Given, pair of linear equations is
kx-5y-2=0 and 6x+2y-7=0
Here a₁=k, b₁=-5, c₁=-2
and a₂=6, b₂=2, c₂=-7
For no solution,
\(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \neq \frac { -2 }{ -7 } \)
\(\Rightarrow \quad \frac { k }{ 6 } =\frac { -5 }{ 2 } \)
\(\Rightarrow \quad k=-15\)

Given pair of linear equations is
41x+53y=135   ..(i)
and 53x+41y=147  ...(ii)
On adding Eqs. (i) and (ii), we get
94x+94y=282
\(\Rightarrow\) x+y=3 [dividingboth sides by 94] ...(iii)
On subtracting Eq. (i) from Eq. (ii), we get
12x-12y=12
\(\Rightarrow\) x-y=1 [dividing both sides by 12]   ...(iv)
Now, on adding Eqs.(iii) and (iv), we get
2x=4 \(\Rightarrow\) x=2
On substituting x=2 in Eq. (iii), we get
y=3-2=1
Hence, x=2 and  y=1 is the required solution.

Given, a and β are the zeroes of the polynomial
p(x)=ax²+bx+c.
Sum of zeroes, a+β=\(-b\over a\) and product of zeroes, aβ=\(c\over a\)
Now, a²β+aβ²=aβ(a+β)
\(=\frac { c }{ a } \times \frac { (b) }{ a } =\frac { -bc }{ a^{ 2 } } \)

Let us assume to the contrary that \(3+2\sqrt { 5 } \)  is a rational number. Then, it can be expressed in the form \(\frac{a}{b}\), where a, b are coprime integers and \(b\neq 0\)
Now,  \(3+2\sqrt { 5 } \) = a/b, where a,b are integers and \(b\neq 0\)
On rearranging, we get
\(2\sqrt { 5 } =\frac { a }{ b } -3\quad or\quad \sqrt { 5 } =\frac { a }{ 2b } -\frac { 3 }{ 2 } \)
Since, a, b are integers and \(b\neq 0\) , therefore \(\frac{a}{2b}\) is rational number and so \(\frac{a}{2b}\) - \(\frac{3}{2}\) is a rational number.
[since, difference of two rational numbers is also a rational number]
\(\Rightarrow \sqrt { 5 } \) is a rational number. But \(\sqrt { 5 } \) is an irrational number.
This shows that our assumption is incorrect.
So, \(3+2\sqrt { 5 } \) is irrational.

Let the original speed of train be x km/hr. Then,
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 360 km = \(\frac{360}{x} \mathrm{hr}\)
Time taken by the train under increased speed to cover 360 km = \(\frac{360}{x+5} \mathrm{hr}\)
Therefore,
\( \frac{360}{x}-\frac{360}{x+5}=1 \)
\(\frac{360(x+5)-360 x}{x(x+5)}=1 \)
\(\frac{360 x+1800-360 x}{x^{2}+5 x}=1 \)
\(\frac{1800}{x^{2}+5 x}=1\)
1800 = x² + 5x
x² + 5x - 1800 = 0
x² - 40x + 45x - 1800 = 0
x(x - 40) + 45(x - 40) = 0
(x - 40)(x + 45) = 0
x - 40 = 0
x = 40
Or
x + 45 = 0
x = -45
But, the speed of the train can never be negative.
Hence, the original speed of train is x = 40 km/hr.

We have
x² + 7x + 10 = (x + 2)(x + 5)
So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,
sum of zeroes = \(-2+(-5)=-(7)=\frac{-(7)}{1}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\) 
product of zeroes = \((-2) \times(-5)=10=\frac{10}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)

Let side of bigger square be x m and side of smaller square be y m
ATQ x²+y² = 640   ...(i)
and 4x-4y=64
\(\Rightarrow x-y=16\Rightarrow x=16+y\)
Substituting the value of x in equation (i) we get 
(16+y²)+y²=640
\(\Rightarrow 256+y^{ 2 }+32y+y^{ 2 }=640\)
\(\Rightarrow 2y^{ 2 }+32y-384=0\)
\(\Rightarrow y^{ 2 }+16y-192=0\)
\(\Rightarrow (y+24)(y-8)\)
\(\Rightarrow \) y=8 or y=-24 (Rejecting)
When y=8, then x=16+8=24 
\(\therefore \) Sides of squares are 8m and 24 m

Given, equation is \( \sqrt2x^2+7x+5\sqrt2=0\) 
\(\Rightarrow\) \( \sqrt2x^2+5x+2x+5\sqrt2=0\) [\(\because\) 5 \(\times\)2 = 10 and 5 + 2 = 7]
\(\Rightarrow\)\(x(\sqrt2x+5)+\sqrt2(\sqrt2x+5)=0\)
\(\Rightarrow\) (\(\sqrt{2}\)x + 5) (x + \(\sqrt{2}\)) = 0
\(\Rightarrow\) \(\sqrt{2}\)x+5 = 0 or x + \(\sqrt{2}\) = 0
\(\Rightarrow\) x=\(\frac{-5}{\sqrt{2}}\) or x = -\(\sqrt{2}\)
Hence, the roots of the equation
\(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \text { are } \frac{-5}{\sqrt{2}} \text { and }-\sqrt{2} \text {. }\)

4s² – 4s + 1 = 4s² – 2s - 2s + 1
=2s(2s - 1) - 1(2s - 1)
=(2s - 1)(2s -1)
\(=2\left(s-\frac{1}{2}\right) 2\left(s-\frac{1}{2}\right)\) 
Therefore the zeroes of the polynomial are 1/2, 1/2 
Relationship between the zeroes and the coefficient of the polynomial: 
Also sum of the zeroes = \(\frac{1}{2}+\frac{1}{2}=1\) 
Also product of the zeroes = \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\) 
hence verfied.

D=b²-4ac=0
Here a=1+m²,b=2mc,c=(c²-a²)=0
(2mc)²-4(1+m²)(c²-a²)=0
\(\Rightarrow 4m^{ 2 }c^{ 2 }-4(1+m^{ 2 })(c^{ 2 }-a^{ 2 })\)
\(\Rightarrow m^{ 2 }c^{ 2 }-4(1+m^{ 2 })(c^{ 2 }-a^{ 2 })=0\)
\(\Rightarrow m^{ 2 }c^{ 2 }-(c^{ 2 }-a^{ 2 }+m^{ 2 }c^{ 2 }-m^{ 2 }a^{ 2 })=0\)
\(\Rightarrow m^{ 2 }c^{ 2 }-(c^{ 2 }+a^{ 2 }+m^{ 2 }c^{ 2 }+m^{ 2 }a^{ 2 })=0\)
\(\Rightarrow -c^{ 2 }+a^{ 2 }+m^{ 2 }a^{ 2 }=0\)
c²=a²(1+m²)
Hence proved.

\(\frac { 3 }{ x+1 } +\frac { 4 }{ x-1 } =\frac { 29 }{ 4x-1 } \)
\(\Rightarrow \frac { 7x+1 }{ x^{ 2 }-1 } =\frac { 29 }{ 4x-1 } \)
\(\Rightarrow (7x+1)(4x-1)=29x^{ 2 }-29\)
\(\Rightarrow 28x^{ 2 }-7x+4x-1=29x^{ 2 }-29\)
\(\Rightarrow 28x^{ 2 }-3x-1=29x^{ 2 }-29\)
\(\Rightarrow x^{ 2 }+3x-28\)
\(\Rightarrow x^{ 2 }+7x-4x-28=0\)
\(\Rightarrow x(x+7)-4(x+7)=0\)
\(\Rightarrow x(x+7)(x-4)=0\)
x=4,-7

(c) If Assertion is correct but Reason is incorrect.

(d) If Assertion is incorrect but Reason is correct.

(i) (b): Let two consecutive integers be x, x + 1.
Given, x² + (x + 1)² = 650
\(\begin{array}{l} \Rightarrow 2 x^{2}+2 x+1-650=0 \\ \Rightarrow 2 x^{2}+2 x-649=0 \end{array}\)
(ii) (c): Let the two numbers be x and 15 - x.
Given, \(\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\)
\(\begin{array}{l} \Rightarrow 10(15-x+x)=3 x(15-x) \\ \Rightarrow 50=15 x-x^{2} \Rightarrow x^{2}-15 x+50=0 \end{array}\)
(iii) (d): Let the numbers be x and x + 3.
Given, x(x + 3) = 504
\(\Rightarrow\) x² + 3x - 504 = 0
(iv) (c): Let the number be x.
According to question, x² - 84 = 3(x + 8)
\(\Rightarrow x^{2}-84=3 x+24 \Rightarrow x^{2}-3 x-108=0\)
(v) (d): Let the number be x.
According to question, x + 12 = \(\frac {160}{x}\)
\(\Rightarrow x^{2}+12 x-160=0\)