Q.

If equation ∫dt3at-2t2  = axsin-1(t2a2-1), the value of  x is

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a

32

b

0

c

12

d

-12

answer is .

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Detailed Solution

(t2a2-1) is dimensionless∴[a] = [t]As, (3at-2t2) = [t]∴ [dt3at-t2]= [t][t]= [M0L0T0]ax should be dimensionless, so x = 0.
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