The given equation is,

$=(\cos 2p\pi +i\sin 2p\pi )(\cos 2q\pi +i\sin 2q\pi )$

$=\cos 2(p+q)\pi +i\sin 2(p+q)\pi$

$=(\cos \pi +i\sin \pi {)}^{2(p+q)}$

$=(-1+0{)}^{2(p+q)}$

$=(-1{)}^{2(p+q)}=1$

Sample space contains \(2^{4}=16\) elements.

The favorable events are:

\(\{ H , H , H , T \},\{ H , H , T , H \},\{ H , T , H , H \},\{ T , H , H , H \},\{ H , H , H , H \}\)

There are 5 possibilities,

\(\therefore\) The required probability is \(\frac{5}{16}\).

Given,

\(|z-1|=|z+i|\)

Let \(x=x+\) iy

\(|x+i y-1|=|x+i y+1|\)

\(|(x-1)+i y|=|x+i(y+1)|\)

Squaring both sides

\(2(x-1)({iy})+({x}-1)^{2}+{y}^{2} {i}^{2}={x}^{2}+{i}^{2}({y}+1)^{2}+2 {xi}({y}+1)\)

\(\Rightarrow({x}-1)^{2}-{y}^{2}+2 {xyi}-2 {yi}={x}^{2}-{y}^{2}-1-2 {y}+2 {xyi}+2 {xi}\)

\(\Rightarrow x^{2}+1-2 x-y^{2}-2 y i=x^{2}-y^{2}-2 y+2 x i-1\)

\(\Rightarrow 2 y-2 x+2=2 x i+2 y i\)

\(\Rightarrow y-x+1=i(x+y)=0\)

\(\Rightarrow {y}-{iy}-{x}-{xi}+1=0\)

\(\Rightarrow y(1-i)-x(1+i)+1=0\)

Therefore it can be written as:

by \(-a x+1=0 \quad((1-i)=b\) and \((1+i)=a\) as they are constant)

\(\therefore\) It represent the straight line passing through origin.

Let set A and B have m and n elements, respectively.

2^m−2ⁿ = 56

2ⁿ(2^m−n−1) = 56  = 8×7 = 2³×7

Comparing both sides, we get

2ⁿ = 2³ and 2^m−n = 7

⇒ n = 3 and 2^m−n = 8

⇒ 2^m−n = 2³ ⇒ m-n = 3

⇒ m−3 = 3 ⇒ m = 6

Number of the elements in A is 6.

Let the position vector of \(A\) and \(C\) be \(\vec{a}\) and \(\vec{c}\) respectively. Therefore, Position vector of \(B=\vec{b}=\vec{a}+\vec{c} ~~\dots\)(i)

Also Position vector of \(E=\frac{\vec{b}+2 \vec{c}}{3}=\frac{\vec{a}+3 \vec{c}}{3}~~\dots\)(ii)

Now point \(P\) lies on angle bisector of \(\angle AOC\). Thus,

Position vector of point \(P=\lambda\left(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{b}}{|\vec{b}|}\right)~~\dots\)(iii)

Also let \(P\) divides \(EA\) in ration \(\mu: 1\). Therefore,

Position vector of \(P\)

\(=\frac{\mu \vec{a}+\frac{\vec{a}+3 \vec{c}}{3}}{\mu+1}=\frac{(3 \mu+1) \vec{a}+3 \vec{c}}{3(\mu+1)}~~\dots\)(iv)

Comparing (iii) and (iv), we get

\(\lambda\left(\frac{\vec{a}}{|\vec{a}|}+\frac{\vec{c}}{|\vec{c}|}\right)=\frac{(3 \mu+1) \vec{a}+3 \vec{c}}{3(\mu+1)}\)

\(\Rightarrow \frac{\lambda}{|\vec{a}|}=\frac{3 \mu+1}{3(\mu+1)}\) and \(\frac{\lambda}{|\vec{c}|}=\frac{1}{\mu+1}\)

\(\Rightarrow \frac{3|\vec{c}|-|\vec{a}|}{3|\vec{a}|}=\mu\)

\(\Rightarrow \frac{\lambda}{|\vec{c}|}=\frac{1}{\frac{3|\vec{c}|-\vec{a}}{3|\vec{a}|}+1}\)

\(\Rightarrow \lambda=\frac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\)

Given,

\(\tan \left(2 \tan ^{-1}(\cos x)\right)\)

As we know, \(2 \tan ^{-1} x =\tan ^{-1} \frac{2 x }{1- x ^{2}}\)

Therefore,

\(2 \tan ^{-1} \cos x=\tan ^{-1} \frac{2 \cos x}{1-\cos ^{2} x}\)

\(=\tan ^{-1} \frac{2 \cos x}{\sin ^{2} x}\)\(\quad(\because 1 - cos^2{x} =sin^2{x})\)

\(=\tan ^{-1}(2 \cot x \operatorname{cosec} x)\)

\(\tan \left(\tan ^{-1}(2 \cot x \operatorname{cosec} x)\right)=2 \cot x \operatorname{cosec} x \quad\left(\because \tan \left(\tan ^{-1} x\right)=x\right)\)

As we know,

If \(\alpha\) and \(\beta\) are the roots of the equation, \(a x^{2}+b x+c=0\)

Sum of roots \((\alpha+\beta)=\frac{-b}{a}\)

Product of roots \((\alpha \beta)=\frac{c}{a}\)

Given,

\(x^{2}-6 x+3=0\)

\(\alpha+\beta=-\left(\frac{-6}{1}\right)=6\)

\( \alpha \beta=\frac{3}{1}=3 \)

Now,

\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}\)...(1)

\((\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta\)

\(\Rightarrow(6)^{2}=\alpha^{2}+\beta^{2}+2 \times 3 \)

\(\Rightarrow \alpha^{2}+\beta^{2}=36-6=30\)

Putting the value of \((\alpha^{2}+\beta^{2})\) in equation (1), we get

\(\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=\frac{30}{3}\)

\(\therefore \frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=10\)

So, the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \cdot \beta}=10\)

Given:

\(\left(1+3 x+3 x^{2}+x^{3}\right)^{2 n}\)

\(=(1+x)^{6 n}\)

Here, \(n\) is an even number.

So the middle terms is \((\frac{6 n}{2} { +1})=(3 n+1)^{\text {th }}\) term

So terms \((3 n+1)^{\text {th }}\) term is:

\(T_{3 n+1}={ }^{6 n} C_{3 n} x^{3 n} \)

\(=\frac{(6 n) !}{(3 n!)^{2}} x^{3 n}\)

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)

\(\therefore\) vector perpendicular to \(\overrightarrow{\mathrm{a}}\) and \(\overrightarrow{\mathrm{b}}\) is \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\)

\(\therefore \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{array}\right|=\hat{i}-2 \hat{j}+\hat{k}\)

Now, projection of vector \(\vec{c}=2 \hat{i}+3=\hat{j}+\hat{k}\) on \(\vec{a} \times \vec{b}\) is

\(=\left|\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|}\right|=\left|\frac{2-6+1}{\sqrt{6}}\right|=\frac{3}{\sqrt{6}}=\sqrt{\frac{3}{2}}\)

There are \(8\) chair on each side of the table.

Let the sides be represented by \({A}\) and \({B}\).

Let four persons sit on side \({A}\), then number of ways of arranging \(4\) persons on \(8\) chairs on side \({A}={ }^{8} {P}_{4}\)

And two persons sit on side \({B}\).

The number of ways of arranging \(2\) persons on \(8\) chairs on side \({B}={ }^{8} {P}_{2}\)

The remaining \(10\) persons can be arranged in remaining \(10\) chairs in \(10 !\) ways.

Hence, the total number of ways in which the persons can be arranged is \({ }^{8} {P}_{4} \times{ }^{8} {P}_{2} \times 10 !=\frac{8 ! 8 ! 10 !}{4 ! 6 !}\)