MHTCET 23 Apr 2025 Shift 2 Paper

Concept:For $(f \circ g)(x)$, first evaluate $g(x)$ on each proper sub-interval, then apply $f$ to those outputs and take the union of the ranges.Explanation:The given $g(x)$ has overlapping rules on $0 \le x \le 1$, so rewrite it as a single-valued function:$$g(x)=\begin{cases}-x, & -2\le x\le 0\\ x, & 0\le x<1\\ -x, & 1\le x\le 3\end{cases}$$All outputs of $g(x)$ lie in $[-3,2]$, so on these values we use $f(t)=1+\frac{5t}{3}$.Case 1: $-2\le x\le 0$$g(x)=-x$, hence$$(f\circ g)(x)=1+\frac{5(-x)}{3}=1-\frac{5x}{3}$$At $x=-2$, value is $\frac{13}{3}$; at $x=0$, value is $1$.So range is $\left[1,\frac{13}{3}\right]$.Case 2: $0\le x<1$$g(x)=x$, hence$$(f\circ g)(x)=1+\frac{5x}{3}$$Range is $\left[1,\frac{8}{3}\right)$.Case 3: $1\le x\le 3$$g(x)=-x$, hence$$(f\circ g)(x)=1+\frac{5(-x)}{3}=1-\frac{5x}{3}$$At $x=1$, value is $-\frac{2}{3}$; at $x=3$, value is $-4$.Range is $\left[-4,-\frac{2}{3}\right]$.Combining the three ranges, the minimum value is $-4$ and the maximum value is $\frac{13}{3}$.Thus the required range interval is $\left[-4,\frac{13}{3}\right]$.Answer:Option C: $\left[-4,\frac{13}{3}\right]$