Learn

Friday, 9 October

JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 6 September 2020, Shift 2, Maths Q51

Question 51 of 75 in this shift, Maths question 1 of 25.

For a suitably chosen real constant a, let a function, f:R−{−a}→Rf:\mathbf{R}-\{-a\}\to\mathbf{R} be defined by f(x)=a−xa+xf(x)=\frac{a-x}{a+x}. Further suppose that for any real number x≠−ax\neq -a and f(x)≠−af(x)\neq -a, (f∘f)(x)=x(f\circ f)(x)=x. Then f(−12)f\left(-\frac{1}{2}\right) is equal to :
  1. (1)33Official answer
  2. (2)−3-3
  3. (3)13\frac{1}{3}
  4. (4)−13-\frac{1}{3}

Official answer

Option 1

NTA final key (Sep 2020).

Chapter
Relations and Functions
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.