Let f ( x ) = a x 2 + 2 a x + 3 4 x 2 + 4 x - 3 , x - 3 2 , 1 2 b , x = - 3 2 , 1 2 be continuous at x = - 3 2 . If f o f ( x ) = 7 5 , then x is equal to: [2026]

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Solution

Q.
Let $f (x)$ $=$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{4 x^{2} + 4 x - 3}, & x \neq - \frac{3}{2}, \frac{1}{2} \\ b, & x = - \frac{3}{2}, \frac{1}{2} \end{matrix}$

be continuous at $x = - \frac{3}{2}$ . If $f o f (x) = \frac{7}{5}$ , then $x$ is equal to: [2026]

1 1.4

2 0

3 1

4 2

Ans.
(3)

$f (x) =$ ${\begin{matrix} \frac{a x^{2} + 2 a x + 3}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$

$For continuity at x = \frac{- 3}{2}$

$LHL = RHL$

$\Rightarrow \lim_{x \to \frac{- 3}{2}} \frac{(a x^{2} + 2 a x + 3)}{(2 x - 1) (2 x + 3)}$

$At x = \frac{- 3}{2} \Rightarrow Numerator = 0$

$a {(\frac{- 3}{2})}^{2} + 2 a (\frac{- 3}{2}) + 3 = 0$

$\frac{9}{4} a - 3 a + 3 = 0$

$\frac{3 a}{4} = 3 \Rightarrow a = 4$

$∴$ $f (x) =$ ${\begin{matrix} \frac{4 x^{2} + 8 x + 3}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$

$f (x) =$ ${\begin{matrix} \frac{(2 x + 1) (2 x + 3)}{(2 x - 1) (2 x + 3)}; & x \neq \frac{- 3}{2}, \frac{1}{2} \\ b; & x = \frac{- 3}{2}, \frac{1}{2} \end{matrix}$
$f o f (x)$ $= f (\frac{2 x + 1}{2 x - 1})$ $= \frac{2 (\frac{2 x + 1}{2 x - 1}) + 1}{2 (\frac{2 x + 1}{2 x - 1}) - 1}$

$= \frac{\frac{4 x + 2}{2 x - 1} + 1}{\frac{4 x + 2}{2 x - 1} - 1}$ $= \frac{\frac{6 x + 1}{2 x - 1}}{\frac{2 x + 3}{2 x - 1}}$

$= \frac{6 x + 1}{2 x + 3}$ $= \frac{7}{5}$

$\Rightarrow 5 (6 x + 1) = 7 (2 x + 3)$

$30 x + 5 = 14 x + 21$

$16 x = 16$

$\Rightarrow x = 1$

$Ans. x = 1$

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