Question

Let g(x) = $\{\begin{array}{cc}-x& x\le 1\\ x+1& x\ge 1\end{array}$ and f(x) = $\{\begin{array}{cc}1-x& x\le 0\\ x^2& x\ge 0\end{array}$

Consider the composition of f and g. i.e, (f $\circ$ g)(x) -f(g(x)). The number of discontinties is (f $\circ$ g)(x) present in the interval (-$\infty$,0) is.

• A. 0
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is A 0

g(x) = $\{\begin{array}{cc}-x& x\le 1\\ x+1& x\ge 1\end{array}$

f(x) = $\{\begin{array}{cc}1-x& x\le 0\\ x^2& x\ge 0\end{array}$

f $\circ$ g(x) = f(g(x)) = $\{\begin{array}{cc}1-g\left(x\right),& g\left(x\right)\le 0\\ \left(g\right(x){)}^2,& g\left(x\right)\ge 0\end{array}$

=$\{\begin{array}{cc}1-(-x)& -x\le 0,x\le 1\\ 1-(x+1),& x+1\le 0,x\ge 1\\ (-x{)}^2& -x>0,x\le 1\\ (x+1{)}^2,& x+1>0,x\ge 1\end{array}$

= $\{\begin{array}{cc}1+x,& 0\le x\le 1\\ x^2& x<0\\ (x+1{)}^2& x>1\end{array}$

f(g(x)) = $\{\begin{array}{cc}{x}^2& x<0\\ x+1,& 0\le x\le 1\\ (x+1{)}^2& x>1\end{array}$

f $\circ$ g(x) is continuous at every point in (-$\infty$, 0).

$\Rightarrow$ Number of points of discontinuities = 0

Overall we have two points of discontinuities at x = 0 and x = 1 but these do not belong to the given interval (-$\infty$, 0).