Question

If [.] denotes the greatest integer function, then match the following columns:

Answer 1

A)

$I={\int }_{-2}^2(\alpha x^3+\beta x+\gamma )dx={\int }_{-2}^2(\alpha x^3+\beta x)dx+{\int }_{-2}^2\left(\gamma \right)dx$
As $\alpha x^3+\beta x$ is odd function, then ${\int }_{-2}^2(\alpha x^3+\beta x)dx=0$
$I={\int }_0^2\gamma dx=2\times 2\gamma =4\gamma$
B)

$B=\frac{\gamma }{2}{\int }_0^{1}2\sin \alpha x\sin \beta xdx=\frac{\gamma }{2}{\int }_0^1(\cos (\alpha -\beta )x-\cos (\alpha +\beta )x)dx=\frac{\gamma }{2}{[\frac{\sin (\alpha -\beta )x}{\alpha -\beta }-\frac{\sin (\alpha +\beta )x}{\alpha +\beta }]}_0^1=\frac{\gamma }{2}[\frac{\sin (\alpha -\beta )}{\alpha -\beta }-\frac{\sin (\alpha +\beta )x}{\alpha +\beta }]$
As $\alpha ,\beta$ are roots of $\tan x=2x$
Then, $2\alpha =\tan \alpha$ and $2\beta =\tan \beta$
Therefore,
$2(\alpha -\beta )=(\tan \alpha -\tan \beta )=\frac{\sin (\alpha -\beta )}{\cos \alpha \cos \beta }$ ...(1)
$2(\alpha +\beta )=(\tan \alpha +\tan \beta )=\frac{\sin (\alpha +\beta )}{\cos \alpha \cos \beta }$ ...(2)
Substituting (1) and (2) in B, we get
$B=\frac{\gamma }{2}{\int }_0^1(\cos \alpha \cos \beta -\cos \alpha \cos \beta )=0$
C)

As $f(x+\alpha )+f\left(x\right)=0\Rightarrow f(x+2\alpha )+f(x+\alpha )=0\Rightarrow f(x+2\alpha )=f\left(x\right)$
Hence $f\left(x\right)$ is periodic with period $2\alpha$
Therefore,
${\int }_{\beta }^{\beta +2\gamma \alpha }(\alpha x^3+\beta b+\gamma )dx=\gamma {\int }_0^{2\alpha }f\left(x\right)dx$
D)

$I={\int }_0^{2\beta \pi }\left[\sin x\right]dx+{\int }_{2\beta \pi }^{(2\beta +1)\pi }\left[\sin x\right]dx+{\int }_{(2\beta +1)\pi }^{\alpha }\left[\sin x\right]dx$

$=\beta {\int }_0^{2\pi }\left[\sin x\right]dx+0+{\int }_{(2\beta +1)\pi }^{\alpha }(-1)dx$

$=-\beta \pi +(2\beta +1)\pi -\alpha =(\beta +1)\pi -\alpha$