Question

Two functions are defined as under
$f(x)=\{\begin{array}{ll}x+1,& \text{ x≤1 }\\ 2x+1,& \text{ 1<x≤2}\end{array}$ and $g(x)=\{\begin{array}{ll}{x}^2,& \text{ −1≤x<2}\\ x+2,& \text{ 2≤x≤3}\end{array}$
then $(fog)(x)$ equals

• A. $\{\begin{array}{ll}{x}^2+1,& \text{ ∣x∣≤1}\\ 2x^2+1,& \text{ 1<x≤2}\end{array}$
• B. $\{\begin{array}{ll}{x}^2+1,& \text{ ∣x∣≤1}\\ 2x^2+1,& \text{ 1≤x<2 }\end{array}$
• C. $\{\begin{array}{ll}{x}^2+1,& \text{ ∣x∣>1}\\ 2x^2+1,& \text{ ∣x∣≥2}\end{array}$
• D. None of these

Answer 1

Answer: (A)

$(fog)(x)=f(g(x))=\{\begin{array}{ll}g(x)+1,& \text{ g(x)≤1}\\ 2g(x)+1,& \text{ 1<g(x)≤2}\end{array}$
Now, $g(x)\le 1$
$(i)x^2\le 1$, $-1\le x<2$
$\Rightarrow |x|\le 1$ , $-1$, $\le x<2$
$\Rightarrow |x|\le 1$
$\left(ii\right)x+2\le 1,2\le x\le 3$
$\Rightarrow x\le -1,2\le x\le 3$
$\Rightarrow x=\varphi$
Now, consider $1<g\left(x\right)\le 2$
$\left(iii\right)1<x^2\le 2,-1\le x<2$
$\Rightarrow x\in [-\sqrt{2}$, $-1]\cup (1$, $\sqrt{2}]$, $-1\le x<2$
$\Rightarrow 1<x\le \sqrt{2}$
$\left(iv\right)1<x+2\le 2$, $2\le x\le 3$
$\Rightarrow -1<x\le 0$, $2\le x\le 3$
$\Rightarrow x=\varphi$
So $(fog)(x)=f(g(x))=\{\begin{array}{ll}{x}^2+1,& \text{ ∣x∣≤1}\\ 2x^2+1,& \text{ 1<x≤2}\end{array}$