Prasanna

Answer: (d) {n(n + 1)/2}²
Hint:
Given, series is 1³ + 2³ + 3³ + ……….. n³
Sum = {n(n + 1)/2}²

Answer: (b) a + b
Hint:
Given number = an + bn
Let n = 1, 3, 5, ……..
an + bn = a + b
an + bn = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)

Answer: (b) n/(n + 1)
Hint:
Let the given statement be P(n). Then,
P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} = n/(n + 1).
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.
LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i)
Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}
[1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}
= k/(k + 1)+1/{ (k + 1)(k + 2)}.
{k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)
= {k(k + 2) + 1}/{(k + 1)(k + 2}
= {(k + 1)² }/{(k + 1)(k + 2)}
= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)
⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}
= (k + 1)/(k + 1 + 1)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Answer: (a) 1/(n + 1) for all n ∈ N.
Hint:
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (c) 5
Hint:
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 7¹ – 2¹ = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Hint:
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} .
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. ……. (i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 4)}/{4(k + 2)(k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 2)}/{4(k + 2)(k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (b) n² + n + 1
Hint:
Let S = 3 + 7 + 13 + 21 +……….a_n-1 + a_n …………1
and S = 3 + 7 + 13 + 21 +……….a_n-1 + a_n …………2
Subtract equation 1 and 2, we get
S – S = 3 + (7 + 13 + 21 +……….a_n-1 + a_n) – (3 + 7 + 13 + 21 +……….a_n-1 + a_n)
⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_n – a_n-1) – a_n
⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – a_n
⇒ a_n = 3 + {4 + 6 + 8 + ……(n-1)terms}
⇒ a_n = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}
⇒ a_n = 3 + (n – 1)/2 × {8 + (n – 2)2}
⇒ a_n = 3 + (n – 1) × {4 + n – 2}
⇒ a_n = 3 + (n – 1) × (n + 2)
⇒ a_n = 3 + n² + n – 2
⇒ a_n = n² + n + 1
So, the nth term is n² + n + 1

Answer: (b) 3
Hint:
Let P(n) : n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k) : k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1)(k + 5) = 3m for some natural number m, … (i)
Now, (k + 1)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2)
= k(k + 1)(k + 2) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification]
= 3m + 3(k + 1 )(k + 4) [using (i)]
= 3[m + (k + 1)(k + 4)], which is a multiple of 3
⇒ P(k + 1) : (k + 1 )(k + 2)(k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2)×{∑n² + ∑n}
= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6

Answer: (c) 5
Hint:
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 7¹ – 2¹ = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Answer: (a) Even
Hint:
Let P(n): (n² + n) is even.
For n = 1, the given expression becomes (1² + 1) = 2, which is even.
So, the given statement is true for n = 1, i.e., P(1)is true.
Let P(k) be true. Then,
P(k): (k² + k) is even
⇒ (k² + k) = 2m for some natural number m. ….. (i)
Now, (k + 1)² + (k + 1) = k² + 3k + 2
= (k² + k) + 2(k + 1)
= 2m + 2(k + 1) [using (i)]
= 2[m + (k + 1)], which is clearly even.
Therefore, P(k + 1): (k + 1)² + (k + 1) is even
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n)is true for all n ∈ N.

Answer: (b) 17
Hint:
Given, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹
Let n = 1,
3 × 5^2×1+1 + 2^3×1+1 = 3 × 5²⁺¹ + 2³⁺¹ = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391
Which is divisible by 17
Let n = 2,
3 × 5^2×2+1 + 2^3×2+1 = 3 × 5⁴⁺¹ + 2⁶⁺¹ = 3 × 5⁵ + 2⁷ = 3 × 3125 + 128 = 9375 + 128
= 9503
Which is divisible by 17
Hence, For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2) × {∑n² + ∑n}
= (1/2) × {n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6

Answer: (a) 1/(n + 1) for all n ∈ N.
Hint:
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (a) 1 + nx
Hint:
Let P(n): (1 + x) )ⁿ ≥ (1 + nx).
For n = 1, we have LHS = (1 + x))¹ = (1 + x), and
RHS = (1 + 1 ∙ x) = (1 + x).
Therefore LHS ≥ RHS is true.
Thus, P(1) is true.
Let P(k) is true. Then,
P(k): (1 + x)¹ ≥ (1 + kx). …….. (i)
Now,(1 + x)^k+1 = (1 + x)^k (1 + x)
≥ (1 + kx)(1 + x) [using (i)]
=1 + (k + 1)x + kx²
≥ 1 + (k + 1)x + x [Since kx² ≥ 0]
Therefore P(k + 1) : (1 + x)k + 1 ≥ 1 + (k + 1)x
⇒ P(k +1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (c) 11
Hint:
Let P (n): (10^2n-1 + 1) is divisible by 11.
For n=1, the given expression becomes {10^(2×1-1) + 1} = 11, which is divisible by 11.
So, the given statement is true for n = 1, i.e., P (1) is true.
Let P(k) be true. Then,
P(k): (10^2k-1 + 1) is divisible by 11
⇒ (10^2k-1 + 1) = 11 m for some natural number m.
Now, {10^2(k-1)-1 – 1 + 1} = (10^2k+1 + 1) = {10² ∙ 10^(2k+1)+ 1}
= 100 × {10^2k-1 + 1 } – 99
= (100 × 11 m) – 99
= 11 × (100 m – 9), which is divisible by 11
⇒ P (k + 1) : {10^2(k-1) – 1 + 1} is divisible by 11
⇒ P (k + 1) is true, whenever P(k) is true.
Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Answer: (b) 2304
Hint:
Given number = 7²ⁿ − 48n − 1
Let n = 1, 2 ,3, 4, ……..
7²ⁿ − 48n − 1 = 7² − 48 − 1 = 49 – 48 – 1 = 49 – 49 = 0
7²ⁿ − 48n − 1 = 7⁴ − 48 × 2 − 1 = 2401 – 96 – 1 = 2401 – 97 = 2304
7²ⁿ − 48n − 1 = 7⁶ − 48 × 3 − 1 = 117649 – 144 – 1 = 117649 – 145 = 117504 = 2304 × 51
Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..
So, the given number is divisible by 2304

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Answer: (c) n/{3(2n + 3)}
Hint:
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1) : 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.

Answer: (b) sin² (x + y)
Hint:
cos² x + cos² y – 2cos x × cos y × cos(x + y)
{since cos(x + y) = cos x × cos y – sin x × sin y }
= cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y)
= cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y
= cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y
= (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y
= cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y
= sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 )
= sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y
= (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y
= (sin x × cos y + sin y × cos x)²
= {sin (x + y)}²
= sin² (x + y)

Answer: (d) a² + b² – c²
Hint:
We have
(a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Answer: (c) -1
Hint:
Given, cos 5π = cos (π + 4π) = cos π = -1

Answer: (c) 1
Hint:
Given cosec A (sin B cos C + cos B sin C)
= cosec A × sin(B+C)
= cosec A × sin(180 – A)
= cosec A × sin A
= cosec A × 1/cosec A
= 1

Answer: (a) 4 : (√5 – 1)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)

Answer: (c) -1
Hint:
180 is a standard degree generally we all know their values but if we want to go theoretically then
cos(90 + x) = – sin(x)
So, cos 180 = cos(90 + 90)
= -sin 90
= -1 {sin 90 = 1}
So, cos 180 = -1

Answer: (b) 90°
Hint:
Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2

Answer: (b) 45
Hint:
Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45

Answer: (c) 2π/3
Hint:
Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b² + c² – a²)/2bc
⇒ Cos A = (7² + 8² – 13²)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is
= 2π/3

Answer: (b) tan 60
Hint:
Given, tan 20 × tan 40 × tan 80
= tan 40 × tan 80 × tan 20
= [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20)
= [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20)
= [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20)
= [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20)
= [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20)
= [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20)
= [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20)
= [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}]
= (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)
= sin 60/cos 60
= tan 60
So, tan 20 × tan 40 × tan 80 = tan 60

Answer: (a) 4 : (√5 – 1)
Hint:
Given, the angles of a triangle be in the ratio 1 : 4 : 5
⇒ x + 4x + 5x = 180
⇒ 10x = 180
⇒ x = 180/10
⇒ x = 18
So, the angle are: 18, 72, 90
Since a : b : c = sin A : sin B : sin C
⇒ a : b : c = sin 18 : sin 72 : sin 90
⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1
⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4
Now, c /a = 4/(√5 – 1)
⇒ c : a = 4 : (√5 – 1)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6)
Hint:
√3 cos x-sin x=1
⇒ (√3/2)cos x – (1/2)sin x = 1/2
⇒ sin 60 × cos x – cos 60 × sin x = 1/2
⇒ sin (x – 60) = 1/2
⇒ sin (x – π/3) = sin 30
⇒ sin (x – π/3) = sinπ/6
⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z}
⇒ x = π/3 + n × π + (-1)n × (π/6)

Answer: (d) (2 – e²)^3/2
Hint:
Given, tan² θ = 1 – e²
⇒ tan θ = √(1 – e²)

From the figure and Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – e²)}² + 12
⇒ AC² = 1 – e² + 1
⇒ AC² = 2 – e²
⇒ AC = √(2 – e²)
Now, sec θ = √(2 – e²)
cosec θ = √(2 – e²)/√(1 – e²)
and tan θ = √(1 – e²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – e²) + {(1 – e²)^3/2 × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + {(1 – e²) × (1 – e²) × √(2 – e²)/√(1 – e²)}
= √(2 – e²) + (1 – e²) × √(2 – e²)
= √(2 – e²) × (1 + 1 – e²)
= √(2 – e²) × (2 – e²)
= (2 – e²)^3/2
So, sec θ + tan³ θ × cosec θ = (2 – e²)^3/2

Answer: (b) 1
Hint:
Given, cos 20 + 2sin² 55 – √2 sin65
= cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x}
= 1 + cos 20 – cos 110 – √2 sin65
= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula}
= 1 – 2 × sin 65 × sin (-45) – √2 sin65
= 1 + 2 × sin 65 × sin 45 – √2 sin65
= 1 + (2 × sin 65)/√2 – √2 sin65
= 1 + √2 ( sin 65 – √2 sin 65
= 1
So, cos 20 + 2sin² 55 – √2 sin65 = 1

Answer: (a) 2π/3
Hint:
Let S be the center of the circumcircle of triangle PQR.
So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii.
Thus SPQ & SPR are equilateral triangles.
⇒ ∠QSP = 60°;
Similarly ∠RQP = 60°
⇒ Angle at the center QSP = 120°
So, SRPQ is a rhombus, since all the four sides are equal.
Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Answer: (a) 2b = a + c
Hint:
Given, cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2)
⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2)
⇒ cos((A-C)/2) = 2sin (B/2)
⇒ cos((A-C)/2) = 2cos((A+C)/2)
⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
⇒ 2sin(A/2)sin(C/2) = sin(B/2)
⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2(s – b) = b
⇒ a + b + c – 2b = b
⇒ a + c – b = b
⇒ a + c = 2b

Answer: (c) sin 3x
Hint:
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4}
= sin x × {-sin 2x + 3 × cos 2x}
= sin x × {-sin 2x + 3 × (1 – sin 2x)}
= sin x × {-sin 2x + 3 – 3 × sin 2x}
= sin x × {3 – 4 × sin 2x}
= 3 × sin x – 4 sin 3x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Answer: (d) 1/x + 1/y
Hint:
Given,
tan A – tan B = x ……………. 1
and cot B – cot A = y ……………. 2
From equation,
1/cot A – 1/cot B = x
⇒ (cot B – cot A)/(cot A × cot B) = x
⇒ y/(cot A × cot B) = x {from equation 2}
⇒ y = x × (cot A × cot B)
⇒ cot A × cot B = y/x
Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A)
⇒ cot (A – B) = (y/x + 1)/y
⇒ cot (A – B) = (y/x) × (1/y) + 1/y
⇒ cot (A – B) = 1/x + 1/y

Answer: (b) 2 tan 6x
Hint:
Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)
⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}]
⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}]
⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x)
⇒ tan 6x + tan 6x
⇒ 2 tan 6x

Answer: (b) 4i
Hint:
Given, √(-16) = √(16) × √(-1)
= 4i {since i = √(-1) }

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1) × 144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

Answer: (c) 17i
Hint:
Given, √(-25) + 3√(-4) + 2√(-9)
= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}
= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}
= 5i + 3×2i + 2×3i {since √(-1) = i}
= 5i + 6i + 6i
= 17i
So, √(-25) + 3√(-4) + 2√(-9) = 17i

Answer: (a) a circle
Hint:
Let w = 2/z
Now, |w| = |2/z|
=> |w| = 2/|z|
=> |w| = 2
This shows that w lies on a circle with center at the origin and radius 2 units.

Answer: (d) -128 ω²
Hint:
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω – ω²)⁷ = (-ω² – ω²)⁷
⇒ (1 + ω – ω²)⁷ = (-2ω²)⁷
⇒ (1 + ω – ω²)⁷ = -128 ω¹⁴
⇒ (1 + ω – ω²)⁷ = -128 ω¹² × ω²
⇒ (1 + ω – ω²)⁷ = -128 (ω³)⁴ ω²
⇒ (1 + ω – ω²)⁷ = -128 ω²

Answer: (b) 2
Hint:
Given, {(1 + i)/(1 – i)}ⁿ
= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ
= [{(1 + i)²}/{(1 – i²)}]ⁿ
= [(1 + i² + 2i)/{1 – (-1)}]ⁿ
= [(1 – 1 + 2i)/{1 + 1}]ⁿ
= [2i/2]ⁿ
= iⁿ
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2

Answer: (a) -2√3 + 2i
Hint:
Let z = r(cos θ + i × sin θ)
Then r = 4 and θ = 5π/6
So, z = 4(cos 5π/6 + i × sin 5π/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i

Answer: (c) i
Hint:
Given, i^-999
= 1/i⁹⁹⁹
= 1/(i⁹⁹⁶ × i³)
= 1/{(i⁴)²⁴⁹ × i³}
= 1/{1²⁴⁹ × i³} {since i⁴ = 1}
= 1/i³
= i⁴/i³ {since i⁴ = 1}
= i
So, i^-999 = i

Answer: (c) a² = 3b
Hint:
Given, z₁ and z₂ be two roots of the equation z² + az + b = 0
Now, z₁ + z₂ = -a and z₁ × z₂ = b
Since z₁ and z₂ and z₃ from an equilateral triangle.
⇒ z₁² + z₂² + z₃² = z₁ × z₂ + z₂ × z₃ + z₁ × z₃
⇒ z₁²+ z₂² = z₁ × z₂ {since z₃ = 0}
⇒ (z₁ + z₂)² – 2z₁ × z₂ = z₁ × z₂
⇒ (z₁ + z₂)² = 2z₁ × z₂ + z₁ × z₂
⇒ (z₁ + z₂)² = 3z₁ × z₂
⇒ (-a)² = 3b
⇒ a² = 3b

Answer: (d) no value of x
Hint:
Given complex number = sin x + i cos 2x
Conjugate of this number = sin x – i cos 2x
Now, sin x + i cos 2x = sin x – i cos 2x
⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}
⇒ tan x = 1 and tan 2x = 1
Now both of them are not possible for the same value of x.
So, there exist no value of x

Answer: (d) a hyperbola
Hint:
Let z = x + iy
Now, z² = (x + iy)²
⇒ z² = x² – y² + 2xy
Given, Im(z²) = k
⇒ 2xy = k
⇒ xy = k/2 which is a hyperbola.

Answer: (a) x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
⇒ y = 2/3
and 7 – 2x = 0
⇒ x = 7/2
So, the value of x = 7/2 and y = 2/3

Answer: (b) θ = nπ ± π/3 where n is an integer
Hint:
Given,
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1
Now, equation 1 is imaginary if
3 – 4sin² θ = 0
⇒ 4sin² θ = 3
⇒ sin² θ = 3/4
⇒ sin θ = ±√3/2
⇒ θ = nπ ± π/3 where n is an integer

Answer: (d) 4
Hint:
Given, {(1 + i)/(1 – i)}ⁿ = 1
⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]ⁿ = 1
⇒ [{(1 + i)²}/{(1 – i²)}]ⁿ = 1
⇒ [(1 + i² + 2i)/{1 – (-1)}]ⁿ = 1
⇒ [(1 – 1 + 2i)/{1 + 1}]ⁿ = 1
⇒ [2i/2]ⁿ = 1
⇒ iⁿ = 1
Now, iⁿ is 1 when n = 4
So, the least value of n is 4

Answer: (a) π
Hint:
Given, arg (z) < 0
Now, arg (-z) – arg (z) = arg(-z/z)
⇒ arg (-z) – arg (z) = arg(-1)
⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}

Answer: (c) -1
Hint:
Given x + 1/x = 1
⇒ (x² + 1) = x
⇒ x² – x + 1 = 0
⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1)
⇒ x = {1 ± √(1 – 4)}/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(-1)×√3}/2
⇒ x = {1 ± i√3}/2 {since i = √(-1)}
⇒ x = -w, -w²
Now, put x = -w, we get
x²⁰⁰⁰ + 1/x²⁰⁰⁰ = (-w)²⁰⁰⁰ + 1/(-w)²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= w²⁰⁰⁰ + 1/w²⁰⁰⁰
= {(w³)⁶⁶⁶ × w²} + 1/{(w³)⁶⁶⁶ × w²}
= w² + 1/w² {since w³ = 1}
= w² + w³ /w²
= w² + w
= -1 {since 1 + w + w² = 0}
So, x²⁰⁰⁰ + 1/x²⁰⁰⁰ = -1

Answer: (a) 12i
Hint:
Given, √(-144) = √{(-1)×144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i

Answer: (c) – 1, 1 – 2ω, 1 – 2ω²
Hint:
Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this)
If we let u = x – 1, then the equation becomes
u³ + 8 = (u + 2)(u² – 2u + 4) = 0.
So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when:
u² – 2u + 4 = 0,
which has roots, by the Quadratic Formula, to be u = 1 ± i√3
So, x – 1 = 1 ± i√3
⇒ x = 2 ± i√3
Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and
x² + x + 1 = 0
⇒ x = (-1 ± i√3)/2
If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2
then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0
So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²

Answer: (b) 2²ⁿ
Hint:
Given, (1 – w + w²)×(1 – w² + w⁴)×(1 – w⁴ + w⁸) × …………… to 2n factors
= (1 – w + w²)×(1 – w² + w )×(1 – w + w²) × …………… to 2n factors
{Since w⁴ = w, w⁸ = w²}
= (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors
= (2² w³)×(2² w³)×(2² w³) …………… to 2n factors
= (2²)ⁿ {since w³ = 1}
= 2²ⁿ

Answer: (c) √41
Hint:
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = √(5² + 4²)
⇒ |Z| = √(25 + 16)
⇒ |Z| = √41
So, the modulus of 5 + 4i is √41

Answer: (b) x
Hint:
Given, f(x) = (a – x)^1/n
Now, f(f(x)) = [(a – f(x))ⁿ]^1/n
⇒ f(f(x)) = [(a – {(a – xⁿ)^1/n }ⁿ ]^1/n
⇒ f(f(x)) = [a – (a – xⁿ)]^1/n
⇒ f(f(x)) = [a – a + xⁿ)]^1/n
⇒ f(f(x)) = (xⁿ)^1/n
⇒ f(f(x)) = x

Answer: (b) |x| ≥ 1
Hint:
We have f(x) = √(log₁₂ x²)
Since, log_a k ≥ 0 if a > 1, k ≥ 1
or 0 < a < 1 and 0 < k ≤ 1
So, the function f(x) exists if
log₁₂ x² ≥ 0
⇒ x² ≥ 1
⇒ |x| ≥ 1

Answer: (b) 1
Hint:
Given, f(x) = e^x
and g(x) = log x
fog(x) = f(g(x))
= f (log x)
= e^{log x}
= x
So, fog(1) = 1

Answer: (d) all of above
Hint:
Two functions f and g are said to be equal if f
1. the domain of f = the domain of g
2. the co-domain of f = the co-domain of g
3. f(x) = g(x) for all x

Answer: (b) f(-x) = -f(x)
Hint:
A function f(x) is said to be an odd function if
f(-x) = -f(x) for all x

Answer: (a) Even function
Hint:
Given, f(x) is an odd differentiable function on R
⇒ f(-x) = -f(x) for all x ∈ R
differentiate on both side, we get
⇒ -df(-x)/dx = -df(x)/dx for all x ∈ R
⇒ df(-x)/dx = df(x)/dx for all x ∈ R
⇒ df(x)/dx is an even function on R.

Answer: (a) 4
Hint:
Period of sin (‎πx/2) = 2π/(π/2) = 4
Period of cos (πx/2) = 2π/(π/2) = 4
So, period of f(x) = LCM (4, 4) = 4

Answer: (c) (1, 3)
Hint:
Since f(x) = log₃ x is an increasing function
So, f(A) = (log₃ 3, log₃ 27) = (1, 3)

Answer: (a) R
Hint:
Since tan^-1 x exists if x ∈ (-∞, ∞)
So, tan^-1 (2x + 1) is defined if
-∞ < 2x + 1 < ∞
⇒ -∞ < x < ∞
⇒ x ∈ (-∞, ∞)
⇒ x ∈ R
So, domain of tan^-1 (2x + 1) is R.

Answer: (b) 1
Hint:
Let T is a positive real number.
Let f(x) is periodic with period T.
Now, f(x + T) = f(x), for all x ∈ R
⇒ x + T – [x + T] = x – [x], for all x ∈ R
⇒ [x + T] – [x] = T, for all x ∈ R
Thus, there exist T > 0 such that f(x + T) = f(x) for all x ∈ R
Now, the smallest value of T satisfying f(x + T) = f(x) for all x ∈ R is 1
So, f(x) = x – [x] has period 1

Answer: (a) x
Hint:
Given, f(x) = (3x – 2)/(2x – 3)
Now, f(f(x)) = f{(3x – 2)/(2x – 3)}
= {(3×(3x – 2)/(2x – 3) – 2)}/{(2(3x – 2)/(2x – 3) – 3)}
= {(9x – 6)/(2x – 3) – 2)}/{((6x – 4)/(2x – 3) – 3)}
= [{(9x – 6) – 2(2x – 3)}/(2x – 3)]/[{(6x – 4) – 3(2x – 3)}/(2x – 3)]
= {(9x – 6) – 2(2x – 3)}/{(6x – 4) – 3(2x – 3)}
= (9x – 6 – 4x + 6)/(6x – 4 – 6x + 9)
= 5x/5
= x
So, f(f(x)) = x

Answer: (b) 2x² + 1
Hint:
Given, f(x) = x² and g(x) = 2x + 1
Now gof(x) = g(f(x)) = f(x²) = 2x² + 1

Answer: (b) (0, -4, 4)
Hint:
Given that:
(x, y) ∈ R ⇔ x² + y² = 16
⇔ y = ±√(16 – x² )
when x = 0 ⇒ y = ±4
(0, 4) ∈ R and (0, -4) ∈ R
when x = ±4 ⇒ y = 0
(4, 0) ∈ R and (-4, 0) ∈ R
Now for other integral values of x, y is not an integer.
Hence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}
So, Domain(R) = {0, -4, 4}

Answer: (d) 16
Hint:
Let S is a finite set containing n elements.
Since binary operation on S is a function from S×S to S, therefore total number of
binary operations on S is the
total number of functions from S×S to S = (nⁿ)²
Given Set = {a, b}
Total number of elements = 2
Total number of binary operations = (2²)² = 2⁴ = 16

Answer: (a) Even function
Hint:
Given, f is an even function and g is an odd function.
Now, fog(-x) = f{g(-x)}
= f{-g(x)} {since g is an odd function}
= f{g(x)} for all x {since f is an even function}
So, fog is an even function.

Answer: (c) R – {1, 2}
Hint:
Given, function is f(x) = 1/(x² – 3x + 2)
Clearly, f(x) is not defined when x² – 3x + 2 = 0
⇒ (x – 1)×(x – 1) = 0
⇒ x = 1, 2
So, f(x) is not defined when x = 1, 2
So, domain of function is R – {1, 2}

Answer: (b) nπ – π/4 ≤ x ≤ nπ + π/4
Hint:
sin^-1 (tan x) is defined for -1 ≤ tan x ≤ 1
= -π/4 ≤ x ≤ π/4
The general solution of the above inequality is
nπ -π/4 ≤ x ≤ nπ + π/4

Answer: (d) {-7, -5, -3}
Hint:
Given, A = {-2, -1, 0}
and f(x) = 2x – 3
Now, f(-2) = 2 × (-2) – 3 = -4 – 3 = -7
f(-1) = 2 × (-1) – 3 = -2 – 3 = -5
f(0) = 2 × 0 – 3 = -3
So, range of f = {-7, -5, -3}

Answer: (d) {1, 2, 3}
Hint:
The function f(x) = ^7-xP_x-3 is defined only if x is an integer satisfying the following inequalities:
1. 7 – x ≥ 0
2. x – 3 ≥ 0
3. 7 – x ≥ x – 3
Now, from 1, we get x ≤ 7 ……… 4
from 2, we get x ≥ 3 ……………. 5
and from 2, we get x ≤ 5 ………. 6
From 4, 5 and 6, we get
3 ≤ x ≤ 5
So, the domain is {3, 4, 5}
Now, f(3) = ^7-3P_3-3 = ⁴P₀ = 1
⇒ f(4) = ^7-4P_4-3 = ³P₁ = 3
⇒ f(5) = ^7-5P_5-3 = ²P₂ = 2
So, the range of the function is {1, 2, 3}

Answer: (d) π/6
Hint:
Since g(x) = sin⁴ x + cos⁴ x is periodic with period π/2
So, f(x) = sin⁴ 3x + cos⁴ 3x is periodic with period (π/2)/3 = π/6

Answer: (d) (A – B) ∩ (A – C)
Hint:
Given A, B and C are any three sets.
Now, A – (B ∪ C) = (A – B) ∩ (A – C)

Answer: (d) A
Hint:
(A’)’ = A

Answer: (a) Difference of A and B of B and A
Hint:
A – B will read as difference of A and B of B and A
Ex: Let A = {1, 2, 3, 4, 5} and B = {1, 3, 5, 7}
Now, A – B = {2, 4}

Answer: (a) (A × B) ∪ (A × C)
Hint:
Given A, B and C are any three sets.
Now, A × (B ∪ C) = (A × B) ∪ (A × C)

Answer: (a) [5, 6, 7, 8, 9]
Hint:
Given A = [5, 6, 7] and B = [7, 8, 9]
then A ∪ B = [5, 6, 7, 8, 9]

Answer: (b) 2 and 3
Hint:
2 and 3 is the null set.

Answer: (b) Range R = {1, 3, 5}
Hint:
Given R = {(2, 1),(4, 3),(4, 5)}
then Range(R) = {1, 3, 5}

Answer: (b) {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
Hint:
The set S consists of the square of an integer less than 100
So, S = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81}

Answer: (b) 41
Hint:
The number of students who took at least one of the three subjects can be found by finding out A ∪ B ∪ C, where A is the set of those who took Physics, B the set of those who took Chemistry and C the set of those who opted for Math.
Now, A ∪ B ∪ C = A + B + C – (A ∩ B + B ∩ C + C ∩ A) + (A ∩ B ∩ C)
A is the set of those who opted for Physics = 120/2 = 60 students
B is the set of those who opted for Chemistry = 120/5 = 24
C is the set of those who opted for Math = 120/7 = 17
The 10th, 20th, 30th….. numbered students would have opted for both Physics and Chemistry.
Therefore, A ∩ B = 120/10 = 12
The 14th, 28th, 42nd….. Numbered students would have opted for Physics and Math.
Therefore, C ∩ A = 120/14 = 8
The 35th, 70th…. numbered students would have opted for Chemistry and Math.
Therefore, B ∩ C = 120/35 = 3
And the 70th numbered student would have opted for all three subjects.
Therefore, A ∪ B ∪ C = 60 + 24 + 17 – (12 + 8 + 3) + 1 = 79
Number of students who opted for none of the three subjects = 120 – 79 = 41

Answer: (b) symmetric
Hint:
Given {(a, b) : a² + b² = 1} on the set S.
Now a² +b² = b² + a² = 1
So, the given relation is symmetric.

Answer: (b) 6, 4
Hint:
Let A and B be two sets having m and n numbers of elements respectively
Number of subsets of A = 2m
Number of subsets of B = 2n
Now, according to question
2m – 2n = 48
⇒ 2n(2m – n – 1) = 24(22 – 1)
So, n = 4
and m – n = 2
⇒ m – 4 = 2
⇒ m = 2 + 4
⇒ m = 6

Answer: (a) (- ∞, ∞)
Hint:
Let the given function is
y = 3x – 2
⇒ y + 2 = 3x
⇒ x = (y + 2)/3
Now x is saisfied by all values.
So, Range{f(x)} = R = (-∞, ∞)

Answer: (a) B = C
Hint:
Given A, B, C be three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C then B = C

Answer: (b) X < 0, Y > 0
Hint:
In the second quadrant,

X < 0, Y > 0

Answer: (c) Infinite
Hint:
There are infinite many rational and irrational numbers are possible between 0 and 1
This is because between any two numbers, there are infinite numbers.

Answer: (a) Finite Set
Hint:
In mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality (count of elements in a set) is zero.
So, an empty set is a finite set.

Answer: (a) [5, 6, 7, 8, 9]
Hint:
Given A = [5, 6, 7] and B = [7, 8, 9]
then A U B = [5, 6, 7, 8, 9]

Answer: (c) A = {1, 2, 3} and B = {2, 1, 3}
Hint:
Two sets are equal if and only if they have the same elements.
So, A = {1, 2, 3} and B = {2, 1, 3} are equal sets.

Answer: (d) 27
Hint:
Total students = 50
Students who did not opt for math = 10
Students who did not opt for Science = 15
Students who did not opt for either maths or science = 2
Total of 40 students in math and 13 did not opt for science but did for math = 40 – 13 = 27
So, students of the class opted for both math and science is 27

Answer: (d) X > 0, Y > 0
Hint:
In the last quadrant,

X > 0, Y > 0