Question

Match the entries of Column-I with Column-II.

Column - I	Column - II
(A) ${\int }_0^1\sqrt[4]{4x}(\sqrt{x}+1{)}^{3}dx$	(P) is greater than 4
(B) ${\int }_0^1{x}^{-\frac{1}{2}}{(1+x^{\frac{1}{3}})}^{-1}dx$	(Q) is less than 5
(C) ${\int }_0^1{x}^{-\frac{3}{4}}{(1+x^{\frac{1}{4}})}^{\frac{1}{3}}dx$	(R) is less than $\frac{3}{2}$
(D) ${\int }_{\frac{1}{\sqrt{2}}}^{\sqrt{3}}{x}^{-4}(1-x^2{)}^{-\frac{1}{2}}dx$	(S) is greater than $\frac{1}{2}$
	(T) is greater than 5

• A. (A)(P, Q, S),(B)(Q, R, S),(C)(P, Q, S),(D)(Q, R, S)
• B.

Answer 1

The correct option is A (A)(P, Q, S),(B)(Q, R, S),(C)(P, Q, S),(D)(Q, R, S)

$I_1={\int }_0^1{x}^{1/4}(x^{3/2}+1+3x+3x^{1/2})dx$

$={\int }_0^1(x^{7/4}+x^{1/4}+3x^{5/4}+3x^{3/4})dx$

$\Rightarrow 4<I_1<5$

$I_2={\int }_0^1{x}^{-1/2}(1+x^{1/3}{)}^{-1}dx,\text{Put}x=t^6,dx=6t^{6}dt$

$I_2={\int }_{x=0}^1\left(t^{-3}\right)\left(6t^5\right)(1+t^2{)}^{-1}dt=6{\int }_{x=0}^1\frac{t^2}{1+t^2}dt$

$=[6t-6{\tan }^{-1}t{]}_{x=0}^1=[6(x{)}^{1/6}-6{\tan }^{-1}(x{)}^{1/6}{]}_0^1$

$=6(1-\frac{\pi }{4})$

$\Rightarrow 1<I_2<\frac{3}{2}$

$I_3={\int }_0^1{x}^{-3/4}(1+x^{1/4}{)}^{1/3}dx,\text{Put}1+x^{1/4}=t^3$

$\Rightarrow x^{-3/4}dx=12t^{2}dt$

$I_3={\int }_{x=0}^{1}12t^{3}dt=[3t^4{]}_{x=0}^1={\left[3\right(1+x^{1/4}{)}^{4/3}]}_0^1$

$=3(\sqrt[3]{316}-1)$

$\Rightarrow 4<I_3<5$

$I_4=\int x^{-4}(1+x^2{)}^{-1/2}dx,\text{Put}1+x^2=t^2{x}^2$

$\Rightarrow x=\frac{1}{(t^2-1{)}^{1/2}},dx=\frac{-tdt}{(t^2-1{)}^{3/2}}$

$I_4=-\int (t^2-1)dt=-[\frac{t^3}{3}-t]$

$={[-\frac{1}{3x^3}\sqrt{1+x^2}(1-2x^2\left)\right]}_{-\frac{1}{\sqrt{2}}}^{\sqrt{3}}$

$=\frac{1}{9\sqrt{3}}\times 2\times 5=\frac{10}{9\sqrt{3}}$

$\Rightarrow \frac{1}{2}<I_4<1$