JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 10)
Let $$f:[0,\infty ) \to [0,\infty )$$ be defined as $$f(x) = \int_0^x {[y]dy} $$
where [x] is the greatest integer less than or equal to x. Which of the following is true?
where [x] is the greatest integer less than or equal to x. Which of the following is true?
f is continuous at every point in $$[0,\infty )$$ and differentiable except at the integer points.
f is both continuous and differentiable except at the integer points in $$[0,\infty )$$.
f is continuous everywhere except at the integer points in $$[0,\infty )$$.
f is differentiable at every point in $$[0,\infty )$$.
Explanation
$$f:[0,\infty ) \to [0,\infty ),f(x) = \int_0^x {[y]dy} $$
Let $$x = n + f,f \in (0,1)$$
So, $$f(x) = 0 + 1 + 2 + ... + (n - 1) + \int\limits_n^{n + f} {n\,dy} $$
$$f(x) = {{n(n - 1)} \over 2} + nf$$
$$ = {{[x]([x] - 1)} \over 2} + [x]\{ x\} $$
Note $$\mathop {\lim }\limits_{x \to {n^ + }} f(x) = {{n(n - 1)} \over 2},\mathop {\lim }\limits_{x \to {n^ - }} f(x) = {{(n - 1)(n - 2)} \over 2} + (n - 1)$$
$$ = {{n(n - 1)} \over 2}$$
$$f(x) = {{n(n - 1)} \over 2}(n \in {N_0})$$
so f(x) is cont. $$\forall$$ x $$\ge$$ 0 nd diff. except at integer points
Let $$x = n + f,f \in (0,1)$$
So, $$f(x) = 0 + 1 + 2 + ... + (n - 1) + \int\limits_n^{n + f} {n\,dy} $$
$$f(x) = {{n(n - 1)} \over 2} + nf$$
$$ = {{[x]([x] - 1)} \over 2} + [x]\{ x\} $$
Note $$\mathop {\lim }\limits_{x \to {n^ + }} f(x) = {{n(n - 1)} \over 2},\mathop {\lim }\limits_{x \to {n^ - }} f(x) = {{(n - 1)(n - 2)} \over 2} + (n - 1)$$
$$ = {{n(n - 1)} \over 2}$$
$$f(x) = {{n(n - 1)} \over 2}(n \in {N_0})$$
so f(x) is cont. $$\forall$$ x $$\ge$$ 0 nd diff. except at integer points
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