Solved Examples on Sets Relations Functions

Illustration 2: Suppose, f(x) = (x+1)² for x ≥ -1. If g(x) is the function whose graph is a reflection of the graph of f(x) with respect to the line y = x, then what is the value of g (x)? (2002)

Solution: The question just requires us to find the inverse.

Let us assume y = f(x) = (x+1)² for x ≥ -1

Then ± √y = (x+1) for x ≥ -1

Hence, √y = x + 1 which gives y ≥ 0 and x+1 ≥ 0.

Hence, x = √y - 1 

So, f ^-1 (y) = √y - 1 

Hence, f ^-1 (x) = √x – 1, x ≥ 0.

Illustration 3: Let f(x) = x² and g(x) = sin x for all x ∈ R. Then, the set of all x satisfying (fogogof)(x) = (gogof)(x), where (fog)(x) = f(g(x)), is                                    (2011)

1. ±√nπ, n ∈ {0, 1, 2, ….}                 

2. ±√nπ, n ∈ {1, 2, …. }

3. π/2 + 2nπ, n ∈ {….. -2, -1, 0, 1, 2, ….}

4. 2nπ, n ∈ {….. -2, -1, 0, 1, 2, ….}

Solution: Let f(x) = x² and g(x) = sin x

Then (gof) (x) = sin x²

Hence, go (gof) (x) = sin (sin x²)

Therefore, (fogogof) (x) = (sin (sin x²))²

Again, (gof) (x) = sin x²

So, (gogof) (x) = sin (sin x²)

Given, (fogogof) (x) = (gogof) (x)

Also, (sin (sin x²))² = sin (sin x²)

Hence, we get, sin (sin x²).{sin (sin x²) – 1} = 0

This gives either sin (sin x²) = 0 or sin (sin x²) = 1

Hence, sin x² = 0 or sin x² = π /2

Therefore, x² = nπ (but this is not possible since -1 ≤ sin θ ≤ 1.

So, x = ± √ nπ.