Given : $f(x)=\left\{\begin{array}{cc}x, & 0 \leq x<\frac{1}{2} \\ \frac{1}{2}, & x=\frac{1}{2} \\ 1-x, & \frac{1}{2}<x \leq 1\end{array}\right.$ and $g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R$. Then the area (in sq. units) of the region bounded by the curves, $y=f(x)$ and $y=g(x)$ between the lines $2 \mathrm{x}=1$ and $2 \mathrm{x}=\sqrt{3}$, is:

(C) $\frac{\sqrt{3}}{4}-\frac{1}{3}$

(A) $\frac{1}{3}+\frac{\sqrt{3}}{4}$

(B) $\frac{1}{2}+\frac{\sqrt{3}}{4}$

(D) $\frac{1}{2}-\frac{\sqrt{3}}{4}$

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Area Under Curves

Given :

f (x) = ⎩ ⎨ ⎧ x, \frac{1}{2}, 1 - x, 0 \leq x < \frac{1}{2} x = \frac{1}{2} \frac{1}{2} < x \leq 1

and

g (x) = (x - \frac{1}{2})^{2}, x \in R

. Then the area (in sq. units) of the region bounded by the curves,

y = f (x)

and

y = g (x)

between the lines

2 x = 1

and

2 x = 3

, is:

Answer: (C)

Between

x = \frac{1}{2}

and

x = \frac{3}{2}

the upper curve is

f (x) = 1 - x

and the lower is

g (x) = (x - \frac{1}{2})^{2}

, so the required region has area

\int_{1/2}^{3 /2} [(1 - x) - (x - \frac{1}{2})^{2}] d x

. Evaluating the straight-line part as a trapezium and subtracting

\int_{1/2}^{3 /2} (x - \frac{1}{2})^{2} d x = \frac{1}{3} (\frac{3 - 1}{2})^{3}

, the enclosed area simplifies to

\frac{3}{4} - \frac{1}{3}

JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Single correct option