If K is the coefficient of x4 in the expansion of (1+x+ax2)10, then the absolute value of a for which K is minimum, is
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Solution
We know that, (1+x+ax2)10=10∑r=010Cr(1+x)10−r(ax2)r The coefficient of x4 is, K=10C010C4+10C19C2a+10C28C0a2⇒K=45a2+45×8a+45×143⇒K=45[a2+8a+143] Minimum value of K occurs at −B2A=−82=−4