JEE Main 2021 · 27 July · morning · Single correct · +4 / −1
If the coefficients of x⁷ in ( x² + 1/bx )¹¹ and x⁻⁷ in ( x - 1/bx² )¹¹, b ≠ 0, are equal, then the value of b is equal to :
- (a) 2
- (b) -1
- (c) 1
- (d) -2
Answer and solution
Answer (c) 1
Solution
Maths sir · 4 steps
Find r in each general term, then notice C(11, 5) = C(11, 6) and b falls out in one line.
- First expansion. The general term of (x² + 1/bx)¹¹ is C(11, r) x²⁽¹¹⁻ʳ⁾ (1/bx)ʳ = C(11, r) 1/bʳ x²²⁻³ʳ. For x⁷ we need 22 - 3r = 7, so r = 5 and the coefficient is C(11, 5)/b⁵.
- Second expansion. The general term of (x - 1/bx²)¹¹ is C(11, r) x¹¹⁻ʳ (-1/bx²)ʳ = C(11, r) (-1)ʳ/bʳ x¹¹⁻³ʳ. For x⁻⁷ we need 11 - 3r = -7, so r = 6. Since (-1)⁶ = 1, the coefficient is C(11, 6)/b⁶.
C(11, 5)/b⁵ = C(11, 6)/b⁶
- Cancel and solve. Because C(11, 5) = C(11, 6) = 462, the equation becomes 1/b⁵ = 1/b⁶. Multiply both sides by b⁶ (allowed since b ≠ 0) to get b = 1.
- So. The value of b is 1, which is option (c).
The idea to keep. For a particular power of x, write the general term, collect the powers of x into one exponent, and set it equal to the target. Watch the sign (-1)ʳ, and remember C(n, r) = C(n, n-r).
Answer from the NTA final answer key. Solution by PrepWiser.