Question ID: #789
Let $f(x) = \begin{cases} \frac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x\ne-\frac{3}{2}, \frac{1}{2} \\ b, & x=-\frac{3}{2} \end{cases}$ be continuous at $x=-\frac{3}{2}$. If $f(f(x)) = \frac{7}{5}$, then $x$ is equal to:
- (1) 2
- (2) 1
- (3) 0
- (4) 1.4
Solution:
Factorize the denominator:
$4x^2 + 4x – 3 = 4x^2 + 6x – 2x – 3 = 2x(2x+3) – 1(2x+3) = (2x-1)(2x+3)$.
For $f(x)$ to be continuous at $x = -3/2$, the limit must exist.
Since the denominator tends to 0 at $x = -3/2$, the numerator must also be 0 for the limit to be finite.
$$ \lim_{x \to -3/2} (ax^2 + 2ax + 3) = 0 $$
$$ a\left(-\frac{3}{2}\right)^2 + 2a\left(-\frac{3}{2}\right) + 3 = 0 $$
$$ a\left(\frac{9}{4}\right) – 3a + 3 = 0 $$
Multiply by 4:
$$ 9a – 12a + 12 = 0 $$
$$ -3a = -12 \Rightarrow a = 4 $$
Now substitute $a=4$ into $f(x)$ and simplify:
$$ f(x) = \frac{4x^2 + 8x + 3}{(2x-1)(2x+3)} $$
Factorize numerator: $4x^2 + 8x + 3 = 4x^2 + 6x + 2x + 3 = 2x(2x+3) + 1(2x+3) = (2x+1)(2x+3)$.
$$ f(x) = \frac{(2x+1)(2x+3)}{(2x-1)(2x+3)} = \frac{2x+1}{2x-1} \quad \text{for } x \ne -3/2 $$
We are given $f(f(x)) = \frac{7}{5}$.
First, find $f(f(x))$:
$$ f(f(x)) = \frac{2f(x)+1}{2f(x)-1} = \frac{2\left(\frac{2x+1}{2x-1}\right)+1}{2\left(\frac{2x+1}{2x-1}\right)-1} $$
Multiply numerator and denominator by $(2x-1)$:
$$ = \frac{2(2x+1) + (2x-1)}{2(2x+1) – (2x-1)} $$
$$ = \frac{4x+2+2x-1}{4x+2-2x+1} = \frac{6x+1}{2x+3} $$
Set this equal to $\frac{7}{5}$:
$$ \frac{6x+1}{2x+3} = \frac{7}{5} $$
$$ 5(6x+1) = 7(2x+3) $$
$$ 30x + 5 = 14x + 21 $$
$$ 16x = 16 $$
$$ x = 1 $$
Ans. (2)
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