QJEEM24Maaa3f: JEE Main 2024 Mathematics - Class 11 JEEM24Maaa3f Solution: Let f : R − { − 1 2 } → R and g : R − { − 5 2 } → R be defined…

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QJEEM24Maaa3f: JEE Main 2024 Mathematics - Class 11 JEEM24Maaa3f Solution: Let f : R − { − 1 2 } → R and g : R − { − 5 2 } → R be defined…

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 :

Multiple Choice Options

A R−{−74}
B R
C R−{−52,−74}
D R−{−52}

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Accepted Answer

$\begin{aligned} f (x) = \frac{2 x + 3}{2 x + 1}; x \neq - \frac{1}{2} \\ g (x) = \frac{| x | + 1}{2 x + 5}, x \neq - \frac{5}{2} \end{aligned}$

Domain of $f (g (x))$

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

$x \neq - \frac{5}{2}$ and $\frac{| x | + 1}{2 x + 5} \neq - \frac{1}{2}$

$x \in R - {- \frac{5}{2}}$ and $x \in R$

$∴$ Domain will be $R - {- \frac{5}{2}}$

QJEEM24Maaa3f: JEE Main 2024 Mathematics - Class 11 JEEM24Maaa3f Solution: Let f : R − { − 1 2 } → R and g : R − { − 5 2 } → R be defined…

Multiple Choice Options

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