First, we will find the HCF of 65 and 117 by using Euclid's division algorithm, then put 65 m -117 =HCF and simplify.
= 2

Let three consecutive integers are n, (n + 1) and (n + 2).
Then, one of these three must be divisible by 2 and another one must be divisible by 3.
Hence, the product of numbers is divisible by 6.
e.g.lf n = 1, then the three consecutive numbers are 1,2,3.
∴ Product = 1 x  2 x 3 := 6;
So, it is divisible by 6.
Given statement is true.

The less than type frequency distribution of given table is

Hence the required number of athletes, who completed the race in less than 14.6 is 82.

Here, the highest frequency is 30, which lies in the interval 30-40. So, it is the modal class, i.e. 30-40.

LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.

1

No, here degree of (x-1)=degree of (2x+3)=1.
We know that, degree of remainder  (x) is always less than the degree of divisior g(x).
i.e. degree r(x) So, (x-1) cannot be remainder of a polynomial p(x), when divided by (2x+3).

Let α and α are the same zeroes of the given polynomial ax²+bx+c. Then, we have
\(\alpha .\alpha =\frac { c }{ a } \Rightarrow a^{ 2 }=\frac { c }{ a } \Rightarrow \frac { c }{ a } >0\quad \left[ \because \alpha ^{ 2 } \text{is always positive} \right] \)
⇒ c and a have same sign.

True, since, \(-\frac { b }{ a } \)= sum of the zeroes <0, so \(\frac { b }{ a } \) >0. Also, the product of the zeroes =\(\frac { c }{ a } \) >0
∴ a,b and c all have same sign.

No, because 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
If the number 6ⁿ ends with digit 5, then it will be divisible by 5, i.e. its one factor will be 5. But the only primes in the factorisation of 6ⁿ are 2 and 3, but not 5.
Hence, it cannot end with digit 5.

False

We know that, the nth term of an AP from the end is 
\({a}_{n}=l-(n-1)d\)       ...(i)
Here, l = last term = 49
and common difference, d = -8 - (-11)
= - 8 + 11 = 3
From Eq. (i), \({ a }_{ 4 }=49-(4-1)3=49-9=40\) 

Given, d = - 4, n = 7 and \(a_n=4\) 
We know that,
\({a}_{n}=a + (n-1)d\)
\(\Rightarrow\) 4 = a + (7 - 1) (-4)                  
\(\Rightarrow\) 4 = a + 6 (-4) 
\(\Rightarrow\)   4+24 = a
\(\Rightarrow\) a = 28

Here,   \(OP\ \bot \ PR\)            [radius is perpendicular to the tangent at the point of contact]
\(\Rightarrow \)    \(\angle OPR={ 90 }^{ \circ }\)
\(\Rightarrow \)    \(\angle OPQ+\angle QPR={ 90 }^{ \circ }\)
\(\Rightarrow \)    \(\angle OPQ+{ 50 }^{ \circ }={ 90 }^{ \circ }\)       [ \(\because\) \(\angle OPQ+{ 50 }^{ \circ }\), given]
\(\Rightarrow \)     \(\angle OPQ={ 90 }^{ \circ }-{ 50 }^{ \circ }={ 40 }^{ \circ }\)
In \(\triangle OPQ\), we have
OP = OQ             [radii of circle]
\(\therefore\) \(\angle OPQ+\angle OQP={ 40 }^{ \circ }\)     [ \(\because\) angles corresponding to equal sides are equal]
Now,   \(\angle POQ={ 180 }^{ \circ }-\left( \angle OPQ+\angle OQP \right)\)
\(={ 180 }^{ \circ }-\left( { 40 }^{ \circ }+{ 40 }^{ \circ } \right) ={ 100 }^{ \circ }\quad \)

Elementary events associated to the given random experiment, throwing an ordinary die are 1,2,3,4,5,6.
∴  n(S)=6
Let E be the events of getting multiple of 3, i.e.3,6.
∴    n(E)=12
Now, \(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
Hence, the probability of getting multiple of 3 in a single throw of an ordinary die is \(\frac { 1 }{ 3 } \) .

Total number of outcomes, n(S)=100
Let E=Event of getting a prime number
={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,71,73,79,83,89,97}
∴ n(E)=25
Hence, P (getting a prime number)
\(=\frac { n(E) }{ n(S) } =\frac { 25 }{ 100 } =\frac { 1 }{ 4 } \)

an isosceles triangle

\(5\sqrt { 2 }\ \ units\)

38

0

\({3\over16}\)

No, for example equation 5x² - 7x + 2 = 0 has integral coefficient but its roots are 1 and \(\frac { 2 }{ 5 } and \frac { 2 }{ 5 } \) is not an integer.

Yes, (x-1) (x+2) +2=0
\(\Rightarrow x^{ 2 }+x-2+2=0\)
\(\Rightarrow x^{ 2 }+x=0\)
\(\therefore \) b²-4ac=1>0
Distinct real roots

Yes \(\sqrt { 2 } x^{ 2 }-\frac { 3 }{ \sqrt { 2 } } x+\frac { 1 }{ \sqrt { 2 } } =0\)
\(\Rightarrow 2x^{ 2 }-3x+1=0\)
b²-3x+1=0
Distinct real roots.