If `p` is the probability that a man aged `x` will die in a year, then the probability that out of `n` men `A_1,A_2, A_n` each aged `x ,A_1` will die

Correct Answer - C Given that 5 and 6 have appeared on two of the dice, the sample space reduces to `6^(4) - 2 xx 5^(4) + 4^(4)` (inclusion-exclusion principle). Also,...

Correct Answer - A::D `A_(n+1)-A_(n)=int_(0)^(pi//2)(sin(2n+1)xsin(2n-1)x)/(sinx) dx` `=int_(0)^(pi//2)2 cos 2nx dx=0` or `A_(n+1)=A_(n)` `B_(n+1)-B_(n)=int_(0)^(pi//2)(sin^(2)(n+1)x-sin^(2)nx)/(sin^(2)x)dx` `=int_(0)^(pi//2) (sin(2n+1)x)/(sinx)dx=A_(n+1)`

If elements are not repeated , then number of elements in `A_1 cup A_2 cup A_3 ……cup A_30 xx5 =150 .` but each elements of S is used 10 times...

Here, `tan^(_1) ((a_(1) x- y)/(a + a_(1) y)) = tan^(-1) ((a_(1) - (y)/(x))/(1 + a_(1) (y)/(x))) = tan^(-1) a_(1) - tan^(-1) (y)/(x)` `tan^(-1) ((a_(2) -a_(1))/(1 + a_(2) a_(1))) = tan^(-1)...

`P_1` is midpoint of `A_1A_2`. `therefore" "P_1-=((x_1+x_2)/(2),(y_1+y_2)/(2))` `P_2` divides `P_1A_3` in `1:2`. `therefore" "P_2-=((2((x_1+x_2)/2)+x_3)/(2+1),(2((y_1+y_2)/2)+y_3)/(2+1))` `-=((x_1+x_2+x_3)/(3),(y_1+y_2+y_3)/(3))` Now, `P_3` divides `P_2A_4` in ` 1:3` `therefore" "P_3-=((3.((x_1+x_2+x_3)/3)+x_4)/(3+1),(3.((y_1+y_2+y_3)/3)+y_4)/(3+1))` `-=((x_1+x_2+x_3+x_4)/(4),(y_1+y_2+y_3+y_4)/(4))` Proceeding in this manner, we...

Correct Answer - A We have `(1)/(2)|{:(x_1,,y_1,,1),(x_2,,y_2,,1),(x_3,,y_3,,1):}|=|{:(a_1,,b_1,,1),(a_2,,b_2,,1),(a_3,,b_3,,1):}|or (1)/(2)|{:(x_1,,y_1,,1),(x_2,,y_2,,1),(x_3,,y_3,,1):}|=(1)/(2)|{:(a_1,,b_1,,1),(a_2,,b_2,,1),(a_3,,b_3,,1):}|` Hence, the area of triangle with vertices `(x_1,y_1),(x_2,y_2),(x_3,x_3)` is the same as the area of triangle with vertices `(a_1,b_1),(a_2,b_2),(a_3,b_3)`. Hence, the two triangles...

Correct Answer - D Let (h,k) be the point on the locus. Then by the given conditions, `(h-a_1)^1+(k-b_1)^2=(h-a_2)^2+(k-b_2)^2` or `2h(a_1-a_2)+2k(b_1-b_2)+a_2^2+b_2^2-a_1^2=0` `h(a_1-a_2 )+k(b_1-b_2)+(1)/(2)(a_2^2+b_2^2-a_1^2-b_1^2)=0` .....(1) Also, since (h,k) lies on the given locus,...

Correct Answer - B `(f(x))/(1-x) = b_(0) + b_(1)x+b_(2)x^(2)+"…."+b_(n)x^(n)+"….."` `rArr a_(0) + a_(1)x +a_(2)x^(2) + "….." + a_(n)x^(n) + "…."` `= (1-x)(b_(0) + b_(1)x+b_(2)x^(2) + "….." + b_(n)x^(n)+ "…..")` Comparing the...

Correct Answer - C `(c )` Let `d` be common difference of `A.P.impliesd=a_(i)-a_(i-1)` Now `(1)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(a_(i+1)^(1//3)-a_(i)^(1//3))/(a_(i+1)-a_(i))` `=(1)/(d)[a_(i+1)^(1//3)-a_(i)^(1//3)]` Thus `sum_(i=1)^(n)(n)/(a_(i+1)^(2//3)+a_(i+1)^(1//3)*a_(i)^(1//3)+a_(i)^(2//3))=(1)/(d)sum_(i=1)^(n-1)(a_(i+1)^(1//3)-a_(i)^(1//3))` `=(1)/(d)(a_(n)^(1//3)-a_(1)^(1//3))` `=(1)/(d)((a_(n)-a_(1)))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))` `=((n-1))/(a_(n)^(2//3)+a_(n)^(1//3)*a_(1)^(1//3)+a_(1)^(2//3))`

Correct Answer - A `(a)` First arrange `12` persons `A_(4),A_(5),….A_(15)` in `"^(15)P_(12)` ways There remains `3` places. Keep `A_(1)` in the first place and arrange `A_(2)`, `A_(3)` in the remaining two...