Questions
If the coefficient of x7 in the expansion of and the coefficient of x– 7 is the expansion of are equal, then
detailed solution
Correct option is A
Tr+1,the(r+1) th term in the expansion ax2+1bx11is Tr+1=11Crax211−r1bxr=11Cra11−rbrx22−3rFor the coefficient of x7,set 22−3r=7 ⇒ r=5∴Coefficient of x7 in ax2+1bx11 is 11C5a6b5Next, tr + 1, the (r + 1)th term in the expansion of ax−1bx211is tr+1=11Cr(ax)11−r−1bx2r=11Cra11−rbr(−1)rx11−3rFor the coefficient of x−7, set 11−3r=−7⇒r=6∴Coefficient of x−7 in the expansion of =11C6a5b6We are given 11C5a6b5=11C6a5b6⇒ab=1Talk to our academic expert!
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