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प्रश्न
\[\int\frac{dx}{\sqrt{2ax - x^2}} = a^n \sin^{- 1} \left[ \frac{x}{a} - 1 \right]\]
The value of n is
पर्याय
0
-1
1
none of these.
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उत्तर
0
[ax] = [x2]
⇒ [a] = [x] ...(1)
Dimension of LHS = Dimension of RHS
\[\Rightarrow \left[ \frac{dx}{\sqrt{x^2}} \right] = \left[ a^n \right]\]
\[ \Rightarrow \left[ \frac{L}{L} \right] = \left[ a^n \right] . . . (2) \]
\[[ L^0 ] = [ a^n ]\]
\[n = 0\]
Notes
You may use dimensional analysis to solve the problem.
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