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Question

a20adx(xa)(x2a) equals to

A
ln23
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B
ln32
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C
ln6
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D
none
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Solution

The correct option is A ln32
a20a(xa)(x2a)dx
a(xa)(x2a)=Axa+Bx2a
a=A(x2a)+B(xa)
a=(A+B)x2aAaB
on comparing both sides we get
A+B=0A=B --(i)
and 2aAaB=a
2aBaB=a (from (i))
B=1
So, A=B=1
Now, a20a(xa)(x2a)dx
=a201xadx+a201x2adx
=[ln|xa|+ln|x2a|]a20
=[lnx2axa]a20
=ln∣ ∣ ∣a22aa2a∣ ∣ ∣ln2aa
=ln3ln2
=ln32

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