We have, vertices of hyperbola lies on X-axis. So, the equation of hyperbola is
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\) --- (i)
Vertices \(\equiv \left( \pm a,0 \right) =\left( \pm 7,0 \right) \Rightarrow a=7\)
\(Now,\quad e=\sqrt { 1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } } \Rightarrow \quad { e }^{ 2 }=1+\frac { { b }^{ 2 } }{ { a }^{ 2 } } \)
\(\Rightarrow \left( \frac { 4 }{ 3 } \right) ^{ 2 }=1+\frac { { b }^{ 2 } }{ { 7 }^{ 2 } } \quad [\because e=4/3\quad and\quad a=7]\)
\(\Rightarrow \frac { { b }^{ 2 } }{ 49 } =\frac { 16 }{ 9 } -1\Rightarrow b^{ 2 }=\frac { 343 }{ 9 } \)
\(Thus,\quad { a }^{ 2 }=49,\quad { b }^{ 2 }=\frac { 343 }{ 9 } \)
Hence, the equation of hyperbola is
\(\frac { { x }^{ 2 } }{ 49 } -\frac { { 9y }^{ 2 } }{ 343 } =1\ \Rightarrow \ { 7x }^{ 2 }-{ 9y }^{ 2 }=343\quad \quad [from\quad Eq.(i)]\)

\(\left( { x }^{ 2 }+3+2\sqrt { 3x } \right) ^{ 3 }=\left( x+\sqrt { 3 } \right) ^{ 6 }\)
\( =^{ n }{ C }_{ o }\left( x \right) ^{ 6 }\times \left( \sqrt { 3 } \right) ^{ o }+^{ n }{ C }_{ 1 }{ x }^{ 5 }\left( \sqrt { 3 } \right) ^{ 1 }+^{ n }{ C }_{ 2 }{ x }^{ 4 }\left( \sqrt { 3 } \right) ^{ 2 }+^{ n }{ C }_{ 3 }{ x }^{ 3 }\left( \sqrt { 3 } \right) ^{ 3 }+^{ n }{ C }_{ 4 }{ x }^{ 2 }\left( \sqrt { 3 } \right) ^{ 4 }+^{ n }{ C }_{ 5 }{ x }^{ 1 }\left( \sqrt { 3 } \right) ^{ 5 }+^{ n }{ C }_{ 6 }\left( \sqrt { 3 } \right) ^{ 6 }\)
\(= { x }^{ 6 }+6\sqrt { 3x^{ 5 } } +45{ x }^{ 4 }+60\sqrt { 3x^{ 3 } } +135{ x }^{ 2 }+54\sqrt { 3x } +27\)

\({ x }^{ 5 }-\frac { 5x^{ 3 } }{ 2 } +\frac { 5x }{ 2 } -\frac { 5 }{ 4x } +\frac { 5 }{ { 16x }^{ 3 } } -\frac { 1 }{ { 32x }^{ 5 } } \)

\(\left( 1-{ x }^{ 2 } \right) ^{ \frac { 7 }{ 3 } }=(1-\frac { 7 }{ 3 } { x }^{ 2 }+\frac { 14 }{ 9 } { x }^{ 4 }-\frac { 14 }{ 81 } { x }^{ 6 }+...),when\left| x \right| <1\)

Given \(\sqrt [ 3 ]{ 126 } =\left( 126 \right) ^{ 1/3 }=\left( 125+1 \right) ^{ 1/3 }\)
\(=\left( 126 \right) ^{ 1/3 }\left( 1+\frac { 1 }{ 125 } \right) ^{ 1/3 }=5\left( 1+\frac { 1 }{ { 5 }^{ 3 } } \right) ^{ 1/3 }\)
\(=5\left[ 1+\frac { 1 }{ 3 } .\frac { 1 }{ { 5 }^{ 3 } } +\frac { \frac { 1 }{ 3 } \left( \frac { 1 }{ 3 } -1 \right) }{ 2! } .\left( \frac { 1 }{ { 5 }^{ 3 } } \right) ^{ 2 }+....... \right] \)
\(=\left( 5+\frac { 1 }{ 75 } -\frac { 1 }{ 28125 } \right) =\left[ 5+\frac { \left( 375-1 \right) }{ 28125 } \right] \)
\(=\left( 5+\frac { 374 }{ 28125 } \right) \)
= (5+0.01329)
= 5.01329

(-1)ⁿ (1-x²)ⁿ
Ans:ⁿC₂

Given expression is \(\left( 1+\sqrt { 5x } \right) ^{ 7 }+\left( 1-\sqrt { 5x } \right) ^{ 7 }\)
Here, n = 7, which is odd.
\(\therefore \) Total number of terms 
\(=\frac { n+1 }{ 2 } =\frac { 7+1 }{ 2 } =\frac { 8 }{ 2 } =4\)

Here, n=10 (even) So there will be one one middle term i.e.\(\left( \frac { 10+2 }{ 2 } \right) \) th term or 6th term
T₆=T₅₊₁ =¹⁰C₅ \(\left( \frac { x }{ 3 } \right) ^{ 10-5 }\)(9y)⁵ =61236 x⁵y⁵

Here, given number without index is 96. Since, it is less than 100, So it can be written as 100-4.
Now,\(​​\left( 96 \right) ^{ 3 }=\left( 100-4 \right) ^{ 3 }\)
\(=^{ 3 }C_{ 0 }\left( 100 \right) ^{ 3 }-^{ 3 }C_{ 1 }\left( 100 \right) ^{ 2 }4+^{ 3 }C_{ 2 }\left( 100 \right) 4^{ 2 }-^{ 3 }C_{ 3 }4^{ 3 }\)
\(\left[ by\quad binomial\quad theorem \right] \)
\( =1000000-3\left( 10000 \right) 4+3\left( 100 \right) \left( 16 \right) -64\)
\(=1000000-120000+4800-64\)
\(=884736\)​​​

\({ \left( a+b \right) }^{ 4 }-{ \left( a-b \right) }^{ 4 }=2\left[ ^{ 4 }{ { C }_{ 1 }{ a }^{ 3 }b^{ 1 }+^{ 4 }{ { C }_{ 3 }{ ab }^{ 3 } } } \right] \)
\(=2\left[ { 4{ a }^{ 3 }b^{ 1 }+4{ { ab }^{ 3 } } } \right] \)
\(=8\left( { a }^{ 3 }b+{ { ab }^{ 3 } } \right) \) 
Put \(a=\sqrt { 3 } ,b=\sqrt { 2 }\) to get the required answer.

The general term in the expansion of \(\left( y-\frac { 1 }{ y } \right) ^{ 20 }\) is
\({ T }_{ r+1 }=^{ 20 }{ C }_{ r }(y)^{ 20-r }\left( -\frac { 1 }{ y } \right) ^{ r }[\because { T }_{ r+1 }=^{ n }{ C }_{ r }{ a }^{ n-r }{ b }^{ r }]\)
\(=^{ 20 }{ C }_{ r }{ y }^{ 20-2r }({ -1 }^{ r })\)
For determining 6th term, put r=5 we get
\({ T }_{ 5+1 }=^{ 20 }{ C }_{ 5 }y^{ 20-10 }(-1)^{ 5 }=-^{ 20 }{ C }_{ 5 }{ y }^{ 10 }\)

Given expression is (1-z)^4. Here, n= 4
\(\therefore \) The number of terms in the expansion is (n+1)
i.e. 4+1 = 5

11ⁿ - 10ⁿ = (1+10)ⁿ - 10n
= 1 + 10n + ⁿC₂(10)² + ⁿC₃(10)³ + -10n
= 1 + 100 {ⁿC₂ + ⁿC₃ 10+...+ ⁿC_n10^n-2}
= 100 x an integer + 1
\(\Rightarrow \) 11ⁿ - 10n leaves remainder 1 when divided by 100.

T_r+1=¹⁰C_r x^10-2r (sin \(\alpha \))^10-r (cos \(\alpha \))^r
For the term independent of x, put 10-2r=0 =>r=5
 Term independent of x=T₆=¹⁰C₅ (sin\(\alpha \) cos\(\alpha \))⁵
=¹⁰C₅ \(\times \)2^-5 (sin 2\(\alpha \))⁵
Clearly, it is greatest when 2\(\alpha \) =\(\frac { \pi }{ 2 } \)
Ans: \(\frac { 10! }{ { 2 }^{ 5 }(5!)^{ 2 } } \)

Let two successive terms be (r+1)th and (r+1)th terms. Then,
T_r+1=²⁴C_r x^r and T_r+2 =²⁴C_r+1 x^r+1
Now, according to the given condition, we have
\(\therefore \)   \(\frac { { ^{ 24 }C }_{ r } }{ { ^{ 24 }C }_{ r+1 } } =\frac { 1 }{ 4 } \)
Ans:5th and 6th terms

Coefficient of (2m + 1)th term = ¹⁰C_2m
and coefficient of (4m + 5)th term = ¹⁰C_{4m + 4}
Now, ¹⁰C_2m = ¹⁰C_{4m + 4}
\(\Rightarrow \frac { 10! }{ \left( 2m \right) !\left( 10-2m \right) ! } =\frac { 10! }{ \left( 4m+4 \right) !\left( 10-4m-4 \right) ! } \)
\(\Rightarrow \frac { 1! }{ \left( 2m \right) !\left( 10-2m \right) ! } =\frac { 1 }{ \left( 4m+4 \right) !\left( 6-4m \right) ! } \)
solve it.