Let for $\mathrm{x} \in \mathrm{R}$ $$f(x)=\frac{x+|x|}{2} \text { and } g(x)=\begin{cases}x, & x<0 \\ x^{2} & x \geq 0\end{cases}$$ Then area bounded by the curve $\mathrm{y}=(\mathrm{fog})(\mathrm{x})$ and the lines $\mathrm{y}=0,2 \mathrm{y}-\mathrm{x}=15$ is equal to

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Area Under Curves

Let for

x \in R

f (x) = \frac{x + ∣ x ∣}{2} and g (x) = {x, x^{2} x < 0 x \geq 0

Then area bounded by the curve

y = (fog) (x)

and the lines

y = 0, 2 y - x = 15

is equal to

Answer: 72

Composing the functions,

f (x) = max (x, 0)

while

g (x) = x^{2}

for

x \geq 0

and

g (x) = x

for

x < 0

, so

(f \circ g) (x) = x^{2}

for

x \geq 0

and

0

for

x < 0

. The line

2 y - x = 15

, i.e.

y = \frac{x + 15}{2}

, meets

y = x^{2}

at the point

(3, 9)

and meets the

x

-axis at

(- 15, 0)

. The enclosed region is the triangle from

x = - 15

0

of area

\frac{1}{2} \cdot 15 \cdot \frac{15}{2} = \frac{225}{4}

together with the part from

x = 0

3

of area

\int_{0}^{3} (\frac{x + 15}{2} - x^{2}) d x = \frac{63}{4}

. Their sum is

\frac{288}{4} = 72

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Integer answer