Let `f(x0=a_0+a_1x+a_2x^2++a_n x^n+` and `(f(x))/(1-x)=b_0+b_1x+b_2x^2++b_n x^n+` , then `b_n+b_(n-1)=a_n` b. `b_n-b_(n-1)=a_n` c. `b_n//b_(n-1)=a_n`

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)`

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)...

`(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 - 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 `a_(1)=` coefficient of `x` in `(1+2n+3x^(2))^(10)` `=` coefficient of x in `((1+2x)+3x^(2))^(10)` `=` coefficient of x in `(.^(10)C_(0)(1+2x)^(10) + .^(10)C_(1)(1+2x)^(9)(3x^(2))+"…..")` `=` coefficient of x in `.^(10)C_(0)(1+2x)^(10)` `=...

Correct Answer - C `a_(0) + a_(1)x+a_(2)x+"...."+a_(99)x+x^(100) = 0` has root `.^(99)C_(0), .^(99)C_(1), .^(99)C_(2), "....", .^(99)C_(99)`. or `a_(0) + a_(1)x+a_(2)x^(2) + "...."+a_(99)x^(99) + x^(100)` `= (x-.^(99)C_(0))(x-.^(99)C_(1)) (x-.^(99)C_(2))"....."(x-.^(99)C_(99))` Now, sum of root...

Correct Answer - 6 `(1+2x+5x^(2)-10x^(3))[.^(n)C_(0) + .^(n)C_(1)x+.^(n)C_(2)x^(2)+ "…."]` `= 1+a_(1)x+a_(2)x^(2) +"….."` `rArr a_(1)= n-2` and `a_(2) = (n(n-1))/(2)-2n+5` Given that `a_(1)^(2) = 2a_(2)` `rArr (n-2)^(2) = n(n-1)-4n + 10` or `n^(2)...

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