Here f(x)=\(\frac{x^{n}-a^{n}}{x-a}\)
∴ \(f^{'}(x)=\frac{d}{dx}[\frac{x^{n}-a^{n}}{x-a}]\)
=\(\frac{(x-a)\frac{d}{dx}(x^{n}-a^{n})-(x^{n}-a^n)\frac{d}{dx}(x-a)}{(x-a)^{2}}\)
=\(\frac{(x-a)\times nx^{n-1}-(x^{n}-a^{n}\times 1)}{(x-a)^{2}}\)
=\(\frac{nx^{n}-anx^{n-1-(x^{n}+a^{n})}}{(x-a)^{2}}\)
 

Here, h=5, k=-2 and r=3
(x - h)² + (y - k)² = r²
∴ (x - 5)² + (y + 2)² = (3)²
⇒ x² + 25 - 10x + y² + 4 + 4y = 9
⇒ x² + y² - 10x + 4y + 29 - 9 = 0
⇒ x² + y² -10x + 4y + 20 =0
which is required equation of circle.

Use P(A∪B) = P(A) + P(B),for mutually exclusive events.
Ans.0.3

Given limit = \(\lim_ { x\rightarrow \pi /2 }{ lim } \frac { 1-{ sin }^{ 2 }x }{ 1-sinx } =\lim_ { x\rightarrow \pi /2 }{ lim } 1+sin \ x\)
Ans. 2

Given, foci lies on X-axis, So, the equation of ellipse will be of the form
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1,\quad a>b\)
Also, given that, b = 3 and c = 4
\(\therefore{ c }^{ 2 }={ a }^{ 2 }-{ 3 }^{ 2 }\Rightarrow { \left( 4 \right) }^{ 2 }={ a }^{ 2 }-9 \Rightarrow 16={ a }^{ 2 }-9\)
\(\Rightarrow { a }^{ 2 }=16+9 \Rightarrow { a }^{ 2 }=25\)
On putting the values of \({ a }^{ 2 }=25\) and \({ b }^{ 2 }=9\) in Eq. (i) we get
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Given,GP is \(3,\frac { 3 }{ 2 } ,\frac { 3 }{ 4 } \),...
Here,  a=3,\(r=\frac { 3 }{ 2 } \div 3=\frac { 1 }{ 2 } \)
Let n be the number of terms needed
Thus S_n=\(\frac { 3069 }{ 512 } \)\(​​\Rightarrow \frac { a(1-{ r }^{ n }) }{ 1-r } =\frac { 3069 }{ 512 }\quad [\because r<1]\)
\(\Rightarrow \frac { 3\left\{ 1-\frac { 1 }{ { 2 }^{ n } } \right\} }{ 1-\frac { 1 }{ 2 } } =\frac { 3069 }{ 512 } \)
\( \Rightarrow 6\left( 1-\frac { 1 }{ { 2 }^{ n } } \right) =\frac { 3069 }{ 512 } \Rightarrow 1-\frac { 1 }{ { 2 }^{ n } } =\frac { 3069 }{ 3072 }\)
\(\Rightarrow \frac { 1 }{ { 2 }^{ n } } =1-\frac { 3069 }{ 3072 } =\frac { 3072-3069 }{ 3072 } \\ \)
\(\Rightarrow \frac { 1 }{ { 2 }^{ n } } =\frac { 3 }{ 3072 } =\frac { 1 }{ 1024 } \Rightarrow 2^{ v }=1024\Rightarrow { 2 }^{ n }={ 2 }^{ 10 }\) 
On comparing the powers from both sides, we get 
n=10
hence,10 terms are needed to give the sum\(\frac { 3069 }{ 512 } \).

There are 7 letters in the word ALGEBRA. In these letters, 'A' occurs twice and rest all are different. 
(i) Since two A's are always together, so let us consider two 'A' s as one letter.
\({ AA }\) , L, G, E, B, R Now, we have 6 letters, which can be arranged in ⁶P6 = 6! ways
Hence, required number of ways = 6! = 720 
[\(\because \) two A's can be arranged among themselves in one way only]

\(Given,\quad cos\theta =\frac { cos\alpha -cos\beta }{ 1-cos\alpha cos\beta } \)
\(\Rightarrow \frac { 1-tan^{ 2 }\frac { \theta }{ 2 } }{ 1+tan^{ 2 }\frac { \theta }{ 2 } } =\frac { cos\alpha -cos\beta }{ 1-cos\alpha cos\beta } \quad \left[ \because cosx=\frac { 1-tan^{ 2 }\frac { x }{ 2 } }{ 1+tan^{ 2 }\frac { x }{ 2 } } \right] \)
\(\Rightarrow \frac { 1-tan^{ 2 }\frac { \theta }{ 2 } +1+tan^{ 2 }\frac { \theta }{ 2 } }{ 1-tan^{ 2 }\frac { \theta }{ 2 } -1-tan^{ 2 }\frac { \theta }{ 2 } }\)
\( =\frac { cos\alpha -cos\beta +1-cos\alpha cos\beta }{ cos\alpha -cos\beta -1+cos\alpha cos\beta } \)
\([applying\quad componendo\quad and\quad dividendo]\)
\(\Rightarrow -\frac { 2 }{ 2tan^{ 2 }\frac { \theta }{ 2 } } =\frac { (cos\alpha +1)(1-cos\beta ) }{ (cos\alpha -1)(cos\beta +1) } \)
\(\Rightarrow \frac { 1 }{ 2tan^{ 2 }\frac { \theta }{ 2 } } =\frac { (1+cos\alpha )(1-cos\beta ) }{ (1-cos\alpha )(1+cos\beta ) } \)
\(\Rightarrow \frac { 1 }{ tan^{ 2 }\frac { \theta }{ 2 } } =\frac { \left( 1+2cos^{ 2 }\frac { \alpha }{ 2 } -1 \right) \left( 1-1+2sin^{ 2 }\frac { \beta }{ 2 } \right) }{ \left( 1-1+2sin^{ 2 }\frac { \alpha }{ 2 } \right) \left( 1+2cos^{ 2 }\frac { \beta }{ 2 } -1 \right) } \)
\(\Rightarrow \frac { 1 }{ tan^{ 2 }\frac { \theta }{ 2 } } =\frac { 4cos^{ 2 }\frac { \alpha }{ 2 } sin^{ 2 }\frac { \beta }{ 2 } }{ 4sin^{ 2 }\frac { \alpha }{ 2 } cos^{ 2 }\frac { \beta }{ 2 } } \Rightarrow tan^{ 2 }\frac { \theta }{ 2 } =\frac { sin^{ 2 }\frac { \alpha }{ 2 } cos^{ 2 }\frac { \beta }{ 2 } }{ cos^{ 2 }\frac { \alpha }{ 2 } sin^{ 2 }\frac { \beta }{ 2 } } \)
\(\Rightarrow tan^{ 2 }\frac { \theta }{ 2 } =tan^{ 2 }\frac { \alpha }{ 2 } cot^{ 2 }\frac { \beta }{ 2 } \)
\(\therefore \quad tan\frac { \theta }{ 2 } =\pm tan\frac { \alpha }{ 2 } cot\frac { \beta }{ 2 } \)
Hence proved.

If \((a,a)\in R\)  , then  \(a={ a }^{ 2 }\). It is true for \(a=1\epsilon N\) only \(\therefore (1,1)\in R\) but \((2,2),(3,3),(4,4)\).....does not belongs to R.So, is not true

B = {....., -6, -4, -2, 0, 2, 4, 6,......}, so B is an infinite set.

We have
L.H.S. = sin 3x + sin 2x - sin x.
= (sin 3x - sin x) + sin 2x
\(=[2cos({3x+x\over2}) sin({3x+x\over2})]+2sin x cos x\)
\([\because \ sin C- sin D=2cos {C+D\over2}.sin{C-D\over2}]\)
= 2 cos 2x sin x + 2 sin x cos x
= 2 sin x [cos 2x + cos x]
= 2 sin x\([2cos({2cos ({2x+x\over2})}cos({2x+x\over2})]\)
\(=2sin x[2cos({3x\over2})cos({x\over2})]\)
= 4 sin x cos \({x\over2}\) cos \({3x\over2}\) =R.H.S

Required probability = \(\frac { 12 }{ 52 } \)
\(\frac { 3 }{ 13 } \)

Arranging the data in ascending order, we have 44, 48, 52, 54, 56, 58, 64, 65, 73, 80
Here, n=10. So, median is the mean of 5th and 6th terms.
\(\therefore \) Median (M) = \(\left( \frac { 56+58 }{ 2 } \right) =57\)
We make the table from the given data.

\(\therefore \) Mean deviation = \(\frac { \sum { |{ x }_{ i }-M| } }{ n } =\frac { 86 }{ 10 } =8.6\)
Hence, the mean deviation from the median is 8.6.

\(PQ=QR\quad and\quad { PQ }^{ 2 }+{ QR }^{ 2 }={ PR }^{ 2 }\)

Slope of the required perpendicular line=1/3
\(\therefore \) Equation of the required perpendicular line is
y-2=(1/3)(x-2)
For y-intercept put x=0 ,y = 4/3

We first expand each of the factors of the given product using Binomial
Theorem. We have
(1+2a)⁴ \(\left[ { ^{ 4 }C }_{ 0 }+{ ^{ 4 }C }_{ 1 }(2a)+{ ^{ 4 }C }_{ 2 }(2a)^{ 2 }+{ ^{ 4 }C }_{ 3 }(2a)^{ 3 }+{ ^{ 4 }C }_{ 4 }(2a)^{ 4 } \right] \)
= 1 + 4 (2a) + 6(4a²) + 4 (8a³) + 16a⁴.
= 1 + 8a + 24a² + 32a³ + 16a⁴
and (2-a)⁵ \(\left[ { ^{ 5 }C }_{ 0 }({ 2 }^{ 5 })-{ ^{ 5 }C }_{ 1 }(2)^{ 4 }a+{ ^{ 5 }C }_{ 2 }(2)^{ 3 }(a)^{ 2 }-{ ^{ 5 }C }_{ 3 }(2)^{ 2 }(a)^{ 3 }+{ ^{ 5 }C }_{ 4 }(2)(a)^{ 4 }-{ ^{ 5 }C }_{ 5 }(a)^{ 5 } \right] \)
= 32 – 80a + 80a² – 40a³ + 10a⁴ – a⁵
Thus (1 + 2a)⁴ (2 – a)⁵
\(\left[ 1+8a+{ 24 }a^{ 2 }+{ 32a }^{ 3 }+{ 16a }^{ 4 } \right] \times \left[ 32-80a+{ 80a }^{ 2 }-{ 40a }^{ 3 }+{ 10a }^{ 4 }-{ a }^{ 5 } \right] \)The complete multiplication of the two brackets need not be carried out. We write only
those terms which involve a⁴. This can be done if we note that a^r. a^{4 – r} = a⁴. The terms containing a⁴ are
1 (10a⁴) + (8a) (–40a³) + (24a²) (80a²) + (32a³) (– 80a) + (16a⁴) (32) = – 438a⁴
Thus, the coefficient of a⁴ in the given product is – 438.

We have, z₁ = 3+2i and z₂ = 2-4i
Now, LHS= \({ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }+ { \left| { z }_{ 1 }-{ z }_{ 2 } \right| }^{ 2 }\)
On substituting the value of z₁ and z_{2, we get
LHS =} \({ \left| { z }_{ 1 }+{ z }_{ 2 } \right| }^{ 2 }+ { \left| { z }_{ 1 }-{ z }_{ 2 } \right| }^{ 2 }=2({ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 })\)
\(={ \left| 3+2i+2-4i \right| }^{ 2 }+\left| 3+2i+2-4i \right| \)
\( ={ \left| 5-2i \right| }^{ 2 }+{ \left| 1+6i \right| }^{ 2 }={ (5) }^{ 2 }+{ (-2) }^{ 2 }+{ (1) }^{ 2 }+{ (6) }^{ 2 }\)
\( [if\quad z=a+ib,\quad then\quad { \left| z \right| }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }]\)
\(=25+4+1+36\)
\(\because \quad LHS=66\quad ......(i)\)
\(and\ RHS=2({ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 } =2({ \left| 3+2i \right| }^{ 2 }+{ \left| 2-4i \right| }^{ 2 })\)
\(=2[{ (3) }^{ 2 }+{ (2) }^{ 2 }+{ (2) }^{ 2 }+{ (-4) }^{ 2 }]\)
\(=2(9+4+4+16) =2\times 33\)
\(\because \quad RHS=66\quad ....(ii)\)
From Eq (i) and (ii), we get LHS = RHS.
Hence proved.

\(​​LHS=\frac { co{ s }^{ 2 }(\frac { B-C }{ 2 } ) }{ (b+c{ ) }^{ 2 } } +\frac { si{ n }^{ 2 }(\frac { B-C }{ 2 } ) }{ (b-c{ ) }^{ 2 } }\)
\(=\frac { co{ s }^{ 2 }(\frac { B-C }{ 2 } ) }{ (ksinB+ksinC{ ) }^{ 2 } } +\frac { si{ n }^{ 2 }(\frac { B-C }{ 2 } ) }{ (ksinB-KsinC{ ) }^{ 2 } } \)
\(=\frac { co{ s }^{ 2 }(\frac { B-C }{ 2 } ) }{ { k }^{ 2 }(sinB+sinC{ ) }^{ 2 } } +\frac { si{ n }^{ 2 }(\frac { B-C }{ 2 } ) }{ { K }^{ 2 }(sinB-sinC{ ) }^{ 2 } } \)
\(=\frac { 1 }{ { K }^{ 2 } } \left[ \frac { co{ s }^{ 2 }(\frac { B-C }{ 2 } ) }{ 4si{ n }^{ 2 }\left( \frac { B+C }{ 2 } \right) co{ s }^{ 2 }\left( \frac { B-C }{ 2 } \right) } +\frac { si{ n }^{ 2 }\left( \frac { B-C }{ 2 } \right) }{ 4co{ s }^{ 2 }(\frac { B+C }{ 2 } )si{ n }^{ 2 }\left( \frac { B-C }{ 2 } \right) } \right] \)
\(=\frac { 1 }{ { k }^{ 2 } } \left[ \frac { 1 }{ 4si{ n }^{ 2 }\left( \frac { B+C }{ 2 } \right) } +\frac { 1 }{ 4co{ s }^{ 2 }\left( \frac { B+C }{ 2 } \right) } \right] \)
\( =\frac { 1 }{ { k }^{ 2 } } \left[ \frac { co{ s }^{ 2 }(\frac { B+C }{ 2 } +si{ n }^{ 2 }\left( \frac { B+C }{ 2 } \right) }{ \{ 2si{ n }\left( \frac { B+C }{ 2 } \right) cos\left( \frac { B+C }{ 2 } \right) { \} }^{ 2 } } \right] =\frac { 1 }{ { k }^{ 2 }si{ n }^{ 2 }(B+C) } \)
\(=\frac { 1 }{ { k }^{ 2 }si{ n }^{ 2 }A } ={ a }^{ -2 }=RHS\)

\(X\cap Y=\{ b,d\} \)

p: All the angles of a triangle are equal.
q: the triangles is an obtuse angled triangled.
We have to show that if p then \(\sim \) q. Take each angle equal to \({ 60 }^{ \circ }\), it means they are acute angle.
Hence, we conclude that the given statement is false.

Let P(n) = (xⁿ-yⁿ) is divisible by (x - y) for all n ∊ N.
For n =1
P(1) = (x¹-y¹) is divisible by (x-y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k)=(x^k-y^k) is divisible by (x-y)
⇒ (x^k-y^k) =m(x-y) for some m ∊ Z     ..(i)
For n = k+1
∴  P(k+1)=x^k+1-y^k+1 is divisible by (x-y)
x^k+1-y^k+1 = x^k+1-x^ky+x^ky-y^k+1
= x^k(x-y)+y(x^k-y^k)
= x^k(x -y)+y.m(x-y)
= (x-y)[x^k+my]
which is divisible by (x-y)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, pen) is true for all n∊N.

Put x=e+h
then \(x\rightarrow e \Rightarrow h\rightarrow 0\)
ஃ \(\overset{Lt}{h\rightarrow 0}\frac{log x-1}{x-e}=\overset{Lt}{h\rightarrow 0}\frac{log(e+h)-log e}{e+h-e}\)   [∵ log e=1]
=\(\overset{Lt}{h\rightarrow 0} \frac{log[\frac{e+h}{e}]}{h} \Rightarrow \overset{Lt}{h\rightarrow 0}\frac{log[1+\frac{h}{e}]}{\frac{h}{e}\times e}\)
∴ \(\frac{h}{e}\rightarrow 0 \Rightarrow \frac{1}{e}.\frac{1}{1} \Rightarrow \frac{1}{e} \)
∴ \(\overset{Lt}{x\rightarrow e}\frac{log x-1}{x-e}=\frac{1}{e}\).

When n is even
1² -2² +3² - 4² + 5² - 6² - 7² -8² + ....+(n -1)² - n²
= (1² - 2²) + (3²- 4²) + (5² - 6²) + (7² - 8²) + ...+ [(n - 1)² - n²]
= (1 + 2) (1 -2) + (3 -4)(3 +4) + (5 - 6) (5 +6)+...+ (n - 1 - n) (n - 1 + n)
= -[1 + 2 + 3 + 4 + 5 + 6 + ...+ (n - 1) + n]
= \(\frac { n(n+1) }{ 2 } \)
When n is odd:
(1² - 2²) + (3² - 4²) + (5² - 6²) + ... [(n - 2)2 - (n- 1)2] + n2
= (1 + 2) (1 - 2) + (3 + 4) (3 - 4) + (5 + 6)
(5 - 6) + ...+ (n - 2 - n + 1) (n - 2 + n - 1) + n2
= - [1 + 2 + 3 + 4 + 5 + 6 + ...+ (n - 2) + (n -1)] + n²
= \(\frac { \left( n-1 \right) \left( n-1+1 \right) }{ 2 } +{ n }^{ 2 }\)
= \(\frac { n(n-1) }{ 2 } +{ n }^{ 2 }\)
= \(\frac { -{ n }^{ 2 }+n+{ 2n }^{ 2 } }{ 2 } =\frac { { n }^{ 2 }+n }{ 2 } =\frac { n(n+1) }{ 2 } \)

The equation of lines are
y - x = 0.....(i)
x + y = 0....(ii)
x - k = 0....(iii)

By solving (i) and (ii), we get the coordinates of point C.
\(\therefore\) Coordinate of Care (0, 0).
By solving (ii) and (iii), we get the coordinates of point A.
\(\therefore\) Coordinate of A are (k, - k).
By solving (i) and (iii), we get the coordinates of point B.
\(\therefore\) Coordinates of B are (k, k).
\(\therefore\) Area of \(\Delta ABC=\frac { 1 }{ 2 } \left| \begin{matrix} k & -k & 1 \\ k & k & 1 \\ 0 & 0 & 1 \end{matrix} \right| \)
= \(\frac{1}{2}\)[(k² + k²) + (0 - 0) + (0 - 0)]
= \(\frac{1}{2}\times\) 2k² = k² sq.units.

The given inequality ix x + y \(\ge\) 4
Draw the gtaph of the line x + y = 4

Total of value satisfying the equation x +y = 4

Putting (0, 0) in the given inequation, we have 0 + 0 \(\ge\) 4 \(\Rightarrow\) 0 \(\ge\) 4 which is false
\(\therefore\) Half plane of x + y \(\ge\) 4 is away from origin
Also the given inequality is 2x - y > o.
Draw the graph of the line 2x - y = o.
Table of values satisfying the equation
2x -y =0

Putting (3, 0) in the given inequation, we have
2 x 3 - 0 > 0 => 6 > 0, which is true.
\(\therefore\) Half plane of 2x - y > 0 containing (3,0)

Arrange the data in ascending order, we have
10,11,12,13,13,14,16,16,17,17,18
Here n = 12 (which is even)
So median is average of 6th and 7th observations
∴ Median =\(\frac { 13+14 }{ 2 } =\frac { 27 }{ 2 } =13.5\)

M.D. about median =\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{12}\times28=2.33\)