The term independent of in expansion of is
- A.4
- B.120
- C.210
- D.310
Show answer and solution
Correct option: C
Solution
Simplify each fraction first.
First term: (x+1)/(x^(2/3) - x^(1/3) + 1). Note x+1 = (x^(1/3)+1)(x^(2/3)-x^(1/3)+1), so this simplifies to x^(1/3)+1.
Second term: (x-1)/(x - x^(1/2)) = (x-1)/(x^(1/2)(x^(1/2)-1)) = (x^(1/2)+1)(x^(1/2)-1) / (x^(1/2)(x^(1/2)-1)) = (x^(1/2)+1)/x^(1/2) = 1 + x^(-1/2).
So the expression inside the brackets becomes:
x^(1/3) + 1 - 1 - x^(-1/2) = x^(1/3) - x^(-1/2).
Now expand (x^(1/3) - x^(-1/2))10 using binomial theorem. General term:
T(r+1) = C(10,r) · (x^(1/3))^(10-r) · (-x^(-1/2))r = C(10,r)·(-1)r · x^((10-r)/3 - r/2).
For term independent of x, set the exponent to 0:
(10-r)/3 - r/2 = 0
2(10-r) - 3r = 0
20 - 2r - 3r = 0
r
First term: (x+1)/(x^(2/3) - x^(1/3) + 1). Note x+1 = (x^(1/3)+1)(x^(2/3)-x^(1/3)+1), so this simplifies to x^(1/3)+1.
Second term: (x-1)/(x - x^(1/2)) = (x-1)/(x^(1/2)(x^(1/2)-1)) = (x^(1/2)+1)(x^(1/2)-1) / (x^(1/2)(x^(1/2)-1)) = (x^(1/2)+1)/x^(1/2) = 1 + x^(-1/2).
So the expression inside the brackets becomes:
x^(1/3) + 1 - 1 - x^(-1/2) = x^(1/3) - x^(-1/2).
Now expand (x^(1/3) - x^(-1/2))10 using binomial theorem. General term:
T(r+1) = C(10,r) · (x^(1/3))^(10-r) · (-x^(-1/2))r = C(10,r)·(-1)r · x^((10-r)/3 - r/2).
For term independent of x, set the exponent to 0:
(10-r)/3 - r/2 = 0
2(10-r) - 3r = 0
20 - 2r - 3r = 0
r
[JEE Main 2013]