Author name: Prasanna

Answer: (b) 2880
Total number of persons are 9 in which there are 5 men and 4 women
So total number of place = 9
Now women seat in even place
So total number of arrangement = 4! (_W_W_W_W_) (W-Woman)
Men sit in odd place
So total number of arrangement = 5! (MWMWMWMWM) (M-Man)
Now Total number of arrangement = 5! × 4! = 120 × 24 = 2880

Answer: (d) x = 12y
Given, six boys and six girls sit along a line alternately in x ways and along a circle
(again alternatively in y ways).
Now, x = 6! × 6! + 6! × 6!
⇒ x = 2 × (6!)2
and y = 5! × 6!
Now, x/y = {2 × (6!)2}/(5! × 6!)
⇒ x/y = {2 × 6! × 6! }/(5! × 6!)
⇒ x/y = {2 × 6!}/5!
⇒ x/y = {2 × 6 × 5!}/5!
⇒ x/y = 12
⇒ x = 12y

Answer: (a) 720
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: (a) 588
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, the committee consists of atleast 3 girls:
⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃
= [{4! / (3! × 1!)} × {9! / (4! × 5!)}] + 9C₃
= [{(4 × 3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}] + 9! /(3! × 6!)
= [4 × {(9 × 8 × 7 × 6) / 4!}] + (9×8×7×6!)/(3! × 6!)
= [{4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)] + (9 × 8 × 7)/3!
= (9 × 8 × 7) + (9 × 8 × 7)/(3 × 2 × 1)
= 504 + (504/6)
= 504 + 84
= 588

Answer: (d) 12!/{3! × (4!)³}
Number of ways in which
m × n”>
m × n distinct things can be divided equally into n
n”> groups
= (mn)!/{n! × (m!)n }
Given, 12(3 × 4) people needs to be divided into 3 groups where 4 persons must be there in each group.
So, the required number of ways = (12)!/{3! × (4!)n}

Answer: (a) 24
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) 1
We know that
if ⁿC_r1 = ⁿC_r2
⇒ n = r₁ + r₂
Given, ⁿC₁₅ = ⁿC₆
⇒ n = 15 + 6
⇒ n = 21
Now, ²¹C₂₁ = 1

Answer: (d) 6
Given, ⁿ⁺¹C₃ = 2 ⁿC₂
⇒ [(n + 1)!/{(n + 1 – 3) × 3!}] = 2n!/{(n – 2) × 2!}
⇒ [{n × n!}/{(n – 2) × 3!}] = 2n!/{(n – 2) × 2}
⇒ n/3! = 1
⇒ n/6 = 1
⇒ n = 6

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

Answer: (c) 5040
The number of ways in which 8 students can be sated in a circle = ( 8 – 1)!
= 7!
= 5040

Answer: (d) 36
Total number of functions = 3⁴ = 81
All the four elements can be mapped to exactly one element in 3 ways, and exactly 3
elements in 3(2⁴ – 2) = 3(16 – 2) = 3 × 14 = 42
Thus the number of onto functions = 81 – 42 -3 = 81 – 45 = 36

Answer: (b) (2n)!/(n! × n!)
Given, (1 + x)ⁿ = C₀ + C₁ x + C₂ x² + ………….. + C_n xⁿ ………. 1
and (1 + x)ⁿ = C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n ……….. 2
Multiply 1 and 2, we get
(1 + x)²ⁿ = (C₀ + C₁ x + C₂ x² + …………..+ C_n xⁿ) × (C₀ xⁿ + C₁ x^n-1 + C₂ x^n-2 + ………….. C_r x^n-r + ………. + C_n-1 x + C_n)
Now, equating the coefficient of xn on both side, we get
C₀² + C₁² + C₂² + …………..+ C_nⁿ = ²ⁿC_n = (2n)!/(n! × n!)

Answer: (d) 9 × 9!
Given digit in the number = 9
1st place can be filled = 9 ways = 9 (from 1-9 any number can be placed at first position)
2nd place can be filled = 9 ways (from 0-9 any number can be placed except the number which is placed at the first position)
3rd place can be filled = 8 ways
4th place can be filled = 7 ways
5th place can be filled = 6 ways
6th place can be filled = 5 ways
7th place can be filled = 4 ways
8th place can be filled = 3 ways
9th place can be filled = 2 ways
So total number of ways = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2
= 9 × 9!

Answer: (b) 5 ! × 5 !
Again, 6 girls can be arranged among themselves in 5! ways in a circle.
So, the number of arrangements where boys and girls sit attentively in a circle = 5! × 5!

Answer: (c) 451
The required number of triangle = ¹⁵C₃ – ⁴C₃ = 455 – 4 = 451

Answer: (b) 120
A number is divisible by 10 if the unit digit of the number is 0.
Given digits are 0, 1, 3, 5, 7, 9
Now we fix digit 0 at unit place of the number.
Remaining 5 digits can be arranged in 5! ways
So, total 6-digit numbers which are divisible by 10 = 5! = 120

Answer: (a) 604800
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: (c) 504
Given number of boys = 9
Number of girls = 4
Now, A committee of 7 has to be formed from 9 boys and 4 girls.
Now, If in committee consist of exactly 3 girls:
⁴C₃ × ⁹C₄
= {4! / (3! × 1!)} × {9! / (4! × 5!)}
= {(4×3!) /3!} × {(9 × 8 × 7 × 6 × 5!) / (4! × 5!)}
= 4 × {(9 × 8 × 7 × 6) / 4!}
= {4 × (9 × 8 × 7 × 6)} / (4 × 3 × 2 × 1)
= 9 × 8 × 7
= 504

Answer: (a) 24
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: (c) n!
Given, 2 × P(n, n – 2)
= 2 × {n!/(n – (n – 2))}
= 2 × {n!/(n – n + 2)}
= 2 × (n!/2)
= n!
So, 2 × P(n, n – 2) = n!

Answer: (b) (7, -2)
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 coordiantes 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 cooridnates of B are (7, -2)

Answer: (b) y – x – 1 = 0
Given straight line is: x + y + 1 = 0
⇒ y = -x – 1
Slope = -1
Now, required line is perpendicular to this line.
So, slope = -1/-1 = 1
Hence, the line is
y – 2 = 1 × (x – 1)
⇒ y – 2 = x – 1
⇒ y – 2 – x + 1 = 0
⇒ y – x – 1 = 0

Answer: (c) collinear
Let the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a², ab)
Now,
m₁ = slope of OP = b/a
m₂ = slope of OQ = b/a
m₃ = slope of OR = b/a
Since m₁ = m₂ = m₃
So, the points O, P, Q, R are collinear.

Answer: (c) 2x + y – 7 = 0
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: (b) -3/4
Given, points are (3, 2) and (-1, 5)
Now, slope m = (5 – 2)/(-1 – 3)
⇒ m = -3/4
So, the slope of the line is -3/4

Answer: (d) 14
Let the A.P. are 1, A₁, A₂, A₃ …… A_m, 31
a = 1, a_n = 31 and n = m + 2
Now, a_n = a + (n – 1)d
⇒ 31 = 1 + (m + 2 – 1)d
⇒ 30 = (m + 1)d
⇒ d = 30/(m + 1)
Again, A₇ = a + 7d = 1 + 7[30/(m + 1)] …………….. 1
and A_m-1 = a + (m – 1)d = 1 + (m – 1)[30/(m + 1)] ………. 2
From equation 1 and 2, we get
A₇/A_m-1 = 5/9
⇒ 1 + 7[30/(m + 1) / 1 + (m – 1)[30/(m + 1)] = 5/9
⇒ [m + 1 + 7(30)] / [m + 1 + 30 m – 30] = 5/9
⇒ [m + 211] / [31 m – 29] = 5/9
⇒ 9[m + 211] = 5[31 m – 29]
⇒ 9 m + 1899 = 155 m – 145
⇒ 146 m = 2044
⇒ m = 2044/146
⇒ m = 14
So, the value of m is 14

Answer: (a) (0, 0)
Given lines are:
xy = 0 and x + y = 1
⇒ x = 0, y = 0 and x + y = 1

In a triangle OAB, OA and OB are the altitudes which intersect at O.
So, the required orthocentre is (0, 0)

Answer: (a) a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂

Answer: (d) a = −1 and b = -1
Given equation of the line is x/a + y/b = 1
⇒ bx + ay = ab
It is given that this line passes through (2, -3)
⇒ b(2) + a(-3) = ab
⇒ 2b – 3a = ab ——– (1)
It also passes through (4, -5)
⇒ 4b – 5a = ab ——– (2)
On solving equation (1) and (2), we get
a = -1 and b = -1

Answer: (c) tan^-1 (5/4)
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: (d) (0, 32/3) and (0, -8/3)
Given equation of line is (x/3) + (y/4) = 1
⇒ 4x + 3y = 12
⇒ 4x + 3y – 12 = 0 ……………. 1
Let (0, b) is the point of the y-axis whose distance from given line is 4 unit.
When we compare equation 1 with general form of the equation Ax + By + C = 0, we get
A = 4, B = 3, C = -12
Now perpendicular distance of a line Ax + By + C = 0 from a point (x₁, y₁) is
d = |Ax₁ + By₁ + C|/√(A² + B²)
So perpendicular distance of a line 4x + 3y – 12 = 0 from a point (0 ,b) is
4 = |4×0 + 3×b – 12|/√(4² + 3²)
⇒ 4 = |3b – 12|/√(16 + 9)
⇒ 4 = |3b – 12|/√25
⇒ 4 = |3b – 12|/5
⇒ 4 × 5 = |3b – 12|
⇒ |3b – 12| = 20
Now
3b – 12 = 20 and 3b – 12 = -20
⇒ 3b = 20 12 and 3b = -20 + 12
⇒ 3b = 32 and 3b = -8
⇒ b = 32/3 and b = -8/3
So the points are (0, 32/3) and (0, -8/3)

Answer: (c) y = mx
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: (a) 3/10
Given equations are:
3x + 4y = 9
⇒ 3x + 4y – 9 = 0 and
6x + 8y = 15
⇒ 6x + 8y – 15 = 0
⇒ 3x + 4y – 15/2 = 0
Now, compare these lines with a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0, we get
a₁ = 3, b₁ = 4, c₁ = -9 and
a₂ = 3, b₂ = 4, c₂ = -15/2
Now, distance between two parallel line = |c₁ – c₂|/√(a₁² + b₁²)
= |-9 + 15/2|/√(3² + 4²)
= |(-18 + 15)/2|/√25
= |(-3/2)|/5
= (3/2)/5
= 3/10

Answer: (b) θ is an obtuse angle
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) a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Answer: (d) 1/√3
Here inclination of the line is 30°
So, slope of the line m = tan 30° = 1/√3

Answer: (c) 2
The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2)
d = |4 × (-1) + 3 × 3 + 5|/√(4² + 3²)
⇒ d = |-4 + 9 + 5|/√(16 + 9)
⇒ d = 10/√(25)
⇒ d = 10/5
⇒ d = 2

Answer: (b) 45°
Given line is: 5x – 5y + 8 = 0
⇒ 5y = 5x + 8
⇒ y = (5/5)x + 8/5
⇒ y = x + 8/5
Now tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°
So, the inclination of the line is 45°

Answer: (c) collinear
Let the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a², ab)
Now,
m₁ = slope of OP = b/a
m₂ = slope of OQ = b/a
m₃ = slope of OR = b/a
Since m₁ = m₂ = m₃
So, the points O, P, Q, R are collinear.

Answer: (c) concurrent
Since the determinant of these lines is equal to zero
i.e.
|5 4 0|
|1 2 -10| = 0
|2 1 -5|
So, these three lines are concurrent.

Answer: (a) 0
Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 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, a – d = 1/mn – 1/mn
⇒ a – d = 0

Answer: (c) 3
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: (c) a × r^n-1
Given, a is the first term and r is the common ratio.
Now, nth term of GP = a × r^n-1

Answer: (d) 1002001
The odd numbers from 1 to 2001 are:
1, 3, 5, ………, 2001
This froms an AP
where first term a = 1
Common difference d = 3 – 1 = 2
last term l = 2001
Let number of terms = n
Now, l = a + (n – 1)d
⇒ 2001 = 1 + (n – 1)2
⇒ 2001 – 1 = (n – 1)2
⇒ 2(n – 1) = 2000
⇒ n – 1 = 2000/2
⇒ n – 1 = 1000
⇒ n = 1001
Now, sum = (n/2) × (a + 1)
= (1001/2) × (1 + 2001)
= (1001/2) × 2002
= 1001 × 1001
= 1002001
So, the sum of odd integers from 1 to 2001 is 1002001

Answer: (b) 1
Given, a, b, c are in AP
⇒ 2b = a + c ………. 1
and x, y, z are in GP
⇒ y² = xz ……….. 2
Now, x^b-c × y^c-a × z^a-b = x^b-c × (√xz)^c-a × z^a-b
= x^b-c × x^(c-a)/2 × z^(c-a)/2 × z^a-b
= x^b-c + x^(c-a)/2 × z^{(c-a)/2+ a -b}
= x^2b+(c+a) × z^(c+a)-2b
= x° × z°
= 1
So, the value of x^b-c × y^c-a × z^a-b is 1

Answer: (b) 1, 2, 4, 8
1, 2, 4, 8 is the example of geometric series
Here common ratio = 2/1 = 4/2 = 8/4 = 2

Answer: (c) 2 + √3
Given that three numbers from an increasing GP
Let the 3 number are: a, ar, ar² (r > 1)
Now, according to question,
a, 2ar, ar² are in AP
So, 2ar – a = ar² – 2ar
⇒ a(2r – 1) = a(r² – 2r)
⇒ 2r – 1 = r² – 2r
⇒ r² – 2r – 2r + 1 = 0
⇒ r² – 4r + 1 = 0
⇒ r = [4 ± √{16 – 4 × 1 × 1}]/2
⇒ r = [4 ± √{16 – 4}]/2
⇒ r = {4 ± √12}/2
⇒ r = {4 ± 2√3}/2
⇒ r = {2 ± √3}
Since r > 1
So, the common ratio of the GP is (2 + √3)

Answer: (c) 402
Let ais the first term and d is the common difference of the AP
Given,
a₅ = a + (5 – 1)d = 22
⇒ a + 4d = 22 ………….1
and a15 = a + (15 – 1)d = 62
⇒ a + 14d = 62 ………2
From equation 2 – 1, we get
62 – 22 = 14d – 4d
⇒ 10d = 40
⇒ d = 4
From equation 1, we get
a + 4 × 4 = 22
⇒ a + 16 = 22
⇒ a = 6
Now,
a100 = 6 + 4(100 – 1 )
⇒ a100 = 6 + 4 × 99
⇒ a100 = 6 + 396
⇒ a100 = 402

Answer: (d) 1/2 – 1/√2
Given, a, b, c are in AP
⇒ 2b = a + c
⇒ b = (a + c)/2 ………….. 1
Again given, a², b², c² are in GP then b⁴ = a² c²
⇒ b² = ± ac ………… 2
Using 1 in a + b + c = 3/2, we get
3b = 3/2
⇒ b = 1/2
hence a + c = 1
and ac = ± 1/4
So a & c are roots of either x2 −x + 1/4 = 0 or x² − x − 1/4 = 0
The first has equal roots of x = 1/2 and second gives x= (1 ± √2)/2 for a and c
Since a < c,
we must have a = (1−√2)/2
⇒ a – 1/2 – √2/2
⇒ a – 1/2 – √2/(√2×√2)
⇒ a – 1/2 – 1/√2

Answer: (d) in H.P.
Given, the positive numbers a, b, c, d are in A.P.
⇒ 1/a, 1/b, 1/c, 1/d are in H.P.
⇒ 1/d, 1/c, 1/b, 1/a are in H.P.
Now, Multiply by abcd, we get
abcd/d, abcd/c, abcd/b, abcd/a are in H.P.
⇒ abc, abd, acd, bcd are in H.P.

Answer: (a) 0
Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 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, a – d = 1/mn – 1/mn
⇒ a – d = 0

Answer: (d) 2055
Before 2024, there are 44 squares,
So, 1980th term is 2024
Hence, 2011th term is 2055

Answer: (a) 1
Let first term = a
and common ratio = r
Given, a – ar² = 768
⇒ a(1 – r²) = 768
and ar² – ar⁶ = 240
⇒ ar² (1 – r⁴) = 240
Dividing the above 2 equations, we get
ar² (1 – r⁴)/a(1 – r²) = 240/768
⇒ {ar² (1 – r²) × (1 + r²)}/a(1 – r²) = 240/768
⇒ 1 + r² = 0.3125
⇒ r² = 0.25
⇒ r² = 25/100
⇒ r² = √(1/4)
⇒ r = ± 1/2
Now, a(1 – r²) = 768
⇒ a(1 – 1/4 ) = 768
⇒ 3a/4 = 768
⇒ 3a = 4 × 768
⇒ a = (4 × 768)/3
⇒ a = 4 × 256
⇒ a = 1024
⇒ a = 2¹⁰
Now product of first 21 terms = (a² × r²⁰)¹⁰ × a × r¹⁰
= a²¹ × r²¹⁰
= (2¹⁰)²¹ × (1/2)²¹⁰
= 2²¹⁰ /2²¹⁰
= 1

Answer: (c) 11
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: (a) AP
Given, a, b, c are in GP
⇒ b² = ac
⇒ (b²)ⁿ = (ac)ⁿ
⇒ (b2 )ⁿ= aⁿ × cⁿ
⇒ log (b²)ⁿ = log(an × cn )
⇒ log b²ⁿ = log aⁿ + log cⁿ
⇒ log (bⁿ)² = log aⁿ + log cⁿ
⇒ 2 × log bⁿ = log aⁿ + log cⁿ
⇒ log aⁿ, log bⁿ, log cⁿ are in AP

Answer: (c) 28
Given, a_n = 3n – 2
Put n = 10, we get
a₁₀ = 3 × 10 – 2
⇒ a₁₀ = 30 – 2
⇒ a₁₀ = 28
So, the 10th term of AP is 28

Answer: (c) 740
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: (b) 2b = a + c
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Answer: (b) a², b², c² are in AP
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) Arithmetic Series
3, 5, 7, 9, …….. is an example of Arithmetic Series.
Here common difference = 5 – 3 = 7 – 5 = 9 – 7 = 2

Answer: (c) √{a² × (1 + m²)} > c
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
√{a² × (1 + m²)} > c

Answer: (c) y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

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

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

Answer: (c) 9
Given equation of the circle is
x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
⇒ -3 × 2 – 3 + λ = 0
⇒ -6 – 3 + λ = 0
⇒ -9 + λ = 0
⇒ λ = 9

Answer: (b) 1
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: (b) g² + f² – c ≥ 0
Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0
This equation can be written as
{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²
So, the circle is real is g² + f² – c ≥ 0

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
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: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Answer: (d) 0 < e < 1
The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Answer: (c) (x – 2)² + (y – 3)² = 3²
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) 2√2/√3
Given, the length of the major axis of an ellipse is three times the length of the minor axis
⇒ 2a = 3(2b)
⇒ 2a = 6b
⇒ a = 3b
⇒ a² = 9b²
⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}
⇒ 1 = 9(1 – e²)
⇒ 1/9 = 1 – e²
⇒ e² = 1 – 1/9
⇒ e² = 8/9
⇒ e = √(8/9)
⇒ e = 2√2/√3
So, the eccentricity of the ellipse is 2√2/√3

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Answer: (c) 3/5
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: (b) x + 3y = 0
The coordinate of the centre of the circle x² + y² – 12x + 4y + 6 = 0 are (6, -2)
Clearly, the line x + 3y passes through this point.
Hence, x + 3y = 0 is a diameter of the given circle.

Answer: (a) (2, -3)
Given, equation 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)

Answer: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Answer: (b) x² /25 + y² /9 = 1
Given focus is (4, 0)
⇒ ae = 4
and e = 4/5
a × (4/5) = 4
⇒ a = 5
Now, b² = a² (1 – e²)
⇒ b² = 5² {1 – (4/5)²}
⇒ b² = 25{1 – 16/25}
⇒ b² = 25{(25 – 16)/25}
⇒ b² = 9
Hence, the equation of the ellipse is x²/a² + y²/b² = 1
⇒ x²/5² + y²/9 = 1
⇒ x²/25 + y²/9 = 1

Answer: (a) (2, 0)
Given, y² = 8x
General equation is y² = 4ax
Now, 4a = 8
⇒ a = 2
Now, focus = (a, 0) = (2, 0)

Answer: (c) 12/13, 4/13, 3/13
Let AB be the given line and the DCs of AB be l, m, n. Then
Projection on x-axis = AB . l = 12 (Given)
Projection on y-axis = AB . m = 4 (Given)
Projection on z-axis = AB . n = 3 (Given)
⇒ (AB²) (l² + m² + n²) = 144 + 16 + 9
⇒ (AB²) = 169 {since l² + m² + n² = 1}
⇒ AB = 13
Hence, DCs of AB are 12/13, 4/13, 3/13

Answer: (c) cos θ = (n₁ . n₂)/{|n₁| × |n₂|}
The angle between the planes r . n₁ = d₁ and r . n₂ = d₂ is defined as
cos θ = (n₁ . n₂)/{|n₁| × |n₂|}

Answer: (c) z = 0
The perpendicular distance of P(x, y, z) from xy-plane is zero.

Answer: (b) Line parallel to x-axis
Since x = 0 and y = 0 together represent x-axis, therefore x = a and y = b represent a line parallel to x-axis.

Answer: (d) x + 3y + 6z – 7 = 0
Let the equation of the plane is
(2x – 5y + z – 3) + λ(x + y + 4z – 5) = 0
⇒ (2 + λ)x + (λ – 5)y + (4λ + 1)z – (3 + 5λ) = 0
Since the plane is parallel to x + 3y + 6z – 1 = 0
⇒ (2 + λ)/1 = (λ – 5)/3 = (1 + 4λ)/6
⇒ 6 + 3λ = λ – 5
⇒ 2λ = -11
⇒ λ = -11/2
Again,
6λ – 30 = 3 + 12λ
⇒ -6λ = -33
⇒ λ = -33/6
⇒ λ = -11/2
So, the required equation of plane is
(2x – 5y + z – 3) + (-11/2) × (x + y + 4z – 5) = 0
⇒ 2(2x – 5y + z – 3) + (-11) × (x + y + 4z – 5) = 0
⇒ 4x – 10y + 2z – 6 – 11x – 11y – 44z + 55 = 0
⇒ -7x – 21y – 42z + 49 = 0
⇒ x + 3y + 6z – 7 = 0

Answer: (a) (5/3, 7/3, 17/3)
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: (c) (0, 17/2, -13/2)
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: (c) 6√2i + 6j ± 6k
Let l, m, n be the DCs of OP.
Then it is given that l = cos 45 = 1/√2
m = cos 60 = 1/2
Now, l² + m² + n² = 1
⇒ 1/2 + 1/4 + n² = 1
⇒ n² = 1/4
⇒ n = ±1/2
Now, r = |r|(li + mj + nk)
⇒ r = 12(i/√2 + j/2 ± k/√2)
⇒ r = 6√2i + 6j ± 6k

Answer: (a) (-3, 5, 2)
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: (a) 4 points not in the same plane
Sphere is referred to its center and it follows a quadratic equation with 2 roots. The mid-point of chords of a sphere and parallel to fixed direction lies in the normal diametrical plane.
Now, general equation of the plane depends on 4 constants. So, one sphere passes through 4 points and they need not be in the same plane.

Answer: (d) either (0, -1, 0) or (0, 7, 0)
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: (c) (0, 17/2, -13/2)
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) r . (2i – j + 2k) = 3
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: (a) 1/3, 1/6, 1
Give 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: (b) u² + v² + w² > d
Equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere if
u² + v² + w² – d > 0
⇒ u² + v² + w² > d

Answer: (c) plane
In an x-y-z cartesian coordinate system, the general form of the equation of a plane is
ax + by + cz + d = 0
It is an equation of the first degree in three variables.

Answer: (a) (-3, 5, 2)
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) √(b² + c²)
The coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).
So, the required distance = √{(a – a)² + (b – 0)² + (c – 0)²}
= √(b² + c²)

Answer: (a) |r| = 5
We know that the vector equation of a sphere having center at the origin and radius R
= |r| = R
Here R = 5
Hence, the equation of the required sphere is |r| = 5

Answer: (b) 3 : 5
Let the points are P(1, 2, 3) and Q(-3, 4, -5)
Let the line joining the points P(1, 2, 3) and Q(-3, 4, -5) is divided by the xy-plane at point R in the ratio k : 1
Now, the coordinate of R is
{(-3k + 1)/(k + 1), (4k + 2)/(k + 1), (-5k + 3)/(k + 1)}
Since R lies on the xy-plane.
So, z-coordinate is zero
⇒ (-5k + 3)/(k + 1) = 0
⇒ k = 3/5
So, the ratio = 3/5 : 1 = 3 : 5

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

Answer: (b) = (a × cos a – sin a)/a²
Given,
Lim_x→a (a × sin x – x × sin a)/(ax² – xa²)
When we put x = a in the expression, we get 0/0 form.
Now apply L. Hospital rule, we get
Lim_x→a (a × cos x – sin a)/(2ax – a²)
= (a × cos a – sin a)/(2a × a – a²)
= (a × cos a – sin a)/(2a² – a²)
= (a × cos a – sin a)/a²
So, Lim_x→a (a × sin x – x × sin a)/(ax² – xa²) = (a × cos a – sin a)/a²

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

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

Answer: (d) -1/2
Given, Lim_x→1 (1 + log x – x)}/(1 – 2x + x²)
= Lim_x→1 (1/x – 1)}/(-2 + 2x) {Using L. Hospital Rule}
= Lim_x→1 (1 – x)/{2x(x – 1)}
= Lim_x→1 (-1/2x)
= -1/2

Answer: (d) x × tan x × sec x + sec x
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: (d) 3/2
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: (a) a^x = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
a^x = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Answer: (c) e^ab
Given, Lim_n→0 (1 + an)^b/n
= e^{Lim_n→0(an × b/n)}
= e^{Lim_n→0(ab)}
= e^ab

Answer: (c) 1
Given, Lim_x→0 cos x/(1 + sin x)
= cos 0/(1 + sin 0)
= 1/(1 + 0)
= 1/1
= 1

Answer: (c) 1/2
Given, Lim tan_x→π/4 tan 2x × tan(π/4 – x)
= Lim tan_h→0 tan 2(π/4 – x) × tan(-h)
= Lim tan_h→0 -cot 2h/(-cot h)
= Lim tan_h→0 tan h/tan 2h
= (1/2) × Lim tan_h→0 (tan h/h)/(2h/tan 2h)
= (1/2) × {Lim tan_h→0 (tan h/h)}/{Lim tan_h→0 (2h/tan 2h)}
= (1/2) × 1
= 1/2

Answer: (c) 1/2
When x = 2, the expression
(x³ – 6x² + 11x – 6)/(x² – 6x + 8) assumes the form 0/0
Now,
Lim_x→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) = Lim_x→2 {(x – 1) × (x – 2) × (x – 3)}/{(x – 2) × (x – 4)}
= Lim_x→2 {(x – 1) × (x – 3)}/(x – 4)
= {(2 – 1) × (2 – 3)}/(2 – 4)
= 1/2

Answer: (a) 0
Given, Lim_x→2 (x – 2)/√(2 – x)
= Lim_x→2 -(2 – x)/√(2 – x)
= Lim_x→2 -{√(2 – x) × √(2 – x)}/√(2 – x)
= Lim_x→2 -√(2 – x)
= -√(2 – 2)
= 0

Answer: (d) 18
Given, function f(x) = 3x³ – 2x² + 5x – 1
Differentiate w.r.t. x, we get
df(x)/dx = 3 × 3 × x² – 2 × 2 × x + 5
⇒ df(x)/dx = 9x² – 4x + 5
⇒ {df(x)/dx}_{x =-1} = 9 × (-1)² – 4 × (-1) + 5
⇒ {df(x)/dx}_{x =-1} = 9 + 4 + 5
⇒ {df(x)/dx}_{x =-1} = 18

Answer: (d) 1/9
Given, Lim_x→0 sin² (x/3)/ x²
= Lim_x→0 [sin² (x/3)/ (x/3)² × {(x/3)² /x²}]
= Lim_x→0 [{sin (x/3)/ (x/3)}² × {(x² /9)/x²}]
= 1 × 1/9
= 1/9

Answer: (a) a^x = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Answer: (d) -sin √x /2√x
Let y = cos √x
Put u = √x
du/dx = 1/2√x
Now, y = cos u
dy/du = -sin u
dy/dx = (dy/du) × (du/dx)
= -sin u × (1/2√x)
= -sin √x /2√x

Answer: (b) cot x
Let y = log(sin x)
Again let u = sin x
du/dx = cos x
Now, y = log u
dy/du = 1/u = 1/sin x
Now, dy/dx = (dy/du) × (du/dx)
⇒ dy/dx = (1/sin x) × cos x
⇒ dy/dx = cos x/sin x
⇒ dy/dx = cot x

Answer: (c) e⁶
Given, Lim_x→∞ {(x + 5)/(x + 1)}x
= Lim_x→∞ {1 + 6/(x + 1)}x
= e^{Lim_x→∞6x/(x + 1)}
= e^{Lim_x→∞ 6/(1 + 1/x)}
= e^{6/(1 + 1/∞)}
= e^{6/(1 + 0)}
= e⁶