\(x^{2}-(2+\sqrt{3}) x+(2-\sqrt{3})\)

a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then least prime factor of (a + b) is 2.

Let the man finishes the work in x days and the boy in y
According to question,
\(\frac { 8 }{ x } +\frac { 12 }{ y } =\frac { 1 }{ 10 } \quad ...(i)\)
and \(\frac { 6 }{ x } +\frac { 8 }{ y } =\frac { 1 }{ 14 } \quad \quad ...(ii)\)  
Solve Eq. (i) and Eq. (ii) to get value of x.
Man alone finish the work in 140 days.

Let coordinates of A, B, C and D are A(6,3), B(-3,5), C(4,-2) and D(x,3x).
Now, area of △DBC
⇒ \(\frac{1}{2}\)|x(5+2)-3(-2-3x)+4(3x-5)|
= \(\frac{1}{2}\)|7x+6+9x+12x-20|
= \(\frac{1}{2}\)|28x-14|
Again, area of △ABC
= \(\frac{1}{2}\)|6(5+2)-3(-2-3)+4(3-5)|
=\(\frac{1}{2}\)|42+15-8|=\(\frac{1}{2}\)|49|
=\(\frac {49}{2}\)sq.units.
\(\frac {ar(\Delta DBC)} {ar(\Delta ABC)} \)=\(\frac{1}{2}\)
⇒ \(\frac{|14x-7|}{49} = \frac{1}{2}\)
⇒ |14x-7| = \(\frac{49}{4}\)
⇒14x -7 = \(\frac{49}{4}\) or 14x-7 = -\(\frac{49}{4}\)
14x=\(\frac{49}{4}\)+7 or 14x = 7-\(\frac{49}{4}\) 
14x = \(\frac {77}{4}\) or 14x=-\(\frac{21}{4}\)
x=\(\frac {77}{4 \times14}\) or x=\(- \frac {21}{4\times14}\)
14x = \(\frac {77}{4}\) or 14x=\(-\frac {21}{4}\)
x=\(\frac {11}{8}\) or x=\(\frac {-3}{8}\)

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram.
Since, ABCD is a parallelogram.
\(\therefore\) Diagonals AC and BD will bisect each other. So, the mid-point of AC and mid-point of BD will be same 

Thus mid-point of AC = Mid-point of BD
\(\begin{aligned} \Rightarrow \quad & \left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right) \\ \end{aligned}\)
\(\begin{aligned} & {\left[\because \text { coordinates of mid-point }=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right] } \end{aligned}\)
On comparing the coordinate from both sides,we get
\(\frac{1+x}{2}=\frac{4+3}{2} \text { and } \frac{2+6}{2}=\frac{5+y}{2}\)
\(\Rightarrow\) 1 + x = 7 and 8 = 5 + y
\(\therefore\)   x = 6 and y = 3

Hint Sum of zeroes = - Coefficient of x/ Coefficient of x²
⇒ -1 = \(\frac{-(2a+3)}{a+1}\) = -2

 Let a = 7\(\sqrt5\) be a rational number.
\(\Rightarrow \frac{a}{7}=\sqrt{5}\)
Now, \(\frac{a}{7}\) is a rational number since product of two rational number is a rational number.
The above will imply that \(\sqrt5\) is a rational number. But \(\sqrt5\)  is an irrational number.
his contradicts our assumption. Therefore we can conclude that 7\(\sqrt5\) is an irrational number and hence the result.

Given, a pair of linear equations is :
s - t = 3          
\(\Rightarrow\) s = t + 3        ....(i)
and \(\frac { s }{ 3 } +\frac { t }{ 2 } =6\) ...(ii)
On substituting s = t + 3, from egn. (i) in eqn. (ii),
we get
\(\\ \frac { t+3 }{ 3 } +\frac { t }{ 2 } =6\)
\(\Rightarrow\) 2(t + 3) + 3t = 36
\(\Rightarrow\) 5t + 6 = 36
\(\Rightarrow\) 5t = 30
\(\Rightarrow\) t = 6
From eqn., (i), s = 6 + 3 = 9
Hence, s = 9, t = 6

p(y) = 6y² - 7y + 2
\(\alpha +\beta =-\left( -\frac { 7 }{ 6 } \right) =\frac { 7 }{ 6 } \)
and \(\alpha\beta=\frac{2}{6}=\frac{1}{3}\)
Now \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { { 7 }/{ 6 } }{ { 2 }/{ 6 } } =\frac { 7 }{ 2 } \)
and \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =3\)
The required polynomial is \({ y }^{ 2 }-\frac { 7 }{ 2 } y+3=\frac { 1 }{ 2 } \left[ 2{ y }^{ 2 }-7y+6 \right] \)

Given points are A(k, 2-2k), B(-k+1, 2k) and C(-4-k, 6-2k).Here, k is an unknown.
Since given points are collinear.
So, area of \(\Delta \)ABC=0
i.e. x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)=0
\(\Rightarrow \)  k(2k-6+2k)+(-k+1)(6-2k-2+2k)+(-4-k)(2-2k-2k)=0
\(\Rightarrow \) 4k²-6k+(-k+1)(4)+(-4-k)(2-4k)=0
\(\Rightarrow \)  4k²-6k-4k+4-8+16k-2k+4k²=0
\(\Rightarrow \)  8k²+4k-4=0              
\(\Rightarrow \) 2k²+k-1=0  [divide by 4]                
\(\Rightarrow \)  2k²+2k-k-1=0                
\(\Rightarrow \) 2k(k+1)-1(k+1)=0             
\(\Rightarrow \)  2k-1=0      and   k+1=0                    
\(\Rightarrow \)    k=\(\frac { 1 }{ 2 } \quad \) and   k=-1                
Hence, for k=\(\frac { 1 }{ 2 } \)  and  k=-1, given points are collinear.

Since the given points are collinear, the area of the triangle formed by them must be 0, i.e.,
\(\frac{1}{2}[2(k+3)+4(-3-3)+6(3-k)]=0\) 
i.e., \(\frac{1}{2}(-4 k)=0\) 
Therefore, k=0
Let us verify our answer.
area of \(\Delta \mathrm{ABC}=\frac{1}{2}[2(0+3)+4(-3-3)+6(3-0)]=0\) 

For infinitely many solutions,\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{q}{c_{2}}\) 
\(\Rightarrow \frac{4}{k+2}=\frac{7}{21}=\frac{-10}{-3 k} \Rightarrow \frac{4}{k+2}=\frac{7}{21} \Rightarrow k=10\)

Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10x + y in the expanded form (for example, 56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5).
According to the given condition.
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66
i.e., x + y = 6  ...(1)
We are also given that the digits differ by 2, therefore,
either, x – y = 2   ...(2)
or y – x = 2 ...(3)
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.
In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.

Since \(\alpha\) and \(\beta\) are the zeroes of polynomial p(x) = 3x² + 2x + 1
Hence, \(\alpha+\beta=-\frac{2}{3}\)
and \(\alpha\beta=\frac{1}{3}\)
Now for the new polynomial,
Sum of the zeroes = \(\frac { 1-\alpha }{ 1+\alpha } +\frac { 1-\beta}{ 1+\beta} \)
\(=\frac { \left( 1-\alpha +\beta -\alpha \beta \right) +\left( 1+\alpha -\beta -\alpha \beta \right) }{ \left( 1+\alpha \right) \left( 1+\beta \right) } \)
\(=\frac { 2-2\alpha \beta }{ \left( 1+\alpha +\beta +\alpha \beta \right) } =\frac { 2-\frac { 2 }{ 3 } }{ 1-\frac { 2 }{ 3 } +\frac { 1 }{ 3 } } \)
\(\therefore\)  Sum of zeroes \(=\frac { { 4 }/{ 3 } }{ { 2 }/{ 3 } } =2\)
Product of zeroes \(=\left[ \frac { 1-\alpha }{ 1+\alpha } \right] \left[ \frac { 1-\beta }{ 1+\beta } \right] \)
\(=\frac { \left( 1-\alpha \right) \left( 1-\beta \right) }{ \left( 1+\alpha \right) \left( 1+\beta \right) } \)
\(=\frac { 1-\alpha -\beta +\alpha \beta }{ 1+\alpha +\beta +\alpha \beta } =\frac { 1-\left( \alpha +\beta \right) +\alpha \beta }{ 1+\left( \alpha +\beta \right) +\alpha \beta } \)
\(\therefore\)  Product of zeroes \(=\frac { 1+\frac { 2 }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 2 }{ 3 } +\frac { 1 }{ 3 } } =\frac { \frac { 6 }{ 3 } }{ \frac { 2 }{ 3 } } =3\)
Hence, Required polynomial = x²  - (Sum of zeroes)x + Product of zeroes
= x² - 2x + 3

We have, 12, 15 and 21
\(\therefore\) 12 = 2 \(\times\) 2 \(\times\) 3 = 2² \(\times\) 3
15 = 3 \(\times\) 5 and 21 = 3 \(\times\) 7
Here, 3 is a common prime factor of the given numbers and has smallest power 1.
\(\therefore\) HCF of 12, 15 and 21 = 3
and LCM = Product of each prime factor, with greatest power
= 2² \(\times\)3 \(\times\) 5 \(\times\) 7
= 2 \(\times\) 2 \(\times\) 3 \(\times\) 5 \(\times\) 7 = 420

(i) (c): Mid-point of J (6,17) and I (9,16) is
\(x=\frac{6+9}{2} \text { and } y=\frac{17+16}{2}\)
\(x=\frac{15}{2} \text { and } y=\frac{33}{2}\)
(ii) (a): In the top view figure, when we draw perpendicular from point P to the Y-axis, its distance is 4 units.
(iii) (c): In the  front view  figure, it  is clear  the coordinates of A and S are A(1, 8) and (17, 8) The distance between the points A and S is
\(\sqrt{(17-1)^{2}+(8-8)^{2}}\)
\(=\sqrt{16^{2}}=16\)
(iv) (d): In the front view figure, the coordinates of A and B are A(1 ,8) and B(5, 10)

Let C be point, which divides the line joining A and B in the ratio 1:3 
By using internal division formula
\(C=\left(\frac{1 \times 5+3 \times 1}{1+3}, \frac{1 \times 10+3 \times 8}{1+3}\right)=\left(\frac{5+3}{4}, \frac{10+24}{4}\right)\)
\(=\left(\frac{8}{4}, \frac{34}{4}\right)=(2,8.5)\)
(v) (b): Let Point be P(x,y)
PQ² = PS²
or, \((x-9)^{2}+(y-8)^{2}=(x-17)^{2}+(y-8)^{2}\)
\(\Rightarrow x^{2}-18 x+81=x^{2}-34 x+289\)
\(\Rightarrow\) 34x - 18x = 289 - 81
\(\Rightarrow\) 16x - 208
\(\Rightarrow\) x - 13 = 0

(i) (b): Given, a and \(\beta\) are the zeroes of 
\(p(x)=x^{2}-24 x+128\)
\(\text { Putting } p(x)=0 \text { , we get }\)
\( x^{2}-8 x-16 x+128=0 \)
\(\Rightarrow x(x-8)-16(x-8)=0 \)
\(\Rightarrow (x-8)(x-16)=0 \Rightarrow x=8 \text { or } x=16 \)
\(\therefore \alpha=8, \beta=16\)
(ii) (c) : \(\alpha+\beta+\alpha \beta =8+16+(8)(16) =24+128=152 \)
(iii) (d) : \(p(2)=2^{2}-2 4(2)+128=4-48+128=84\)
(iv) (a): Since a and \(\beta\) are zeroes of \(x^{2}+x-2\)
\(\therefore \quad \alpha+\beta=-1 \text { and } \alpha \beta=-2 \)
\(\text { Now, } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-2}=\frac{1}{2}\)
(v) (c): Sum of zeroes \(=\frac{-2}{k}\)
Product of zeroes \(=\frac{3 k}{k}=3\)
According to question, we have \(\frac{-2}{k}=3\)
\(\Rightarrow \quad k=\frac{-2}{3}\)