Given set $=\{1,2,3,\dots \dots ..50\}$

$\mathrm{P}(\mathrm{A})=$ Probability that number is multiple of 4

$P(B)=$ Probability that number is multiple of 6

$\mathrm{P}(\mathrm{C})=$ Probability that number is multiple of 7

Now,

$\mathrm{P}(\mathrm{A})=\frac{12}{50},\mathrm{P}(\mathrm{B})=\frac{8}{50},\mathrm{P}(\mathrm{C})=\frac{7}{50}$

again

$\begin{array}{rl}& P(A\cap B)=\frac{4}{50},P(B\cap C)=\frac{1}{50},P(A\cap C)=\frac{1}{50}\\ & P(A\cap B\cap C)=0\end{array}$

Thus

$P(A\cup B\cup C)=\frac{12}{50}+\frac{8}{50}+\frac{7}{50}-\frac{4}{50}-\frac{1}{50}-\frac{1}{50}+0$

$=\frac{21}{50}$

$|P^{-1}AP-2I|=|P^{-1}AP-2P^{-1}P|$

$=|P^{-1}(A-2I)P|$

$=\left|P^{-1}\right||A-2I||P|$

$=|A-2I|$

$=\left|\begin{array}{ccc}0& 1& 2\\ 6& 0& 11\\ 3& 3& 0\end{array}\right|=69$

So, Prime factor of 69 is $3\mathrm{\&}23$

So, sum $=26$

Given function is,

$\begin{array}{rl}& f(x)=\{\begin{array}{cc}|x|+[x],& -1\le x<1\\ x+|x|,& 1\le x<2\\ x+[x],& 2\le x\le 3\end{array}\\ & =\{\begin{array}{cc}-x-1,& -1\le x<0\\ x,& 0\le x<1\\ 2x,& 1\le x<2\\ x+2,& 2\le x<3\\ 6,& x=3\end{array}\\ & \end{array}$

$\Rightarrow \mathrm{f}(-1)=0,\mathrm{f}(-1^{+})=0$

$\mathrm{f}\left(0^{-}\right)=-1,\mathrm{f}(0)=0,\mathrm{f}\left(0^{+}\right)=0$

$\mathrm{f}\left(1^{-}\right)=1,\mathrm{f}(1)=2,\mathrm{f}\left(1^{+}\right)=2$

$\mathrm{f}\left(2^{-}\right)=4,\mathrm{f}(2)=4,\mathrm{f}\left(2^{+}\right)=4$

$\mathrm{f}\left(3^{-}\right)=5,\mathrm{f}(3)=6$

$\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=\{0,1,3\}$

Hence, $f(x)$ is discontinuous at only three points.

$\begin{array}{rl}& \overrightarrow{a}=\lambda \hat{i}+2\hat{j}-3\hat{k}\\ & \overrightarrow{b}=\hat{i}-\lambda \hat{j}+2\hat{k}\\ & \Rightarrow (\overrightarrow{b}-\overrightarrow{a})\times \overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}\times \overrightarrow{b}=8\hat{i}-40\hat{j}-24\hat{k}\\ & \Rightarrow \overrightarrow{a}-\overrightarrow{b})\cdot (\overrightarrow{a}+\overrightarrow{b}(\overrightarrow{a}\times \overrightarrow{b})=8\hat{i}-40j-24\hat{k}\\ & \Rightarrow 8(\overrightarrow{a}\times \overrightarrow{b})=8\hat{i}-40\hat{j}-24\hat{k}\end{array}$

Now, $\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \lambda & 2& -3\\ 1& -\lambda & 2\end{array}\right|$

$\begin{array}{rl}& =(4-3\lambda )\hat{i}-(2\lambda +3)\hat{j}+(-{\lambda }^2-2)\hat{k}\\ & \Rightarrow \lambda =1\\ & \therefore \overrightarrow{a}=\hat{i}+2\hat{j}-3\hat{k}\\ & \overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k}\\ & \Rightarrow \overrightarrow{a}+\overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k},\overrightarrow{a}-\overrightarrow{b}=3\hat{j}-5\hat{k}\end{array}$

$\Rightarrow (\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -1\\ 0& 3& -5\end{array}\right|=2\hat{i}+10\hat{j}+6\hat{k}$

∴ required answer =4+100+36=140

Given:

$P(n):1\times 3+2\times 3^2+3\times 3^3+\dots +n\times 3^n$ $=\frac{(2n-1)3^{n+1}+3}{4}$

For $n=1,$

$\text{L H.S }=1\times 3=3$ and R.H.S. $=\frac{(2\times 1-1)3^{1+1}+3}{4}=\frac{3^2+3}{4}=\frac{12}{4}=3$

Thus $\mathrm{P}(1)$ is true.

$P(n)$ be true for some $n=k$.

$1\times 3+2\times 3^2+3\times 3^3+\dots +k\times 3^k$ $=\frac{(2k-1)3^{k+1}+3}{4}$

Now, $P(n)$ is true for $n=k+1$.

$1\times 3+2\times 3^2+3\times 3^3+\dots +(k+1)\times 3^{k+1}$ $=\frac{(2k+1)3^{k+2}+3}{4}$

Adding $(k+1)\times 3^{k+1}$ on both sides, we get

$1\times 3+2\times 3^2+3\times 3^3+\dots +k\times 3^k+(k+1)\times 3^{k+1}$ $=\frac{(2k-1)3^{k+1}+3}{4}+(k+1)\times 3^{k+1}$

$=\frac{(2k-1)3^{k+1}+3\mid 4(k+1)3^{k+1}}{4}$

$=\frac{3^{k+1}[2k-1+4(k+1)]+3}{4}$

$=\frac{3^{k+1}(6k+3)+3}{4}$

$=\frac{3^{(k+1)+1}(2k+1)+3}{4}$

$=\frac{(2k+1)3^{k+2}+3}{4}$

Thus, $P(k+1)$ is true whenever $P(k)$ is true.

By the principle of mathematical induction, statement P(n) is true for all natural numbers.

Given,

$4x+3<6x+7$

Subtracting 3 from both sides.

$4x+3-3<6x+7-3$

$\Rightarrow 4x<6x+4$

Subtracting $6x$ from both sides,

$4x-6x<6x+4-6x$

$\Rightarrow -2x<4\text{ or }$

$\Rightarrow x>-2$ i.e., all the real numbers greater than $-2$, are the solutions of the given inequality.

So, the solution set is $(-2,\mathrm{\infty })$, i.e. $x\in (-2,\mathrm{\infty })$

Given, $\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{\frac{(1+\cos \theta )\times (1+\cos \theta ))}{(1-\cos \theta )\times (1+\cos \theta )})}\text{ (By rationalization) }$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{\frac{(1+\cos \theta {)}^2}{1-{\cos }^2\theta }}$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{\frac{(1+\cos \theta {)}^2}{{\sin }^2\theta }}$$(\because 1-{\cos }^{2}x={\sin }^{2}x)$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\sqrt{{\left(\frac{1+\cos \theta }{\sin \theta }\right)}^2}$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}=\frac{1+\cos \theta }{\sin \theta }$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}$ = $\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }$

$\Rightarrow \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}$ = $\mathrm{cosec}\theta +\cot \theta$ $(\because \mathrm{cosec}x=\frac{1}{\sin x},\cot x=\frac{\cos x}{\sin x})$

B={x:x∈(−∞,−1]∪[5,∞)}
A∩B={x:x∈(−2,−1]}
A∪B={x:x∈(−∞,2)∪[5,∞)}
A−B={x:x∈(−1,2)}
B−A={x:x∈(−∞,−2]∪[5,∞)}