Mathematics Test - 28

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

Let, \(y^{2}=9-x^{2}\)

\(\Rightarrow x^{2}=9-y^{2}\)

\(x=\sqrt{9-y^{2}} \quad\quad\ldots\)(1)

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

From equation (1).

\(9-y^{2} \geq 0\)

\(y^{2}-9 \leq 0\)

\(\Rightarrow(y+3)(y-3) \leq 0\)

\(\Rightarrow-3 \leq y \leq 3\)

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

Range \(=\) the value of \({y}=[0,3]\)

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

Given inequality is,

\(2 \sqrt{\sin ^2 x-2 \sin x+5} \leq 2 \sin ^2 y \\\)

\( \Rightarrow \sqrt{\sin ^2 x-2 \sin x+5} \leq 2 \sin ^2 y \\\)

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

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

Therefore, \(\sin x=|\sin y|\)

\(\mathrm{I}=\int_{1}^{4} \frac{\sqrt{x}+3}{\sqrt{x}} d x\)

Let \(\sqrt{x}+3=t\)\( \quad \ldots \ldots(i)\)

Differentiating w.r.t \(x\), we get,

\(\frac{1}{2 \sqrt{x}} d x=d t\)

\(\frac{1}{\sqrt{x}} d x=2 d t\)

The new limits to \(eq^{n}(i)\)

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

\(t=4\)

Similarly,

When \(x=4, t=5\)

\(\therefore \int_{1}^{4} \frac{\sqrt{x}+3}{\sqrt{x}} d x=\int_{4}^{5} 2t d t\)

\(={2}\times\left[\frac{t^{2}}{2}\right]_{4}^{5}\)

\(=[t^{2}]_{4}^{5}\)

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 \(1 \leq \mathrm{y} \leq 2\)

\(\mathrm{y}^2-\mathrm{y}^2+2=\mathrm{y}-1-\mathrm{y}+2 \\\)

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

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

Case 3: when \(\mathrm{y} \leq 1\)

\(\mathrm{y}^2-\mathrm{y}+2=-\mathrm{y}+1-\mathrm{y}+2 \\\)

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

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

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

\(\because\) 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:\(p(n, r)=\frac{n !}{n_{1} ! n_{2} ! n_{3} ! \ldots n_{k} !}\)

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.