Question

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

for continuous at $x = \frac{- 3}{2}$

$${LHL = RHL }{\Rightarrow \lim_{x \rightarrow \frac{- 3}{2}}\mspace{2mu}\frac{\left( ax^{2} + 2ax + 3 \right)}{(2x - 1)(2x + 3)} }$$at $x = \frac{- 3}{2} \Rightarrow$ Numerator $= 0$

$${a\left( \frac{- 3}{2} \right)^{2} + 2a\left( \frac{- 3}{2} \right) + 3 = 0 }{\frac{9}{4}a - 3a + 3 = 0 }{\frac{3a}{4} = 3 \Rightarrow a = 4 }{\therefore f(x) = \left\{ \begin{matrix} \frac{4x^{2} + 8x + 3}{(2x - 1)(2x + 3)} & ;\ x \neq \frac{- 3}{2},\frac{1}{2} \\ b & ;\ x = \frac{- 3}{2},\frac{1}{2} \end{matrix} \right.\ }{f(x) = \left\{ \begin{matrix} \frac{(2x + 1)(2x + 3)}{(2x - 1)(2x + 3)} & ;x \neq \frac{- 3}{2},\frac{1}{2} \\ b & ;x = \frac{- 3}{2},\frac{1}{2} \end{matrix} \right.\ }{f \circ f(x) = f\left( \frac{2x + 1}{2x - 1} \right) = \frac{2\left( \frac{2x + 1}{2x - 1} \right) + 1}{2\left( \frac{2x + 1}{2x - 1} \right) - 1} }{= \frac{6x + 1}{2x + 3} = \frac{7}{5} }$$So $x = 1$ Ans.