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JEE Advanced 2019 Paper 1 · Mathematics · Differentiation and Application of Derivatives · medium

Let f(x)=lfloor xrfloorsin (π x) for real x, where lfloor xrfloor is the greatest integer not exceeding x.

One or more options correct

Let f(x)=⌊x⌋sin⁡(πx)f(x)=\lfloor x\rfloor\sin(\pi x) for real xx, where ⌊x⌋\lfloor x\rfloor is the greatest integer not exceeding xx. Which of the following statements is/are correct?
  1. A. ff is continuous at every real number
  2. B. ff is differentiable at exactly one integer
  3. C. f′ ⁣(52)=2πf'\!\left(\dfrac52\right)=2\pi
  4. D. ∫03f(x) dx=−2π\displaystyle\int_0^{3}f(x)\,dx=-\dfrac{2}{\pi}
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Answer: A

Near an integer nn, ⌊x⌋\lfloor x\rfloor jumps, but sin⁡(πx)→0\sin(\pi x)\to0, so f(n±)=0=f(n)f(n^\pm)=0=f(n); elsewhere ff is a product of continuous functions. Hence ff is continuous everywhere (A true). For differentiability at nn: the left derivative is (n−1)πcos⁡(nπ)(n-1)\pi\cos(n\pi) and the right derivative is nπcos⁡(nπ)n\pi\cos(n\pi). Their difference is πcos⁡(nπ)=±π≠0\pi\cos(n\pi)=\pm\pi\ne0, so ff is non-differentiable at every integer, and (B) is false. On (2,3)(2,3), f=2sin⁡πxf=2\sin\pi x so f′(5/2)=2πcos⁡5π2=0f'(5/2)=2\pi\cos\frac{5\pi}2=0, not 2π2\pi (C false). Finally ∫03f=0+∫12sin⁡πx dx+2∫23sin⁡πx dx=−2π+4π=2π\int_0^3f=0+\int_1^2\sin\pi x\,dx+2\int_2^3\sin\pi x\,dx=-\frac2\pi+\frac4\pi=\frac2\pi, not −2π-\frac2\pi (D false).
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