JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

The inverse function of
f(x) = (8²ˣ - 8⁻²ˣ)/(8²ˣ + 8⁻²ˣ), x ∈ (-1, 1), is :

  1. (a) 1/4log ₑ( (1 - x)/(1 + x) )
  2. (b) 1/4( log ₈e )log ₑ( (1 - x)/(1 + x) )
  3. (c) 1/4( log ₈e )log ₑ( (1 + x)/(1 - x) )
  4. (d) 1/4log ₑ( (1 + x)/(1 - x) )
Answer and solution

Answer (c) 1/4( log ₈e )log ₑ( (1 + x)/(1 - x) )

Solution

Maths sir · 4 steps

Multiply top and bottom by 8²ˣ, solve for 8⁴ˣ, then take log base 8 and switch to ln.

  1. Clear the negative power. Put y = f(x) and multiply the numerator and denominator by 8²ˣ. This gives y = (8⁴ˣ - 1)/(8⁴ˣ + 1).
  2. Isolate the exponential. Cross-multiply: y · 8⁴ˣ + y = 8⁴ˣ - 1. Collect the 8⁴ˣ terms to get 8⁴ˣ(1 - y) = 1 + y.

    8⁴ˣ = (1+y)/(1-y) ⇒ x = 1/4 log₈((1+y)/(1-y))

  3. Change the base. The options use ln, so write log₈ t = (log₈ e)ln t. Then x = 1/4(log₈ e)ln((1+y)/(1-y)).
  4. So. Rename y as x: f⁻¹(x) = 1/4(log₈ e)logₑ((1+x)/(1-x)), which is option (c).

The idea to keep. For (aᵗ - a⁻ᵗ)/(aᵗ + a⁻ᵗ), multiply through by aᵗ, solve for a²ᵗ, and take logₐ. A base other than e leaves a logₐ e factor when you switch to ln.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Functions (Advanced)

Asked about 70 times in JEE Main since 2020. This question is from 2020.