Question

$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{(I)}{\int }_0^2\frac{x^4+1}{(x^2+2{)}^{\frac{3}{4}}}dx=& \text{(P)}\sqrt{3\sqrt{2}}\\ \text{(II)}{\int }_1^3(\sqrt{1+(x-1{)}^3}+\sqrt[3]{3x^2-1})dx=& \text{(Q)}\sqrt{2\sqrt{3}}\\ \text{(III)}{\int }_0^2\sqrt[3]{3x^3-3x^2+8x-6}dx=& \text{(R)}6\\ \text{(IV)}{\int }_e^{\pi }\frac{2^{\frac{x}{e}}-2^{\frac{\pi }{x}}}{x}dx=& \text{(S)}0\\ & \text{(T)}\frac{\pi -e}{2}\end{array}$
Which of the following is only CORRECT combination?

• A. (I)-(Q)
• B. (II)-(R)
• C. (III)-(Q)
• D. (IV)-(T)

Answer 1

The correct option is B (II)-(R)

(I)
$I={\int }_0^2\frac{x^4+1}{(x^2+2{)}^{\frac{3}{4}}}dx$
$\text{Multiplying numerator and denominator by}{x}^3$
$={\int }_0^2\frac{x^7+x^3}{(x^8+2x^4{)}^{\frac{3}{4}}}dx\text{Put}(x^8+2x^4)=tI=\frac{1}{8}{\int }_0^{288}\frac{dt}{t^{\frac{3}{4}}}=3^{\frac{1}{2}}\times 2^{\frac{1}{4}}=\sqrt{3\sqrt{2}}$

(II) $\sqrt{1+(x-1{)}^3}\&1+\sqrt[3]{3x^2-1}\text{are inverse of each other in}[1,3]I={\int }_1^{3}f\left(x\right)dx+{\int }_1^3{f}^{-1}\left(x\right)dx{f}^{-1}\left(x\right)=tx=f\left(t\right)dx=f^{\prime }\left(t\right)dt{\int }_1^{3}f\left(x\right)dx+{\int }_1^{3}t{f}_1^{1}tdt{\int }_1^{3}f\left(x\right)dx+\left[tf\right(t){]}_1^3-{\int }_1^{3}f(t)dt=3f(3)-1.f(1)=3\times 3-1\times 1=8\text{So,}{\int }_1^3\underset{}{\left(\right(\sqrt{1+(x-1{)}^3}+(1+\sqrt[3]{3x^2-1}))}-1)dx=8-2=6$

(III) $2I={\int }_0^2(\sqrt[3]{3(x-1{)}^3+5(x-1)}+\sqrt[3]{3(1-x{)}^3+5(1-x)})dxI=0$

(IV) $\text{Put}x=\frac{\pi e}{t}$
$I={\int }_{\pi }^e(-(\frac{2^{\frac{\pi }{t}}-2^{\frac{t}{e}}}{t}\left)\right)dt=-II=0$