PrepRung

JEE Advanced 2020 Paper 1 · Mathematics · Sets, Relations and Functions · medium

Let f(x)=|x^2-1|+|x-2| for x∈mathbb R.

One or more options correct

Let f(x)=∣x2−1∣+∣x−2∣f(x)=|x^{2}-1|+|x-2| for x∈Rx\in\mathbb R. Which of the following statements is/are correct?
  1. A. ff is not differentiable at exactly three points
  2. B. The minimum value of ff on R\mathbb R is 11
  3. C. ∫03f(x) dx=10\displaystyle\int_0^3 f(x)\,dx=10
  4. D. ff is differentiable at x=0x=0 and f′(0)=1f'(0)=1
Show answer and solution

Answer: A, B

The kinks come from x=±1x=\pm1 (zeros of x2−1x^2-1, simple zeros) and x=2x=2, so there are exactly three points of non-differentiability; (A) is true. On [−1,1][-1,1], f=3−x−x2f=3-x-x^2 (decreasing to 11 at x=1x=1); on [1,2][1,2], f=x2−x+1≥1f=x^2-x+1\ge1 with minimum at x=1x=1; for x≥2x\ge2 and x≤−1x\le-1 the values exceed 33 or are larger. So the minimum is f(1)=1f(1)=1 and (B) is true. Integrating piecewise, ∫01(3−x−x2)dx=136\int_0^1(3-x-x^2)dx=\tfrac{13}{6}, ∫12(x2−x+1)dx=116\int_1^2(x^2-x+1)dx=\tfrac{11}{6}, ∫23(x2+x−3)dx=356\int_2^3(x^2+x-3)dx=\tfrac{35}{6}, total 596≠10\tfrac{59}{6}\ne10; (C) is false. Near 00, f=3−x−x2f=3-x-x^2, so f′(0)=−1f'(0)=-1, not 11; (D) is false.
Report a mistake

Original question written in the JEE Advanced 2020 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake