Verified Solution • 2025–26 Session moderate

QJEEM21Ma10fd: JEE Main 2021 Mathematics - Class 12 JEEM21Ma10fd Solution: Let be defined as where [x] is the greatest integer less than…

Let f:[0,∞)→[0,∞)f:[0,\infty ) \to [0,\infty ) be defined as f(x)=∫0x[y]dyf(x) = \int_0^x {[y]dy}

where [x] is the greatest integer less than or equal to x. Which of the following is true?

Multiple Choice Options

  • A f is continuous at every point in [0,∞)[0,\infty ) and differentiable except at the integer points.
  • B f is both continuous and differentiable except at the integer points in [0,∞)[0,\infty ) .
  • C f is continuous everywhere except at the integer points in [0,∞)[0,\infty ) .
  • D f is differentiable at every point in [0,∞)[0,\infty ) .

Step-by-Step Verified Solution

f:[0,∞)→[0,∞),f(x)=∫0x[y]dyf:[0,\infty ) \to [0,\infty ),f(x) = \int_0^x {[y]dy}

Let x=n+f,f∈(0,1)x = n + f,f \in (0,1)

So, f(x)=0+1+2+...+(n−1)+∫nn+fn dyf(x) = 0 + 1 + 2 + ... + (n - 1) + \int\limits_n^{n + f} {n\,dy}

f(x)=n(n−1)2+nff(x) = {{n(n - 1)} \over 2} + nf

=[x]([x]−1)2+[x]{x}= {{[x]([x] - 1)} \over 2} + [x]\{ x\}

Note lim⁡x→n+f(x)=n(n−1)2,lim⁡x→n−f(x)=(n−1)(n−2)2+(n−1)\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)2= {{n(n - 1)} \over 2}

f(x)=n(n−1)2(n∈N0)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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