Mathematics - Limits Question with Solution | TestHub

Given,

$f\left(x\right)=a\sin \left(\frac{\pi \left[x\right]}{2}\right)+\left[2-x\right],a\in \mathrm{ℝ}$

Now given $\underset{x\to -1}{\lim }f\left(x\right)$ exists,

So $\underset{x\to -1^{+}}{\lim }a\sin \left(\pi \frac{\left[x\right]}{2}\right)+\left[2-x\right]=-a+2$

And $\underset{x\to -1^{-}}{\lim }asin\left(\pi \frac{\left[x\right]}{2}\right)+\left[2-x\right]=0+3=3$

So, $\underset{x\to -1}{\lim }f\left(x\right)$ exist when $-a+2=3\Rightarrow a=-1$

Now,

${\int }_0^{4}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_1^{2}f\left(x\right)dx+{\int }_2^{3}f\left(x\right)dx+{\int }_3^{4}f\left(x\right)dx$

$\Rightarrow {\int }_0^{4}f\left(x\right)dx={\int }_0^1-\sin \left(\frac{\pi \left[x\right]}{2}\right)+\left[2-x\right]dx+{\int }_1^2-\sin \left(\frac{\pi \left[x\right]}{2}\right)+\left[2-x\right]dx+{\int }_2^3-\sin \left(\frac{\pi \left[x\right]}{2}\right)+\left[2-x\right]dx+{\int }_3^4-\sin \left(\frac{\pi \left[x\right]}{2}\right)+\left[2-x\right]dx$

$={\int }_0^1\left(0+1\right)dx+{\int }_1^2\left(-1+0\right)dx+{\int }_2^3\left(0-1\right)dx+{\int }_3^4\left(1-2\right)dx$

$=1-1-1-1=-2$