Let $f: \mathbb{R}-\left\{\frac{-1}{2}\right\} \rightarrow \mathbb{R}$ and $g: \mathbb{R}-\left\{\frac{-5}{2}\right\} \rightarrow \mathbb{R}$ be defined as $f(x)=\frac{2 x+3}{2 x+1}$ and $g(x)=\frac{|x|+1}{2 x+5}$. Then the domain of the function fog is :

(A) $\mathbb{R}-\left\{-\frac{5}{2}\right\}$

(C) $\mathbb{R}-\left\{-\frac{7}{4}\right\}$

(D) $\mathbb{R}-\left\{-\frac{5}{2},-\frac{7}{4}\right\}$

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Sets Relations and Functions

Let

f : R - {\frac{- 1}{2}} \to R

and

g : R - {\frac{- 5}{2}} \to R

be defined as

f (x) = \frac{2 x + 3}{2 x + 1}

and

g (x) = \frac{∣ x ∣ + 1}{2 x + 5}

. Then the domain of the function fog is :

Answer: (A)

f (x) = \frac{2 x + 3}{2 x + 1}

is defined for

x \neq = - \frac{1}{2}

;

g (x) = \frac{∣ x ∣ + 1}{2 x + 5}

is defined for

x \neq = - \frac{5}{2}

.

For

f (g (x))

to be defined,

g (x)

must also avoid

- \frac{1}{2}

\frac{∣ x ∣ + 1}{2 x + 5} = - \frac{1}{2} \Rightarrow 2 (∣ x ∣ + 1) = - (2 x + 5) \Rightarrow 2∣ x ∣ + 2 x + 7 = 0.

For

x \geq 0

2 x + 2 x + 7 = 4 x + 7 = 0 \Rightarrow x = - \frac{7}{4}

, which contradicts

x \geq 0

— no solution.

For

x < 0

- 2 x + 2 x + 7 = 7 \neq = 0

— no solution either.

So

g (x) = - \frac{1}{2}

has no real solution, and the domain of

f \circ g

is limited only by the domain of

g

Domain (f \circ g) = R - {- \frac{5}{2}} .

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option