Since it is a 4 digit number, first digit cannot be 0. Case 1: (When 0's are at 3rd and 4th place) Possible cases are (7,1), (6,2), (5,3), (4,4), (3,5), (2,6), (1.7), i.e. there are 7 possible cases. Similar cases are possible for CASE 2: (When 0's lie at 2nd and 3rd place). Similarly, there are 7 possible cases for CASE3 (When 0's lie at 2nd and 4th place). Total number of cases is 21. Hence, this is the correct option.

First pen can be put in 6 ways. Similarly, each of the second, third and fourth pens can be put in 6 ways. Hence, total number of ways= 6*6*6*6=1296 (Correct Answer)

Since there are 5 Indian couples and 5 American couples, so there are a total of 20 persons in the party. So, the total number of handshakes possible = ²⁰C₂ It comes out to be 190 (20!/(2!*18!). Now if no wife shakes hand with her own husband, then = 190-10=180 It is also given that no Indian wife shakes hand with any male and there are 5 Indian wives.  Any of them cannot shake hand with 10 males in the party, but we have already included the case in which they do not shake hands with their own husband. So, each of them cannot shake hands with 9 males in the party So, the number of rejected handshakes = 5*9 = 45. Total number of handshakes = 180-45 = 135 (Correct Answer)

Since point B is on x-axis, therefore the equation of line OB is y=0 as every point on x-axis has its y coordinate as 0. (Correct Answer)

First equation, x² - 4x + A = 0 Discriminant D = 16 - 4A Roots are (4 - (16 - 4A)^1/2)/2 and (4 + (16 - 4A)^1/2)/2, i.e. the roots are (2-(4-A)^1/2)and (2+(4-A)^1/2). Obviously, q = (2+(4-A)^1/2) and p = (2-(4-A)^1/2) because it is written that q>p. So, q-p = 2(4-A)^1/2 Second equation, x² - 12x + B =0 Discriminant D = 144 - 4B Roots are (12 - (144 - 4B)^1/2)/2 and (12 + (144 - 4B)^1/2)/2, i.e. the roots are (6-(36-B)^1/2)and (6+(36-B)^1/2). Obviously s = (6+(36-B)^1/2) and r = (6-(36-B)^1/2) because it is written that s>r. So, s-r = 2(36-B)^1/2 Since p, q, r and s are in AP, therefore q - p = s - r (In AP, common difference is the same) 2(4-A)^1/2 = 2(36-B)^1/2 Solving, we get 4-A = 36-B, i.e. B-A = 32 Now, possible combinations of (B,A) are (36,4), (35,3), (34,2), (33,1) because A<=4 as per the roots of first equation {(4-A)^1/2}, otherwise roots will be complex. Also, B<=36 as per the roots of the second equation. Hence, this option is correct.  

From this equation, there are four possible equations of line: 2x+3y<=6 -2x+3y<=6 -2x-3y<=6 2x-3y<=6 All these lines will enclose an area of a triangle = (1/2*2*3) = 3 units square  Since there are four such triangles and figure is symmetric, total area enclosed will be = 3*4 = 12 square units (Correct Answer)

According to the question, a chord of length 2a makes an angle θ at the centre. So, in the triangle, side opposite to angle θ is 2a. r (r = radius of the circle) In that triangle, applying the Cosine Law, Cos θ = (r² + r² - (2a)²)/(2r²) Cos θ = (2r² - 4a²)/(2r²) Solving, we get r² = (2a²)/(1-Cosθ)------------(1) Now, Cos θ can be written as Cos 2(θ/2) (Cos 2x = (1 - tan²x)/(1 + tan²x)) Cos 2(θ/2) = (1 - tan²(θ/2))/(1 + tan²(θ/2)) Cos θ = (1 - tan²(θ/2))/(1 + tan²(θ/2)) 1 - Cos θ = 2tan²(θ/2)/(1 + tan²(θ/2)) It can be written as:1 - Cos θ = 2/(1 + Cot²(θ/2)) Divinding numerator and denominator by tan²(θ/2) and substituting the value of 1 - Cos θ in (1), we get r² = a²(1 + Cot²(θ/2)) Now, area of circle is πr². So, area = πa²(1+Cot²(θ/2)) (Correct Answer)

Since the insect turns and moves due south and cuts the x-axis at point B, the coordinates of point B are (6,0). (As x= 6 units from origin in the projection of A) Equation of the line is  x= 6 (Correct Answer)

Let the coordinates of the third vertex be (x,y). Applying the Pythagoras theorem in the right angled triangle, we get 8 = (x-2)² + y² + x² + (y-2)² 8 = x²+4-4x+y²+x²+y²+4-4y Cancelling 8, we get x²+y²-2x-2y=0 (Correct Answer)

Coordinates of origin are (0,0). Since it moves at point A at a distance of 4√3 from origin in the direction of 60° east of north, hence the coordinates of point A are = (6, 2√3). Now, we are having the coordinates of O and A, we can easily find the equation of line. y-y1=m(x-x1), where m is the slope of the line. m = (y2-y1)/(x2-x1) y1=0, x1=0, y2=2√3, x2=6 m = (2√3/6) = 1/√3 Putting in the equation, we get x-√3y=0 (Correct Answer)  

Now 1^st son has x children, 2^nd has x+1 children. So, 3^rd will be having (24-(2x+1))children, i.e. (23-2x) children. Total number of handshakes is ²⁴C₂ = 276 But it includes cases in which children of same parents shake hands among themselves. So, possible number of handshakes = 276 - x (Handshakes among children of same parents) According to question, it should be maximum, i.e. handshakes among children of same parents should be minimum. Handshakes among children of same parents (y) y = ^xC₂ + ^(x+1)C₂ + ^(23-2x)C₂ y= 253+3x²-45x Differentiating it w.r.t. x and putting 0, we get x = 7.5 Since the number of children cannot be in fraction, lets take x=8 x+1=9 and remaining children of 3^rd son come out to be 7. So putting the values and solving, we get  y=21+28+36=85 So, maximum possible handshakes are 276-85 = 191 (Correct Answer)  

e^iθ = (Cosθ + i Sinθ) (According to De Moiver's theorem) Now, it will become e^{(Cosθ + i Sinθ)} = (e^Cosθ)(e^iSinθ) Now, e^iSinθ = Cos(Sinθ) + iSin (Sinθ) So, it will become e^Cosθ (Cos(Sinθ) + iSin (Sinθ)) e^Cosθ (Cos(Sinθ)) + ie^Cosθ Sin (Sinθ) Real part is e^Cosθ (Cos(Sinθ)) (Correct Answer)

Coordinates of point A (6,2√3).   Coordinates of point at x-axis through which median through A passes are (3,0). (As median divides the line joining O and B into two equal parts) Equation of line y-y1=m(x-x1), where m = (y2-y1)/(x2-x1) y2=2√3, y1=0, x2=6, x1=3 m=(2/√3) Putting m in the equation, we get y=(2/√3) (x-3) √3y-2x+6=0 (Correct Answer)

x³-3x²+3x+7=0 (x-1)³+8=0 (x-1)³=-8 (x-1)³=(-2)³ {(x-1)/(-2)}³=(1) Taking the cube root, we get (x-1)/(-2)=1,ω,ω² Solving, we get three different values of x or α, β, γ = -1, 1-2ω, 1-2ω² Putting the values of α, β, γ in the asked equation, (1/ω)+(1/ω)+ω²=3ω² (As 1 can be written as ω³)(Correct Answer)

For Log_a^b to be integer, b>a. So, obviously b cannot be 2 because a and b are distinct. So, if b=2, a should be greater than that, and then it will not be an integer. So if a=2 ,then possible values of b are {2²,2³,2⁴,------2²⁴,2²⁵} There are 24 possible values. Similarly, if a=2², then possible values of b are {2⁴,2⁶,2⁸,------2²²,2²⁴}, i.e. there are 11 possible values. Similarly, if a=2³, then possible values of b are {2⁶,2⁹,2¹²,------2²¹,2²⁴}, i.e. there are 7 possible values. Similarly, if a=2⁴, then possible values of b are {2⁸,2¹²,2¹⁶,------2²⁰,2²⁴}, i.e. there are 5 possible values. Similarly, if a=2⁵, then possible values of b are {2¹⁰,2¹⁵,2²⁰,------2²⁵}, i.e. there are 4 possible values. Similarly, if a=2⁶,then possible values of b are {2¹²,2¹⁸,2²⁴}, i.e. there are 3 possible values. Similalry, if a=2⁷, then possible values of b are {2¹⁴,2²¹}, i.e. there are 2 possible values. Similalry, if a=2⁸, then possible values of b are {2¹⁶,2²⁴}, i.e. there are 2 possible values. Similarly, if a=2⁹, then possible values of b are {2¹⁸), i.e. only 1 possible value. For a=2¹⁰,2¹¹,2¹², the possible values are 2²⁰,2²²,2²⁴ respectively. After that, there is no case possible. So, total possibilities are (24+11+7+5+4+3+2+2+1+1+1+1)= 62 cases Total number of possibilities = a can be 25 numbers and b can be 24 numbers (As a and b should be distinct) Total possibilities are 25*24. Probability = 62/(25*24) = 31/300 (Correct Answer)