Given: $f(x)=\sqrt{9-x^2}$

Let, $y^2=9-x^2$

$\Rightarrow x^2=9-y^2$

$x=\sqrt{9-y^2}\dots$(1)

We know that for any function $f(x)=\sqrt{9-x^2},y\ge 0\dots$. (A)

From equation (1).

$9-y^2\ge 0$

$y^2-9\le 0$

$\Rightarrow (y+3)(y-3)\le 0$

$\Rightarrow -3\le y\le 3$

But from equation (A) y must be positive, then, $y=[0,3]$.

Range $=$ the value of $y=[0,3]$

$\begin{array}{rl}& {\mathrm{S}}_{\mathrm{k}}=\frac{\mathrm{k}+1}{2}\\ & {\mathrm{S}}_{\mathrm{k}}{}^2=\frac{{\mathrm{k}}^2+1+2\mathrm{k}}{4}\\ & \therefore \sum _{\mathrm{j}-1}^{\mathrm{n}}{\mathrm{S}}_{\mathrm{j}}{}^2=\frac{1}{4}[\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+\mathrm{n}+\mathrm{n}(\mathrm{n}+1)]\\ & =\frac{\mathrm{n}}{4}[\frac{(\mathrm{n}+1)(2\mathrm{n}+1)}{6}+1+\mathrm{n}+1]\\ & =\frac{\mathrm{n}}{4}[\frac{2{\mathrm{n}}^2+3\mathrm{n}+1}{6}+\mathrm{n}+2]\\ & =\frac{n}{4}\left[\frac{2n^2+9n+13}{6}\right]=\frac{n}{24}[2n^2+9n+13]\\ & A=24,B=2,C=9,D=13\\ & \end{array}$

Given inequality is,

$2\sqrt{{\sin }^{2}x-2\sin x+5}\le 2{\sin }^{2}y$

$\Rightarrow \sqrt{{\sin }^{2}x-2\sin x+5}\le 2{\sin }^{2}y$

$\Rightarrow \sqrt{(\sin x-1{)}^2+4}\le 2{\sin }^{2}y$

It is true if $\sin x=1$ and $|\sin y|=1$

Therefore, $\sin x=|\sin y|$

$\mathrm{I}={\int }_1^4\frac{\sqrt{x}+3}{\sqrt{x}}dx$

Let $\sqrt{x}+3=t$$\dots \dots (i)$

Differentiating w.r.t $x$, we get,

$\frac{1}{2\sqrt{x}}dx=dt$

$\frac{1}{\sqrt{x}}dx=2dt$

The new limits to $e{q}^n(i)$

When $x=1,\sqrt{1}+3=t$

$t=4$

Similarly,

When $x=4,t=5$

$\therefore {\int }_1^4\frac{\sqrt{x}+3}{\sqrt{x}}dx={\int }_4^{5}2tdt$

$=2\times {\left[\frac{t^2}{2}\right]}_4^5$

$=[t^2{]}_4^5$

Put the value of limit,

$=[5^2-4^2]$

$=9$

Given:

$A=\{2,3,4\};B=\{5,6\}$

We know that:

For any two non-empty sets $A$ and $B$, we have:

I. $A\times B=\{(a,b)\mid a\in A$ and $b\in B\}$

II. $B\times A=\{(b,a)\mid a\in A$ and $b\in B\}$

III. Any two ordered pairs $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$.

Then,

$A\times B=\{(2,5),(2,6),(3,5),(3,6),(4,5),(4,6)\}$

The number of elements in $A\times B$ i.e.,

$n(A\times B)=6$

Therefore, the number of subsets of$A\times B=2^n$

$=2^6$

$=64$

Therefore, there are 64 subsets for $A\times B$.

Intersection:

Let A and B be two sets. The intersection of A and B is the set of all those elements which are present in both sets A and B.

The intersection of A and B is denoted by A ∩ B

i.e., A ∩ B = {x : x ∈ A and x ∈ B}

The Venn diagram for intersection is as shown below:

Union:

Let A and B be two sets. The union of A and B is the set of all those elements which belong to either A or B or both A and B.

The union of A and B is denoted by A ∪ B.

i.e., A ∪ B = {x : x ∈ A or x ∈ B}

The Venn diagram for the union of any two sets is shown below:

A ∪ B = A + B - A ∩ B

As we know, 

P ∪ Q = P + Q - P ∩ Q

Putting the values given in the question,

P = P + Q -  P ∩ Q

P ∩ Q = Q

Let $3^x=y$

$\therefore \mathrm{y}(\mathrm{y}-1)+2=|\mathrm{y}-1|+|\mathrm{y}-2|$

Case 1: when $\mathrm{y}>2$

${\mathrm{y}}^2-\mathrm{y}+2=\mathrm{y}-1+\mathrm{y}-2$

${\mathrm{y}}^2-3\mathrm{y}+5=0$

$\because \mathrm{D}<0[\therefore \text{ Equation not satisfy. ] }$

Case 2 : when $1\le \mathrm{y}\le 2$

${\mathrm{y}}^2-{\mathrm{y}}^2+2=\mathrm{y}-1-\mathrm{y}+2$

${\mathrm{y}}^2-\mathrm{y}+1=0$

$\because \mathrm{D}<0[\therefore \text{ Equation not satisfy. }]$

Case 3: when $\mathrm{y}\le 1$

${\mathrm{y}}^2-\mathrm{y}+2=-\mathrm{y}+1-\mathrm{y}+2$

${\mathrm{y}}^2+\mathrm{y}-1=0$

$\therefore \mathrm{y}=\frac{-1+\sqrt{5}}{2}$

$=\frac{-1-\sqrt{5}}{2}[\therefore \text{ Equation not Satisfy }]$

$\because$ Only one $-1+\frac{\sqrt{5}}{2}$ satisfy equation

Given word is : ASHUTOSH

Total 8 letters are there, in which letter $S$ and $H$ are repeated twice.

We know that:Number of Permutations of '$n$' things taken '$r$' at a time:$p(n,r)=\frac{n!}{(n-r)!}$

Number of Permutations of ‘$n$’ objects where there are $n_1$ repeated items, $n_2$ repeated items, $n_k$ repeated items taken ‘$r$’ at a time:$p(n,r)=\frac{n!}{n_1!n_2!n_3!\dots n_k!}$

Therefore,The number of arrangements will be:

$p(8,2)=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2!}{(2\times 1)2!}$$=10080$

Therefore, the number of arrangements of letters in the word ASHUTOSH will be $10080$

If the constraints in a linear programming problem are changed the problem is to be re-evaluated.The linear inequalities or equations or restrictions on the variables of a linear programming problem are called constraints. The conditions x ≥ 0, y ≥ 0 are called non-negative restrictions.