\(\underset { x\rightarrow \frac { \pi }{ 2 } }{ Lim } \frac { { a }^{ cotx }-{ a }^{ cosx } }{ cotx-cosx } \)
\(=\underset { x\rightarrow \frac { \pi }{ 2 } }{ Lim } { a }^{ cosx }\left[ \frac { { a }^{ cotx }-^{ cosx }-1 }{ cotx-cosx } \right] \)
⇒Put cot x - cos x = t
\(\therefore \quad Lim\quad x\rightarrow \frac { \pi }{ 2 } \Rightarrow cot\frac { \pi }{ 2 } -cos\frac { \pi }{ 2 } =t\)
\(\therefore \quad t\rightarrow 0\Rightarrow \underset { t\rightarrow 0 }{ Lim } \frac { { a }^{ t }-1 }{ t } \Rightarrow { log }_{ e }a\) 

Let a be the first term and r be the common ratio of given G.P. then
a₄ = 4 \(\Rightarrow\) ar³ = 4
Now product of first seven terms
= a₁· a₂· a₃· a₄· a₅· a₆· a₇
= (a)(ar)(ar²)(ar³)(ar⁴)(ar⁵)(ar⁶)
= a⁷r^2I = (ar³)7
= (4)⁷  [\(\therefore\)  ar³ = 4]
= 16384.

Origin is shifted to (-3, 2) by a translation
h = -3 and k = 2
Let (x', y') be the new coordinates of the point (x, y)
\(\therefore\) x = x' + h = x' - 3
and y = y' + k = y' + 2
Now substituting the values of x and y in the given of the parabola y² = 4ax we get
(y' + 2)²= 4a (x'-3)
\(\Rightarrow\) y'² + 4 + 4y' = 4ax' -12a
\(\Rightarrow\) y'² + 4y' - 4ax' + 12a + 4 =0
\(\therefore\) the equation of the parabola in new system is y² + 4y'-4ax+12a + 4 = 0.

There are 7 letters in the word
\(\therefore \) Total events = 7!
Now there are 3 vowels in the word 'SOCIETY' when we put three vowels altogether and consider as one letter. Then favourable events = 5! x 3!
\(\text{thus required probability }=\frac { 5!\times 3! }{ 7! } =\frac { 1 }{ 7 } \)

Here 2x² - 4x + 3 = 0
Comparing the given quadratic equation with ax² + bx + c= 0, we have a = 2, b = - 4 and c = 3
\(\therefore x=\frac{-(-4)\pm\sqrt {(-4)^2-4\times2\times3}}{2\times2}\)
\(=\frac{4\pm\sqrt {16-24}}{4}=\frac{4\pm\sqrt {-8}}{4}\)
\(=\frac{4\pm2\sqrt {2i}}{4}=\frac{2\pm\sqrt {2i}}{2}\)
Thus \(x=\frac{2+\sqrt {2i}}{2}\ and\ x=\frac{2-\sqrt {2i}}{2}.\)

Domain = {-3, -2, -1,0, 1,2, 3};Range = {0,1,2,3, 4}

Domain = [2,\(\infty\));Range = [0,\(\infty\))

It is given that P(n) is true i.e.3ⁿ>n.
we have prove that P(n + 1) is true i.e. 3^{n + 1} >(n + 1)
Now 3ⁿ> n
⇒ 3·3ⁿ> 3n
⇒ 3ⁿ⁺¹>n+2n
⇒ 3ⁿ⁺¹>n+1         [∵ 2n> 1for all n∊N]
which shows that P(n+1) is true.

A card is drawn at random from a pack of 52 cards, therefore there are 52 sample points in the sample space.

When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}

The first n natural numbers are 1,2,3...,n.
\(\because\) Standard deviation,
\(SD=\sqrt { \frac { \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } }{ n } -({ \frac { \sum _{ i=1 }^{ n }{ { x }_{ i } } }{ n } ) }^{ 2 } } \)
\(\therefore \ SD=\sqrt { \frac { (n(n+1)(2n+1) }{ 6n } -({ \frac { n(n+1) }{ 2n } ) }^{ 2 } } \)
\([\because \ \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 } and\ \sum _{ i=1 }^{ n }{ { x }_{ i } } =\frac { n(n+1) }{ 2 } ]\)
\(=\sqrt { (n+1)(\frac { 2n+1 }{ 6 } -\frac { n+1 }{ 4 } ) } \)
\(=\sqrt { (n+1)(\frac { 4n+2-3n-3 }{ 12 } ) } \)
\(=\sqrt { \frac { (n+1)(n-1) }{ 12 } } =\sqrt { \frac { { n }^{ 2 }-1 }{ 12 } } \)
\(\therefore \ Variance={ (SD) }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 12 } \)

The inverse of the given statement is 'If you do not get a job, then your credentials are not good'.

Here, \(f(x)=\frac{x^{2}-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=(x+3)\) 
and \(g(x)=x+3\Rightarrow f(x)=g(x)=x+3\).
But \(f(x)\neq g(x)\) as domain of f(x) is \(R-\left\{ 3 \right\} \) and
domain of g(x) is R.
\(\Rightarrow \) Domain of \(f(x)\neq \) Domain of g(x) \(\Rightarrow \) \(f\neq g\)

We are sure that members of this collection are January, June, and July. So, this collection is well-defined. Hence, it is a set.