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

Given \(\frac { PS }{ SQ } =\frac { PT }{ TR } \) and \(\angle PRQ\)
To prove \(\triangle PQR\) is an isosceles triangle, i.e. PQ = PR.
Proof Since, \(\frac { PS }{ SQ } =\frac { PT }{ TR } \)
\(\therefore ST\parallel QR\)
[ by converse of basic proportionality theorem]
Then, \(\angle PST=\angle PQR\)    [corresponding angles] ... (i)
Also,  \(\angle PST=\angle PRQ\)    [given]   ... (ii)
From Eqs. (i) and (ii), 
\(\angle PRQ=\angle PQR\Rightarrow PQ=PR\)
[since, sides opposite to equal angles of a triangle are also equal]
Hence, \(\triangle PQR\)  is an isosceles triangle.

Let  \(3\sqrt { 2 } \)  be a rational number. Then, it will be of the form  \(\frac { p }{ q } \) , where p, q are coprime integers and  \(q\neq 0\).
Now, \(\frac { p }{ q } \) = \(3\sqrt { 2 } \)   \(\Rightarrow \quad \frac { p }{ 3q } =\sqrt { 2 } \)
Since, p is an integer and 3q is also an integer \(\left( 3q\neq 0 \right) \).
So, \(\frac { p }{ 3q } \) is a rational number.
\(\Rightarrow \quad \sqrt { 2 } \) is a rational number.
But this contradicts the fact that \(\sqrt { 2 } \) is an irrational number.
Hence, \(3\sqrt { 2 } \) is an irrational number.

On applying Euclid's division lemma for 960 and 432, we get
960 = (432 x 2) + 96
Here, remainder = 96 \(\neq \) 0,
so take new dividend as 432 and divisor as 96.
Then, we get 432 = (96 x 4) + 48
Here, remainder = 48 \(\neq \) 0,
so take new dividend as 96 and divisor as 48.
Then. we get 96 = (48 x 2) + 0
Here, the remainder is 0 (zero) and last divisor is 48.
Hence, HCF of 960 and 432 is 48.

Let a be any odd positive integer, then on dividing a by b, we have a = bq + r,\( \ 0\le r ..\). (i) [by Euclid's division lemma]
On putting b = 2 in Eq. (i), we get
a = bq + r,          \(0\le\) r   \(\Rightarrow \)  r = 0 or 1
If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.
If r = 1, then a = 2q + 1, which is not divisible by 2.
\(\therefore \quad \left( 2q+1 \right) \) is odd.
Now, as a is odd, so it cannot be of the form 2q. Thus, any odd positive integer a is of the form (2q + 1).
Now, consider \({ a }^{ 2 }=\left( 2q+1 \right) ^{ 2 }={ 4q }^{ 2 }+1+4q\quad \left[ \because \quad \left( x+y \right) ^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy \right] \) 
= 4(q² + q) + 1 = 4 m + 1,   where m = q² + q
Hence, for some integer m, the square of any odd integer is of the form 4m + 1.

a x 13 = 39
= 3

First, write the given number in the form of p/q and then factorise the denominator.
(i) Denominator is of the form 2ⁿ 5^m
(ii) Denominator contains a factor other than 2 or 5.

Irrational

 We have, 7 x 11 x 13 + 13 = 13(7 x 11 + 1)
= 13 x 78
= 2 x 3 x 13²
As it is the product of prime factors 2, 3 and 13.
Hence, it is a composite number.

a is any positive integer

\({ x }^{ 2 }-4\sqrt { 3 } x+3=0\)
If \(\alpha\) and \(\beta\) are the zeroes of \({ x }^{ 2 }-4\sqrt { 3 } x+3\)
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha +\beta =-\frac { \left( -4\sqrt { 3 } \right) }{ 1 } \)
\(\Rightarrow \quad \alpha +\beta=4\sqrt { 3 }\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{3}{1}\)
\(\Rightarrow \quad \alpha\beta=3\)
\(\therefore \quad \alpha +\beta -\alpha \beta =4\sqrt { 3 } -3\)

f(x) = x² - 7x - 8
Let other zero be k, then
Sum of zeroes \(-1+k=-\left( \frac { -7 }{ 1 } \right) =7\)
\(\Rightarrow k = 8\)

p(x) = 3x² - kx + 6
Sum of the zeroes = 3 \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad 3=-\frac { \left( -k \right) }{ 3 } \)
k = 9

The smallest prime number is 2 and the smallest composite number is 2² Hence, required HCF (2²,2) = 2.

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.

The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number

Let α and \(\frac { 1 }{ \alpha } \) be two zeroes of the given polynomial, which are reciprocal of each other.
On comparing the given polynomial with Ax²+Bx+C, we get
A=a²+9, B=13 and C=6a
Now, product of zeroes,
\(\alpha \times \frac { 1 }{ \alpha } =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \) 
\(\\ \Rightarrow \ 1=\frac { 6\alpha }{ a^{ 2 }+9 }\) 
\( \\ \Rightarrow \ a^{ 2 }+9=6a\) 
\(\\ \Rightarrow \ a^{ 2 }-6a+9=0\) 
\(\\ \Rightarrow \ (a-3)^{ 2 }=0\ \ \left[ \because \quad (x-y)^{ 2 }=x^{ 2 }+y^{ 2 }-2xy \right] \) 
\(\\ \therefore \ a=3\)

We know that a polynomial in one variable x, is an algebraic expression of the form,
\(p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{1} x+a_{0}\)  where n is a positive integer and a₀, a₁, a₂,..., a_n are constants.
(i) No, because powers of any polynomial is always positive integer.
(ii) Yes, because it satisfies the definition of polynomial.
(iii) No, because powers of any polynomial can not be negative integer.
(iv) Yes, because it satisfies the condition of polynomial.

Rational

\(\because \) smallest odd composite number = 9
and smallest odd prime number = 3.
\(\therefore \) HCF of 9 and 3 = 3
and LCM of 9 and 3 = 9
Now, if an odd number p divides q², then p is one of the factors of q², i.e. q² = pm, for some integer m.    .... (i) 
Now, q³ = q² . q \(\Rightarrow \) q³ = pm . q   [from Eq.(i)]
\(\Rightarrow \)  q³ = p (mq)
\(\Rightarrow \)  p is a factor of q³ also \(\Rightarrow \)  p divides q³