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Question

If the term independent of x in the expansion of (xmx2)10 is 405, the value of m is :

A
m=2
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B
m=2
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C
m=3
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D
m=3
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Solution

The correct options are
C m=3
D m=3
The general term in the given expansion is given by
Tr+1=(1)r× 10Cr×(x)10r×(mx2)r
=(1)r× 10Cr×x(5r2)×mrx2r
=(1)r× 10Cr×x(5r22r)×mr
=(1)r× 10Cr×x(55r2)×mr.

Let Tr+1 be free from x.

Then, the power of x in Tr+1 must be 0.

55r2=05r2=5r=2r+1=3.
So, T3 will be free from x.

Now, T3=T(2+1)
=(1)2× 10C2×x0×m2
=(10×92×m2)=45m2.
But, it is given that the term independent of x is 405.

45m2=405m2=9m=±3.

Hence, m=±3.

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