Let the events be defined as:

A: Obtaining a sum of 8

B: Getting an even number on both dice

Now cases favourable to A are \((3,5)(5,3)(2,6)(6,2)(4,4)\)

So, \(P(A)=\frac{5}{36}\)

Cases favourable to B: \((2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)\)

P(B)=\(\frac{9}{36}\)

Now, \((2,6)(6,2)\) and \((4,4)\) are common to both events \(A\) and \(B\)

So, \(P(A \cap B)=\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )=\frac{5}{36}+\frac{9}{36}-\frac{3}{36}\)

\(\Rightarrow P ( A \cup B )\)\(=\frac{11}{36}\)

Given,

\(\int_{1}^{4}\left(5 x^{2}-8 x+5\right) d x\)

Let, \(I=\int_{1}^{4}\left(5 x^{2}-8 x+5\right) d x\)

\(I=\int_{1}^{4}\left(5 x^{2}\right) d x-\int_{1}^{4} 8 x . d x+\int_{1}^{4} 5 d x\)

Using, \(\int^{n} d x=\frac{1}{n+1} x^{n+1}+c\)

\(I=5 \int_{1}^{4} x^{2} d x-8 \int_{1}^{4} x d x+5 \int_{1}^{4} d x\)

\(I=5\left[\frac{x^{3}}{3}\right]_{1}^{4}-8\left[\frac{x^{2}}{2}\right]_{1}^{4}+5[x]_{1}^{4}\)

\(I=5\left[\frac{4^{3}-1^{3}}{3}\right]-8\left[\frac{4^{2}-1^{2}}{2}\right]+5[4-1]\)

\(=5 \times \frac{63}{3}-8 \times \frac{15}{2}+5 \times 3\)

\(=60\)

From figure, we have

\(P=5, O Q=6\)

and \(O M=\frac{5}{2}, C M=3\)

\(\therefore\) ln \(\Delta O M C, O C^{2}=O M^{2}+M C^{2}\)

\(\Rightarrow  O C^{2}=\left(\frac{5}{2}\right)^{2}+(3)^{2} \Rightarrow O C=\frac{\sqrt{61}}{2}\)

Thus, the required circle has its centre \(\left(\frac{5}{2}, 3\right)\)

and radius \(\frac{\sqrt{61}}{2}\).

So, its equation is \(\left(x-\frac{5}{2}\right)+(y-3)^{2}=\left(\frac{61}{4}\right)\).

Thus, \(\lambda=\frac{61}{4}\)

Choose \(P\) as the origin of the reference.

\(\overrightarrow{ PR }=\vec{a}+\vec{b}\)

\(\overrightarrow{ PS }=-(\vec{a}-\vec{b})=\vec{b}-\vec{a}\)

\(\overrightarrow{ PM }=\overrightarrow{ a }+\frac{\overrightarrow{ b }}{2}\)

Given:\(\overrightarrow{ SX } =\frac{4}{5}\overrightarrow{ SM }\)

\(\overrightarrow{ PX }-(\overrightarrow{ b }-\overrightarrow{ a })=\frac{4}{5}\left\{\overrightarrow{ a }+\frac{\overrightarrow{ b }}{2}-(\overrightarrow{ b }-\overrightarrow{ a })\right\}\)

\(\overrightarrow{ PX }=\frac{3}{5}\overrightarrow{ a }-(-\frac{6}{10}\overrightarrow{ b })\)

\(\Rightarrow \overrightarrow{ PX } =\frac{3}{5}(\overrightarrow{ a }+\overrightarrow{ b })\)

\(=\frac{3}{5} \overrightarrow{ PR }\)

Let line L₁: x – 3y = 0 …(i)

L₂: 4x + 3y = 5 …(ii)

L₃: 3x + y = 0 …(iii)

The slope of L₁, m₁ =$\frac{1}{3}$

The slope of L₂, m₂ =$\frac{-4}{3}$

The slope of L₃, m₃ = -3

Here, m₁m₃ = -1

So, L₁ is perpendicular to L₃.

Let L: 3x – 4y = 0

Slope, m =$\frac{3}{4}$

Here, m₂m = -1

So, L is perpendicular to L₂.

The line L passes through the origin and it is perpendicular to L₂.

So, L will pass through the orthocentre.

Number lying between \(100\) and \(1000\) formed by \(3\) digits.

And every digit have \(5\) option \((1,2,3,4,5)\) to select a number.

But repetition is not allowed.

So, number \(={ }^{5} P_{3}\)

\(=\frac{5 !}{(5-3) !}\)

\(=\frac{5 !}{2 !}\)

\({ }^{5} P_{3}\)\(=60\)

Given,\(P(A)=2 P(B)=6 P(C)\)

We know that when events are mutually exclusive and exhaustive,

\(P(A)+P(B)+P(C)=1\)

Let, \(P(A)\) = \(k\)

Then, \(P(B)= \frac{P(A)}{2}=\frac{k}{2}\)

\(P(B)=\frac{k}{2}\)

\(\Rightarrow P(C)= \frac{k}{6}\)

So, according to the concept\(k+\frac{k}{2}+\frac{k}{6}=1\)

\(\Rightarrow \frac{10 k}{6}=1\)

\(\Rightarrow k=\frac{3}{5}\)

Therefore,\(P(B)\) = \(\frac{k}{2}=\frac{3}{5 \times 2}\)\(=0.3\)

Let the parallel plane to \(2 x-2 y+z=0\) is \(2 x-2 y+z+\lambda=0\)

It passes through \((1,-2,3)\)

\(\therefore 2+4+3+\lambda=0\)

\(\Rightarrow \lambda=-9\)

The distance of \((-1,2,0)\) from the plane

\(2 x-2 y+z-9=0\) is \(\left|\frac{-2-4-9}{\sqrt{5+4+1}}\right|=\left|\frac{-15}{3}\right|=5\)

Given:

\(\mathrm{P}(\mathrm{n})=3^{(2 \mathrm{n}+2)}-8 \mathrm{n}-9=8 \mathrm{~m}\)

Where \(\mathrm{a} \in \mathrm{N}\), i.e. a is a natural number.

For \(\mathrm{n}=1\)

\(\text { LHS }=3^{(2 \times 1+2)}-8 \times 1-9 \)

\(=3^{4}-17 \)

\(=81-17 \)

\(=64=8 \times 8\)

Therefore, \(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}=1\)

Suppose, \(\mathrm{P}(\mathrm{k})\) is true.

\(3^{(2 k+2)}-8 k-9=8 a\).....(1) [where \(\mathrm{a} \in \mathrm{N}\)]

Now, we will prove that \(\mathrm{P}(\mathrm{k}+1)\) is true.

\(\text { LHS }=3^{2(\mathrm{k}+1)+2}-8(\mathrm{k}+1)-9 \)

\(=9\left[3^{(2 \mathrm{k}+2)}\right]-8 \mathrm{k}-17\)

From equation (1), we get

\(=9[8 \mathrm{a}+9+8 \mathrm{k}]-8 \mathrm{k}-17 \)

\(=72 \mathrm{a}+64 \mathrm{k}+64 \)

\(=8(9 \mathrm{a}+8 \mathrm{k}+8) \)

\(=8 \mathrm{~b}\)

Where \(\mathrm{b}=9 \mathrm{a}+8 \mathrm{k}+8\) and \(\mathrm{b}\) is natural number.

Therefore, \(\mathrm{P}(\mathrm{k}+1)\) is true whenever \(\mathrm{P}(\mathrm{k})\) is true.

Therefore, \(\mathrm{P}(\mathrm{n})\) is true for \(\mathrm{n}\), where \(\mathrm{n}\) is a natural number.

So, the given equation is divisible by 8.

Given function is,

\(f(x)=\left\{\begin{array}{cc} |x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3 \end{array}\right. \)

\(=\left\{\begin{array}{cc} -x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ 2 x, & 1 \leq x<2 \\ x+2, & 2 \leq x<3 \\ 6, & x=3 \end{array}\right. \)

\(\Rightarrow\)\(\begin{array}{c} f(-1)=0, f\left(-1^{+}\right)=0 \end{array} \)

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

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

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

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

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

Hence, \(\mathrm{f}(\mathrm{x})\) is discontinuous at only three points.