If `0<a_1<a_2<ddot<a_n ,` then prove that `tan^(-1)((a_1x-y)/(x+a_1y))+tan^(-1)((a_2-a_1)/(1+a_2a+1))+tan^(-1)((a_3-a_2)/(1+a_3a_2))++tan^(-1)((a_n-a_

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

`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 - 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,...

`(1+x+x^(2)+"….."+x^(p))^(n)=a_(0) + a_(1)x+"…."+a_(np)x^(np)` Differentiating both side w.r.t. , we get `n(1+x+x^(2)+"……"+x^(p))^(n-1)(1+2x+"….."+px^(p-1))` `= a_(1)+2a_(2)x+"….."+npa_(np)x^(np-1)` Now put `x = 1` `:. a_(1) + 2a_(2) + "......." np a_(np) = n(p+1)^(n-1)(1+2+"...."+p)` `= (n(p+1)^(n).p)/(2)`

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 - 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 - A::C `(1-y)^(m)(1+y)^(n)` `= (1-.^(m)C_(1)y+.^(m)C_(2)y^(2) - "……") (1+.^(n)C_(1)y+.^(n)C_(2)y^(2) + "…..")` `= 1+(n-m)y+{(m(m-1))/(2) + (n(n-1))/(2)-mn} y^(2) + "….."` Given, `a_(1) = 10` `rArr a_(1) = n - m = 10"...

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...

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...