If `(1+x)^5=a_0+a_1x+a2x2+a_3x^3=a_4x^4+a_5x^5,` then the value of `(a_0-a_2+a_4)^2+(a_1-a_3+a_5)^2` is equal to `243` b. `32` c. `1` d. `2^(10)`

`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 - 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-2x^(2))^(20) = a_(0) + a_(1)x+a_(2)x^(2) + "….." +a_(40)x^(40)` Putting `x = 1`, we get `a_(0) + a_(1) + a_(2) + a_(3) +"……."+a_(40) = 0" "(1)` Putting, `x = - 1`,...

`(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 - 31 `(1+x-2x^(2))^(6) = 1+a_(1)x+a_(2)x^(2)+"…."+a_(12)x^(12)` Putting `x = 1` and `x = - 1` and adding the results, we get `64 = 2(1+a_(2)+a_(4)+"….")` `:. a_(2)+a_(4)+a_(6)+"…."a_(12) = 31`

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 - A `(a)` Putting `x=1` and `-1` and adding `a_(0)+a_(2)+…+a_(50)=(3^(25)+1)/(2)` `=((1+2)^(25)+1)/(2)` `=("^(25)C_(0)+^(25)C_(1)*2+^(25)C_(2)*2^(2)+^(25)C_(25)*2^(25)+1)/(2)` `=(2[1+^(25)C_(1)+^(25)C_(2)*2+...+^(25)C_(25)*2^(24)])/(2)` `=2[13+^(25)C_(2)+...+^(25)C_(25)*2^(23)]`

Correct Answer - Inconsistent