Subject: Mathematics — Differential Calculus
Question: Consider the function
$f:\left( - \frac{\pi}{2},\frac{\pi}{2} \right) \rightarrow ( - \infty,\infty)$
defined by
$$f(x) = (|x| + |x - 1|)sinx + \lbrack xsinx\rbrack$$
where $\lbrack xsinx\rbrack$ is the greatest integer less than or
equal to $xsinx$.\
Let $\alpha$ be the total number of points in the interval
$\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$ at which $f$ is NOT
continuous, and let\
$\beta$ be the total number of points in the interval
$\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$ at which $f$ is NOT
differentiable.\
Then the value of $\alpha + \beta$ is
Answer: 5.00\
Solution: Observe $\lbrack xsinx\rbrack$
let $g(x) = xsinx$\
$${g^{'}(x) = sinx + xcosx \geq 0
}{g^{'}(x) \geq 0\forall x \in \left\lbrack 0,\frac{\pi}{2} \right)
}{g^{'}(x) \leq 0\forall x \in \left( - \frac{\pi}{2},0 \right\rbrack}$$
[IMAGE] $\lbrack xsinx\rbrack$ discontinuous at
$x = x_{1}, - x_{1};x_{1} \in \left( 0,\frac{\pi}{2} \right)$\
$\lbrack xsinx\rbrack$ not diff. at
$x = x_{1}, - x_{1},x_{1} \in \left( 0,\frac{\pi}{2} \right)$\
Now $y = (|x| + |x - 1|)sinx$\
(i) continuous
$\forall x\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$\
(ii) Not differentiable at $x = 1$, differentiable at $x = 0$
Here $x_{1} \neq 1$\
Now\
(i) for diffrentiability\
$$f(x) = \underset{f_{1}(x)}{\overset{(|x| + |x - 1|)sinx}{︸}} + \underset{f_{2}(x)}{\overset{\lbrack xsinx\rbrack}{︸}}
$$at $x = x_{1}, - x_{1}$\
$f_{1}(x)$ is differentiable & $f_{2}(x)$ is not differentiable\
Hence $f(x)$ is not differentiable\
at $x = 1;f_{1}(x)$ is not differentiable & $f_{2}(x)$ is
differentiable\
$\Rightarrow f(x)$ is not differentiable\
at $x = 0$\
$f_{1}(x)$ & $f_{2}(x)$ both differentiable\
Hence $\beta = 3$\
(ii) for continuity, we can clearly say\
at $x = x_{1}, - x_{1}$,\
$f(x)$ will be discontinuous\
$${\alpha = 2
}{\alpha + \beta = 5}$$