Let for x R , f ( x ) = x + | x | 2 and g ( x ) = { x , x 0 x 2 , x 0 . Then area bounded by the curve y = ( f o g ) ( x ) and the lines y = 0 , 2 y - x = 15 is equal to _______ . [2023]

Question

Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]

Smart Achievers · Accepted Answer

(72)

Here, $f (x) = \frac{x + | x |}{2} =$ ${\begin{matrix} x, & x \geq 0 \\ 0, & x < 0 \end{matrix}$

$g (x) =$ ${\begin{matrix} x^{2}, & x \geq 0 \\ x, & x < 0 \end{matrix}$

$f o g (x) = f (g (x)) =$ ${\begin{matrix} g (x), & g (x) \geq 0 \\ 0, & g (x) < 0 \end{matrix}$ $=$ ${\begin{matrix} x^{2}, & x \geq 0 \\ 0, & x < 0 \end{matrix}$

$Required area = A_{1} + A_{2}$

$= \int_{0}^{3} (\frac{x + 15}{2} - x^{2}) d x + \frac{1}{2} \times \frac{15}{2} \times 15$

$= {[\frac{x^{2}}{4} + \frac{15 x}{2} - \frac{x^{3}}{3}]}_{0}^{3} + \frac{225}{4} = \frac{9}{4} + \frac{45}{2} - 9 + \frac{225}{4} = \frac{288}{4} = 72$

Solution

Q.
Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]

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Solution

Q. Let for x∈R, f(x)=x+|x|2 and g(x)={x,x<0x2,x≥0. Then area bounded by the curve y=(fog)(x) and the lines y=0, 2y-x=15 is equal to _______ . [2023]

Ans. (72) Here, f(x)=x+|x|2={x,x≥00,x<0 g(x)={x2,x≥0x,x<0 fog(x)=f(g(x))={g(x),g(x)≥00,g(x)<0={x2,x≥00,x<0 Required area =A1+A2 =∫03(x+152-x2)dx+12×152×15 =[x24+15x2-x33]03+2254=94+452-9+2254=2884=72

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Q.
Let for $x \in R,$ $f (x) = \frac{x + | x |}{2}$ and $g (x) =$ ${\begin{matrix} x, & x < 0 \\ x^{2}, & x \geq 0 \end{matrix}$ . Then area bounded by the curve $y = (f o g) (x)$ and the lines $y = 0, 2 y - x = 15$ is equal to _______ . [2023]