MCQ
Given : $f(x)=\left\{\begin{array}{ccc}{x} & {,} & {0 \leq x < \frac{1}{2}} \\ {\frac{1}{2}} & {,} & {x=\frac{1}{2}} \\ {1-x} & {,} & {\frac{1}{2} < x \leq 1}\end{array}\right.$

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

  • A
    $\frac{1}{3}+\frac{\sqrt{3}}{4}$
  • $\frac{\sqrt{3}}{4}-\frac{1}{3}$
  • C
    $\frac{1}{2}+\frac{\sqrt{3}}{4}$
  • D
    $\frac{1}{2}-\frac{\sqrt{3}}{4}$

Answer

Correct option: B.
$\frac{\sqrt{3}}{4}-\frac{1}{3}$
b
Required area $=$ Area of trepezium ABCD -

Area of parabola between $x=\frac{1}{2}$ and $ x=\frac{\sqrt{3}}{2}$

$A=\frac{1}{2}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)\left(\frac{1}{2}+1-\frac{\sqrt{3}}{2}\right)-\int_{1 / 2}^{\sqrt{3} / 2}\left(x-\frac{1}{2}\right)^{2} d x$$=\frac{\sqrt{3}}{4}-\frac{1}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free