MCQ
The function $f(x) = {[x]^2} - [{x^2}]$, (where $[y]$ is the greatest integer less than or equal to $y$),is discontinuous at
- AAll integers
- BAll integers except $0$ and $1$
- CAll integers except $0$
- ✓All integers except $1$
$ - 1 < x < 0,\,\,f(x) = {( - 1)^2} - 0 = 1$
$x = 0,\,\,f(x) = {0^2} - 0 = 0$
$0 < x < 1,\,\,f(x) = {0^2} - 0 = 0$
$x = 1,\,\,f(x) = {1^2} - 1 = 0$
$1 < x < \sqrt 2 ,\,\,f(x) = {1^2} - 1 = 0$
$x = \sqrt 2 ,\,\,f(x) = {1^2} - 2 = - 1$
$\sqrt 2 < x < \sqrt 3 ,\,\,f(x) = {1^2} - 2 = - 1$
$x = \sqrt 3 ,\,\,f(x) = {1^2} - 3 = - 2$
$\sqrt 3 < x < 2,\,\,f(x) = {1^2} - 3 = - 2$
$x = 2,\,\,f(x) = 4 - 4 = 0$;
$2 < x < \sqrt 5 ,\,\,f(x) = 4 - 4 = 0$
$x = \sqrt 5 ,\,\,f(x) = 4 - 5 = - 1$
Hence function is discontinuous at all integers except $1$.
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$f(t)=\left\{\begin{array}{cc}(-1)^{n+1} 2, & \text { if } t=2 n-1, n \in N , \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in N \end{array}\right.$
Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim _{x \rightarrow 1+} \frac{g(x)}{x-1}$. Then the value of $\alpha+\beta$ is equal to. . . . . .