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

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

20,400,80

More than 230 litres but less than 920 litres]

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.

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

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

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

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

(1, 1, -2)

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

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

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 

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.

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, \(\infty \))

0

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

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.

Use the formula of cos (A+B).

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}

{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).

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 \(n\in N\)

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

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

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 \(z=\frac{1+2i}{1-3i}\)
\(=\frac{1+2i}{1-3i}\times\frac{1+3i}{1+3i}=\frac{1+5i+6i^2}{1-9i^2}\)
\(=\frac{-5+5i}{10}=-\frac{1}{2}+\frac{1}{2}i\)
= 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 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.