If coefficient of x3 and x4 in the expansion of 1+ax+bx2(1−2x)18 in powers of x are both zeros, then (a,b) is equal to
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
14,2513
b
14,2723
c
16,2723
d
16,2513
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
S=1+ax+bx2(1−2x)18=1+ax+bx21+18C1(−2x)+ 18C2(−2x)2+18C3(−2x)3+18C4(−2x)4+…Coefficient of x3 in the expansion of S is 18C3(−2)3+a 18C2(−2)2+18C1(−2)b=0Divide by 18C1(−2) to obtain5443−17a+b=0 (1)Similarly, coefficient of x4 is 18C4(−2)4+a 18C3(−2)3+18C2(−2)2b=0Divide by 18C2(−2)2 to obtain80−323a+b=0 (2)Subtract (2) from (1) to obtain3043−193a=0⇒a=16From b=17×16−5443=2723