MCQMediumJEE 2024Composition & Inverse Functions

JEE Mathematics 2024 Question with Solution

Let f:R{1/2}Rf: \mathbb{R} - \{-1/2\} \to \mathbb{R} and g:R{5/2}Rg: \mathbb{R} - \{-5/2\} \to \mathbb{R} be defined as f(x)=2x+32x+1f(x) = \frac{2x + 3}{2x + 1} and g(x)=x+12x+5g(x) = \frac{|x| + 1}{2x + 5}. What is the domain of fgf \circ g?

  • A

    R{5/2}\mathbb{R} - \{-5/2\}

  • B

    R\mathbb{R}

  • C

    R{7/4}\mathbb{R} - \{-7/4\}

  • D

    R{5/2,7/4}\mathbb{R} - \{-5/2, -7/4\}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

  • f(x)=2x+32x+1f(x) = \frac{2x + 3}{2x + 1} with domain R{12}\mathbb{R} \setminus \left\{-\frac{1}{2}\right\}
  • g(x)=x+12x+5g(x) = \frac{|x| + 1}{2x + 5} with domain R{52}\mathbb{R} \setminus \left\{-\frac{5}{2}\right\}

Find: The domain of f(g(x))f(g(x)).

For the composite function f(g(x))f(g(x)) to be defined, two conditions are needed:

  1. g(x)g(x) itself must be defined.
  2. The value of g(x)g(x) must belong to the domain of ff, so g(x)12g(x) \ne -\frac{1}{2}.

From g(x)=x+12x+5g(x) = \frac{|x| + 1}{2x + 5}, we first require

2x+502x + 5 \ne 0

so that

x52.x \ne -\frac{5}{2}.

Now check whether g(x)=12g(x) = -\frac{1}{2} is possible:

x+12x+5=12\frac{|x| + 1}{2x + 5} = -\frac{1}{2}

Cross-multiplying,

2(x+1)=(2x+5)2(|x| + 1) = -(2x + 5)

so

2x+2=2x52|x| + 2 = -2x - 5

which gives

2x+2x=7.2|x| + 2x = -7.

Case Analysis

Consider cases for x|x|.

If x0x \ge 0, then x=x|x| = x. So

2x+2x=72x + 2x = -7 4x=74x = -7 x=74x = -\frac{7}{4}

But this contradicts x0x \ge 0. Hence no solution arises from this case.

If x<0x < 0, then x=x|x| = -x. So

2(x)+2x=72(-x) + 2x = -7 0=7,0 = -7,

which is impossible. Hence there is no solution in this case either.

Therefore, g(x)=12g(x) = -\frac{1}{2} is never attained for any valid real xx. So no extra restriction is added beyond the domain of gg.

Hence the domain of fgf \circ g is

R{52}.\mathbb{R} \setminus \left\{-\frac{5}{2}\right\}.

Therefore, the correct option is A.

Common mistakes

  • Students often exclude x=74x = -\frac{7}{4} by solving g(x)=12g(x) = -\frac{1}{2} incorrectly. The equation must be checked with the cases of x|x|; otherwise an invalid value may be accepted. Use proper case analysis before excluding any point.

  • Another mistake is to consider only the domain of ff or only the domain of gg. For a composite function, both conditions are necessary: g(x)g(x) must be defined and its output must lie in the domain of ff.

  • Some students cross-multiply without first noting that x52x \ne -\frac{5}{2}. Since g(x)g(x) itself is undefined there, that value must be excluded immediately before any further analysis.

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