Consider a positive odd integer as a.
On dividing a by b, let q be the quotient and r be the remainder.
Then, a = bq + r,    .... (i)
[by Euclid's division lemma]
On putting b= 6 in Eq.(i) we get
a = 6q + r, \(0\le r<6\)     .... (ii)
So, possible values of r = 0, 1, 2, 3, 4, 5
If r = 0, then from Eq.(ii), a = 6q
Here, 6q is divisible by 2, so 6q is even.
If r = 1, then from Eq.(ii), a = 6q + 1
Here, 6q + 1 is not divisible by 2, so 6q + 1 is odd.
If r = 2, then from Eq.(ii), a = 6q + 2
Here, 6q + 2 is divisible by 2, so 6q+ 2 is even.
If r = 3, then from Eq.(ii), a = 6q + 3
Here, 6q + 3 is not divisible by 2, so 6q+ 3 is odd.
If r = 4, then from Eq.(ii), a = 6q + 4
Here, 6q+ 4  is divisible by 2, so 6q + 4 is even.
If r = 5, then from Eq.(ii), a = 6q + 5
Here, 6q + 5 is not divisible by 2, so 6q + 5 is odd.
Since, a is odd, so a cannot be 6q, 6q + 2, 6q + 4
Hence, any positive odd integer is of the form 6q + 1, 6q + 3 and 6q + 5.

By division algorithm, p(x)=g(x).q(x)+r(x)=6x²-7x-5

2, 2

Fundamental theorem of Arithmetic: Every composite number can be expressed as a product of primes and decomposition is unique, apart from the order in which the prime factors occur.
HCF = 24
LCM = 540
Let's calculate \(\frac { LCM }{ HCF } =\frac { 540 }{ 24 } =22.5\) not an integer
Since LCM is always a multiple of HCF, hence, two numbers cannot have HCF and LCM as 44 and 540 respectively.

In the polynomial \(7q^{ 6 }+4q^{ 2 }+\frac { 3 }{ 2 } +q-8\) , the highest power of the variable q is 6.

3x²+8x+4;-2 and \(-\frac { 2 }{ 3 } \)

\(\frac { 15 }{ 4 } +\frac { 5 }{ 40 } =\frac { 15 }{ 4 } \times \frac { 25 }{ 25 } +\frac { 5 }{ 40 } \times \frac { 25 }{ 25 } \)
​​​​\(=\frac { 375 }{ 100 } +\frac { 125 }{ 1000 } \)
= 3.75 + 0.125 = 3.875

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

We have, \(\frac{51}{1500}\) \(=\frac{17}{500}\)
Prime factors of 500 = 2 x 2 x 5 x 5 x 5 = 2² x 5³
which is in the form 2ⁿ x 5^m
So, it has terminating decimal expansion.
Now, \(\frac { 51 }{ 1500 } =\frac { 17 }{ { { 2 }^{ 2 }\times 5 }^{ 3 } } \times \frac { { 2 } }{ { 2 } } =\frac { 34 }{ { 10 }^{ 3 } } =0.034\)
which is the required decimal expansion.

(i) Let the zeroes of p(x) be \(\alpha\) and \(\beta\) and \(\alpha=3 \beta\).
Sum of zeroes \(=\alpha+\beta=\frac{-(-8)}{a}\)
\(\Rightarrow  3 \beta+\beta=\frac{8}{a} \Rightarrow \beta=\frac{2}{a}\)
Product of zeroes \(=\alpha \beta=\frac{3}{a} \Rightarrow \beta^2=\frac{1}{a}\)
From Eqs. (i) and (ii), we get
a=4
(ii) Since, a = 4 > 0,
Therefore The graph of $p(x)$ is an open upward parabola.

Quotient, q(x) = x² - 3x + 2 and remainder, r(x) = 2.

\(\therefore\) a is smaller than b
\(\therefore\)  a = b x 0 + a
Comparing with a = bq + r we get q = 0

Number of zeroes of a polynomial p(x) is equal to numbers of the points of intersection of the graph of P(x) with x-axis
∵ the graph of p(x) does not intersects the x-axis
∴ It has no zero

This is non-terminating decimal expansion.

\(\frac{35}{50}=\frac{7}{10}=0.70\)

(i) (c): Here, the simplest form of given options are
125/441 = 5³/(3² x 7²), 77/210 = 11/(2 x 3 x 5),
15/1600 = 3/(2⁶ x 5) Out of all the given options, the denominator of option (c) alone has only 2 and 5 as factors. So, it is a terminating decimal.
(ii) (b): 23/(2³ x 5²) = 23/200 = 0.115
(iii) (a): 441/(2² x 5⁷ x 7²) = 9/(2² x 5⁷), which is a terminating decimal.
(iv) (d): The fraction form of a non-terminating recurring decimal will have at least one prime number other than 2 and 5 as its factors in denominator. So, p can take either of 3, 7 or 15.
(v) (a): Here denominator has only two prime factors i.e., 2 and 5 and hence it is a terminating decimal.

(i) (b): Graph of a quadratic polynomial is a parabolic in shape.
(ii) (c): Since the graph of the polynomial cuts the
x-axis at (-6,0) and (6, 0). So, the zeroes of polynomial are -6 and 6.
\(\therefore\) Required polynomial is p(x) = x² - (-6 + 6)x + (-6)(6) = x² - 36
(iii) (c) : We have, p(x) = x² - 36
Now, p( 6) = 62 - 36 = 36 - 36 = 0
(iv) (b): Letf (x) = x² + 2x - 3. Then,
\(\text { Sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}=-\frac{(2)}{1}=-2\)
(v) (d): The given polynomial is at²+ 5t + 3a Given, sum of zeroes = product of zeroes.
\(\Rightarrow \quad \frac{-5}{a}=\frac{3 a}{a} \Rightarrow a=\frac{-5}{3}\)

(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 \(\frac{-2}{k}=3\)
\(\Rightarrow \quad k=\frac{-2}{3}\)