Question

Let $f\left(x\right)$ is a real valued function defined as :
$f\left(x\right)=\{\begin{array}{cc}2x+3,& -3\le x<-2\\ x+1,& -2\le x<0\\ x+2,& 0\le x\le 1\end{array}$, then

• A. $f\left(x\right)=0$ has one real solution in $(-1,1)$
• B. $f\left(x\right)=\frac{5}{2}$ has one real solution in $[0,1]$
• C. $f\left(x\right)=-2$ has one real solution in $(-3,0)$
• D. $f\left(x\right)=1$ has one real solution in $(-2,1)$

Answer 1

Answer: (B), (C)

$f\left(x\right)=-2$ has one real solution in $(-3,0)$

Clearly, $f\left(x\right)$ is continuous in $[-3,1]-\left\{0\right\}$
So, $I.V.T.$ is not applicable in $(-1,1)$ and in $(-2,1)$
Now, for $x\in (-1,1)$: $f\left(x\right)=0$ gives $x=-1,-2$
$\Rightarrow$ No solution.

For $x\in [0,1]$ $\Rightarrow f\left(x\right)\in [2,3]$and is continuous.
From $I.V.T.,$ there exist atleast one value of $c\in [a,b]$ for $f\left(x\right)=f\left(c\right)$, if $f\left(c\right)\in \left[f\right(a),f(b\left)\right]$
and $f\left(c\right)=\frac{5}{2}\in [2,3]\Rightarrow$ atleast one solution.

For $(-3,0)$: $f\left(x\right)=0$ gives $x=-\frac{3}{2},-1$
$\Rightarrow$ Two solutions.

when $x\in [-3,-2),f\left(x\right)=-2$ gives $2x+3=-2,\Rightarrow x=-\frac{5}{2}$
So, one solution for $f\left(x\right)=-2$ in $[-3,-2)$

Now, when $x\in [-2,0),f\left(x\right)=-2\Rightarrow x+1=-2$
$\Rightarrow x=-3\notin [-2,0)$
So, no solution in $[-2,0)$
$\therefore f\left(x\right)=-2$ has one solution in $(-3,0)$