If I1=∫0114−x2−x3dx,I2=∫01/211−x4dx then .LIST-ILIST-III) I1 is less thanP) 1II) I1 is more thanQ) 12III) I2 is less thanR) π4IV) I2 is more than S) 14 T) π2

For 0≤x≤1,4−x2≥4−x2−43≥4−2x2∫0114−x2dx≤I1≤∫0114−2x2dx π6≤I1≤π42111 I2

If I1=∫0114−x2−x3dx,I2=∫01/211−x4dx then .LIST-ILIST-III) I1 is less thanP) 1II) I1 is more thanQ) 12III) I2 is less thanR) π4IV) I2 is more than S) 14 T) π2

For 0≤x≤1,4−x2≥4−x2−43≥4−2x2∫0114−x2dx≤I1≤∫0114−2x2dx π6≤I1≤π42111 I2

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If $I_{1} = \int_{0}^{1} \frac{1}{\sqrt{4 - x^{2} - x^{3}}} d x, I_{2} = \int_{0}^{1 / 2} \frac{1}{\sqrt{1 - x^{4}}} d x$ then .
LIST-I LIST-II
I) $I_{1}$ is less than P) 1
II) $I_{1}$ is more than Q) $\frac{1}{2}$
III) $I_{2}$ is less than R) $\frac{π}{4}$
IV) $I_{2}$ is more than S) $\frac{1}{4}$
T) $\frac{π}{2}$

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$I \to P, R I I \to P I I I \to S I V \to T$

$I \to Q, S I I \to P, R I I I \to T I V \to P$

$I \to Q, S I I \to R, T I I I \to Q, T I V \to R$

$I \to P, R I I \to Q, S I I I \to P, R I V \to Q, S$

answer is B.

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Detailed Solution

For $0 \leq x \leq 1, 4 - x^{2} \geq 4 - x^{2} - 4^{3} \geq 4 - 2 x^{2}$
$\int_{0}^{1} \frac{1}{\sqrt{4 - x^{2}}} d x \leq I_{1} \leq \int_{0}^{1} \frac{1}{\sqrt{4 - 2 x^{2}}} d x \frac{π}{6} \leq I_{1} \leq \frac{π}{4 \sqrt{2}}$

111 $I_{2}$

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LIST-I	LIST-II
I) $I_{1}$ is less than	P) 1
II) $I_{1}$ is more than	Q) $\frac{1}{2}$
III) $I_{2}$ is less than	R) $\frac{π}{4}$
IV) $I_{2}$ is more than	S) $\frac{1}{4}$
	T) $\frac{π}{2}$

If I1=∫0114−x2−x3dx,I2=∫01/211−x4dx then .LIST-ILIST-III) I1 is less thanP) 1II) I1 is more thanQ) 12III) I2 is less thanR) π4IV) I2 is more than S) 14 T) π2

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