Prasanna

Answer: (b) 210
Hint:
Total number of triangles that can be formed with 12 points (if none of them are collinear)
= ¹²C₃
(this is because we can select any three points and form the triangle if they are not collinear)
With collinear points, we cannot make any triangle (as they are in straight line).
Here 5 points are collinear. Therefore we need to subtract ⁵C₃ triangles from the above count.
Hence, required number of triangles = ¹²C₃ – ⁵C₃ = 220 – 10 = 210

Answer: (d) ⁿC_r
Hint:
The number of combination of n distinct objects taken r at a time be x is given by
ⁿC_r = n!/{(n – r)! × r!}
Let the number of combination of n distinct objects taken r at a time be x.
Now consider one of these n ways. There are e objects in this selection which can be arranged in r! ways.
So, each of the x combinations gives rise to r! permutations. So, x combinations will give rise to x×(r!).
Consequently, the number of permutations of n things, taken r at a time is x×(r!) and it is equal to ⁿP_r
So, x×(r!) = ⁿP_r
⇒ x×(r!) = n!/(n – r)!
⇒ x = n!/{(n – r)! × r!}
⇒ ⁿC_r = n!/{(n – r)! × r!}

Answer: (b) 671
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671

Answer: (c) 450
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450

Answer: (a) 5⁸
Hint:
Total number of toys = 8
Total number of children = 5
Now, each toy can be distributed in 5 ways.
So, total number of ways = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5
= 5⁸

Answer: (c) n!
Hint:
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!

Answer: (b) 5⁴
Hint:
Here, both balls and boxes are different.
Now, 1st ball can be placed into any of the 5 boxes.
2nd ball can be placed into any of the 5 boxes.
3rd ball can be placed into any of the 5 boxes.
4th ball can be placed into any of the 5 boxes.
So, the required number of ways = 5 × 5 × 5 × 5 = 5⁴

Answer: (a) 1
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1

Answer: (c) 1008
Hint:
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x₁ is for apple, x₂ is for mango and x₃ is for orange.
Now, first we have to select total 15 fruits out of them.
x₁ + x₂ + x₃ = 15 (where 0 ⇐ x₁ ⇐ 5, 0 ⇐ x₂ ⇐ 10, 0 ⇐ x₃ ⇐ 13)
= (x⁰ + x¹ + x² +………+ x⁵)×(x⁰ + x¹ + x² +………+ x1¹⁰)×(x⁰ + x¹ + x² +………+ x¹³)
= {(1- x⁶)/(1 – x)}×{(1- x¹¹)/(1 – x)}×{(1- x¹⁴)/(1 – x)}
= {(1- x⁶)×(1- x¹¹)×{(1- x¹⁴)}/(1 – x)³
= {(1- x⁶)×(1- x¹¹)×{(1- x¹⁴)} × ∑^3+r+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷)×{(1- x¹⁴)} × ∑^3+r+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷ – x¹⁴ + x²⁵ + x²⁰ – x³¹)} × ∑^2+rC_r × x^r
= 1 × ∑^2+rC_r × x^r – x¹¹ × ∑^2+rC_r × x^r – x⁶ × ∑^2+rC_r × x^r + x¹⁷ × ∑^2+rC_r × x^r – x¹⁴ × ∑^2+rC_r × x^r + x²⁵ × ∑^2+rC_r × x^r + x²⁰ × ∑^2+rC_r × x^r – x³¹ × ∑^2+rC_r × x^r
= ∑^2+rC_r × x^r – ∑^2+rC_r × x^r+11 – ∑^2+rC_r × x^r+6 + ∑^2+rC_r × x^r+17 – ∑^2+rC_r × x^r+14 + ∑^2+rC_r × x^r+25 + ∑^2+rC_r × x^r+20 – ∑^2+rC^r × x^r+25
Now we have to find co-efficeient of x¹⁵
= ²⁺¹⁵C₁₅ – ²⁺⁴C₄ – ²⁺⁹C⁹ – ²⁺¹C¹ (rest all terms have greater than x¹⁵, so its coefficients are 0)
= ¹⁷C₁₅ – ⁶C₄ – ¹¹C₉ – ³C₁
= ¹⁷C₂ – ⁶C₂ – ¹¹C₂ – ³C₁
= {(17×16)/2} – {(6×5)/2} – {(11×10)/2} – 3
= (17×8) – (3×5) – (11×5) – 3
= 136 – 15 – 55 – 3
= 136 – 73
= 63
Again we have to distribute 15 fruits between 2 persons.
So x₁ + x₂ = 15
= ^2-1+15C₁₅
= ¹⁶C₁₅
= ¹⁶C₁
= 16
Now total number of ways of distribution = 16 × 63 = 1008

Answer: (a) 604800
Hint:
6 men can be sit as
× M × M × M × M × M × M ×
Now, there are 7 spaces and 4 women can be sit as ⁷P₄ = ⁷P₃ = 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800

Answer: (d) 151200
Hint:
Given word is : ASSASSINATION
Total number of words = 13
Number of A : 3
Number of S : 4
Number of I : 2
Number of N : 2
Number of T : 1
Number of O : 1
Now all S are taken together. So it forms a single letter.
Now total number of words = 10
Now number of ways so that all S are together = 10!/(3!×2!×2!)
= (10×9×8×7×6×5×4×3!)/(3! × 2×2)
= (10×9×8×7×6×5×4)/(2×2)
= 10×9×8×7×6×5
= 151200
So total number of ways = 151200

Answer: (c) 450
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450

Answer: (b) 7
Hint:
The number of triangles that can be formed using the vertices of a regular polygon = ⁿC₃
Given, T_n+1 – T_n = 21
⇒ ⁿ⁺¹C₃ – ⁿC₃ = 21
⇒ ⁿC₂ + ⁿC₃ – ⁿC₃ = 21 {since ⁿ⁺¹C_r = ⁿC_r-1 + ⁿC_r}
⇒ ⁿC₂ = 21
⇒ n(n – 1)/2 = 21
⇒ n(n – 1) = 21×2
⇒ n² – n = 42
⇒ n² – n – 42 = 0
⇒ (n – 7)×(n + 6) = 0
⇒ n = 7, -6
Since n can not be negative,
So, n = 7

Answer: (c) 360
Hint:
Given word is GARDEN.
Total number of ways in which all letters can be arranged in alphabetical order = 6!
There are 2 vowels in the word GARDEN A and E.
So, the total number of ways in which these two vowels can be arranged = 2!
Hence, required number of ways = 6!/2! = 720/2 = 360

Answer: (a) 24
Hint:
Any factors of 2⁵ × 3⁶ × 5² which is a perfect square will be of the form 2^a × 3^b × 5^c
where a can be 0 or 2 or 4, So there are 3 ways
b can be 0 or 2 or 4 or 6, So there are 4 ways
a can be 0 or 2, So there are 2 ways
So, the required number of factors = 3 × 4 × 2 = 24

Answer: (b) 196
Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement = ⁵C₄ × ⁸C₆
= ⁵C₁ × ⁸C₂
= 5 × (8×7)/(2×1)
= 5 × 4 × 7
= 140
2. When all 5 is selected from the first 5 and rest 5 from remaining 8
Total arrangement = ⁵C₅ × ⁸C₅
= 1 × ⁸C₃
= (8×7×6)/(3×2×1)
= 8×7
= 56
Now, total number of choices available = 140 + 56 = 196

Answer: (b) 671
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671

Answer: (b) 40320
Hint:
The number of ways in which 8 students can be sated in a line = ⁸P₈
= 8!
= 40320

Answer: (d) 204
Hint:
A chess board contains 9 lines horizontal and 9 lines perpendicular to them.
To obtain a square, we select 2 lines from each set lying at equal distance and this equal
distance may be 1, 2, 3, …… 8 units, which will be the length of the corresponding square.
Now, two lines from either set lying at 1 unit distance can be selected in ⁸C₁ = 8 ways.
Hence, the number of squares with 1 unit side = 8²
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7², 6², …… 1²
Hence, total number of square = 8² + 7² + ……+ 1² = 204

Answer: (a) 720
Hint:
The word LOGARITHMS has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= ¹⁰P₃
= 10 ×9 ×8
= 720

Answer: (b) 143/11760
Hint:
Total number of cards = 52
Two cards are lost.
So remaining cards = 50
Now one card is drawn.
Probability that it is a diamond card = 13/50
Now probability that both lost cards are heart = 13/50 ×(¹¹C₂ / ⁴⁹C₂)
= 13/50 ×[{(11×10)/2}/{(49×48/2)}]
= 13/50 ×{(11×10)/(49×48)}
= {(13×11×10)/(50×49×48)}
= {(13×11)/(5×49×48)}
= 143/11760
So probability that both lost card are heart = 143/11760

Answer: (c) 16/625
Hint:
The last digit of the four whole number can be
0, 1, 2, 3, 4, 5, 6, 7, 8, 9
The chance that any of the four numbers is divisible by 2 or 5 = 6/10 = 3/5
Hence, the chance that any of the four numbers is not divisible by 2 or 5 = 1 – 3/5 = 2/5
So, the chance that all of the four numbers are divisible by 2 or 5 = (2/5)×(2/5)×(2/5)×(2/5)
= 16/625
This is the chance that the last digit in the product will not be 0, 2, 4, 5, 6, 8 and this is also the chance that the last digit in the product is 1, 3, 7 or 9

Answer: (b) 1/36
Hint:
Total number of cases = 6³ = 216
The same number can appear on each of the dice in the following ways:
(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)
So, favourable number of cases = 6
Hence, required probability = 6/216 = 1/36

Answer: (b) 1/6
Hint:
First, we choose 1 machine out of given 4.
The probability that it is fault = 2/4 = 1/2
Now, we have to pick the second fault machine.
The probability that it is fault = 1/3
So, required probability = (1/2)×(1/3) = 1/6

Answer: (d) 7/9
Hint:
When two dice are throw, then Total outcome = 36
A doublet: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Favourable outcome = 6
Sum is 10: {(4, 6), (5, 5), (6, 4)}
Favourable outcome = 3
Again, A doublet and sum is 10: (5, 5)
Favourable outcome = 1
Now, P(either dublet or a sum of 10 appears) = P(A dublet appear) + P(sum is 10) – P(A dublet appear and sum is 10)
⇒ P(either dublet or a sum of 10 appears) = 6/36 + 3/36 – 1/36
= (6 + 3 – 1)/36
= 8/36
= 2/9
So, P(neither dublet nor a sum of 10 appears) = 1 – 2/9 = 7/9

Answer: (d) 21
Hint:
When two dice are thrown, then total outcome = 6×6 = 36
A: Getting an odd number on the first die.
A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}
Total outcome = 18
B: Getting a total of 7 on the two dice.
B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Total outcome = 6
C: Getting a total of greater than or equal to 8 on the two dice.
C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Total outcome = 15
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A ∪ B) = 18 + 6 – 3
⇒ n(A ∪ B) = 21

Answer: (a) 4/5
Hint:
Total number of ways of choosing two numbers out of six = ⁶C₂ = (6×5)/2 = 3×5 = 15
If smaller number is chosen as 3 then greater has choice are 4, 5, 6
So, total choices = 3
If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6
So, total choices = 4
If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6
So, total choices = 5
Total favourable case = 3 + 4 + 5 = 12
Now, required probability = 12/15 = 4/5

Answer: (c) 9/1547
Hint:
Total number of cards = 52
Number of king card = 4
Now, 7 cards are drawn from 52 cards.
P (3 cards are king) = {⁴C₃ × ⁴⁸C₄}/⁵²C₇
= {4×(48×47×46×45)/(4×3×2×1)}/{(52×51×50×49×48×47×46)/(7×6×5×4×3×2×1)}
= {4×(48×47×46×45)×(7×6×5×4×3×2×1)}/{(4×3×2×1)×{(52×51×50×49×48×47×46)}
= (7×6×5×4×45)/(52×51×50×49)
= (6×5×4×45)/(52×51×50×7)
= (6×4×45)/(7×52×51×10)
= (6×45)/(7×13×51×10)
= (6×3)/(7×13×17×2)
= (3×3)/(7×13×17)
= 9/1547

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2/(4×2) = e^-1/2/8

Answer: (d) 0.0545
Hint:
Given word: INSTITUTION
Total letters = 11
The word contains 3 I, 2 N, 1 S, 3 T, 1 U and 1 O
Total number of arrangement = 11!/(3!×2!×3!) = 554400
Now, taken 3 T are together.
So total latter = 9
The number of favorable cases = 9!/(3!×2!) = 30240
Now, P(3 T are together) = 30240/554400 = 0.0545

Answer: (b) 1/9
Hint:
One person can select one house out of 3 = ³C₁ = 3
So, three persons can select one house out of three = 3×3×3 = 27
Thus, probability that all the three can apply for the same house = 3/27 = 1/9

Answer: (d) 4/9
Hint:
Total number of shocks = 5 + 4 = 9
Two shocks are pulled.
Now, P(Both are same color) = (⁵C₂ + ⁴C₂)/⁹C₂
= {(5×4)/(2×1) + (4×3)/(2×1)}/{(9×8)/(2×1)}
= {(5×4) + (4×3)/}/{(9×8)
= (5 + 3)/(9×2)
= 8/18
= 4/9

Answer: (c) 247/256
Hint:
Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
The general form for probability of random variable x is
P(X = x) = ⁿC_x × p^x × q^n-x
Now, in the question, we want at least two heads
Now, p = q = 1/2
So, P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)^8-2
⇒ P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)⁶
⇒ 1 – P(X < 2) = ⁸C₀ × (1/2)⁰ × (1/2)⁸ + ⁸C₁ × (1/2)¹ × (1/2)^8-1
⇒ 1 – P(X < 2) = (1/2)⁸ + 8 × (1/2)¹ × (1/2)⁷
⇒ 1 – P(X < 2) = 1/256 + 8 × (1/2)⁸
⇒ 1 – P(X < 2) = 1/256 + 8/256
⇒ 1 – P(X < 2) = 9/256
⇒ P(X < 2) = 1 – 9/256
⇒ P(X < 2) = (256 – 9)/256
⇒ P(X < 2) = 247/256

Answer: (c) 1/2
Hint:
Given, a couple has two children.
Let A denotes both children are females i.e. {FF}
Now, P(A) = (1/2)×(1/2) = 1/4
and B denotes elder children is a female i.e. {FF, FM}
P(B) = 1/4 + 1/4 = 1/2
Now, P(A ∩ B) = 1/4
Now, P(Both the children are female if elder child is female)
P(A/B) = P(A ∩ B)/P(B)
⇒ P(A/B) = (1/4)/(1/2)
⇒ P(A/B) = 1/2

Answer: (c) e^-1/2/8
Hint:
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2 /(4×2) = e^-1/2/8

Answer: (a) 1 – 3/e²
Hint:
Here m = 2
Now, P(X > 1.5) = ∑_r {(e^-2 × 2^r)/r!} {2 ≤ r ≤ ∞}
= e^-2 {2²/2! + 2³/3! + 2⁴/4! + …}
= e^-2 {(1 + 2 /1! + 2²/2! + 2³/3! + …) – 1 – 2}
= e^-2 (e² – 3)
= 1 – 3e^-2
= 1 – 3/e²

Answer: (d) 0.9
Hint:
Given, A and B are two mutually exclusive events.
So, P(A ∩ B) = 0
Again given P(A) = 0.5 and P(B ̅) = 0.6
P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B)
⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

Answer: (b) 2/7
Hint:
In a leap year, the total number of days = 366 days.
In 366 days, there are 52 weeks and 2 days.
Now two days may be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Now there are total 7 possibilities, So total outcomes = 7
In 7 possibilities, Sunday came two times.
So, favorable case = 2
Hence, the probabilities of getting 53 Sundays in a leap year = 2/7

Answer: (a) 1 – (5/6)⁶
Hint:
Let A is the event that 6 does not occur at all.
Now, the probability of at least one 6 occur = 1 – P(A)
= 1 – (5/6)⁶

Answer: (d) 1/12
Hint:
Total cities are 4 i.e. A, B, C, D
Given, Rahul visit four cities, So, n(S) = 4! = 24
Now, sample space IS:
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CDAD, CDAB,CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}
Let G = Rahul visits A firsta and B last
⇒ G = {ACDB, ADCB}
⇒ n(G) = 2
So, P(G) = n(G)/n(S) = 2/24 = 1/12

Answer: (c) 3/5
Hint:
Given, ∑x = 12 and ∑x² = 18
Now, varience = ∑x²/n – (∑x/n)²
⇒ varience = 18/10 – (12/10)²
⇒ varience = 9/5 – (6/5)²
⇒ varience = 9/5 – 36/25
⇒ varience = (9 × 5 – 36)/25
⇒ varience = (45 – 36)/25
⇒ varience = 9/25
⇒ Standard deviation = √(9/25)
⇒ Standard deviation = 3/5

Answer: (b) 30.1
Hint:
Given, algebraic sum of of the deviation of 20 observations measured from 30 is 2
⇒ ∑(x_i – 30) = 2 {1 ≤ i ≤ 20}
⇒ ∑x_i – 30 × 20 = 2
⇒ (∑x_i)/20 – (30 × 20)/20 = 2/20
⇒ (∑x_i)/20 – 30 = 0.1
⇒ Mean – 30 = 0.1
⇒ Mean = 30 + 0.1
⇒ Mean = 30.1

Answer: (b) S.D./Mean
Hint:
The coefficient of variation = S.D./Mean

Answer: (b) 1446
Hint:
Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5
= 7230/5
= 1446

Answer: (b) 8
Hint:
Given, Mode = Mean + 12
⇒ Mode – 12 = Mean
Now, Mode = 3×Median – 2×Mean
⇒ Mode = 3×Median – 2(Mode – 12)
⇒ Mode = 3×Median – 2×Mode + 24
⇒ Mode + 2×Mode = 3×Median + 24
⇒ 3×Mode = 3×Median = 24
⇒ Mode = Median + 8
So, mode exceeds the median by 8

Answer: (c) 22, 4
Hint:
Since each value is increased by 2, therefore the median value is also increased by
2. So, new median = 22
Again, the variance is independent of the change of origin. So it remains the same.

Answer: (c) 13
Hint:
Give, data are: 4, 7, 8, 9, 10, 12, 13 and 17
Range = Maximum value – Minimum Value
= 17 – 4
= 13

Answer: (a) Symmetric distribution
Hint:
In a symmetric distribution,
Mean = Median = Mode

Answer: (a) 12
Hint:
Given the difference of mode and median of a data is 24
⇒ Mode – Median = 24
⇒ Mode = Median + 24
Now, Mode = 3×Median – 2×Mean
⇒ Median + 24 = 3×Median – 2×Mean
⇒ 24 = 3×Median – 2×Mean – Median
⇒ 24 = 2×Median – 2×Mean
⇒ Median – Mean = 24/2
⇒ Median – Mean = 12

Answer: (a) |r| ≤ 1
Hint:
If r is the correlation coefficient, then |r| ≤ 1

Answer: (b) 11
Hint:
Given, varience of the data = 121
Now, the standard deviation of the data = √(121) = 11

Answer: (b) 20
Hint:
Mean = ∑ f _i× x _i /∑ f _i
⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)
⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50
⇒ 20.6 = (530 + 25p)/50
⇒ 530 + 25p = 20.6 × 50
⇒ 530 + 25p = 1030
⇒ 25p = 1030 – 530
⇒ 25p = 500
⇒ p = 500/25
⇒ p = 20
So, the value of p is 20

Answer: (c) 9
Hint:
Given mean of first n natural number is 5n/9
⇒ (n+1)/2 = 5n/9
⇒ n + 1 = (5n×2)/9
⇒ n + 1 = 10n/9
⇒ 9(n + 1) = 10n
⇒ 9n + 9 = 10n
⇒ 10n – 9n = 9
⇒ n = 9

Answer: (d) 0
Hint:
Let the observations are 0, a, b, c, ……… up to n
Now, geometric mean = (0 × a × b × c × ……… up to n)^1/n
= 0
So, geometric mean is 0

Answer: (d) all of the above
Hint:
Range, Quartile deviation, Mean deviation all are the measure of dispersions method.

Answer: (a) x – 5/4
Hint:
Given, discrete values x + 4, x – 7/2, x – 5/2, x – 3, x – 2, x + 1/2, x – 1/2, x + 5
Now, arrange them in ascending order, we get
x – 7/2, x – 3, x – 5/2, x – 2, x – 1/2, x + 1/2, x + 4, x + 5
Total observations = 8
Now, median = AM of 4th and 5th observations
= AM of (x – 2) and (x – 1/2) observations
= (x – 2 + x – 1/2)/2
= (2x – 5/2)/2
= x – 5/4

Answer: (b) 0
Hint:
The relationship between the correlation coefficient and covariance for two variables as shown below:
r_{(x, y)} = COV(x, y)/{s_x × s_y}
r_{(x, y)} = correlation of the variables x and y
COV(x, y) = covariance of the variables x and y
s_x = sample standard deviation of the random variable x
s_x = sample standard deviation of the random variable y
Now given COV(x, y) = 0
Then r_{(x, y)} = 0

Answer: (b) 20
Hint:
Given mean of 100 observations is 20
Now
∑ xi/100 = 20 (1 = i = 100)
⇒ ∑xi = 100×20
⇒ ∑xi = 2000
3 observations 21, 21 and 18 are recorded in-correctly.
So ∑xi = 2000 – 21 – 21 – 18
⇒ ∑xi = 2000 – 60
⇒ ∑xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So, the new mean is 20

Answer: (a) origin only
Hint:
Varience is independent of change of origin only.

Answer: (c) √{∑(x_i – mean)²/n}
Hint:
Given, x₁, x₂, x₃, ………. , x_n be n observations and X be the arithmetic mean.
Now standard deviation = √{∑(x_i – mean)²/n}

Answer: (d) e^-1/2
Hint:
Given, Lim_x→0 (cos x)^{cot² x}
= Lim_x→0 (1 + cos x – 1)^{cot² x}
= eLim_x→0 (cos x – 1) × cot² x
= eLim_x→0 (cos x – 1)/tan² x
= e^-1/2

Answer: (c) 2 cos a
Hint:
Given, Lim_x→0 {sin (a + x) – sin (a – x)}/x
= Lim_x→0 {2 × cos a × sin x}/x
= 2 × cos a × Lim_x→0 sin x/x
= 2 cos a

Answer: (b) 1
Hint:
Given, Lim_x→-1 [1 + x + x² + ……….+ x¹⁰]
= 1 + (-1) + (-1)² + ……….+ (-1)¹⁰
= 1 – 1 + 1 – ……. + 1
= 1

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2× tan^-1 x)/x
= 2 × 1
= 2

Answer: (a) 0
Hint:
We know that
log(1 – x) = -x – x²/2 – x³/3 – ……..
Now,
Lim_x→0 log(1 – x) = Lim_x→0 {-x – x²/2 – x³/3 – ……..}
⇒ Lim_x→0 log(1 – x) = Lim_x→0 {-x} – Lim_x→0 {x²/2} – Lim_x→0 {x³/3} – ……..
⇒ Lim_x→0 log(1 – x) = 0

Answer: (c) log (a/b)
Hint:
Given, Lim_x→0 {(a^x – b^x)/ x}
= Lim_x→0 {(a^x – b^x – 1 + 1)/ x}
= Lim_x→0 {(a^x – 1) – (b^x – 1)}/ x
= Lim_x→0 {(a^x – 1)/x – (b^x – 1)/x}
= Lim_x→0 (a^x – 1)/x – Lim_x→0 (b^x – 1)/x
= log a – log b
= log (a/b)

Answer: (d) x × tan x × sec x + sec x
Hint:
Given, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y
= lim_y→0 {x sec (x + y) + y sec (x + y) – x×sec x}/y
= lim_y→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= lim_y→0 x{ sec (x + y) – sec x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 x{1/cos (x + y) – 1/cos x}/y + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{cos x – cos (x + y)} × x/{y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y×cos (x + y)×cos x}] + lim_y→0 {y sec (x + y)}/y
= lim_y→0 {sin (x + y/2) × lim_y→0 {sin(y/2)/(2y/2)} × lim_y→0 { x/{y×cos (x + y)×cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, lim_y→0 {(x + y) × sec (x + y) – x×sec x}/y = x × tan x × sec x + sec x

Answer: (c) e⁵
Hint:
Given, Lim_y→∞ {(x + 6)/(x + 1)}(x + 4)
= Lim_y→∞ {1 + 5/(x + 1)}(x + 4)
= e^{Lim_y→∞ 5(x + 4)/(x + 1)}
= e^{Lim_y→∞ 5(1 + 4/x)/(1 + 1/x)}
= e^{5(1 + 4/∞)/(1 + 1/∞)}
= e^{5/(1 + 0)}
= e⁵

Answer: (d) A
Hint:
Let y = [1+(1/x)] /[1-(1/x)]
then dy/dx = [{1-(1/x)}*(-1/x²)]/[{1+(1/x)}*(1/x²)]
= (1/x²) [(1/x) -1 – 1 – (1/x)]/[1-(1/x)]²
= [-2/x²]/[(x-1)/x]²
= -2/(x-1)²

Answer: (d) -x – x²/2 – x³/3 – ……..
Hint:
log(1 – x) = -x – x²/2 – x³/3 – ……..

Answer: (b) 0
Hint:
Given, f(x) = x × sin(1/x)
Now, Lim_x→0 f(x) = Lim_x→0 x × sin(1/x)
⇒ Lim_x→0 f(x) = 0

Answer: (a) 0
Hint:
Given, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³
= Lim_n→∞ [{n×(n + 1)×(2n + 1)}/6]/{n(n + 1)/2}²
= Lim_n→∞ [{n×n×n ×(1 + 1/n)×(2 + 1/n)}/6]/{n × n ×(1 + 1/n)/2}²
= Lim_n→∞ [{n³ ×(1 + 1/n)×(2 + 1/n)}/6]/{n² ×(1 + 1/n)/2}²
= Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n⁴ × {(1 + 1/n)/2}²]
⇒ Lim_n→∞ [{(1 + 1/n)×(2 + 1/n)}/6]/[n × {(1 + 1/n)/2}²]
= [{(1 + 1/∞)×(2 + 1/∞)}/6]/[∞×{(1 + 1/∞)/2}²
= [{(1 + 0)×(2 + 0)}/6]/∞ {since 1/∞ = 0}
= {(1 × 2)/6}/∞
= (2/6)/∞
= (1/3)/∞
= 0
So, Lim_n→∞ {1² + 2² + 3² + …… + n²}/n³ = 0

Answer: (a) 0
Hint:
Lim_n→∞ (sin x/x) = Lim_y→0 {y × sin (1/y)} = 0

Answer: (b) 1
Hint:
We know that
a^x = 1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + ………..
Now,
Lim_x→0 a^x = Lim_x→0 {1 + x/1! × (log a) + x²/2! × (log a)² + x³/3! × (log a)³ + …}
⇒ Lim_x→0 a^x = Lim_x→0 1 + Lim_x→0 {x/1! × (log a)} + Lim_x→0 {x² /2! × (log a)²}+ ………
⇒ Lim_x→0 a^x = 1

Answer: (b) 1
Hint:
Given, Lim_x→0 f(x) exists
⇒ Lim_x→0 – f(x) = Lim_x→0 + f(x)
⇒ Lim_x→0 (x + k) = Lim_x→0 cos x
⇒ k = cos 0
⇒ k = 1

Answer: (c) 2
Hint:
Given, Lim_x→0 (1/x) × sin^-1 {2x/(1 + x²)
= Lim_x→0 (2 × tan^-1 x)/x
= 2 × 1
= 2

Answer: (c) a/b
Hint:
Given, Lim_x→0 sin (ax)/bx
= Lim_x→0 [{sin (ax)/ax} × (ax/bx)]
⇒ (a/b) Lim_x→0 sin (ax)/ax
= a/b

Answer: (c) a
Hint:
Given, Lim_x→0 {log(1 + ax)}/x
= Lim_x→0 {ax – (ax)² /2 + (ax)³ /3 – (ax)⁴ /4 + …….}/x
= Lim_x→0 {ax – a² x² /2 + a³ x³ /3 – a⁴ x⁴ /4 + …….}/x
= Lim_x→0 {a – a² x /2 + a³ x² /3 – a⁴ x³ /4 + …….}
= a – 0
= a

Answer: (c) -1/x²
Hint:
Given, f(x) = (x + 1)/x
Now, df(x)/dx = d{(x + 1)/x}/dx
= {1 × x – (x + 1)×1}/x²
= (x – x – 1)/x²
= -1/x²

Answer: (d) 3/2
Hint:
Given, Lim_x→0 (e^x² – cos x)/x²
= Lim_x→0 (e^x² – cos x -1 + 1)/x²
= Lim_x→0 {(e^x² – 1)/x² + (1 – cos x)}/x²
= Lim_x→0 {(e^x² – 1)/x² + Lim_x→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2

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

Answer: (b) x – y = 0
Hint:
Let the coordinate of the variable point P is (x, y)
Now, the abscissa of this point = x
and its ordinate = y
Given, abscissa = ordinate
⇒ x = y
⇒ x – y = 0
So, the locus of the point is x – y = 0

Answer: (c) y – 2 = 3 × (x – 1)
Hint:
Given straight line is: y = 3x + 1
Slope = 3
Now, required line is parallel to this line.
So, slope = 3
Hence, the line is
y – 2 = 3 × (x – 1)

Answer: (b) θ is an obtuse angle
Hint:
Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.
Then its slope is given by m = tan θ
Given, slope is positive
⇒ tan θ < 0
⇒ θ lies between 0 and 180 degree
⇒ θ is an obtuse angle

Answer: (a) x + y = α + β
Hint:
Let the equation of the line be x/a + y/b = 1 which cuts off intercepts a and b with
the coordinate axes.
It is given that a = b, therefore the equation of the line is
x/a + y/a = 1
⇒ x + y = a …..1
But it is passes through (α, β)
So, α + β = a
Put this value in equation 1, we get
x + y = α + β

Answer: (d) a₁/a₂ = b₁/b₂ = c₁/c₂
Hint:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincedent if
a₁/a₂ = b₁/b₂ = c₁/c₂

Answer: (c) 2x – y – 1 = 0
Hint:
Given, the point (2, 3) and slope of the line is 2
By, slope-intercept formula,
y – 3 = 2(x – 2)
⇒ y – 3 = 2x – 4
⇒ 2x – 4 – y + 3 = 0
⇒ 2x – y – 1 = 0

Answer: (b) -a/b
Hint:
Give, equation of line is ax + by + c = 0
⇒ by = -ax – c
⇒ y = (-a/b)x – c/b
It is in the form of y = mx + c
Now, slope m = -a/b

Answer: (c) y = mx
Hint:
Equation of the line passing through (x₁, y₁) and slope m is
(y – y₁) = m(x – x₁)
Now, required line is
(y – 0 ) = m(x – 0)
⇒ y = mx

Answer: (c) tan^-1 (5/4)
Hint:
Given, lines are:
x – 2y = 5 ………. 1
and y – 2x = 5 ………. 2
From equation 1,
x – 5 = 2y
⇒ y = x/2 – 5/2
Here, m₁ = 1/2
From equation 2,
y = 2x + 5
Here. m₂ = 2
Now, tan θ = |(m₁ + m₂)/{1 + m₁ × m₂}|
= |(1/2 + 2)/{1 + (1/2) × 2}|
= |(5/2)/(1 + 1)|
= |(5/2)/2|
= 5/4
⇒ θ = tan^-1 (5/4)

Answer: (a) a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hint:
Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Answer: (b) x – y = 0
Hint:
Let the coordinate of the variable point P is (x, y)
Now, the abscissa of this point = x
and its ordinate = y
Given, abscissa = ordinate
⇒ x = y
⇒ x – y = 0
So, the locus of the point is x – y = 0

Answer: (b) (7, – 2)
Hint:
The equation of median through B is x + y = 5
The point B lies on it.
Let the coordinates of B are (x₁, 5 – x₁)
Now CF is a median through C,
So co-ordiantes of F i.e. mid-point of AB are
((x₁+1)/2, (5 – x₁+ 2)/2)
Now since this lies on x = 4
⇒ (x₁ + 1)/2 = 4
⇒ x₁ + 1 = 8
⇒ x₁ = 7
Hence, the co-oridnates of B are (7, -2)

Answer: (c) √3x + y = 14
Hint:
Given, The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis.
Now, equation of line is
x × cos 30 + y × sin 30 = 7
⇒ √3x/2 + y/2 = 7
⇒ √3x + y = 7×2
⇒ √3x + y = 14

Answer: (d) (-5, -3)
Hint:
Let the third vertex of the triangle is C(x, y)
Given, two vertices of a triangle are A(3,-2) and B(-2,3)
Now given orthocentre of the circle = H(-6, 1)
So, AH ⊥ BC and BH ⊥ AC
Since the product of the slope of perpendicular lines equal to -1
Now, AH ⊥ BC
⇒ {(-2 – 1)/(3 + 6)} × {(y + 2)/(x – 3)} = -1
⇒ (-3/9) × {(y + 2)/(x – 3)} = -1
⇒ (-1/3)×{(y – 3)/(x + 2)} = -1
⇒ (y – 3)/{3×(x + 2)} = 1
⇒ (y – 3) = 3×(x + 2)
⇒ y – 3 = 3x + 6
⇒ 3x + 6 – y = -3
⇒ 3x – y = -3 – 6
⇒ 3x – 2y = -9 ………… 1
Again, BH ⊥ AC
⇒ {(3 – 1)/(-2 + 6)} × {(y – 3)/(x + 2)} = -1
⇒ (2/4) × {(y – 3)/(x + 2)} = -1
⇒ (1/2)×{(y – 3)/(x + 2)} = -1
⇒ (y – 3)/{2×(x + 2)} = 1
⇒ (y – 3) = 2×(x + 2)
⇒ y – 3 = 2x + 4
⇒ 2x + 4 – y = -3
⇒ 2x – y = -3 – 4
⇒ 2x – y = -7 ………… 2
Multiply equation 2 by 2, we get
4x – 2y = -14 ……… 3
Subtract equation 1 and we get
-x = 5
⇒ x = -5
From equation 2, we get
2×(-5) – y = -7
⇒ -10 – y = -7
⇒ y = -10 + 7
⇒ y = -3
So, the third vertex of the triangle is (-5, -3)

Answer: (c) x² + y² = c² – a²
Hint:
Let P(h, k) be any position of the moving point and let A(a, 0) and B(-a, 0) be the given points. Then
PA² + PB² = 2c²
⇒ (h – a)² + (k – 0)² + (h + a)² + (k – 0)² = 2c²
⇒ h² – 2ah + a² + k² + h² + 2ah + a² + k² = 2c²
⇒ 2h² + 2k² + 2a² = 2c²
⇒ h² + k² + a² = c²
⇒ h² + k² = c² – a²
Hence, the locus of (h, k) is x² + y² = c² – a²

Answer: (c) 2x + y – 7 = 0
Hint:
Given, points are: (1, 5) and (2, 3)
Now, equation of line is
y – y₁ = {(y₂ – y₁)/(x₂ – x₁)} × (x – x₁)
⇒ y – 5 = {(3 – 5)/(2 – 1)} × (x – 1)
⇒ y – 5 = (-2) × (x – 1)
⇒ y – 5 = -2x + 2
⇒ 2x + y – 5 – 2 = 0
⇒ 2x + y – 7 = 0

Answer: (c) Either the line is x-axis or it is parallel to the x-axis.
Hint:
Let θ be the angle of inclination of the given line with the positive direction of x- axis in the anticlockwise sense.
Then its slope is given by m = tan θ
Given, slope is zero
⇒ tan θ = 0
⇒ θ = 0°
⇒ Either the line is x-axis or it is parallel to the x-axis.

Answer: (c) -1
Hint:
Let m₁ is the slope of first line and m₂ is the slope of second line.
Now, two lines are perpendicular if m₁ × m₂ = -1
i.e. the product of their slopes is equals to -1

Answer: (d) 5
Hint:
Given, equation of line is 4x – 3y + 15 = 0
⇒ 4x – 3y = -15
⇒ 4x/(-15) + (-3)y/(-15) = 1
⇒ x/(-15/4) + 3y/15 = 1
⇒ x/(-15/4) + y/(15/3) = 1
⇒ x/(-15/4) + y/5 = 1
Now, compare with x/a + y/b = 1, we get
y-intercept b = 5

Answer: (b) 6x + 4y = 5
Hint:
Let P(h, k) be any point on the locus. Then
Given, PA = PB
⇒ PA² = PB²
⇒ (h – 1)² + (k – 3)² = (h + 2)² + (k – 1)²
⇒ h² – 2h + 1 + k² – 6k + 9 = h² + 4h + 4 + k² – 2k + 1
⇒ -2h – 6k + 10 = 4h – 2k + 5
⇒ 6h + 4k = 5
Hence, the locus of (h, k) is 6x + 4y = 5

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: (a) AP
Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP

Answer: (b) 2b = a + c
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Answer: (a) 2 + √3
Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Answer: (b) (e – 1)² /2 e
Hint:
We know that,
e^x = 1 + x/1! + x² /2! + x³ /3! + x⁴ /4! + ………..
Now,
e¹ = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e^-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e¹ + e^-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Answer: (b) 4⁵
Hint:
here it is given that T₃ = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar⁴
= a⁵r¹⁰
= (ar²)⁵
= 4⁵

Answer: (c) 1
Hint:
Let first term is a and the common difference is d of the AP
Now, T_m = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and T_n = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, T_mn = 1/mn + (mn-1)/mn
⇒ T_mn = 1/mn + 1 – 1/mn
⇒ T_mn = 1

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b
Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Answer: (c) H.P.
Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP

Answer: (b) a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP

Answer: (d) 350
Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a₁ = 1/2 + 1 = 3/2
a₂ = 2/2 + 1 = 2
a₃ = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a₃₅ = a₁ + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b
Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Answer: (c) 3
Hint:
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar⁴
Now
⇒ ar² + ar⁴ = 90
⇒ a(r² + r⁴) = 90
⇒ r² + r⁴ = 90
⇒ r² ×(r² + 1) = 90
⇒ r²(r² + 1) = 3² ×(3² + 1)
⇒ r = 3
So the common ratio is 3

Answer: (b) 3775
Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775

Answer: (b) 31/48
Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Answer: (c) 740
Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3×7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2×4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2)×{2×(-1) + (20-1)×4}
= 10×{-2 + 19×4)}
= 10×{-2 + 76)}
= 10 × 74
= 740

Answer: (c) 11
Hint:
Given,
the sum of the first 2n terms of the A.P. 2, 5, 8, …..= the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}
⇒ n×{4 + 6n – 3} = (n/2)×{114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Answer: (b) 2abc
Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G₁ and G₁ are two GM between b and c,
⇒ b, G₁, G₂, c are in the GP having common ration r, then
⇒ r = (c/b)^1/(2+1) = (c/b)^1/3
Now,
G₁ = br = b×(c/b)^1/3
and G₁ = br = b×(c/b)^2/3
Now,
(G₁)³ + (G₂)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G₁)³ + (G₂)³ = b² × c × (2a/b)
⇒ (G₁)³ + (G₂)³ = b × c × 2a
⇒ (G₁)³ + (G₂)³ = 2abc

Answer: (a) 1/3, 1/6, 1
Hint:
Given 3x + 1 = 6y – 2 = 1 – z
= (3x + 1)/1 = (6y – 2)/1 = (1 – z)/1
= (x + 1/3)/(1/3) = (y – 2/6)/(1/6) = (1 – z)/1
= (x + 1/3)/(1/3) = (y – 1/3)/(1/6) = (1 – z)/1
Now, the direction ratios are: 1/3, 1/6, 1

Answer: (a) (-3, 5, 2)
Hint:
Let image of the point P(1, 3, 4) is Q in the given plane.
The equation of the line through P and normal to the given plane is
(x-1)/2 = (y-3)/-1 = (z-4)/1
Since the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
Now, the coordinate of the mid-point of PQ is
(r + 1, -r/2 + 3, r/2 + 4)
Now, this point lies in the given plane.
2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0
⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0
⇒ 3r + 6 = 0
⇒ r = -2
Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
= (-3, 5, 2)

Answer: (c) meet in a unique point
Hint:
Given, three planes are
x + y = 0 …….. 1
y + z = 0 …….. 2
and x + z = 0 ……… 3
add these planes, we get
2(x + y + z) = 0
⇒ x + y + z = 0 ……… 4
From equation 1
0 + z = 0
⇒ z = 0
From equation 2
x + 0 = 0
⇒ x = 0
From equation 3
y + 0 = 0
⇒ y = 0
So, (x, y, z) = (0, 0, 0)
Hence, the three planes meet in a unique point.

Answer: (a) (5/3, 7/3, 17/3)
Hint:
Let D be the foot of perpendicular and let it divide BC in the ration m : 1
Then the coordinates of D are {(3m + 4)/(m + 1), (5m + 7)/(m + 1), (3m + 1)/(m + 1)}
Now, AD ⊥ BC
⇒ AD . BC = 0
⇒ -(2m + 3) – 2(5m + 7) – 4 = 0
⇒ m = -7/4
So, the coordinate of D are (5/3, 7/3, 17/3)

Answer: (b) Plane
Hint:
Let the position vectors of the given points A and B be a and b respectively and that of the variable point be r.
Now, given that
PA² – PB² = k (constant)
⇒ |AP|² – |BP|² = k
⇒ |r – a|² – |r – b|² = k
⇒ (|r|² + |a|² – 2r.a) – (|r|² + |b|² – 2r.b) = k
⇒ 2r.(b – a) = k + |b|² – |a|²
⇒ r.(b – a) = (k + |b|² – |a|²)/2
⇒ r.(b – a) = C where C = (k + |b|² – |a|²)/2 = constant
So, it represents the equation of a plane.

Answer: (b) 9x² + 25y² + 25z² – 225 = 0
Hint:
Let the point P is (x, y, z)
Now given that
PA + PB = 10
⇒ √{(x-4)² + y² + z²} + √{(x+4)² + y² + z²} = 10
⇒ √{(x-4)² + y² + z²} = 10 – √{(x+4)² + y² + z²}
Now square both side
[√{(x-4)² + y² + z²}]² = (10)² + [{(x+4)² + y² + z²}]² – 2 ×10×√{(x+4)² + y² + z²}
⇒ {(x-4)² + y² + z²} = 100 + {(x+4)² + y² + z²} – 20×√{(x+4)² + y² + z²}
⇒ x² + 16 – 8x + y² + z² = 100 + x² + 16 + 8x + y² + z² – 20×√{(x+4)² + y² + z²}
⇒ – 8x = 100 + 8x – 20×√{(x+4)² + y² + z²}
⇒ -8x -8x – 100 = – 20×√{(x+4)² + y² + z²}
⇒ -16x -100 = – 20×√{(x+4)² + y² + z²}
⇒ 4x + 25 = 5×√{(x+4)² + y² + z²}
Again square both side,
(4x + 25)² = 25 ×[√{(x+4)² + y² + z²}]²
⇒ 16x² + 625 + 200x = 25×{(x+4)² + y² + z²}
⇒ 16x² + 625 + 200x = 25×(x² + 16 + 8x + y² + z²)
⇒ 16x² + 625 + 200x = 25x² + 400 + 200x + 25y² + 25z²
⇒ 25x² + 400 + 200x + 25y² + 25z² – 16x² – 625 – 200x = 0
⇒ 9x² + 25y² + 25z² – 225 = 0

Answer: (c) 5
Hint:
Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)
Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
⇒ distance = √{(4cos θ – 3sin θ)²}
⇒ distance = 4cos θ – 3sin θ ……………. 1
Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}
= √(16 + 9)
= √25
= 5
From equation 1, we get
distance = 5
So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

Answer: (d) 2√3(i + j + k)
Hint:
Let l, m, n are DCs of r.
Given, l = m = n
⇒ l² + m² + n² = 1
⇒ 3l² = 1
⇒ l² = 1/3
⇒ l = m = n = 1/√3
So, r = |r|(li + mj + nk)
⇒ r = 6(i/√3 + j/√3 + k/√3)
⇒ r = 2√3(i + j + k)

Answer: (d) A
Hint:
Given, equation of plane is:
2x – (1 + a)y + 3az = 0
=> (2x – y) + a(-y + 3z) = 0
which is passing through the intersection of the planes
2x – y = 0 and -y + 3z = 0
2x – y = 0 and y – 3z = 0

Answer: (a) √3 unit
Hint:
Let a is the length of the side of a square.
Given, the diagonal of a square are (1,–2,3) and (2, -3, 5)
Now, length of the diagonal of square = √{(1 – 2)² + (-2 + 3)² + (3 – 5)²}
= √{1 + 1 + 4}
= √6
Again length of the diagonal of square is √2 times the length of side of the square.
⇒ a√2 = √6
⇒ a√2 = √3×√2
⇒ a = √3
So, the length of side of square is √3 unit

Answer: (c) (0, 17/2, -13/2)
Hint:
The line passing through the points (5, 1, 6) and (3, 4, 1) is given as
(x-5)/(3-5) = (y-1)/(4-1) = (z-6)/(1-6)
⇒ (x-5)/(-2) = (y-1)/3 = (z-6)/(-5) = k(say)
⇒ (x-5)/(-2) = k
⇒ x – 5 = -2k
⇒ x = 5 – 2k
(y-1)/3 = k
⇒ y – 1 = 3k
⇒ y = 3k + 1
and (z-6)/(-5) = k
⇒ z – 6 = -5k
⇒ z = 6 – 5k
Now, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane
So, 5 – 2k = 0
⇒ k = 5/2
Now, 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 = 17/2
and 6 – 5k = 6 – 5×5/2 = 6 – 25/2 = -13/2
Hence, the required point is (0, 17/2, -13/2)

Answer: (b) π/3
Hint:
Let a is a vector parallel to the vector having direction ratio is 4, -3, 5
⇒ a = 4i – 3j + 5k
Let b is a vector parallel to the vector having direction ratio is 3 ,4, 5
⇒ b = 3i + 4j + 5k
Let θ be the angle between the given vectors.
Now, cos θ = (a . b)/(|a|×|b|)
⇒ cos θ = (12 – 12 + 25)/{√(16 + 9 + 25)×√(9 + 16 + 25)}
⇒ cos θ = 25/{√(50)×√(50)}
⇒ cos θ = 25/50
⇒ cos θ = 1/2
⇒ cos θ = π/3
⇒ θ = π/3
So, the angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is π/3

Answer: (b) r . (2i – j + 2k) = 3
Hint:
The equation of plane parallel to the plane r . (2i – j + 2k) = 5 is
r . (2i – j + 2k) = d
Since it passes through the point i + j + k, therefore
(i + j + k) . (2i – j + 2k) = d
⇒ d = 2 – 1 + 2
⇒ d = 3
So, the required equation of the plane is
r . (2i – j + 2k) = 3

Answer: (d) 2√3(i + j + k)
Hint:
Let l, m, n are DCs of r.
Given, l = m = n
⇒ l² + m² + n² = 1
⇒ 3l² = 1
⇒ l² = 1/3
⇒ l = m = n = 1/√3
So, r = |r|(li + mj + nk)
⇒ r = 6(i/√3 + j/√3 + k/√3)
⇒ r = 2√3(i + j + k)

Answer: (c) 5
Hint:
Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)
Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
⇒ distance = √{(4cos θ – 3sin θ)²}
⇒ distance = 4cos θ – 3sin θ …………….1
Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}
= √(16 + 9)
= √25
= 5
From equation 1, we get
distance = 5
So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

Answer: (a) (-3, 5, 2)
Hint:
Let image of the point P(1, 3, 4) is Q in the given plane.
The equation of the line through P and normal to the given plane is
(x-1)/2 = (y-3)/-1 = (z-4)/1
Since the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
Now, the coordinate of the mid-point of PQ is
(r + 1, -r/2 + 3, r/2 + 4)
Now, this point lies in the given plane.
2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0
⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0
⇒ 3r + 6 = 0
⇒ r = -2
Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
= (-3, 5, 2)

Answer: (d) either (0, -1, 0) or (0, 7, 0)
Hint:
Let the point on y-axis is O(0, y, 0)
Given point is A(2, 3, -1)
Given OA = 3
⇒ OA² = 9
⇒ (2 – 0)² + (3 – y)² + (-1 – 0)² = 9
⇒ 4 + (3 – y)² + 1 = 9
⇒ 5 + (3 – y)² = 9
⇒ (3 – y)² = 9 – 5
⇒ (3 – y)² = 4
⇒ 3 – y = √4
⇒ 3 – y = ±4
⇒ 3 – y = 4 and 3 – y = -4
⇒ y = -1, 7
So, the point is either (0, -1, 0) or (0, 7, 0)

Answer: (d) None of these
Hint:
Let l, m, n be the direction cosines of the given vector.
Then, α, β, γ
l = cos α
m = cos β
n = cos γ
Now, l² + m² + n² = 1
⇒ cos² α + cos² β + cos² γ = 1
⇒ 1 – sin² α + 1 – sin² β + 1 – sin² γ = 1
⇒ 3 – sin² α – sin² β – sin² γ = 1
⇒ 3 – 1 = sin² α + sin² β + sin² γ
⇒ sin² α + sin² β + sin² γ = 2

Answer: (b) 3 : 2
Hint:
Since OP has projections 13/5, 19/5 and 26/5 on the coordinate axes, therefore
OP = 13i/5 + 19j/5 + 26/5k
Let P divides the join of Q(2, 2, 4) and R(3, 5, 6) in the ratio m : 1
Then the position vector of P is
{(3m + 2)/(m + 1), (5m + 2)/(m + 1), (6m + 4)/(m + 1)}
So, 13i/5 + 19j/5 + 26/5k = (3m + 2)/(m + 1)+ (5m + 2)/(m + 1)+ (6m + 4)/(m + 1)
⇒ (3m + 2)/(m + 1) = 13/5
⇒ 2m = 3
⇒ m = 3/2
⇒ m : 1 = 3 : 2
Hence, P divides QR in the ration 3 : 2

Answer: (c) a plane containing Z axis
Hint:
Given, equation is 3x – 4y = 0
Here z = 0
So, the given equation 3x – 4y = 0 represents a plane containing Z axis.

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) 8x – 19 = 0
Hint:
Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0
Now, the required line is the radical axis of the two circles are
(x² + y² – 4) – (x² + y² – 8x + 15) = 0
⇒ x² + y² – 4 – x² – y² + 8x – 15 = 0
⇒ 8x – 19 = 0

Answer: (a) 7
Hint:
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Answer: (d) x²/25 + y²/9 = 1
Hint:

From the question, it is clear that the path traced by the man is an ellipse having its foci at two posts.
Let the equation of the ellipse be
x²/a² + y²/b² = 1
It is given that the sum of the distances of the man from the two flag posts is 10 m
This means that the sum of focal distances of a point on the ellipse is 10 m
⇒ PS + PS₁ = 10
⇒ 2a = 10
⇒ a = 5
Again, given that the distance between the flag posts is 8 meters
⇒ 2ae = 8
⇒ ae = 4
Now, b² = a² (1 – e²)
⇒ b² = a² – a² e²
⇒ b² = a² – (ae)²
⇒ b² = 5² – 4²
⇒ b² = 25 – 16
⇒ b² = 9
⇒ b = 3
Hence, the equation of the path is x²/5² + y²/3² = 1
⇒ x²/25 + y²/9 = 1

Answer: (d) (1, 1)
Hint:
The center of the given ellipse is the point of intersection of the lines
x + y – 2 = 0 and x – y = 0
After solving, we get
x = 1, y = 1
So, the center of the ellipse is (1, 1)

Answer: (c) (a sin²t, -2a sin t)
Hint:
The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all
values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is
(a sin²t, -2a sin t)

Answer: (b) y² = 9x/2
Hint:
A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:
y² = 4ax ………….. 1
Given, point (2,3) lies on the parabola,
⇒ 3² = 4a × 2
⇒ 9 = 4a × 2
⇒ 9/2 = 4a
From equation 1, we get
y² = (9/2)x
⇒ y² = 9x/2
This is the required equation of the parabola.

Answer: (b) (3, 1)
Hint:
Given, parabola is x² = 9y
Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.
So, h = 3k ……… 1
Since P(h, k) lies on the parabola
So, h² = 9k ……… 2
From equation 1 and 2, we get
(3k)² = 9k
⇒ 9k² = 9k
⇒ 9k² – 9k = 0
⇒ 9k(k – 1) = 0
⇒ k = 0, 1
When k = 0, h = 0
So k = 1
Now, from equation 1,
h = 3 × 1 = 3
So, the point is (3, 1)

Answer: (b) 1
Hint:
Given point (1, 2) and equation of circle is x² + y² = 5
Now, x² + y² – 5 = 0
Put (1, 2) in this equation, we get
1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0
So, the point (1, 2) lies on the circle.
Hence, only one tangent can be drawn.

Answer: (c) 3/5
Hint:
Given, distance between foci = 6
⇒ 2ae = 6
⇒ ae = 3
Again minor axis = 8
⇒ 2b = 8
⇒ b = 4
⇒ b² = 16
⇒ a² (1 – e²) = 16
⇒ a² – a² e² = 16
⇒ a² – (ae)² = 16
⇒ a² – 3² = 16
⇒ a² – 9 = 16
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Now, ae = 3
⇒ 5e = 3
⇒ e = 3/5
So, the eccentricity is 3/5

Answer: (c) (x – 2)² + (y – 3)² = 3²
Hint:
Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}
= √(4 + 9 – 4)
= √9
= 3
So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
Hint:
Given focus S(3, 0)
and equation of directrix is: 3x + 4y = 1
⇒ 3x + 4y – 1 = 0
Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
Then, SP = PM
⇒ SP² = PM²
⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}²
⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25
⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0
⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
This is the required equation of parabola.

Answer: (a) a parabola
Hint:
Let x = 2 + t²
⇒ x – 2 = t² ……….. 1
and y = 2t + 1
⇒ y – 1 = 2t
⇒ (y – 1)/2 = t
From equation 1, we get
x – 2 = {(y – 1)/2}²
⇒ x – 2 = (y – 1)²/4
⇒ (y – 1)² = 4(x – 2)
This represents the equation of a parabola.

Answer: (b) x²/a² – y²/b² = 1
Hint:
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² – y²/b² = 1

Answer: (b) 12y = x² + 4x + 16
Hint:
Given, parabola having vertex is (-2, 1) and focus is (-2, 4)
As the vertex and focus share the same abscissa i.e. -2,
parabola axis of symmetry as x = -2
⇒ x + 2 = 0
Hence, the equation of a parabola is of the type
(y – k) = a(x – h)² where (h, k) is vertex
Now, focus = (h, k + 1/4a)
Since, vertex is (-2, 1) and parabola passes through vertex
So, focus = (-2, 1 + 1/4a)
Now, 1 + 1/4a = 4
⇒ 1/4a = 4 -1
⇒ 1/4a = 3
⇒ 4a = 1/3
⇒ a = /1(3 × 4)
⇒ a = 1/12
Now, equation of parabola is
(y – 1) = (1/12) × (x + 2)²
⇒ 12(y – 1) = (x + 2)²
⇒ 12y – 12 = x² + 4x + 4
⇒ 12y = x² + 4x + 4 + 12
⇒ 12y = x² + 4x + 16
This is the required equation of parabola.

Answer: (c) (5, 0)
Hint:

given diameter of the parabola is 20 m.
The equation of parabola is y² = 4ax.
Since this parabola passes through the point A(5,10) then
10² = 4a×5
⇒ 20a = 100
⇒ a = 100/20
⇒ a = 5
So focus of parabola is (a, 0) = (5, 0)

Answer: (c) √77/2
Hint:
Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0
⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0
⇒ x² + y² – 2x + 3y – 25/4 = 0
Now, radius = √{(-2)² + (3)² – (-25/4)}
= √{4 + 9 + 25/4}
= √{13 + 25/4}
= √{(13×4 + 25)/4}
= √{(52 + 25)/4}
= √{77/4}
= √77/2

Answer: (d) 2a = b²
Hint:
Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)
Now, (x + 0)/2 = a
⇒ x = 2a
and (y + 0)/2 = b
⇒ y = 2b
Now, y² = 4x
⇒ (2b)² = 4 × 2a
⇒ 4b² = 8a
⇒ b² = 2a

Answer: (a) x²/81 + y²/9 = 1
Hint:
Given a rod of length 12 cm moves with its ends always touching the coordinate axes.
Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.
It is shown in the figure.

Here AP = 3 cm, AB = 12
Now BP = AB – AP
⇒ BP = 12 – 3
⇒ BP = 9 cm
Again from figure,
∠PAO = ∠BPO = θ (since PQ || OA and are corresponding angles)
Now in ΔBPO,
cosθ = QP/BP
⇒ cosθ = x/9 …………. 1
Again in ΔPAr,
sinθ = PR/PA
⇒ sinθ = y/3 …….. 2
Now square equation 1 and 2 and then add them, we get
cos² θ + sin² θ = x²/81 + y²/9
⇒ x²/81 + y²/9 = 1 (since cos² θ + sin² θ = 1 )
So, the equation of the locus of a point P is x²/81 + y²/9 = 1

Answer: (a) ln = am²
Hint:
Given, lx + my + n = 0
⇒ my = -lx – n
⇒ y = (-l/m)x + (-n/m)
This will touches the parabola y² = 4ax if
(-n/m) = a/(-l/m)
⇒ (-n/m) = (-am/l)
⇒ n/m = am/l
⇒ ln = am²

Answer: (a) (2,-3)
Hint:
Given, equation fo the of the circle is 4x² + 4y² – 8x + 12y – 25 = 0
⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0
⇒ x² + y² – 2x + 3y – 25/4 = 0
Now, center = {-(-2), -3} = (2, -3)