JEE Mathematics 2023 Question with Solution

Given: $f(x) = \frac{x + |x|}{2}$ for $x \geq 0$ and $f(x) = \frac{x}{2}$ for $x < 0$; $g(x) = x^2$ for $x \geq 0$ and $g(x) = x$ for $x < 0$.

Find: The area bounded by $y = f \circ g(x)$, $y = 0$ and $2y - x = 15$.

From the composition shown in the solution,

Hence,

The line is

so

The curve $y = x^2$ intersects this line at $x = 3$ as indicated in the solution figure.

Therefore the required area is taken as

Now evaluate:

Substituting the limits,

Therefore, the required area is $72$.

The plotted region confirms that for $x < 0$, the graph of $y = f \circ g(x)$ lies on $y = 0$, while for $x \geq 0$ it becomes $y = x^2$. The line $y = \frac{x+15}{2}$ forms a triangle with the $x$-axis on the left and a curvilinear part with the parabola on the right. Adding these two parts gives the total area as $72$.