For natural numbers `m`, `n`, if `(1-y)^m(1+y)^n=1+a_1y+a_2y^2+...,` and `a_1=a_2=10` then (A)`mltn` (B)`mgtn` (C)`m+n=80` (D)`m-n=20`

`p^(|)=(pK_(a_2))+pk_(a_3))/2=(8.3+10.8)/2=8.85`

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)...

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 Put `x = i` `(1+i)^(5) = (a_(0) - a_(2) + a_(4)) + i(a_(1)-a_(3)+a_(5))` `rArr |1+i|^(5) = |(a_(0) - a_(2) + a_(4)) + i(a_(1) - a_(3) + a_(5))|`...

Correct Answer - C Let the probability that a man aged x dies in a year p. Thus the probability that a man aged x does not die in a year...

`cos B = (c^(2) + a^(2) - b^(2))/(2ca)` or `a^(2) - (2c cos B) a + c^(2) - b^(2) = 0` This equation has roots `a_(1) and a_(2)` `rArr a_(1)...

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...

Correct Answer - Inconsistent