\({ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) +{ cot }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 2x } \right) =\frac { \pi }{ 3 } \)
\({ cot }^{ -1 }\left( \frac { 1 }{ x } \right) ={ tan }^{ -1 }x\)
Therefore \({ cot }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 2x } \right) ={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) \)
Hence, \(LHS={ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) +{ tan }^{ -1 }\left( \frac { 2x }{ 1-{ x }^{ 2 } } \right) =\frac { \pi }{ 3 } \)
\(\Rightarrow { tan }^{ -1 }\frac { 2x }{ 1-{ x }^{ 2 } } =\frac { \pi }{ 6 } \)
\(\frac { 2x }{ 1-{ x }^{ 2 } } =tan\frac { \pi }{ 6 } =\frac { 1 }{ \sqrt { 3 } } \)
1-x² = 2\(\sqrt3\)x ⇒  x² + 2\(\sqrt3\)x - 1 = 0
\(​​x=\frac { -2\sqrt { 3 } \pm \sqrt { 12+4 } }{ 2 } =\frac { -2\sqrt { 3 } \pm 4 }{ 2 } \)
= 2 -\(\sqrt3\) (or) - 2\(\sqrt3\) 
∴ x = 2\(\sqrt3\) 

The unknown parts are B, a, c,
B = 180 - (A + C) = 180- (35° 17' + 45° 13')
= 99° 30'
To find sides, use sine formula
\(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { bsinA }{ sinB } =\frac { 42.1\times sin{ 35 }^{ 0 }17' }{ sin{ 99 }^{ 0 }30' } \)

log a = log 42.1 + log sin 36° 17' - log sin 99° 30'
= 1.6243 + 1.7616 - 1.9940
= 1.3859 - 1.9940
= 1.3859 - [-1 + 0.9940] = 1.3919
⇒ a = 24.65⁰
Again \(\Rightarrow \frac { bsinC }{ sinB } =\frac { 42.1\times sin{ 45 }^{ 0 }13' }{ sin{ 99 }^{ 0 }30' } \)
log c = log 42.1 + log sin 45° 31' - log sin 99° 30'
= 1.6243 + 1.8511 - 1.9940
= 1.4754 - 1.9940
= 1.4754 - [-1 + 0.9940] = 1.4814
⇒ c = 30.3°
Thus B = 99° 30' ; a = 24.65° ; C = 30.3°

P(n) : (11 + x)ⁿ \(\ge\) 1 + nx
P(1): (1 +x)1\(\ge\)1 +n
\(\Rightarrow\)1 + x\(\ge\)1 + x, which is true.
Hence, P( 1) is true.
Let P(k) be true
(i.e.) (1 +x)^k \(\ge\)1 + kx
We have to prove that P(k + 1) is true.
(i.e.) (1+x)^k+1\(\ge\) 1+(k+l)x
Now, (1 +x)^k+1\(\ge\) 1 + kx [\(\because\)P(k) is true]
Multiplying both sides by (1 + x), we get
(1 + x)^k (1 + x)\(\ge\)(1 + kx) (1 + x)
\(\Rightarrow\)(1 + x)^{k + 1}\(\ge\) 1 + kx + x + kx²
\(\Rightarrow\) (1 + X)^{k+ 1} \(\ge\) 1 + (k + 1) x + kx²....(1)
Now, 1 + (k+ 1)x + kx² \(\ge\)1 + (k+ 1)x ...(2)
From (1) and (2), we get
(1+x)^k+1\(\ge\)  1+(k+1)x
\(\therefore\) P(k+ 1) is true if P(k) is true.
Hence, by the principle of mathematical induction, p(n) is true for all values, of n.

LHS = sin2A - sin2B+ sin2C
= sin2A+ sin2C- sin2B
\(=2sin\frac{2A+2C}{2}cos\frac{2A-2C}{2}-sin2B\)
= 2 sin(A + C) cos (A -C) - 2 sin B cos B
= 2 sin (180° - B) cos (A - C) -2 sin B cos B
= 2 sin B cos (A - C) - 2 sin B cos B
= 2 sin B [cos(A - C) - cos B]
= 2 sin B [cos (A- C) - cos (180° - (A+ C)]
= 2 sin B[cos(A - C)+ cos(A+ C)]
= 2 sin B [2cos A cos C]
= 4 cos A sinB cos C = 1 RHS

\(\frac { { x }^{ 3 }-1 }{ { x }^{ 2 }+x+1 } \)= \(\frac { { x }^{ 3 }-1 }{ (x+2)(x-1) } =\frac { A }{ x+2 } +\frac { B }{ x-1 } \)
\(\frac { { x }^{ 3 }-1 }{ { x }^{ 2 }+x+1 } =\frac { A(x-1) }{ (x+2) } +\frac { B(x+2) }{ (x-1) } \)
Equating numerator on both sides
x³-1=A(x-1)+B(x+2)
put x = 1
0 = 0+B(1+2)
3B =0 ⇒ \(\boxed{B =0}\)
put x = -2
(-2)³-1 =A(-2-1)+B(0)
-8-1 = -3A
-9 = -3A
A= 9/3 ⇒ \(\boxed{A=3}\)
∴ \(\frac { { x }^{ 3 }-1 }{ { x }^{ 2 }+x+1 } =\frac { 3 }{ x+2 } +\frac { 0 }{ x-1 } \)
\(\frac { { x }^{ 3 }-1 }{ { x }^{ 2 }+x+1 } =\frac { 3 }{ x+2 } \)

\(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { A }{ x-1 } +\frac { B }{ x+2 } +\frac { C }{ (\quad ) } \)
= \(\frac { A(x+2)^{ 2 }+B(x-1)(x+2)+C(x-1) }{ (x-1)(x+2)(x+2)^{ 2 } } \)
Equating numerator on b/s
9=A(x+2)²+B(x-1)(x+2)+C(x-1)
put x =-2
9 = A(0)+B(0)+C(-3)
-3C = 9 ⇒ C=-3
put x =1
9 - A(1+2)² +B(0)+C(0)
9A =9
A =1
put x =0
9 =4A-2B-C
9=4(1)-2B+3
9-7 =-2B
2 = -2B
B =-1
∴ \(\frac { 9 }{ (x-1)(x+2)^{ 2 } } =\frac { 1 }{ x-1 } \frac { 1 }{ x+2 } \frac { -3 }{ (x+2)^{ 2 } } \)

LHS = cos 20° cos 40° cos 80°
\(=\frac{1}{2}\)cos 20° (2 cos 80° cos 40°)
\(=\frac{1}{2}\)cos 20° [ cos 80° + 40°) + cos (80° - 40°)]
[\(\because\) 2 cos A cos B = cos (A + B) + cos (A - B)]
\(=\frac{1}{2}\) cos 20° (cos 120° + cos 40°)
\(=-cos 20°(-\frac{1}{2}+cos 40°)\)
\(=-\frac{1}{4}cos 20°+\frac{1}{2}cos40°cos20°\)
\(=-\frac{1}{4}cos20°+\frac{1}{4}[cos(40°+20°)+cos(40°-20°)]\)
\(=-\frac{1}{4}cos20°+\frac{1}{4}(cos60°+cos20°)\)
\(=-\frac{1}{4}cos20°+\frac{1}{4}cos60°+\frac{1}{4}cos20°\)
\(=\frac{1}{4}cos60°=\frac{1}{4}(\frac{1}{2})=\frac{1}{8}=RHS\)

Here  ²ⁿC₃ : ⁿC₃ = 11 : 1
\(\Rightarrow {(2n)!\over 3!(2n-3)!}\times {2!(n-3)!\over n!}={11\over 1}\)
\(\Rightarrow {(2n)(2n-1)(2n-2)(2n-3)!\over 3!(2n-3)!}\times {3!(n-3)!\over n(n-1)(n-2)(n-3)!}={11\over 1}\)
\(\Rightarrow {(2n)(2n-1)(2n-2)\over n(n-1)(n-2) }={11\over 1}\)
\(\Rightarrow {4(2n-1)\over n-2}={11\over1}\)
\(\Rightarrow \) 8n - 4 = 11n - 22
\(\Rightarrow \) 3n = 18 \(\Rightarrow \) n = 6

We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women 
Therefore, Number of ways =¹⁰C₃ \(\times\) ⁷C₃ + ¹⁰C₄\(\times\)⁷C₂
\(={10\times9\times8\over3\times2\times1}\times{7\times6\times5\over3\times2\times1}+{10\times9\times8\times7\over4\times3\times2\times1}\times{7\times6\over2\times1}\)
= (120 x 35) + (210 x 21) = 4200 + 4410 = 8610
If 2 particular women to be always present, then the number of ways
=¹⁰C₄ \(\times\) ⁵C₀ + ¹⁰C₃ \(\times\) ⁵C₄ \(\times\)⁵C₁
\(={10\times9\times8\times7\times\over 4\times3\times2\times1}1+{10\times9\times8\over3\times2\times1}\times5=\) 210 + 120 x 5 = 210 + 600 = 810
\(\therefore\)Total number of committee = 8610 - 810 = 7800
Hence, the value of the filler is 7800

The range of the function is {x, y, z}.

Let p(n): be \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) ...\left( 1+\frac { 1 }{ n } \right) =\left( n+1 \right) \)
Step 1: p(1): \(( 1+\frac { 1 }{ 1 }) =\left( 1+1 \right) \)
\(\Rightarrow\) p(1):2 = 2
\(\Rightarrow\) p(1) is true.
Step 2: Let p(m) be true
\(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) \left( 1+\frac { 1 }{ 3 } \right) ...\left( 1+\frac { 1 }{ m } \right) =\left( m+1 \right) \)
Step 3: To prove that p(m + 1) is true
i.e. to prove that \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) ...\left( 1+\frac { 1 }{ m } \right)\left( 1+\frac { 1 }{ m+1 } \right) =\left( m+2 \right) \)
LHS = \(\left( 1+\frac { 1 }{ 1 } \right) \left( 1+\frac { 1 }{ 2 } \right) ...\left( 1+\frac { 1 }{ m } \right)\left( 1+\frac { 1 }{ m+1 } \right) \)
= (m + 1) \(\left( 1+\frac { 1 }{ m+1 } \right)\) = m + 1 + 1
= (m + 2) = RHS
p(m + 1) is true.
By the principle of mathematical induction, p(n) is true for all \(n\in N\)

Let p(n): be x²ⁿ - y²ⁿ is divisible by (x +y).
Step 1: p(1): x² - y² is divisible by (x +y)
⇒ p(1): (x + y) (x - y) is divisible by (x +y)
⇒ p(1) is true.
Step 2: Let p(m) be true.
x^2m - y^2m is divisible by (x +y)
x^2m - y^2m = ⋋ (x +y) where A is a constant
Step 3: To prove that p(m + 1) is true
i.e. to prove that x^2(m+1) - y^2(m+1) is divisible by (x + y)
Consider x^2(m+1) - y^2(m+1)
= x^2m+2 - y^2m+2
= x^2m . x² - y^2m .y²
x^2m . x² - x^2m .y² + x^2m . y² -y^2m .y²     [Adding and subtracting x^2m . y^2m]
x^2m (x² - y²) +y² (x^2m - y^2m)
x^2m (x² - y²) + y² ⋋ (x +y)          [by (1)]
x^2m (x +y) (x - y) + ⋋ . y^2m (x +y)
(x + y) (x^2m (x - y) + ⋋y²) which is divisible by (x +y)
⇒ p(m + 1) is true.
Hence, by the principle of mathematical induction, p(n) is true for all n E N.

Let the H.P. be \({{1}\over{a}},{1\over a+d},{1\over a+2d},...\)
\(\therefore\)     \({T}_{1}={1\over a}={2 \over 5}\) and \({T}_{2}={ 1 \over a+d}={12 \over 13}\)
\(\Rightarrow\)  \(a={5\over 2}\) and \(a+d={13 \over 12}\)
\(\Rightarrow\)  \(a={5\over 2}\) and \({5\over 2}+d={13 \over 12}\Rightarrow d{13 \over 12}-{5\over 2}\Rightarrow d={13-20\over12 }={-17\over 12}\)
Now, \({T}_{n}-{1 \over {a+(n-1)d}}={{1}\over{{{5}\over{2}}+(n-1)\left( {{-17}\over{12}} \right)}}={12\over30-17n+17}\)
\(={12\over47-17n}\)
Tn is largest when 47 - 17n is least and positive
For n = 2, 47 - 17n is least and positive
\(\therefore\) Tn is largest when n = 2
\(\therefore\) The largest term \(={12\over 47-17(2)}={12\over13}.\)

Let a and r be the first term and common ratio of the G.P.
\(\therefore\)  \({S}_{n}={a(1-r^n)\over1-r},n\epsilon N\)
LHS \(={{S}_{10}-{S}_{20}}^{2}=\left[ {a(1-{r}^{10})\over1-r}-{a(1-{r}^{20})\over1-r} \right]^{2}\)
\(={{a}^{2}\over{{(1-r)}^{2}}}[1-{r}^{10}-1+{r}^{20}]^2\)
\(={{a}^{2}\over{(1-r)}^{2}}.{r}^{20}{({r}^{10}-1)}^{2}={{a^2.{r}^{20}.{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
RHS = S₁₀ (S₃₀ - S₂₀)
\(={a(1-{r}^{10}\over1-r)}\left[ {a(1-{r}^{30})\over1-r}-{{a(1-{r}^{30})}\over{1-r}} \right]\)
\(={{a^2}\over{(1-r^2)}}(1-{r}^{10})[1-{r}^{30}-1+{r}^{20}]={{a^2(1-{r}^{10})}\over{{(1-r)}^{2}}}.{r}^{20}(1-{r}^{10})\)
\(={{a^3.{r}^{20}(1-{r}^{10})^2}\over{{(1-r)}^{2}}}={{{a}^{2}.{r}^{20}{({r}^{10}-1)}^{2}}\over{{(1-r)}^{2}}}\)
\(\therefore\) LHS = RHS.

Given T_p+1 = 2.T_q+1
\(\Rightarrow\)   a+(+1-1)d = 2[a+(q+1-1)d]    [ \(\because\) T_n = a + ( n - 1) d ]
\(\Rightarrow\)  a + pd = 2a +2qd
\(\Rightarrow\) a = ( p - 2q ) d   ...(1)
 Now T_3p+1 = a+ ( 3p + 1 - 1 )d = a + 3pd
= ( p - 2q ) d + 3pd   (using (1))
= 4pd - 2qd
= 2d (2p-q) ...(2)
Also T_p+q+1 = a + (p+q+1-1)d
= a+(p+q)d
= (p-2q)d + (p+q)d   (using (1))
= d(p-2q+p+q) = d(2p-q) ...(3)
From (2) and (3), T_3p+1 = 2. T_p+q+1.

We have
AB² = AC² + BC² - 2AC . BC.cos\(\frac{\pi}{4}\)

⇒ AB² = 250² + 3002 - 2 . 250 \(\times\) 300 \(\times\) \(\frac{1}{\sqrt{2}}\)
⇒ AB² = 62500 + 90000 - 75000√2
⇒ AB² = 152500 - 75000 \(\times\) 1.44
⇒ AB² = 44500
⇒ AB = \(\sqrt{44500}\) = 210.95 m

Given a = 18, b = 24, c = 30 and ㄥc = 90°
By sine formula,
\(\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { 18 }{ sinA } =\frac { 24 }{ sinB } =\frac { 30° }{ sin90° } \)
\(\Rightarrow \frac { 18 }{ sinA } =\frac { 24 }{ sinB } =\frac { 30 }{ 1 } \)
From I and III,\(\frac { 18 }{ sinA } =30°\)
 \(\Rightarrow sinA=\frac { 18 }{ 30 } \)
\(\Rightarrow sinA=\frac { 3 }{ 5 } \)
From II and III, \(\frac { 24 }{ sinB } =30°\)
\(\Rightarrow sinB=\frac { 24 }{ 30 } =\frac { 4 }{ 5 } \)
And sin C = sin 90° = 1
⇒ sin A = \(\frac{3}{5}\) , sin B = \(\frac{4}{5}\), sin C = 1

Given \(2\left(x+{{1}\over{x}} \right)^2-7\left(x+{{1}\over{x}} \right)+5=0.\)
Let \(x+{1\over x}=y\)
2y²-7y + 5 = 0
(2y - 5) (y - 1) = 0
2y = 5 or y = 1
\(y={5\over 2}\ or\ y=1\)
\({x+{1\over x}}{={5\over2}}\ or\ y=1\)

Case(i) when \({x+{1\over x}}{={5\over2}}\)
⇒ \({x^2+1\over x}={5\over2}\)
⇒ 2x²+2 = 5x
⇒ 2x²-5x+ 2 = 0
⇒ (x-2) (2x-1) = 0
⇒ \(x=2\ or\ {1\over2}\) 
Case (ii) \(x+{1\over x}=1\)
⇒ \({x^2+1\over x}=1\)
⇒ x²+1 = x
⇒ x²-x+1 = 0
⇒ \(x={+1\pm\sqrt{1-4(1)(1)}\over 2}\)
⇒ \(x={1\pm\sqrt{-3}\over2}\)
Which is not possible
∴ The real roots of the given equation are 2 and \(\frac{1}{2}\) 

Given polynomial is

x⁴ - 14x²y² - 51y⁴
_​​​​​​​= (x² - 17y²) (x² + 3y²)
= [x² - (17y)²] [x²+ 3y²]
\(=(x-\sqrt{17}y)(x+\sqrt{17}y)(x^2+3y^2)[\because\ a^2-b^2 =(a+b)(a-b)]\)

f +g : R ⟶ R and f - g : R ⟶ R
Now (f + g) (x) = f(x) + g(x)
⇒ ( f+ g)(x) = x + |x|
\(⇒\ \left( f+g \right) x =\begin{cases}x +x\quad if\quad x\ge 0 \\ x-x \ if\quad x<0\quad \end{cases}\)
\(⇒\ \left( f+g \right) x =\begin{cases}2x\quad if\quad x\ge 0 \\ 0 \ if\quad x<0\quad \end{cases}\)
Similarly (f - g) (x) = f(x) - g(x) = x - |x|
\(⇒\ \left( f-g \right) x =\begin{cases}x -x\quad if\quad x\ge 0 \\ x-(-x )\ if\quad x<0\quad \end{cases}\)
\(⇒\ ​​\left( f-g \right) \left( x \right) =\begin{cases} 0 \quad if\quad x\ge 0 \\ 2x \ if\quad x<0\quad \end{cases}\)

Let A= {-1, 2, 3,4, 5}, and B = {0,3, 6}
A \(\times\) B = {(-1, 0) (-1,3) (-1, 6) (2, 0) (2, 3) (2, 6) (3, 0) (3, 3) (3, 6) (4, 0) (4, 3) (4, 6) (5, 0) (5, 3) (5,6)}
The set of all ordered pairs (a, b) such that a + b = 5 is R = {(-1, 6) (2, 3) (5, 0)}

Let A and B be two sets having elements m and n respectively
Then, number of subsets of set A = 2^m
Number of subsets of set B = 2ⁿ
Given 2^m - 2ⁿ = 56
\(⇒\ 2^n\left({2^m\over 2^n}-1\right)=56\)
⇒ 2ⁿ(2^m-n - 1) = 8 \(\times\) 7
⇒ 2ⁿ(2^m-n - 1) = 2³ \(\times\)(2³ - 1)
Equating both sides we get
n = 3 and m - n = 3
⇒ n = 3 and m - 3 = 3
⇒ n = 3 and m = 6

Venn diagram

From (ii) and (iv), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Let \({{{log}_{e}^{x}}\over{b-c}}={{{log}_{e}^{y}}\over{c-a}}={{{log}_{e}^{z}}\over{a-b}}=k\)
\({log}_{e}^{x}= k(b-c),{log}_{e}^{y}=k(c-a)\) and
\({log}_{e}^{z}=k(a-b)\)  ...(1)
\(\Rightarrow\)  x = e^k(b-c),y=e^k(c-a)and z = e^k(a-b) ....(2)
xyz = e^k(b-c).e^k(c-a).e^k(a-b)
= e^{k(b-c+c-a+a-b) = ek(0)} = e⁰ = 1
\(\Rightarrow\) xyz = 1

Given \(\sqrt{x+5}+\sqrt{x+21}=\sqrt{6x+40}\)
Squaring both sides we get
\((\sqrt{x+5}+\sqrt {x+21})^2=(\sqrt{6x+40})^2\)
⇒ \(z+5+z+21+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \( 2x+26+2\sqrt{(x+5)(x+21)}=6x+40\)
⇒ \(2\sqrt{(x+5)(x+21)}=6x+40-2x-26\)
⇒ \(2\sqrt{(x+5)(x+21)}=4x+14\)
 \(\sqrt{(x+5)(x+21)}=2x+7\)
Squaring again we get
(x + 5)(x + 21) = (2x + 7)²
⇒ x² + 21x + 5x + 105 = 4x²+ 49 + 28x
⇒ x² + 26x + 105 = 4x² +49 + 28x
⇒ 3x² + 2x- 56 = 0
\(x = {-2 \pm \sqrt{4-4(3)(-56)} \over 6}\)
\(x = {-2 \pm \sqrt{4+672)} \over 6}\)
\(x={-2\pm26\over 6}⇒x=4,{-14\over 3}\)
⇒ When x = 4
Case (i) :
\(\sqrt{4+5}+\sqrt{4+21}=\sqrt{6(4)}+40\)
\(\sqrt9+\sqrt{25}=\sqrt{64}\)
3 + 5 = 8
8 = 8 which is true  ⇒ x = 4 is a root
Case (ii) : When x = \(-14\over 3\)
\(\sqrt{{1-\over3}+5}+\sqrt{{-14\over 3}+21}=\sqrt{+6\left(-14\over 3\right)+40}\)
\(\sqrt{1\over 3}+\sqrt{49\over 3}=\sqrt{12}\) which is not true
\(\therefore\) x  \(={{-14}\over{3}}\) is not a root.