{x : x \(\in\) R, 0 \(\le\) x < 7} is the set that contain 0 but not 7. So, it can be represented as an interval whose first end is closed and the other end is open.
So, the interval is [0,7).

Given, R = {(x-2,\({ x }^{ 2 }\)) : x is a prime number less than 15}
\(\therefore \) R = {(0,4),(1,9),(3,25),(5,49),(9,121),(11,169)}
Domain(R) = {0,1,3,5,9,11}
Range(R) = {4,9,25,49,121,169}

Use the formula of cos (A+B).

LHS= \(cos\frac { \pi }{ 8 } +cos\frac { 3\pi }{ 8 } +cos\frac { 5\pi }{ 8 } +cos\frac { 7\pi }{ 8 } \)
\(=cos\frac { \pi }{ 8 } +cos\frac { 3\pi }{ 8 } +cos\left( \pi -\frac { 3\pi }{ 8 } \right) +cos\left( \pi -\frac { \pi }{ 8 } \right)\)
\( \\ =cos\frac { \pi }{ 8 } +cos\frac { 3\pi }{ 8 } -cos\frac { 3\pi }{ 8 } -cos\frac { \pi }{ 8 } \quad [\because cos(\pi -\theta )=-cos\theta ]\)
\(=0\)=RHS.
Hence proved.

\(z=\frac { 1-i }{ 1+i } x \frac { 1-i }{ 1-i } =\frac { 1-1-2i }{ 1+1 } =-i\)   = i

0

2(2x+3)−10<6(x−2)
⇒4x+6−10<6x−12
⇒4x−4<6x−12
⇒−4+12<6x−4x
⇒8<2x
⇒4 Thus, all real numbers x, which are greater than 4, are the solution of the given inequality.
Hence, the solution set of the given inequality is (4, \(\infty \))

There will be as many ways as there are ways of filling 2 vacant places  in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.

We have ,\(\frac{^{n-1}{C}{_{r}} } { ^{n}{C}{_{r}}}\) = \(\frac {6}{9}\)  
\(\Rightarrow\) \(\frac { \frac { (n-1)! }{r!(n-1-r)! } }{ \frac { n! }{ r!(n-r)! } }\) =\(\frac {2}{3}\)
\(\Rightarrow\)  \(\frac { (n-1)! }{(n-(r+1))!} \times \frac {(n-r)!}{n!}\) = \(\frac {2}{3}\)
\(\Rightarrow\) \(\frac { (n-1)! }{r!(r+1))!} \times \frac {(n-r)(n-(r=1))!}{n(n-1)!}\) =\(\frac {2}{3}\)
\(\Rightarrow\) \(\frac {n-r}{n}\)= \(\frac {2}{3}\)
\(\Rightarrow\)3n - 3r = 2n
\(\Rightarrow\)n - 3r = 0 ......(i)
Also we have            \( \frac{^{n}{C}{_{r}}}{^{n+1}{C}{_{r}}}\) = \(\frac{9}{13}\) 
\(\Rightarrow\) \(\frac{n!}{r!(n-r)!} \times \frac{r!(n+1-r)!}{(n+1)}\) = \(\frac{9}{13}\)
\(\Rightarrow\)\(\frac{n!}{ (n-r)!} \times \frac{n-(r-1)! }{(n+1)n!}\)  =  \(\frac{9}{13}\)
\(\Rightarrow\)\(\frac {(n-(r-1))(n-r)!}{(n-r!)(n+1)}\) =  \(\frac{9}{13}\) 
\(\Rightarrow\) \(\frac{n-r+1}{n+1} \) = \(\frac{9}{13}\)
\(\Rightarrow\)13n - 13r + 13 = 9n + 9
\(\Rightarrow\) 4n - 13r = -4  .....(ii)
On solving Eqs.(i) and (ii), we get
n = 12 and r = 4 

On dividing both sides of given equation by
\(\sqrt { (\sqrt { 3 } )^{ 2 }+(1)^{ 2 } } \) we get
\(x.\left( \frac { \sqrt { 3 } }{ 2 } \right) +y.\left( \frac { 1 }{ 2 } \right) =2\)
\(xcos30^{ \circ }+ysin30^{ 0 }=2\)
\(\alpha =30^{ \circ }\quad p=2\quad units\)

Given, equation of parabola is .x² = -9y, which is of the form .x² = -4ay i.e focus lies on the negative direction of Y-axis
Here, 4a=9
\(\Rightarrow\)   a =\(\frac{9}{4}\)
\(\therefore\)  Focus = (0, -a) =\(\left( 0,-\frac { 9 }{ 4 } \right) \) 
Axis = Y-axis
Directrix, y = a \(\Rightarrow\) y =\(\frac{9}{4}\) and length of latusrectum = 4a =9

The centroid of a triangle is equal to the centroid of the triangle formed by mid-points of its sides.

(1, 1, -2)

\(-\frac { 4a }{ { x }^{ 5 } } +\frac { 2b }{ { x }^{ 3 } } -sinx\)

\(\frac { 1 }{ 3 } \)

\({1\over 7!}+{1\over 8!}={x\over 9!}\)
⇒ \({1\over 7!}+{1\over 8\times7!}={x\over 9\times8\times7!}\)
⇒ \({1\over 7!}\left[1+{1\over8}\right]={1\over 7!}\left[x\over 9\times8\right]\)
⇒ \({9\over 8}={x\over 9\times8}\)
⇒ x=81

Here h = -2, k = 3 and r = 4
The equation of circle is
(x - h)² + (y - k)² = r²
∴ (x + 2)² + (y - 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 - 6y = 16
⇒ x² + y² + 4x - 6y - 3 = 0
which is required equation of circle.

More than 230 litres but less than 920 litres]

20,400,80

\(\frac{x^{2}}{(x sinx+cos x)^{2}}\)

a⁶x⁶ - 6a⁵x⁴b + 15a⁴x²b² - 20a³b³ + \(\frac { { 15 a }^{ 2 }{ b }^{ 4 } }{ { x }^{ 4 } } -6\frac { { ab }^{ 5 } }{ { x }^{ 4 } } +\frac { { b }^{ 6 } }{ { x }^{ 6 } } \)

Step I  Let P(n) be the given statement,
i.e. P(n): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 }=\left[ \frac { n(n+1) }{ 2 } \right] ^{ 2 }\)
Step II  For n=1, we have, LHS=1³=1
and RHS= \(\left[ \frac { 1(1+1) }{ 2 } \right] ^{ 3 }=\left( \frac { 1.2 }{ 2 } \right) ^{ 3 }=1\)
LHS=RHS;     P(1) is true
Step III  Let us assume that P(n) is true for n=k. Then we have
P(k): \({ 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+.....+{ k }^{ 3 }=\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }...(i)\)
Step IV  Now, we shall prove the statement for n=k+1 For this we have to show
1³+2³+3³+.....+k³+(k+1)³
=\(\left[ \frac { (k+1)(k+1+1) }{ 2 } \right] ^{ 2 }\)
Now, LHS=1³+2³+3³+.....+k³+(k+1)³
= \(\left[ \frac { k(k+1) }{ 2 } \right] ^{ 2 }+(k+1)^{ 3 }\)       [from Eq.(i)]
\(=(k+1)^{ 2 }\left[ \frac { k^{ 2 } }{ 4 } +(k+1) \right] \)
\(\\ =(k+1)^{ 2 }\left[ \frac { k^{ 2 }+4(k+1) }{ 4 } \right] =\frac { (k+1)^{ 2 }\left[ k^{ 2 }+4k+4 \right] }{ 4 } \)
\(=\frac { (k+1)^{ 2 }(k+2)^{ 2 } }{ 4 } =RHS\)
So, P(k+1) is true, whenever P(k) is true. Hence, by principle of mathematical induction, P(n) is true for all \(n\in N\)

The given inequality is 3x + 2y \(\le\)150.
Draw the graph of the line 3x + 2y = 150.
Table of values satisfying the equation 3x + 2y = 150

Putting (0,0) in the given inequation, we have 3 x 0 + 2 x 0 \(\le\)150 \(\Rightarrow\) 0 \(\le\)150, which is true.

\(\therefore\) Half plane of 3x + 2y \(\le\)150 is towards origin.
Also the given inequality is x + 4y \(\le\) 80.
Draw the graph of the line x + 4y = 80.
Table of values satisfying the equation x + 4y = 80

Putting (0, 0) in the given inequation, we have o + 4 x 0 \(\le\) 80 \(\Rightarrow\) 0 \(\le\) 80, which is true.
\(\therefore\) Half plane of x + 4y \(\le\) 80 is towards origin.
The given inequality is x \(\le\)15.
Draw the graph of the line x = 15.
Putting (0, 0) in the given inequation, we have 0 \(\le\)15 which is true
\(\therefore\) Half plane of x \(\le\)15 is towards origin.

Let \(z=\frac{1+2i}{1-3i}\)
\(=\frac{1+2i}{1-3i}\times\frac{1+3i}{1+3i}=\frac{1+5i+6i^2}{1-9i^2}\)
\(=\frac{-5+5i}{10}=-\frac{1}{2}+\frac{1}{2}i\)
= r (cos \(\theta\) + i sin \(\theta\))
\(\Rightarrow r\ cos\theta=-\frac{1}{2}\ and\ r\ sin\theta=\frac{1}{2}\)...... (i)
Squaring both sides of (i) and adding
\(r^2(cos^2\theta+sin^2\theta)=\frac{1}{4}+\frac{1}{4}\)
\(\Rightarrow r^2=\frac{1}{2}\Rightarrow r=\frac{1}{\sqrt 2}\)
\(\therefore\frac{1}{\sqrt 2}\ cos\theta=-\frac{1}{2}\ and\frac{1}{\sqrt 2}\ sin\theta=\frac{1}{2}\)
\(\Rightarrow\ cos\theta=-\frac{1}{\sqrt 2}\ and\ sin\theta=\frac{1}{\sqrt 2}\)
Since sin \(\theta\) is positive and cos \(\theta\) is negative.
\(\therefore\) \(\theta\) lies in second quadrant.
\(\theta=(\pi-\frac{\pi}{4})=\frac{3\pi}{4}\)
Thus polar form of z is \(\frac{1}{\sqrt 2}(cos\frac{3\pi}{4}+isin\frac{3\pi}{4}).\)

The equations of lines AB, BC and AC are 
x + y - 6 = 0   ...(i)
x - 3y - 2 = 0  ...(ii)
5x- 3y + 2 = 0 ...(iii)
On solving the equation (i) and (ii), we have coordinates of point B (5,1).
On solving the equation (ii) and (iii), we have coordinates of point C (-1, -1).
On solving the equation (iii) and (i), we have coordinates of point A (2, 4).
Let D, E, F are mid points of BC, AC and AB respectively.
Coordinates of D are \(\left({5-1\over 2},{1-1\over 2}\right)\ i.e.(2,0)\)
Coordinates of E are \(\left({2-1\over 2},{4-1\over 2}\right)\ i.e.\left({1\over 2},{3\over 2}\right)\)
Coordinates of D are  \(\left({2+5\over 2},{4+1\over 2}\right)\ i.e.\left({7\over 2},{5\over 2}\right)\)
Equation of median AD is
\(y-4={(0-4)\over (2-2)}(y-2)\)
⇒  x - 2 = 0
Equation of median BE is
\(y-1={\left({3\over2}-1\right)\over \left({1\over2}-5\right)}(x-5)\)
\(⇒\ y-1={{1\over 2}\over -{9\over 2}}(x-5)\)
⇒ 9y - 9 = -x + 5
⇒ x + 9y - 14 = 0
Equation of CF is
\(y+1={\left({5\over 2}+1\right)\over \left({7\over 2}+5\right)}(x+1)\)
\(⇒\ y+1={{7\over 2}\over {9\over 2}}(x+1)\)
\(⇒\ y+1={7\over 9}(x+1)\)
⇒ 9y + 9 = 7x + 7 ⇒ 7x - 9y - 2 = 0
Thus equations of medians are x - 2 = 0, x + 9y - 14 = 0 and 7x - 9y - 2 = O.

Let F₁ and F₂ be two points where the flag parts are fixed on the ground. The origin O is the mid point of F₁F₂

∴ OF₁=OF₂=\(\frac { 1 }{ 2 } \)F₁F₂=\(\frac { 1 }{ 2 } \) x 8 =4m
∴ Coordinates of F; are (-4, 0) and F₂ are (4, 0)
Let P(α, β) be any point on the track.
∴ PF₁+PF₂ = 0
∴ \(\sqrt { (\alpha +4)^{ 2 }+(\beta -0)^{ 2 } } +\sqrt { (\alpha -4)^{ 2 }+(\beta -0)^{ 2 } } \)
=10
⇒ \(\sqrt { \alpha ^{ 2 }+16+8\alpha +{ \beta }^{ 2 } } \)
=10-\(\sqrt { { \alpha }^{ 2 }+16-8\alpha +{ \beta }^{ 2 } } \)
Squaring both sides, we have
\(\\ { \alpha }^{ 2 }+{ \beta }^{ 2 }+8\alpha +16=100+{ \alpha }^{ 2 }+{ \beta }^{ 2 }-8\alpha \)+16
=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
⇒ 16\(\alpha \)-100=-20\(\sqrt { { \alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16 } \)
Squaring both sides again, we have
(16\(\alpha \)-100)2=(-20\(\sqrt { { (\alpha }^{ 2 }+{ \beta }^{ 2 }-8d+16)^{ 2 } } \)
⇒ 256\(\alpha \)²+1000-3200\(\alpha \) = 400(\(\alpha \)²+\(\beta \)²+8\(\alpha \)+16)
⇒ 256\(\alpha \)²+10000-3200\(\alpha \)=400\(\alpha \)²+400\(\beta \)²-3200\(\alpha \)+6400
⇒ 144\(\alpha \)²+400\(\beta \)²=3600
⇒ \(\frac { 144\alpha ^{ 2 } }{ 3600 } +\frac { 400{ \beta }^{ 2 } }{ 3600 } \)=1 ⇒ \(\frac { { \alpha }^{ 2 } }{ 25 } +\frac { { \beta }^{ 2 } }{ 9 } \)=1
Thus required equation of locus of point P is
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } \)=1

cos 6\(\theta\) + 6 cos 4\(\theta\) + 15 cos 2\(\theta\) + 10
= (cos 6\(\theta\) + cos 4\(\theta\)) + (5 cos 4\(\theta\) + 5 cos 2\(\theta\)) + (10 cos 2\(\theta\) + 10)
= (cos 6\(\theta\) + cos 4\(\theta\)) + 5 (cos 4\(\theta\) + cos 2\(\theta\)) + 10 (cos 2\(\theta\)+ 1)
= 2cos 5\(\theta\) cos \(\theta\) + 5 x 2 cos 3\(\theta\) cos \(\theta\)+ 10 x 2 cos \(\theta\) cos \(\theta\)
= 2cos \(\theta\) [cos5\(\theta\) + 5 cos 3\(\theta\) + 10 cos \(\theta\)]
\(\therefore {cos \theta + 6 cos 4 \theta + 15 cos 2 \theta + 10 \over cos 5 \theta + 5 cos 3 \theta + 10 cos \theta}=2cos \theta\)