Question

If $[.]$ denotes the greatest function, then answer the following by appropriately matching the lists based ​​​​​​on the information given in $\text{Column I}$ and $\text{Column II}$
$\begin{array}{cc}ColumnI& ColumnII\\ \text{a.}\underset{-1}{\overset{1}{\int }}[x+[x+\left[x\right]\left]\right]dx& \text{p.}3\\ \text{b.}\underset{2}{\overset{5}{\int }}\left(\right[x]+[-x\left]\right)dx& \text{q.}5\\ \text{c.}\underset{-1}{\overset{3}{\int }}\text{sgn}(x-[x\left]\right)dx& \text{r.}4\\ \text{d.}25\underset{0}{\overset{\pi /4}{\int }}\left(\right({\tan }^6(x-[x\left]\right)+{\tan }^4(x-[x\left]\right))dx& \text{s.}-3\end{array}$

• A. $\left(a\right)\to \left(s\right),\left(b\right)\to \left(q\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(s\right)$
• B. $\left(a\right)\to \left(s\right),\left(b\right)\to \left(r\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(q\right)$
• C. $\left(a\right)\to \left(s\right),\left(b\right)\to \left(s\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(q\right)$
• D. $\left(a\right)\to \left(q\right),\left(b\right)\to \left(s\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(p\right)$

Answer 1

The correct option is C $\left(a\right)\to \left(s\right),\left(b\right)\to \left(s\right),\left(c\right)\to \left(r\right),\left(d\right)\to \left(q\right)$

$\left(a\right)$
$I=\underset{-1}{\overset{1}{\int }}[x+[x+\left[x\right]\left]\right]dx$
We know that,
$[x+n]=\left[x\right]+n;$ if $n$ is an integer
$\Rightarrow I=3\underset{-1}{\overset{1}{\int }}\left[x\right]dx\Rightarrow I=3\underset{-1}{\overset{0}{\int }}-1dx+3\underset{0}{\overset{1}{\int }}0dx\Rightarrow I=-3$
$\left(a\right)\to \left(s\right)$

$\left(b\right)$
$I=\underset{2}{\overset{5}{\int }}\left(\right[x]+[-x\left]\right)dx$
We know that,
$\left[x\right]+[-x]=\{\begin{array}{c}-1,\text{if}x\text{is not an integer}\\ 0,\text{if}x\text{is an integer}\end{array}$

$\Rightarrow I=-\underset{2}{\overset{5}{\int }}1dx=-3$
$\left(b\right)\to \left(s\right)$

$\left(c\right)$
$\underset{-1}{\overset{3}{\int }}\text{sgn}(x-[x\left]\right)dx$
We know that,
$\text{sgn}(x-[x\left]\right)=\{\begin{array}{c}1,\text{if}x\text{is not an integer}\\ 0,\text{if}x\text{is an integer}\end{array}$
Hence,
$\underset{-1}{\overset{3}{\int }}\text{sgn}(x-[x\left]\right)dx=4$
$\left(c\right)\to \left(r\right)$

$\left(d\right)$
$I=25\underset{0}{\overset{\pi /4}{\int }}\left(\right({\tan }^6(x-[x\left]\right)+{\tan }^4(x-[x\left]\right))dx$
We know that,
$0<x\le \frac{\pi }{4}\Rightarrow \left[x\right]=0$
$\Rightarrow I=25\underset{0}{\overset{\pi /4}{\int }}{\tan }^{6}x+{\tan }^{4}xdx\Rightarrow I=25\underset{0}{\overset{\pi /4}{\int }}{\tan }^{4}x({\tan }^{2}x+1)dx\Rightarrow I=25\underset{0}{\overset{\pi /4}{\int }}{\tan }^{4}x{\sec }^{2}xdx$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\Rightarrow I=25\underset{0}{\overset{1}{\int }}{t}^{4}dt\Rightarrow I=\frac{25}{5}=5$
$\left(d\right)\to \left(q\right)$