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JEE Advanced 2022 Paper 1 · Mathematics · Differentiation and Application of Derivatives · hard

Let f:mathbb R→mathbb R be defined by f(x)=lfloor xrfloorsin (π x), where lfloor xrfloor is the greatest integer not exceeding x.

One or more options correct

Let f:R→Rf:\mathbb R\to\mathbb R be defined by f(x)=⌊x⌋sin⁡(πx)f(x)=\lfloor x\rfloor\sin(\pi x), 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 on R\mathbb R.
  2. B. ff is differentiable at every integer.
  3. C. f′ ⁣(74)=π2f'\!\left(\dfrac74\right)=\dfrac{\pi}{\sqrt2}.
  4. D. ∫03f(x) dx=2π\displaystyle\int_0^3 f(x)\,dx=\dfrac{2}{\pi}.
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Answer: A, C, D

(A) At an integer nn, sin⁡πx→0\sin\pi x\to0 and ⌊x⌋\lfloor x\rfloor is bounded, so f(n±)=0=f(n)f(n^\pm)=0=f(n): continuous. (B) Left derivative at nn is (n−1)πcos⁡nπ(n-1)\pi\cos n\pi, right derivative is nπcos⁡nπn\pi\cos n\pi; they differ by πcos⁡nπ≠0\pi\cos n\pi\neq0, so ff is not differentiable at any integer: false. (C) On (1,2)(1,2), f=sin⁡πxf=\sin\pi x, f′=πcos⁡πxf'=\pi\cos\pi x, and πcos⁡7π4=π2\pi\cos\frac{7\pi}4=\frac{\pi}{\sqrt2}: true. (D) f=0f=0 on [0,1)[0,1); on (1,2)(1,2), sin⁡πx≤0\sin\pi x\le0 so ∫12sin⁡πx dx=−2π\int_1^2\sin\pi x\,dx=-\frac2\pi; on (2,3)(2,3), f=2sin⁡πx≥0f=2\sin\pi x\ge0 so ∫23=4π\int_2^3=\frac4\pi. Total 2π\frac2\pi: true.
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