  \(I=\int { \frac { sinx }{ sin^{ 3 }x+cos^{ 3 }x } } dx\)
\(=\int { \frac { tanxsec^{ 2 }x }{ { tan }^{ 3 }x+1 } } dx\)
On substituting tanx = t and sec²x = dx = dt, we get
\(I=\int { \frac { t }{ { t }^{ 3 }+1 } } dt\)
\(=\int { \frac { t }{ (t+1)({ t }^{ 2 }-t+1) } } dt\)
\(=-\frac { 1 }{ 3 } \int { \frac { 1 }{ t+1 } dt+\frac { 1 }{ 3 } } \int { \frac { t+1 }{ { t }^{ 2 }-t+1 } dt } \)
\(=-\frac { 1 }{ 3 } log\left| t+1 \right| +\frac { 1 }{ 6 } \int { \frac { (2t-1)+3 }{ { t }^{ 2 }-t+1 } } dt+\frac { 1 }{ 2 } \int { \frac { 1 }{ { t }^{ 2 }-t+1 } } dt\)
\(=-\frac { 1 }{ 3 } log\left| t+1 \right| +\frac { 1 }{ 6 } log\left| { t }^{ 2 }t+1 \right| +\frac { 1 }{ 2 } \int { \frac { 1 }{ \left( t-\frac { 1 }{ 2 } \right) ^{ 2 }+\left( \frac { \sqrt { 3 } }{ 2 } \right) ^{ 2 } } } dt\)
\(=-\frac { 1 }{ 3 } log\left| t+1 \right| +\frac { 1 }{ 6 } log\left| { t }^{ 2 }-t+1 \right| +\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 2t-1 }{ \sqrt { 3 } } \right) \)
\(=-\frac { 1 }{ 3 } log\left| tanx+1 \right| +\frac { 1 }{ 6 } log\left| { tan }^{ 2 }x-tanx+1 \right| +\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 2tanx-1 }{ \sqrt { 3 } } \right) +C\)

f(x) = sin3x - cos3x, 0 < x < π 
\(\Rightarrow f'(x)=3\cos3x+3\sin3x\)
\(=3(\cos3x+\sin3x)\)
Put \(f'(x)=0\)
\(\Rightarrow \cos3x+\sin3x=0\)
\(\Rightarrow \sin3x=-\cos3x\)
\(\Rightarrow -\tan3x=1\)
\(\Rightarrow \tan3x=-1\)
As 0\(\pi\), 0 < 3x > 3\(\pi\)
ஃ tan 3x is negative for the following values:
\(3x=\frac{3\pi}{4}\)
\(\Rightarrow x=\frac{\pi}{4}\)
\(3x=\pi+\frac{3\pi}{4}=\frac{7\pi}{4}\)
\(\Rightarrow x=\frac{7\pi}{12}\)
\(3x=\frac{7\pi}{4}+\pi=\frac{11\pi}{4}\)
\(\Rightarrow x=\frac{11\pi}{12}\)
Hence we have intervals:
Hence, \(f(x)-\sin3x-\cos3x\) is strictly increasing in the intervals \((0,\frac{\pi}{4})\cup(\frac{7\pi}{12},\frac{11\pi}{12}) \) and strictly decreasing in intervals \((\frac{\pi}{4},\frac{7\pi}{12})\cup(\frac{11\pi}{12},\pi)\)

\(=2\tan ^{ -1 }{ \left\{ \tan { \frac { \alpha }{ 2 } } \tan { \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } \right\} } \)
\(=\tan ^{ -1 }{ \frac { 2\tan { \frac { \alpha }{ 2 } } .\tan { \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } }{ 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \tan ^{ 2 }{ \left( \frac { \pi }{ 4 } -\frac { \beta }{ 2 } \right) } } } \)   \(\left[ \because \quad 2\tan ^{ -1 }{ x } =\tan ^{ -1 }{ \frac { 2x }{ 1-{ 2x }^{ 2 } } } \right] \)
\(=\tan ^{ -1 }{ \frac { 2\tan { \frac { \alpha }{ 2 } } \left[ \frac { 1-\tan { \frac { \beta }{ 2 } } }{ 1+\tan { \frac { \beta }{ 2 } } } \right] }{ 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } { \left( \frac { 1-\tan { \frac { \beta }{ 2 } } }{ 1+\tan { \frac { \beta }{ 2 } } } \right) }^{ 2 } } } } \)
\(\left[ \because \tan { (A-B) } =\frac { \tan { A } -\tan { B } }{ 1+\tan { A } \tan { B } } \right] \)
\(=\tan ^{ -1 }{ \left( \frac { 2\tan { \frac { \alpha }{ 2 } } \left[ \frac { \left( 1-\tan { \frac { \beta }{ 2 } } \right) \left( 1+\tan { \frac { \beta }{ 2 } } \right) }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } \right] }{ \frac { { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 }-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } { \left( 1-\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 2\tan { \frac { \alpha }{ 2 } } .\left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ { \left( 1+\tan { \frac { \beta }{ 2 } } \right) }^{ 2 }-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } { \left( 1-\tan { \frac { \beta }{ 2 } } \right) }^{ 2 } } \right) } \)
\(=\tan ^{ -1 }{ \left[ \frac { 2\tan { \frac { \alpha }{ 2 } } .\left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) +2\tan { \frac { \beta }{ 2 } } \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } \right] } \)
\(=\tan ^{ -1 }{ \left[ \frac { \frac { 2\tan { \frac { \alpha }{ 2 } } }{ \left( 1+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } .\frac { \left( 1-\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) } }{ \frac { 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } }{ \left( 1-\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) } +\frac { 2\tan { \frac { \beta }{ 2 } } }{ \left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) } } \right] } \)
\(=\tan ^{ -1 }{ \frac { \sin { \alpha } +\cos { \beta } }{ \cos { \alpha } +\sin { \beta } } } \)
[Dividing N^, and D ^' by \(\left( 1+\tan ^{ 2 }{ \frac { \alpha }{ 2 } } \right) \)\(\left( 1+\tan ^{ 2 }{ \frac { \beta }{ 2 } } \right) \)]
= RHS

We know that:
A square matrix \(A=\left[ { a }_{ ij } \right] \) is said to be skew-symmetric if the (i, j)^th element of
A = -(j, i)^th element of A i.e., \({ a }_{ ij }=-{ a }_{ ji }\) for all i, j.
\(\therefore \ { a }_{ ii }=-{ a }_{ ii }\) for all i
\(\Rightarrow 2{ a }_{ ii }=0 \Rightarrow \quad { a }_{ ii }=0\).
Hence, the elements on the main diagonal of skew-symmetric matrix are all zero.

\(\Rightarrow { y }^{ ' }=\frac { log\quad x }{ \left( 1+log\quad x \right) ^{ 2 } } \)