Suppose the loan in cleared in n months. Clearly the amount forms an. A.P. with a = 20 and d = 15
∴ Sum of the amounts = 3250
S_n = 3250

\(⇒\ {n\over2}[2a + (n -1)d]=3250\)
\(⇒\ {n\over2}[40+(n-1)15]=3250\)
⇒ n(40 + 15n - 15) = 6500
⇒ n (15n + 25) 6500
⇒ 15n² + 25n = 6500
⇒ 15n² + 25n = 6500
⇒ 3n² + 5n - 1300 = 0
⇒ (n - 20) (3n + 65) = 0
⇒ n = 20 or \(n={-65\over 3}\) which is not possible
∴ n = 20
Thus, the amount is cleared in 20 months.

Given A = {a, b, c, d}, B = {a, c, e}, C = {a, e}
B∩C = {a, c}
A∩(B∩C) = {a}
A ∩ B = {a, e}
(A ∩ B) ∩ C = {a}
From (1) and (2), it is clear that A ∩ (B ∩ C) = (A ∩ B) ∩ C.

Given cos(\(\alpha \) - β)+cos(β - \(\gamma \))+cos(\(\gamma \) - \(\alpha \)) = - \(\frac{3}{2}\)
\(cos\alpha cos\beta +sin\alpha sin\beta +cos\beta cos\gamma +sin\beta sin\gamma +cos\gamma cos\alpha +sin\alpha sin\gamma =-\frac { 3 }{ 2 } \)
\(2\left[ cos\alpha cos\beta +cos\beta cos\gamma +cos\gamma cos\alpha +sin\alpha sin\beta +sin\beta sin\gamma +sin\alpha sin\gamma \right] =-3\)
\(\left( 2cos\alpha cos\beta +2cos\beta cos\gamma +2cos\gamma cos\alpha \right) +\left( 2sin\alpha sin\beta +2sin\beta sin\gamma +2sin\alpha sin\gamma \right) +3=0\)
\(\left( 2cos\alpha cos\beta +2cos\beta cos\gamma +2cos\gamma cos\alpha \right) +\left( 2sin\alpha sin\beta +2sin\beta sin\gamma +2sin\alpha sin\gamma \right) +\left( { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha \right) +\left( { cos }^{ 2 }\beta +{ sin }^{ 2 }\beta \right) +\left( { cos }^{ 2 }\gamma +{ sin }^{ 2 }\gamma \right) =0\)
\({ \left( cos\alpha +cos\beta +cos\gamma \right) }^{ 2 }+{ \left( sin\alpha +sin\beta +sin\gamma \right) }^{ 2 }=0\)
\(cos\alpha +cos\beta +cos\gamma =0\) and \(sin\alpha +sin\beta +sin\gamma =0\)

LHS = cos²A + (1-sin²B) - 2cos A cos B.cos(A+B)
= cos²A - sin²B+1 - 2cos A cos B.cos(A+B)
=cos(A+B)cos(A-B) + 1 - 2 cos A cos B.cos(A+B)
= cos(A + B)cos(A - B) + 1 - 2cosA cosB.cos(A+B)
= cos(A+B)[cos(A-B) - 2cos A cos B]+1
= cos(A+B)[cos A cos B + sin A sin B - 2 cos A cos B]+1
= -cos(A+B)[cosA cosB - sinA sinB]+1
= -cos(A+B) cos(A+B)+1
= -cos²(A+B)+1 = 1-cos²(A+B)
= sin²(A+B) = RHS
Hence proved.

Let A, B be the land marks and C be the position of the plane,
Given ㄥACB 45°

Using cosine formula,
c² = a² + b² - 2ab cos C
c² = 2² + 1² - 2(2)(1) cos 45⁰
\(=4+1-4\left(1\over \sqrt3\right)\)

c² = 5 - 2√2
\(c^2=\sqrt{5-2\sqrt2}km\)

Let C be the position of the target and A and B be the positions of the fighter jet

Given ㄥBAC = 30, ㄥABC = 45
ஃ ㄥC = 180 - (30 - 45) = 180 - 75 = 105
Given AB = 100 km
Using sine formula,
\(\frac { a }{ sinA } =\frac { c }{ sinC } \)
\(\Rightarrow \frac { a }{ sin30° } =\frac { 100 }{ sin105° } \)
\(\Rightarrow \frac { a }{ \frac { 1 }{ 2 } } =\frac { 100 }{ sin105° } \Rightarrow 2a=\frac { 100 }{ sin105° } \Rightarrow a=\frac { 50 }{ sin105° } \)
Now, sin 105° = sin (60 + 45) = sin 60 cos 45 + cos 60 sin 45
= \(\frac { \sqrt { 3 } }{ 2 } .\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } .\frac { 1 }{ \sqrt { 2 } } =\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } \)
Substituting (2) in (1) we get,
a = \(\frac { 50 }{ \frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } } } \Rightarrow a=\frac { 50\left( 2\sqrt { 2 } \right) }{ \sqrt { 3 } +1 } =\frac { 100\sqrt { 2 } }{ \sqrt { 3 } +1 } \times \frac { \sqrt { 3 } -1 }{ \sqrt { 3 } -1 } \)
a = \(\frac { 100\left( \sqrt { 6 } -\sqrt { 2 } \right) }{ 3-1 } =50\left( \sqrt { 6 } -\sqrt { 2 } \right) km\)

\(\frac{2}{x+2}+\frac{1}{x-1}=\frac{2 x-2+x+2}{(x+2)(x-1)}=\frac{3 x}{(x+2)(x-1)}= k = 3\)

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

\(\log _{a} b \log _{b} c \log _{c} a=\log _{a} c \log _{c} a=\log _{a} a=1\)

\(\text { Since }|\cos x|<1\)
\(\text { From option (4), }\)
\(\sec \theta=\frac{1}{4}\)
\(\Rightarrow \cos \theta=4 \text { is not possible. }\)

\(\sqrt{2+2 \cos 4 \theta} =\sqrt{2+2\left(2 \cos ^{2} 2 \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} 2 \theta} \)
\(=2 \cos 2 \theta \)
\(\sqrt{2+\sqrt{2+2 \cos 4 \theta}} =\sqrt{2+2 \cos 2 \theta} \)
\(=\sqrt{2+2\left(2 \cos ^{2} \theta-1\right)} \)
\(=\sqrt{4 \cos ^{2} \theta} \)
\(=\pm 2 \cos \theta \)
\(\Rightarrow \pi<2 \theta<\frac{3 \pi}{2} \)
\(\Rightarrow \frac{\pi}{2}<\theta<\frac{3 \pi}{4} \text { in II quadrant. } \)
\(\therefore \sqrt{2+\sqrt{2+2 \cos 4 \theta}}=-2 \cos \theta\)

Let X = (a,b,c)
Then R = Universal relation
= {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}.
It is transitive

LetA = (a, b),B = (a,b,c)
A \(\times\)B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}
B \(\times\)A = {(a, a),(a, b), (b, a), (b, b), (c, a),(c, b)}
(A \(\times\)B) ⋂ (B \(\times\)A) = {(a, a), (a,b), (b, a): (b, b)}
= A \(\times\)A

\(\mathrm{T}: x-y \text { is an integer } \Rightarrow x \mathrm{R} y\)
\(\text { i) } x-x=0 \text { is an integer }\)
\(\therefore \text { T is reflexive. }\)
\(\text { ii) }(x-y) \text { is an integer } \Rightarrow y-x \text { is also an integer }\)
\(\therefore \mathrm{T} \text { is symmetry }\)
iii) If (x - y)is an integer and y- z is also an integer, by adding
x - z is also an integer.
\(\therefore \mathrm{T} \text { is transitive }\)
Thus, T is an equivalence relation.
\(\mathrm{S}: \mathrm{y}=x+1 \Rightarrow x \mathrm{~S} y\)
\(\text { i) } x=x+1 \Rightarrow x S x \text { is not true. }\)
\(\therefore S \text { is not reflexive. }\)
Hence T is an equivalence relation but S is not an equivalence relation

\(n(A \cap B)=1\)

It is not a function since it has two images

f(0) = 0, f(-3) = 9 and f(3) = 9
.'. S is [0,9]

It is onto not one-one

\(\text { Since } \sin 30^{\circ}=\frac{1}{2}\)
\(\sin 150^{\circ}=\frac{1}{2}\)

Let the roots be α, 2α 
sum of the roots α+2α = 3α =\(\frac { -(-a) }{ 2 } \) = \(\frac { a }{ 2 } \)
⇒ \(\alpha =\frac { a }{ 6 } \)      ...(1)
Product of the roots α(2α)= 2α² = \(\frac { 64 }{ 2 } \) = 32
⇒ a² = \(\frac { 32 }{ 2 } \) = 16 ⇒ α =\(\sqrt { 16 } =\pm 4\)
Substituting a value in (1) we get
± 4 = \(\frac { a }{ 6 } \) ⇒ a = ± 6\(\times\)4 = ± 24

Let the roots be α and α²
sum of the roots α + α² = \(\frac { -k }{ 3 } \)   ...(1)
product of the roots α(α²) = (α³) = \(\frac { -81 }{ 3 } \) = -27
α³ = -27=(-3)³ ⇒ α = -3
substituting α value in (1) we get
-3 + (-3)² = \(\frac { -k }{ 3 } \)
(i.e) -3 + 9 = \(\frac { -k }{ 3 } \)
(i.e) \(\frac { -k }{ 3 } \) = 6 ⇒ k = -18

cos 204= cos (180° + 24°) = - cos 24°
cos 125°= cos (180° - 55°) = - cos 55°
\(\therefore\) LHS cos 24° + cos 55° + (- cos 55°) + (- cos 24°) + cos 300°
= cos 24° + cos 55° - cos 55° - cos 24° + cos 300°
\(=cos300°=cos(360°-60°)=cos60°=\frac{1}{2}=RHS\)

LHS sin (270° - \(\theta\)) sin (90° - \(\theta\)) - cos (270° - \(\theta\)) cos (90° + \(\theta\)) + 1
Now, sin (270° - \(\theta\)) = sin {180° + (90° - \(\theta\))}
= - cos (90° - \(\theta\)) = - sin \(\theta\)
LHS =  - cos \(\theta\) . cos \(\theta\) - (- sin \(\theta\)) (- sin \(\theta\)) + 1
= - cos² \(\theta\) - sin²\(\theta\) + 1
= - (cos²\(\theta\) + sin² \(\theta\)) + 1 = -1 + 1 = 0 = RHS

Here T₂-T₁ = \(log\frac { { a }^{ 2 } }{ b } -log\quad a=log\frac { { a }^{ 2 }/b }{ a } \)
= \(log\frac { a }{ b } \)
T₃-T₂ = \(log\frac { a^{ 3 } }{ b^{ 2 } } -log\frac { a^{ 2 } }{ b^{ 2 } } =log\frac { a^{ 3 } }{ b^{ 2 } } +log\frac { a^{ 2 } }{ b^{ 2 } } \)
= \(log\frac { a^{ 3 } }{ b^{ 2 } } \times \frac { b }{ { a }^{ 2 } } =log\frac { a }{ b } \)
∴ T₂-T₁ = T₃-T₂ = \(log\left( \frac { a }{ b } \right) \)
∴ The given sequences is an A.P

In (1 + x)ⁿ, (r + 1)^th term from the beginning.
T_r+1 = nC_r 1^n-r. x^r = nC_rx^r   ....(1)
Its co-efficient is nC_r
In (1 + x)ⁿ, there are (n + 1)terms
So, the (r +1)^th term from the end will have (n + 1) - (r + 1) = n - r terms
∴ T_n-r+1 = nC_n-r 1^n-(n-r).x^n-r = nC_n-rx^n-r  ...(2)
Its Co-efficient is  nC_n-r
From (1) and (2), the Co-efficient of (r + 1)th term from the beginning and from the end are equal

(i) a particular teacher is included?
There are 5 teachers and 20 students 2 teachers out of 5 teachers can be selected in 5C₂ ways.
3 students out of 20 students can be selected in 20C₃ ways
Hence, total number of committees = 20C₃ \(\times \) 5C₂
= \(\frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= \(10\times 19\times 6\times 5\times 2\)
= 11400
Since a particular teacher is included, the committee will have 1 teacher and 3 students.
∴ 1 teacher can be selected from 4'teachers in 4C₁ = 4 ways.
3 students out of 20 students can be selected in 20C₃ ways.
Hence, required number of committees
= 4C₁ \(\times \) 20C₃
= \(4\times \frac { 20\times 19\times 18 }{ 3\times 2\times 1 } \)
= 4560
(ii) 2 teachers can be selected from 5 teachers in 5C₂ ways
Since a particular student is excluded, 3 students can be selected from 19 students in 19C₃ ways
Hence required number of committees = 19C₃ \(\times \) 5C₂
=\(\frac { 19\times 18\times 17 }{ 3\times 2\times 1 } \times \frac { 5\times 4 }{ 2\times 1 } \)
= 19 \(\times \)6\(\times \)17\(\times \)5
= 9690

Let there be n sides of the polygon. We know that the number of diagonals of n sided polygon is \(\frac { n(n-3) }{ 2 } \)
⇒ Given \(\frac { n(n-3) }{ 2 } =90\)
⇒ n²-2n = 180
⇒ n²-3n-180 = 0
⇒ (n-15) (n+12) = 0
⇒  n = 15 or n = -12
⇒ There are 15 sides for the polygon which has 90 diagonals.

Given (n+2)! = 60(n-1)!

\(\Rightarrow \) (n+2) (n+1)(n) = 60
\(\Rightarrow \) (n+2)(n+1)(n) = 5 \(\times\)4 \(\times\)3
Equating the terms both sides we get, n = 0

LHS = n!(n + 2)
RHS = n! + (n+1)!
= n!+(n+1)(n)!....(1)
= n!(1+n+1)
= n!(n+20)...(2)
From (1) and (2), LHS = RHS
Hence Proved.

Any line parallel to 2x + 3y = 10 is of the form 2x + 3y + k = 0 ....(1)
2x + 3y = -k
Dividing by - k we get,
\(\frac { 2x }{ -k } +\frac { 3y }{ -k } =1\)
\(\Rightarrow \frac { x }{ \left( -\frac { k }{ 2 } \right) } +\frac { y }{ \left( -\frac { k }{ 3 } \right) } =\pm \left( \frac { 5+k }{ 5 } \right) \)
Its intercepts are \(\frac { -k }{ 2 } \) and \(\frac { -k }{ 3 } \)
Also, it is given that sum of the intercepts on the axes is 15.
\(\therefore \ \frac { -k }{ 2 } +\left( \frac { -k }{ 3 } \right) =15\)
\(\Rightarrow \frac { -3k-2k }{ 6 } =15\)
\(\Rightarrow \frac { -5k }{ 6 } =15\)
\(\Rightarrow -5k=15\times 6\)
\(\Rightarrow k=\frac { -15\times 6 }{ 5 } =-18\)
\(\therefore\) Equation the required lines is 2x + 3y - 18 = 0 \(\Rightarrow\) 2x + 3y = 18.

Set of concentric circles in a plane is an infinite set since number of concentric circles in a plane is infinite

Given A = {0}  ...(1)
B = {x : x > 15 and x < 5} ⇒ B = Ф  ...(2)
C = {x : x-5 = 0} ⇒ B = {5}   ...(3)
D = { x : x² = 25} ⇒ D = {-5, 5}  ...(4)
E = {x : x is an integral positive root of x² - 2x - 15 = 0}
⇒ E = {5}  ...(5)
From (3) and (5), clearly C = E.
Hence C and E are equal sets

Given number is \(\frac { 1 }{ { 2 }^{ 1000 } } \)
we know 1000 < 1001
\(\Rightarrow\) 2¹⁰⁰⁰ < 2¹⁰⁰¹
\(\Rightarrow\) \(\frac { 1 }{ { 2 }^{ 1000 } } <\frac { 1 }{ { 2 }^{ 1001 } } \)
\(\therefore\)  A positive number smaller than \(\frac { 1 }{ { 2 }^{ 1000 } } is\frac { 1 }{ { 2 }^{ 1001 } } \)

Let the two irrational numbers be \(5+\sqrt { 7 } \) and \(7-\sqrt { 7 } \)
Their sum = \(\left( 5+\sqrt { 7 } \right) +\left( 7-\sqrt { 7 } \right) \ = \ 5+\sqrt { 7 } +7-\sqrt { 7 } \)
= 5 + 7 = 12 which is a rational number.
Consider  the two irrational numbers \(4+\sqrt { 6 } \) and \(4-\sqrt { 6 } \)
Their product = \(\left( 4+\sqrt { 6 } \right) +\left( 4-\sqrt { 6 } \right) ={ 4 }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 }\)= 16 - 6 = 10 which is a rational number.

(i) [-2, 0],
(ii) (0, 8),
(iii) (-8, -2],
(iv) [-5, -9]

A = { x | x is a whole number less than or equal to 7 } is the set-bilder form of A.

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m