Question

Let $f\left(x\right)=\frac{1}{x},g\left(x\right)=\frac{1}{4x^2-1}$ and $h\left(x\right)=\frac{5x}{x+2}$ be three functions and $k\left(x\right)=h\left(g\right(f\left(x\right)\left)\right).$ If domain and range of $k\left(x\right)$ are $R-\{a_1,a_2,a_3,\dots ,a_n\}$ and $R-A$ respectively, where $R$ is the set of real numbers, then

• A. $n+\sum _{i=1}^n{a}_i=5$
• B. $n+\sum _{i=1}^n{a}_i=10$
• C. Number of integers in set $A$ is $5$
• D. Number of integers in set $A$ is $7$

Answer 1

Answer: (A), (D)

The correct options are
A $n+\sum _{i=1}^n{a}_i=5$
D Number of integers in set $A$ is $7$

$k\left(x\right)=h\left(g\right(f\left(x\right)\left)\right)=\frac{5x^2}{8-x^2}$
Domain of $k\left(x\right)$ is $R-\{0,\pm 2,\pm 2\sqrt{2}\}$

Let $y=\frac{5x^2}{8-x^2}$
Then, $(y+5)x^2-8y=0$
As $x$ is real, so $D\ge 0$ and $y+5\ne 0$
$\Rightarrow 8y(y+5)\ge 0$ and $y\ne -5$
$\Rightarrow y\in (-\infty ,-5)\cup [0,\infty )$
But $y=0$ for $x=0$ and $y=5$ for $x=\pm 2$
which are not in the domain of $k\left(x\right)$.

$\therefore$ Range of $k\left(x\right)$ is $(-\infty ,-5)\cup (0,\infty )-\left\{5\right\}$ or $R-\left(\right[-5,0]\cup \{5\left\}\right)$
$\therefore A=[-5,0]\cup \left\{5\right\}$