If I=∫01/2 dx1−x2n for n≥1 the value of I is

For n≥1,0≤x≤1/2,x2n≤x2.⇒0≤1−x2≤1−x2n⇒11−x2n≤11−x2⇒∫01/2 dx1−x2n≤∫01/2 dx1−x2=sin−1⁡x01/2 =π6<1

If I=∫01/2 dx1−x2n for n≥1 the value of I is

For n≥1,0≤x≤1/2,x2n≤x2.⇒0≤1−x2≤1−x2n⇒11−x2n≤11−x2⇒∫01/2 dx1−x2n≤∫01/2 dx1−x2=sin−1⁡x01/2 =π6<1

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If $I = \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2 n}}} for n \geq 1$ the value of $I$ is

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less than 1

more than 1/2

more than 1

less than 1/2

answer is A.

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Detailed Solution

For $n \geq 1, 0 \leq x \leq 1 / 2, x^{2 n} \leq x^{2}$ .

$\begin{array}{l} \Rightarrow 0 \leq 1 - x^{2} \leq 1 - x^{2 n} \Rightarrow \frac{1}{\sqrt{1 - x^{2 n}}} \leq \frac{1}{\sqrt{1 - x^{2}}} \\ \begin{aligned} \Rightarrow \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2 n}}} \leq \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2}}} & {= \sin^{- 1} x]}_{0}^{1 / 2} \\ = \frac{π}{6} < 1 \end{aligned} \end{array}$