If \( f(x) = 3x + 6 \), \( g(x) = 4x + k \), and \( f \circ g(x) = g \circ f(x) \), then find \( k \):

Question

Statement 1 : The function \(f:\mathbb{R}\to\mathbb{R}\) defined by \[ f(x)=\frac{x}{1+|x|} \] is one–one.
Statement 2 : The function \(f:\mathbb{R}\to\mathbb{R}\) defined by \[ f(x)=\frac{x^2+4x-30}{x^2-8x+18} \] is many–one.
Which of the following is correct?

Answer 1

The functions \( f(x) = 3x + 6 \) and \( g(x) = 4x + k \) are given. It is stated that \( f \circ g(x) = g \circ f(x) \), which implies \( f(g(x)) = g(f(x)) \). The compositions are computed as follows:\[f(g(x)) = f(4x + k) = 3(4x + k) + 6 = 12x + 3k + 6.\]\[g(f(x)) = g(3x + 6) = 4(3x + 6) + k = 12x + 24 + k.\]Equating these two expressions yields:\[12x + 3k + 6 = 12x + 24 + k.\]After canceling the \( 12x \) terms, we get:\[3k + 6 = 24 + k.\]Solving for \( k \):\[3k - k = 24 - 6,\]\[2k = 18,\]\[k = 9.\]

If \( f(x) = 3x + 6 \), \( g(x) = 4x + k \), and \( f \circ g(x) = g \circ f(x) \), then find \( k \):

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The Correct Option is C

Solution and Explanation

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