<p>For what values of \(p\) does \(\displaystyle\int_0^1\frac{dx}{x^p}\) converge? [JEE Advanced 2010]</p>
Step-by-Step Solution
Key Concept: \int_0^1 x^(-p) dx = [x^(1-p)/(1-p)]_0^1. Converges when 1-p > 0, i.e., p < 1.
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<p>$\int_0^1\frac{dx}{x^p}=\int_0^1 x^{-p}dx=\left[\frac{x^{1-p}}{1-p}\right]_\epsilon^1\to\frac{1}{1-p}$ if $1-p>0$, i.e., $p<1$.</p>
<p>If $p=1$: integral = $[\ln x]_0^1\to\infty$. Diverges.</p>
<p>If $p>1$: integral $\to\infty$. Diverges.</p>
<p>Converges iff $\boxed{p<1}$.</p>
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Correct Answer: A