Question

$\begin{array}{cc}\text{List I}& \text{List II}\\ \text{(I)}{\int }_{-1}^1\frac{dx}{(1+x^2)(1+3^x)}\text{is}& \text{(P)}\frac{\pi }{2}\\ \text{(II)}7\pi {\int }_0^1(1-x^2{)}^{\frac{3}{2}}{x}^{3}dx\text{is}& \text{(Q)}\frac{\pi }{4}\\ \text{(III)}\frac{\pi }{4}.\frac{{\int }_0^{10}\left[x\right]dx}{{\int }_0^{10}\left\{x\right\}dx},\text{(where [.] {.} denotes}& \text{(R)}\frac{9\pi }{4}\\ \text{the greatest integer and}& \\ \text{the fractional part function)}& \\ \text{(IV)}{\int }_0^{\frac{\pi }{4}}[\tan x+[\sin x+[\cos x+[\sec x\left]\right]\left]\right]dx,& \text{(S)}\frac{2\pi }{5}\\ \text{(where [.] denotes the greatest integer function)}& \end{array}$

Which of the following is only CORRECT combination?

• A. (I)-(S)
• B. (II)-(R)
• C. (III)-(R)
• D. (IV)-(P)

Answer 1

The correct option is C (III)-(R)

(I)
$I={\int }_{-1}^1\frac{dx}{(1+3^x)(1+x^2)}I={\int }_{-1}^1\frac{dx}{(1+3^{-x})(1+x^2)}2I={\int }_{-1}^1\frac{dx}{1+x^2}=\frac{\pi }{2}I=\frac{\pi }{4}$

(II)
$I=7\pi {\int }_0^1(1-x^2{)}^{\frac{3}{2}}.x^{3}dxx=\sin \theta \Rightarrow dx=\cos \theta d\theta I=7\pi {\int }_0^{\frac{\pi }{2}}{\cos }^4\theta .{\sin }^3\theta d\theta \text{put}\cos \theta =t\Rightarrow -\sin \theta d\theta =dtI=-7\pi {\int }_1^0{t}^4(1-t^2)dtI=7\pi (\frac{1}{5}-\frac{1}{7})=\frac{2\pi }{5}$

(III) ${\int }_0^{10}\left\{x\right\}dx=10{\int }_0^{10}xdx(\because {\int }_a^{a+nT}f(x)dx=n{\int }_0^{T}f(x\left)dx\right),\text{T is the period of the periodic function}I=\frac{\pi }{4}.\frac{({\int }_0^{1}0.dx+{\int }_1^{2}1.dx+{\int }_2^{3}2.dx+...+{\int }_9^{10}9.dx)}{10{\int }_0^{1}xdx}I=\frac{\pi }{4}.\frac{0+1+2+...+9}{10.\frac{1}{2}}=\frac{9\pi }{4}$

(IV) $0<x<\frac{\pi }{4}\Rightarrow \left[\sec x\right]=1\Rightarrow [\cos x+[\sec x\left]\right]=1\Rightarrow [\sin x+[\cos x+\left[\sec x\right]\left]\right]=1\Rightarrow [\tan x+[\sin x+[\cos x+[\sec x\left]\right]\left]\right]=1$
$I={\int }_0^{\frac{\pi }{4}}1.dx=\frac{\pi }{4}$