x² +y² = 25
y = 土\(\sqrt { 25-{ x }^{ 2 } } \)
x = 0 ⇒ y = 士5
y = 0 ⇒ x = 土5
(0, 5), (0, -5) ∈R
x = 3 ⇒ y = 土4
x = -3 ⇒ y = 土4
Domain of R = {0, 3, -3, -4, 4, -5, 5}
Domain of R^-1 {0, 3, -3, -4, 4, -5, 5}

tan^-1 (x + 1) + tan^-1 (x - 1) = tan^-1\((\frac{4}{7})\)
\({ tan }^{ -1 }\left( \frac { x+1+x-1 }{ 1-({ x }^{ 2 }-1) } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 7 } \right) \)
\(tan\left( \frac { 2x }{ 2-{ x }^{ 2 } } \right) ={ tan }^{ -1 }\left( \frac { 4 }{ 7 } \right)\)
\(\frac { 2x }{ 2-{ x }^{ 2 } } =\frac { 4 }{ 7 } \)
8 - 4x² = 14x
4x² + 14x - 8 = 0
⇒ 2x²+ 7x - 4 = 0
2x²+ 8x - x - 4 = 0
2x (x + 4) -1(x + 4) = 0
(x + 4) (2x - 1) = 0
x = -4 or x = \(\frac{1}{2}\)
∴ x = \(\frac{1}{2}\) 

2 tan θ - cot θ = - 1
2 tan θ-\(\frac{1}{tan θ}\) = 1 ⇒ 2tan²θ - 1 = tan θ
2 tan² θ - 1 + tan θ = 0
⇒ (2tan θ-1) (tan θ+1) = 0
2tan θ-1 = 0 (or) tan θ+1​​​​​​​ = 0
tan θ = \(\frac{1}{2}\) (or) tan θ = -1
When tan θ = -1 = -tan\(\frac{\pi}{4}\)
tan θ = tan\((\frac{-\pi}{4})\)
⇒ θ = nπ + \((\frac{-\pi}{4})\)
= nπ ​​​​​​​-\(\frac{\pi}{4}\);n∈Z
When tan θ = \(\frac{1}{2}\)= tan β (say)
∴ θ = nπ + β + nπ + tan^-1\((\frac{1}{2})\)
Hence  θ = nπ - \(\frac{\pi}{4}\) (or) θ = nπ + tan^-1\((\frac{1}{2})\); n∈Z

Let p(n) : 2n < (n + 2)! for all k\(\in\)N.
Step1: P(1) 2.1 < (1+2)!
\(\Rightarrow\) 2 < 3!\(\Rightarrow\) 2 < 6 which is true for P(1)  [\(\because\) 3! = 3 x 2 x 1 = 6]
Step 2: P(k) : 2k < (k + 2)!. Let it be true for P(k)
Step3:P(k+1) 2(k+1) < (k+1+2)!
Since 2k < (k + 2)!  (from Step 2)
\(\Rightarrow\)2k+2 < (k+2)! +2
\(\Rightarrow\) 2(k+1) < (k+2)!+2
Also, (k+ 2)! + 2 < (k+ 3)!
\(\therefore\) 2(k+1) < (k+3)!
\(\Rightarrow\) 2(k + 1) < (k + 2 + 1)! which is true for P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true.

\((x+1)^{1\over 3} = (x-3)^{1\over 2}\)
L.C.M. of 2 and 3 is 6 & Raising to the power 6
\(\left\{ (x+1)^{ \frac { 1 }{ 3 } } \right\} ^{ 6 }=\left\{ (x-3)^{ \frac { 1 }{ 2 } } \right\} ^{ 6 }\)
(x+1)² = (x-3)³
x²+2x+1 =x³-9x²+27x-27
0 = x³-9x²+27x-27-x²-2x-1
x³-10x²+25x-28 =0
since constant term is - 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x - 7) is a factor
Using synthetic division

we get x³-10x²+25x-28=(x-7)(x²-3x+4)
solving x²-3x+4 =0
x=\(\frac { 3\pm \sqrt { 9-16 } }{ 2 } =\frac { 3\pm \sqrt { -7 } }{ 2 } \)
the roots are x =7 x=\(\frac { 3\pm \sqrt { -7 } }{ 2 } \)

The given inequality is x+y≤9
Draw the graph of the line x+y =9

Table of values satisfying the equation
x+y = 9

Putting (0, 0) in the given inequation, we have 0+0≤9⇒0≤9, is towards is origin
∴ Half plane of x+y≤9 is away from origin
Also, the given inequality is x-y < 0
Draw the graph of the line x - y = 0.
Table of values satisfying the equation
 x-y =0

Putting (0, 3) in the given inequation, we have 0-3-0<0⇒-3<0 which is true.
∴  Half plane of x-y < 0 containing the points (0,3).

LHS = tan 70° - tan 20° - 2 tan 40°
\(=(\frac{sin70°}{cos70°}-\frac{sin20°}{cos20°})-2tan40°\)
\(=(\frac{sin70°cos20°-cos70°sin20°}{cos70°cos20°})-2tan40°\)
\(=\frac{sin(70°-20°)}{cos70°cos20°}-2tan40°\)
\(=\frac{2sin50°}{2cos70°cos20°}-2tan40°\)
\(=\frac{2sin50°}{cos90°+cos50°}-2tan40°\)
\(=\frac{2sin50°}{cos50°}-2tan40°\ [\because cos90°=0]\)
\(=2(\frac{sin50°}{cos50°}-\frac{sin40°}{cos40°})\)
\(=2(\frac{sin50°cos40°-cos50°sin40°}{cos50°cos40°})\)
\(=2[\frac{sin(50°-40°)}{cos50°cos40°}]=\frac{4sin10°}{2cos50°cos40°}\)
\(=\frac{4sin10°}{cos90°+cos10°}=\frac{4sin10°}{cos10°}\)
= 4 tan 10° = RHS

Given,⁵P_r = 2 ⁶P_r-1
\(\Rightarrow {5!\over (5-r)!}=2\times{6!\over \{6-(r-1)\}!}\)
\(\Rightarrow {5!\over (5-r)!}=2\times{6.5!\over (7-r)!}\)
\(\Rightarrow {5!\over (5-r)!}={6.5!\over (7-r)(6-r)(5-r)!}\)
\(1={12\over (7-r)(6-r)}\)
\(\Rightarrow\) r² - 13r + 42 = 12 or r² - 13r + 30 = 0
or (r-3)(r-10) = 0
\(\therefore\) r = 3, 10
Now, ⁵p_rand ⁶p_r-1are meaningless when r = 10
[\(\therefore\)ⁿP_r is defined only if r\(\le\)n]
\(\therefore\) Rejecting r = 10, we have r = 3

Given h = 61.4 + 2.3 F
Given h = 160 ⇒160 = 61.4 + 2.3 F
⇒ 2.3 F = 160-61.4 = 98.6
F =\(\frac { 98.6 }{ 2.3 } \) = 42.87
Given h = 170,
170 = 61.4 + 2.3 F
⇒ 170 - 61.4 = 2.3 F
2.3 F = 108.6
⇒ F =\(\frac { 108.6 }{ 2.3 } \)= 47.23
So the ranges of values are
42.8 < x < 47.23

Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2. According to the given conditions.
x < 10, x + 2 < 10
and x + ( x + 2 ) > 11
⇒ x < 10, x < 8
and 2x > 9       (\(\because\) x < 8 automatically smallest of the lesser than) ...(1)
⇒ x < 10
and \(x>\frac { 9 }{ 2 } \) ....(2)
From (1) and (2), we get
\(\frac { 9 }{ 2 } \) Also, x is an odd positive integer
x can take values 5 and 7.
So, the required possible pairs will be (x, x+2) = (5,7),(7, 9)

Given that \(\frac { |x-2|-1 }{ |x-2|-2 } \le 0\)
Put | x - 2 | = y
∴   \(\frac { y-1 }{ y-2 } \le 0\)
⇒ y - 1 > 0, y - 2 < 0 ⇒ y < 2
⇒ 1 < y, 2 ⇒ 1 < | x - 2 | < 2
⇒  1 ⇒ 1<| x - 2 |, | x - 2 | < 2
⇒ x - 2 < -1 (or) x - 2 >1 and -2 ⇒ x<1 (or) x > 3 and -2 + 2 < x < 2 + 2
⇒ x < 1 (or) x > 3 and 0< x < 4
Hence, the required solution is (0, 1) U (3, 4)

Suppose that is rational
then \(\sqrt { 5 } \) = (where p and q are integer which are co-prime)
⇒ p =\(\sqrt { 5 } \)q = p² = 5q²   ....(1)
\(\frac { { p }^{ 2 } }{ 5 } \)= q² ⇒ 5 is a factor of p
So let P = 5c
substituting p = 5c in (1) we get
(5c)² = 5q² ⇒ 25² = 5q²
⇒ c²  = \(\frac { { 5q }^{ 2 } }{ 25 } =\frac { { q }^{ 2 } }{ 5 } \)
⇒ 5 is a factor of q also
⇒ So 5 is a factor of p and q which is a contradiction.
⇒ \(\sqrt { 5 } \) is not a rational number.
⇒\(\sqrt { 5 } \)  is an irrational number.

Given nP_r = nP_{r+ 1}

⇒ 1 = n-r       ...(1)
Also, it is given that nC_r = nC_r-1

\(⇒\ {1\over r}={1\over n-r+1}\)
⇒ n - r + 1 = r
⇒ n -2r = -1 ...(2)
(1) - (2} ⟶ 2 = r
Substituting r = 2 in (1) we get,
n-2 = 1 ⇒ = 3
∴ n = 3 and r = 2
n = 3 and r = 2

There are 12 letters in the given word of which 2 are T's and the remaining are distinct.
Remaining 10 letters between P and S can be arranged in \(\frac{10!}{2!}\) ways.
ஃ Total number of words starting with P and ending is S = \(\frac{10!}{2!}\) 
= \(\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } \)
= 1814400

Clearly a number between 100 and 1000 has 3 digits.
\(\therefore\) Total number of 3 digit numbers having atleast one of their digits as y.
= (Total number of 3 digit numbers) - (Total number of 3 digit numbers in which y does not appear at all)
Total number of 3 digit numbers' we have to form 3 digit numbers by using the digits 0, 1, 2, 3, ... 9. Hundreds' place can be filled in 9 ways and each of the ten's and one's place can be filled in 10 ways. So, total number of 3 digit numbers = 9 \(\times\) 10 \(\times\) 10 = 900.
Total number of 3 digit number in which 7 does not appear. Here, we have to form 3-digit numbers by using the digits 0 to 9, except 7. So, hundreds place can be filled in 8 ways, and each of the ten's and one's place can be filled in 9 ways.
So, total number of 3-digit numbers in which 7 does not appear at all is 8 \(\times\) 9 \(\times\) 9.
Hence, total number of 3-digit numbers having atleast one of their digits as 7 is 9 \(\times\) 10 \(\times\) 10 - 8 \(\times\)9 \(\times\)9
= 900 - 648 = 252

Let sin^-1 \(\left( \frac { 12 }{ 13 } \right) +{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =\pi \)
Then \(sin\quad x=\frac { 12 }{ 13 } ,cosy\frac { 4 }{ 5 } ,\quad and\quad tanz\frac { 63 }{ 16 } \)
\(cos\quad x\sqrt { 1-{ sin }^{ 2 }x } =\sqrt { 1-\frac { 144 }{ 169 } } =\sqrt { \frac { 25 }{ 169 } } =\frac { 5 }{ 13 } \)
\(and\quad tan\quad x=\frac { sin\quad x }{ cos\quad x } =\frac { 12 }{ 13 } /\frac { 5 }{ 13 } =\frac { 12 }{ 5 } \)
\(sin\quad y\sqrt { 1-{ cos }^{ 2 }y } =\sqrt { 1-\frac { 16 }{ 25 } } =\sqrt { \frac { 9 }{ 25 } } =\frac { 3 }{ 5 } \)
\(tan\quad y=\frac { sin\quad y }{ cos\quad y } =\frac { \frac { 3 }{ 5 } }{ \frac { 4 }{ 5 } } =\frac { 3 }{ 4 } \)
i.e have tan (x + y) = \(\frac { tan\quad x+tan\quad y }{ 1-tan\quad x.tan\quad y } \)
\(\frac { \frac { 12 }{ 5 } +\frac { 3 }{ 4 } }{ 1-\frac { 12 }{ 5 } \times \frac { 3 }{ 4 } } =\frac { \frac { 48+15 }{ 20 } }{ \frac { 20-36 }{ 20 } } =-\frac { 63 }{ 16 } \)
From (1) and (2), tan (x+y) = -tan z
\(\Rightarrow tan\quad (x+y)=tan\quad (-z)\)
\(\Rightarrow tan(x+y)=tan(\pi -z)\)
\(\Rightarrow x+y= -z\quad or\quad x+y=\pi -z\)
Since x, y, and z are positive, x + y \(\neq \) -z
\(\therefore\)  x + y + z = p
\(\Rightarrow { sin }^{ -1 }\left( \frac { 12 }{ 13 } \right) +{ cos }^{ -1 }\left( \frac { 4 }{ 5 } \right) +{ tan }^{ -1 }\left( \frac { 63 }{ 16 } \right) =\pi \)  

Given sin 2x + sin 4x + sin 6x = 0
⇒ sin 4x + (sin 2x + sin 6x) = 0
\(\Rightarrow sin4x+2sin\left( \frac { 6x+2x }{ 2 } \right) cos\left( \frac { 6x-2x }{ 2 } \right) =0\)
⇒ sin 4x + 2 sin 4x . cos 2x = 0
⇒ sin 4x (1 + 2 cos 2x) = 0
⇒ sin 4x = 0 or 1 + 2 cos 2x = 0
⇒ sin 4x = 0, cos 2x = \(\frac{-1}{2}\)
case (i) : 
When sin 4x = 0
⇒ 4x = nπ, n∈z
⇒ x = \(\frac { n\pi }{ 4 } ,\ n\epsilon z\)
case(ii) :
When cos 2x = \(\frac{-1}{2}\)
\(\Rightarrow cos2x=-cos\left( \frac { \pi }{ 3 } \right) \)
\(cos2x=-cos\left( \pi -\frac { \pi }{ 3 } \right) \)
\(cos2x=-cos\frac { 2\pi }{ 3 } m\epsilon z\)
\(x=m\pi \pm \frac { \pi }{ 3 } ,\quad m\epsilon z\)
Hence \(x=\frac { n\pi }{ 4 } \) or x = \(m\pi \pm \frac { \pi }{ 3 } \), where m,n∈z

For any two distinct elements x₁, x₂ in \([0,{\pi\over 2}]\)
in x₁ = sin x₂ ⇒ x₁ = x₂
∴ f is one - one...(1)
Also cos x₁ = cos x₂ ⇒ x₁ = x₂
∴ g is also one-one...(2)
Now, (f + g) (x) = f(x) + g(x) = sin x + cos x
⇒ (f+g)(0) = sin0 + cos0 = 0+1=1
and \(f+g\left(\pi\over 2\right)=sin\frac{\pi}{2}+cos{\pi}{2}=1+0=1\)
(f+g)(0) = (f+g)\((\frac{\pi}{2})\) ⇒ 0 ≠ \(\frac{\pi}{2}\)
∴ (f + g) is not one-one...(3)
From (1), (2) and (3), we get f and g are one-one but (f +g) is not one-one.

LHS= \({{1}\over{3-\sqrt{8}}}-{{1}\over{\sqrt{8}-\sqrt{7}}}+{{1}\over{\sqrt{7}-\sqrt{6}}}-{{1}\over{\sqrt{6}-\sqrt{5}}}+{{1}\over{\sqrt{5}-2}}=5\)
Multiplying each term by the conjugate of the denominator we get
\(={3+\sqrt8\over (3-\sqrt8)(3+\sqrt8)}-{\sqrt8+\sqrt7\over(\sqrt8-\sqrt7)(\sqrt8+\sqrt7)}+{\sqrt7+\sqrt6\over( \sqrt7-\sqrt6)(\sqrt7+\sqrt6)}-{\sqrt6+\sqrt5\over(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}+{\sqrt5+2\over(\sqrt5-2)(\sqrt5+2)}\)
\(={3+\sqrt8\over3^2-(\sqrt8)^2}-{\sqrt8+\sqrt7\over (\sqrt8)^2-(\sqrt7)^2}+{\sqrt7+\sqrt6\over (\sqrt7)^2-( \sqrt6)^2}-{\sqrt6+\sqrt5\over(\sqrt6)^2-(\sqrt5)^2}+{\sqrt5+2\over (\sqrt5)^2-2^2}\)
\(={3+8\over 9-8}-{\sqrt8-\sqrt7\over 8-7}+{\sqrt7+\sqrt6\over 7-6}-{\sqrt6+\sqrt5\over 6-5}+{\sqrt5+2\over5-4}\)

Hence proved

Let x = \(\sqrt[3]{(45.4)^2\over (3.2)^2\times(6.5)^3}\)
Taking logarithms of both sides, we get
log x =  \(log\left[ (45.4)^2\over (3.2)^2\times(6.5)^3 \right]^{1/3}\)
\(={1\over 3}log\left(45.4)^2\over (3.2)^2\times(6.5)^2\right)\)
\(={1\over 3}[2log 45.4 - 2 log3.2 - 3 log 6.5]\)
\(={1\over 3}[2(1.6571) - 2(0.5052)- 3 (0.8129)]\)
\(={1\over 3}(-0.1349)= -0.0450\)
= -1 + 1 - 0.0450 = -1.9550
log x  = -1.9550
⇒ x = anti log (-1.9550)
⇒ x = 0.9016

Given R = {(a, b): a < b²} where a, b ∈ R
reflexivity: We know that \(\left(1\over 2\right)\le\left(1\over 2\right)^2\)is not true
\(⇒\ \left({1\over 2},{1\over 2}\right)∉R\)
⇒ R is not reflexive
Symmetry: We know that -1< 3² but 3 ≰ (-1)² is not true
⇒ (-1, 3) ∈ R but (3, -1)∉R
∴ R is not symmetric
Transitive: We observe that 2< (-3)² and -3< (1)² but 2 ≰ (1)² is not true
⇒ (2, -3) ∈ R and (-3, 1) ∈ R but (2, 1) ∉ R
⇒ R is not transitive
∴ R is neither reflexive nor symmetric nor transitive.

Given (-1, 0) ∈ A x A and (0, 1) ∈ A \(\times\) A
(-1,0) ∈ A \(\times\) A ⇒ -1,0 ∈ A and (0, 1) ∈ A \(\times\) A ⇒ 0, 1 ∈ A
∴ -1,0,1 ∈ A
∴ A = {-1, 0, 1}
Also it is given that A\(\times\) A has 9 elements.
∴ A has exactly three elements.
∴ A {-1, 0, 1}
∴  A \(\times\) A = {(-1, -1) (-1, 0) (-1, 1) (0, -1) (0, 0) (0, 1) (1, -1) (1, 0) (1, 1)}

B ⋂ C = {4, 6}
A\(B ⋂ C) = {0, 1, 2, 3}
A\B = {0,2}
A\B = {0, 1, 3}
(A\B) U (A\C)= {a, 1,2, 3}
From (1) and (2), A\(B ∩ C) = (A\B) U (A\C)
Venn diagram:

From (ii) and (v), we get
A\(B ∩ C) = (A\B) U (A\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.