QJEEM20Ma43a0: JEE Main 2020 Mathematics - Class 12 JEEM20Ma43a0 Solution: Given : f ( x ) = { x , 0 ≤ x < 1 2 1 2 , x = 1 2 1 − x , 1 2 < …

Question

Verified Solution • 2025–26 Session moderate

Given :

f (x) = {\begin{matrix} x, & 0 \leq x < \frac{1}{2} \\ \frac{1}{2}, & x = \frac{1}{2} \\ 1 - x, & \frac{1}{2} < x \leq 1 \end{matrix}

and

g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R

Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x =

\sqrt 3

, is :

Accepted Answer

Required area = Area of trepezium ABCD – Area of parabola between x =

{1 \over 2}

and x =

{{\sqrt 3 } \over 2}

= Area of trepezium ABCD -

\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{\left( {x - {1 \over 2}} \right)}^2}dx}

=

{1 \over 2}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)\left( {{1 \over 2} + 1 - {{\sqrt 3 } \over 2}} \right)

-

{1 \over 3}\left[ {{{\left( {x - {1 \over 2}} \right)}^3}} \right]_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}

=

{1 \over 2}\left( {{{\sqrt 3 - 1} \over 2}} \right)\left( {{{3 - \sqrt 3 } \over 2}} \right)

- {1 \over 3}\left[ {{{\left( {{{\sqrt 3 - 1} \over 2}} \right)}^3} - 0} \right]

=

{{\sqrt 3 } \over 4} - {1 \over 3}

Step-by-Step Verified Solution