Let the fraction be \(\frac{x}{y}\), where x is the numerator and y is the denominator.
According to the given condition,
y = 2x + 1
So,the fraction = \(\frac{x}{2 x+1}\)
Reciprocal of the fraction \(=\frac{2 x+1}{x}\)
It is given that the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\).
\(\begin{array}{ll} \therefore & \frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21} \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+(2 x+1)^2}{x(2 x+1)}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+4 x^2+1+4 x}{2 x^2+x}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} \Rightarrow & \frac{x^2+4 x^2+1+4 x}{2 x^2+x}=\frac{58}{21} \\ \end{array}\)
\(\begin{array}{ll} & \frac{5 x^2+4 x+1}{2 x^2+x}=\frac{58}{21} \end{array}\)
\(\Rightarrow\)   21(5x² + 4x + 1) = 58(2x² + x)
\(\Rightarrow\)   105x² + 84x + 21 = 116x² + 58x
\(\Rightarrow\)    -11x² + 26x + 21 = 0
\(\Rightarrow\)      11x² - 26x - 21 = 0
\(\Rightarrow\)  11x² - 33x + 7x - 21 = 0 
\(\Rightarrow\)  11x(x-3) + 7(x - 3) = 0
\(\Rightarrow\)  (x - 3)(11x + 7) = 0
\(\Rightarrow \quad x=3, x \neq-\frac{7}{11}\)
                                 [x cannot be negative]
\(\Rightarrow\)  x = 3
\(\therefore\)   y = 2(3) + 1 = 6 + 1 = 7
Hence, the required fraction = \(\frac{3}{7}\)

6x² - 3 - 7x = 6x² - 7x - 3
=6x² -9x + 2x -3
=3x(2x - 3) + 1(2x - 3)
=(2x - 3)(3x + 1)
\(=2\left(x-\frac{3}{2}\right) 3\left(x+\frac{1}{2}\right)\) 
\(=6\left(x-\frac{3}{2}\right)\left(x+\frac{1}{3}\right)\) 
The zeroes of the polynomial are {3/2, -1/3}
Relationship between the zeroes and the coeffieicients of the polynomial
Also sum of the zeroes = \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6}\) 
Also product of the zeroes = \(\frac{3}{2} \times-\frac{1}{3}=-\frac{1}{2}\) 
Hence Verified.

Let the actual speed of the train be x km/hr and actual time taken be y hr.
Distance = Speed \(\times\) Time
\(\because\) =xy km
According to the given condition,
xy = (x + 10)(y - 2)
\(\Rightarrow\) xy = xy - 2x + 10y-20
\(\Rightarrow\) 2x - 10y + 20 = 0
\(\Rightarrow\) x - 5y + 10 = 0 [divide by 2] ... (i)
and xy = (x - 10)(y + 3)
\(\Rightarrow\) xy = xy + 3x - 10y - 30
\(\Rightarrow\) 3x-10y- 30 = 0 ... (ii)
On multiplying eqn. (i)by 3 and subtracting eqn. (ii) from eqn. (i),
3 \(\times\) (x - 5y + 10) - (3x - 10y - 30) = 0
\(\Rightarrow\) - 5y = - 60
\(\therefore\) y = 12
On substituting y = 12 in eqn. (i),
x - 5 x 12 + 10 = 0
\(\Rightarrow\) x - 60 + 10 = 0
\(\Rightarrow\) x = 50
Hence, the distance covered by the train
= 50 \(\times\) 12
= 600 km.

Let us assume, to the contrary, that \(\sqrt 3\) is rational.
That is, we can find integers a and b (≠ 0) such that \(\sqrt 3\) = a/b ⋅
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b\(\sqrt 3\) = a ⋅
Squaring on both sides, and rearranging, we get 3b² = a ² .
Therefore, a ² is divisible by 3, and by  it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b ² = 9c ² , that is, b ² = 3c ² .
This means that b ² is divisible by 3, and so b is also divisible by 3 (using with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that \(\sqrt 3\) is rational. So, we conclude that \(\sqrt 3\)  is irrational.  we mentioned that :
1. The sum or difference of a rational and an irrational number is irrational and 
2.The product and quotient of a non-zero rational and irrational number is irrational.
We prove some particular cases here.

Given, α and β are the zeroes of the quadratic polynomial p(s)=3s²-6s+4.
\(\therefore \ a+\beta =-\frac { Coefficient \ of \ s }{ Coefficient \ of \ s^{ 2 } } =\frac { -(-6) }{ 3 } =\frac { 6 }{ 3 } =2\)
and \(a\beta =\frac { Constant \ term }{ Coefficient \ of \ s^{ 2 } } =\frac { 4 }{ 3 } \)
Now, \(\frac { a }{ \beta } +\frac { \beta }{ a } +2\left( \frac { 1 }{ a } +\frac { 1 }{ \beta } \right) +3a\beta \)
\(=\frac { a^{ 2 }+\beta ^{ 2 } }{ a\beta } +2\left( \frac { a+\beta }{ a\beta } \right) +3a\beta \)
\(=\frac { (a+\beta )^{ 2 } }{ a\beta } +2\left( \frac { a+\beta }{ a\beta } \right) +3a\beta \)
\(\left[ \because \quad a^{ 2 }+b^{ 2 }=(a+b)^{ 2 }-2ab \right] \)
\(=\frac { (2)^{ 2 }-2(4/3) }{ 4/3 } +2\left( \frac { 2 }{ 4/3 } \right) +3\times \frac { 4 }{ 3 } \)
\(\quad \left[ \because \quad a+\beta =2\quad and\quad a\beta =4/3 \right] \)
\(=\frac { 4-\frac { 8 }{ 3 } }{ \frac { 4 }{ 3 } } +2\times 2\times \frac { 3 }{ 4 } +3+4\)
\(=\frac { 4 }{ 3 } \times \frac { 3 }{ 4 } +7=1+7=8\)

Let us assume that \(2 \sqrt{3}+\sqrt{5}\) is rational. Then, it will be of the form \(\frac{a}{b},\) where a and b are coprime integers and \(b \neq 0\)
Now \(2 \sqrt{3}+\sqrt{5}=\frac{a}{b} \Rightarrow 2 \sqrt{3}=\frac{a}{b}-\sqrt{5}\)
On squaring both sides, we get
\(12=\left(\frac{a}{b}-\sqrt{5}\right)^{2}\)
\(\Rightarrow \quad 12=\frac{a^{2}}{b^{2}}+5-2 \sqrt{5} \frac{a}{b}\)
\(\Rightarrow \quad 2 \sqrt{5} \frac{a}{b}=\frac{a^{2}}{b^{2}}-7\)
\(\Rightarrow \quad \sqrt{5}=\frac{a^{2}-7 b^{2}}{2 a b}\)
Now, \(\frac{a^{2}-7 b^{2}}{2 a b}\) is rational,
\(\Rightarrow \sqrt{5}\) J5is a rational number \(\because\) a, b are integers and b  \(\ne\)0 which is a contradiction to the fact that \(\sqrt{5}\) is irrational. So, our assumption is wrong.
Hence, \(2 \sqrt{3}+\sqrt{5}\) is irrational.
Use the identity (a + b)(a - b) = a² - b²
Rational.

Given polynomial is x² - 8x +  k.
On comparing with ax² + bx + c, we get
a = 1, b = -8 and c = k
Let one of the zeroes be \(\alpha\)
\(\therefore\) Other zero is \(\alpha\) + 2.
We know that
Sum of zeroes = -\(\frac{b}{a}\)
\(\begin{aligned} & \Rightarrow \quad \alpha+\alpha+2=-\frac{b}{a} \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad 2 \alpha+2=-\frac{(-8)}{1} \end{aligned}\)
\(\begin{array}{rrr} \Rightarrow & 2(\alpha+1) & =8 \end{array}\)
\(\begin{array}{rrr} \Rightarrow & \alpha+1 & =4 \end{array}\)
\(\begin{array}{rrr} \Rightarrow & & \alpha=3 \end{array}\)
Therefore, one zero is 3 and other is 3 + 2 = 5
Now, we find the value of k.
\(\because\) Product of zeroes = \(\frac{c}{a}\)
\(\begin{array}{rr} \Rightarrow & \alpha \times(\alpha+2)=\frac{k}{1} \\ \end{array}\)
\(\begin{array}{rr} \Rightarrow & 3 \times 5=\frac{k}{1} \end{array}\)
\(\Rightarrow\)     k = 15
Hence, the value of k is 15.

Let p(t) = t² - 15 = t² - \(\left ( \sqrt{15} \right )^{2}\)
= (t - \(\sqrt{15}\))(t + \(\sqrt{15}\))     [\(\because\) a² - b² = (a - b) (a + b)]
To find zeroes, put p(t) = 0
\(\Rightarrow (t-\sqrt{15})(t+\sqrt{15})=0\)
\(\Rightarrow t-\sqrt{15}=0\) or t + \(\sqrt{15}\) = 0 \(\Rightarrow\) t = \(\sqrt{15}\) or t = -\(\sqrt{15}\)
Hence, zeroes of the given polynomial are -\(\sqrt{15}\) and \(\sqrt{15}\).
Verification
Hence, sum of zeroes = -\(\sqrt{15}\) + \(\sqrt{15}\) = 0 = -(0/1)
\(=-\frac{Coefficient \quad of \quad t}{Coefficient \quad of \quad t^{2}}\)
and product of zeroes = -\(\sqrt{15}\) \(\times\)\(\sqrt{15}\)= -15 = \(\frac{-15}{1}\)
\(=\frac{Constant \quad term}{Coefficient \quad of \quad t^{2}}\)
So, the relationship between the zeroes and its coefficients is verified.

The given pair oflinear equations is
3x-5y+8 = 0
and 7x +6y - 9 = 0
On comparing the given equations with standard form of pair of linear equations i.e. a₁x + b₁y + c₁ = 0 and
a₂x + b₂y + c₂= 0, we get al = 3, b₁ = - 5, c₁ = 8
and  a₂=7, b₂=6, c₂= 9 
Here, \( \frac{a_{1}}{a_{2}}\) = \(\frac{3}{7}\)and \(\frac{b_{1}}{b _{2}}\) = \(\frac{-5}{ 6}\)
  \(\therefore \)  \(\frac{a_{1}}{a_{2}}\neq \frac{b_{1}}{b _{2}}\)
\(\therefore \) The lines representing the given pair of linear equations will intersect at a point.

Hint Let us suppose that \( \sqrt{p} \)+ \( \sqrt{q} \)is a rational
number, Again, let  \( \sqrt{p} \)+ \( \sqrt{q} \) = a, where a is rational.
Therefore, \( \sqrt{q} \)  = a - \( \sqrt{p} \)
On squaring both sides, we get
\({l} q=a^{2}+p-2 a \sqrt{p}\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \)
\(\\ \sqrt {p}=\frac{a^{2}+p-q}{2 a} \)
Since, p and q are primes and a is a rational number, so
\(\frac{a^{2}+p-q}{2 a}\) is rational, therefore \( \sqrt{p} \) is a rational number.
But this contradicts the fact that \( \sqrt{p} \) is irrational
number as p is prime. So, our assumption was incorrect.
Hence, \( \sqrt{p} \) + \( \sqrt{q} \)  is irrational

Smallest number exactly divisible by 520 and 468 = LCM. of 520 and 468 = 4680
\(\therefore\) Required number = 4680 - 17 = 4663

Let the unit's place digit be x and ten's place digit be y.
Original number=10y+x
Number obtained by reversing the order of digits
=10x+y
Sum of both number=99  [given]
(10y+x)+(10x+y)=99
11x+11y=99
x+y=9  .....(i) [dividing both sides by 11]
Also, given that the digits of the numbers differ by 3.
If y>x, then y=-x=3  ..(ii)
and if y On adding Eqs. (i) and (ii), we get
2y=12
\(\Rightarrow\)  \(y= \frac {12}{2}\)=6
Then, from Eq. (i),
x=9-y=9-6=3  [\(\therefore\)y=6]
\(\therefore\)Number=10y+x=10(6)+3=63
Now, adding Eqs. (i) and (iii), we get
2x=12
x=6
Then, from Eq. (i), we get
y=9-x=9-6=3  [\(\therefore\)x=6]
\(\therefore\)Number=10y+x=10(3)+6=36
Hence, the required number 63 and 36

The given pair of linear equations is
5x-4y+8=0   ....(i)
and 7x+6y-9=0    ....(ii)
On comparing with standard form of pair of linear equations, we get
a₁=5, b₁=-4, c₁=8
and a₂=7, b₂=6, c₂=-9
Here, \(\frac { 5 }{ 7 } \neq \frac { -4 }{ 6 } \)    i.e., \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
So, lines (i) and (ii) are intersecting lines.

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.

\(\frac{3}{2} x+\frac{5}{3} y=7,9 x-10 y=14\) 
Here,\(\frac{a_{1}}{a_{2}}=\frac{3}{2 \times 9}=\frac{1}{6}, \frac{b_{1}}{b_{2}}=-\frac{5}{3 \times 10}=\frac{-1}{6}, \frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{1}{2}\) 
\(\because \quad \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) 
\(\therefore\) Pair of equations is consistent with unique solution
 

\(0, \sqrt{5}\) 
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
Given \(\alpha+B=0=-\frac{b}{a}\) 
\(\alpha \beta=\sqrt{5}=\frac{c}{a}\) 
If \(a=1, b=0, c=\sqrt{5}\) 
Therefore, the quadratic polynomial is \(x^{2}+\sqrt{5}\)

Given, HCF of 306 and 657 = 9
We know that
LCM \(\times\) HCF = Product of two numbers
\(\Rightarrow\) LCM \(\times\) 9 = 306 \(\times\) 657
\(\Rightarrow\)   LCM = \( \frac{306 \times 657}{9}\)
= 34 \(\times\) 657 = 22338
\(\therefore\) LCM of 306 and 657 = 22338

If the number 4ⁿ , for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4ⁿ would contain the prime 5. This is not possible because 4ⁿ = (2)²ⁿ ; so the only prime in the factorisation of 4ⁿ is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4ⁿ . So, there is no natural number n for which 4ⁿ ends with the digit zero.
You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it. This method is also called the prime factorisation method. Let us recall this method through an example.

Let HCF = x
ஃ LCM = 14x
A.T,Q. x + 14x = 600 ⇒ x= 40
Now 280 x other number = HCF x LCM : 40 x 560 Other number = 80.

\(x=5,\quad y=\frac { 2 }{ 13 } \)

k=6

(c) Assertion (A) Simplitying the expression \(\sqrt{2}(5-\sqrt{2})=5 \sqrt{2}-(\sqrt{2})^2=5 \sqrt{2}-2\)  It is known that \(\sqrt{2}\) is an irrational number and multiplying an irrational number by a rational number (here, 5) results in an irrational number. Subtracting a rational number (2) from an irrational number still gives us an irrational number.
Thus, \(\sqrt{2}(5-\sqrt{2})\) is an irrational number.
Hence, Assertion (A) is true.
Reason (R) It is known that \(\sqrt{2}\)  is an irrational number. Product of \(\sqrt{2}\) and \(\sqrt{2}\) is equal to 2, which is a rational number.
Therefore, product of two irrational numbers is always an irrational number is false.
Hence, Reason (R) is false.

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

(a) (ii)
96 = 2⁵ x 3
240 = 2⁴ x3  x5
(b) (iii)
Total number of books = 96 +240+336=672
Number of books in each stack = 48
\(\therefore\) Number of stacks formed -= \(\frac{672}{48}=14\) 
(c) (i)
Number of mathmatics books = 336
Number of stacks of mathematics books formed = \(\frac{336}{48}\) 
= 7
(d) (iv)
Number of books in each stack of english books = 48
Thickness of each english book = 3 cm
\(\therefore\)  Height of each stack of english books = (48X3) cm
= 144cm
(e) (iii)
Number of books in a stack of hindi books = 48
Weight of each hindi book = 1.5kg
\(\therefore\) The weight of books in a stack of hindi books
= (48X1.5)kg = 72kg

(i) (b)
(ii) (d): Since, the parabola intersects the x-axis at -3 and 4. So, zeroes of the polynomial are -3 and 4.
(iii) (c): Let \(f(x)=x^{2}-x-12\)
\(=x^{2}-4 x+3 x-12=(x+3)(x-4) \)
\(\text { Consider } f(x)=0 \Rightarrow(x+3)(x-4)=0 \Rightarrow x=4,-3\)
(iv) (b): Sum of zeroes \(=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}} \)
\(=-\frac{(-9)}{4}=\frac{9}{4}\)
(v) (c): Product of zeroes \(=\frac{20}{1}=20\)
\(\therefore\) Reciprocal of product of zeroes \(=\frac{20}{1}\)

(i) (c): For finding \(\alpha,\beta,\) \(\gamma\) consider p(x) = 0
\(\Rightarrow \quad x^{3}-18 x^{2}+95 x-150=0 \)
\(\Rightarrow \quad(x-3)\left(x^{2}-15 x+50\right)=0 \)
\(\Rightarrow \quad(x-3)(x-5)(x-10)=0 \Rightarrow x=10 \text { or } x=5 \text { or } x=3 \)
\(\text { Thus } \alpha=10, \beta=5 \text { and } \gamma=3\)
(ii) (d): Here \(\alpha=10, \beta=5 \text { and } \gamma=3\)
\(\therefore\) Sum of product of zeroes taken two at a time
\(\begin{array}{l} =\alpha \beta+\beta \gamma+\gamma \alpha=(10)(5)+(5)(3)+(3)(10) \\ =50+15+30=95 \end{array}\)
(iii) (a): Product of zeroes of polynomial p(x) = \(\alpha\beta\gamma\)
= (10) (5) (3) = 150
(iv) (b): We have \(p(x)=x^{3}-18 x^{2}+95 x-150\)
\(\begin{array}{l} \text { Now, } p(4)=4^{3}-18(4)^{2}+95(4)-150 \\ =64-288+380-150=6 \end{array}\)
(v) (d): \(g(x)=x^{3}-(\alpha+\beta+\gamma) x^{2} +(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma \)
\(\Rightarrow g(x)=x^{3}-3 x^{2}-16 x-(-48)=x^{3}-3 x^{2}-16 x+48\)