Let f:R - { α/6 } → R be defined by f(x) = (5x + 3)/(6x - α). Then the value of α for which (fof)(x) = x, for all x ∈ R - { α/6 }, is :

Solution

Maths sir · 4 steps

Compose f with itself, set the result equal to x, and let the coefficients choose α.

1. Put f inside f. Replace x by (5x+3)/(6x-α) in f. Then multiply the top and the bottom by (6x-α) to clear the inner fractions.

   f(f(x)) = (5(5x+3) + 3(6x-α))/(6(5x+3) - α(6x-α)) = (43x + 15 - 3α)/((30-6α)x + 18 + α²)

2. Match coefficients. For this to equal x for every x, cross-multiply: (30-6α)x² + (18+α²)x = 43x + 15 - 3α. The x² term must vanish, so α = 5. The constants match only if 15 - 3α = 0, again α = 5, and the x terms need 18 + α² = 43, which α = 5 satisfies.
3. Quick check. With α = 5, the numerator is 43x and the denominator is 43, so f(f(x)) = x. Also f(x) can never equal 5/6, so f(f(x)) is defined for every x in the domain.
4. So. The required value is α = 5, option (b).

The idea to keep. A function (ax+b)/(cx+d) (with c ≠ 0) is its own inverse exactly when a + d = 0. Here that means 5 - α = 0, so α = 5.

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