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Question

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=10r=0 10Cr(1+x)10r(ax2)r
The coefficient of x4 is,
K= 10C0 10C4+ 10C1 9C2a+ 10C2 8C0a2K=45a2+45×8a+45×143K=45[a2+8a+143]
Minimum value of K occurs at B2A=82=4

Hence, the required value of a is 4.

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