JEE Main202229 Jun 2022Morning ShiftMathematicsBinomial TheoremActual
If the constant term in the expansion of 3 x 3 - 2 x 2 + 5 x 5 10 is 2 k . l , where l is an odd integer, then the value of k is equal to
Options
- A6
- B7
- C8
- D9
Correct answer
D. 9
Step-by-step solution
For 3 x 3 - 2 x 2 + 5 x 5 10 the general term is given by T r + 1 = 10 ! r 1 ! r 2 ! r 3 ! 3 r 1 - 2 r 2 5 r 3 x 3 r 1 + 2 r 2 - 5 r 3 For term independent of x the exponent 3 r 1 + 2 r 2 - 5 r 3 = 0           … 1 Also we know r 1 + r 2 + r 3 = 10         ⋯ 2 From 1 and 2 , we get r 1 + 2 10 - r 3 - 5 r 3 = 0 i.e. r 1 + 20 = 7 r 3 So r 1 , r 2 , r 3 = 1 , 6 , 3 Hence the constant term = 10 ! 1 ! 6 ! 3 ! 3 1 - 2 6 5 3 = 2 9 · 3 2 · 5 4 · 7 1