JEE Mains · Maths · STD 12 - 8. Application and integration
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 \(\frac{1}{3}+\frac{\sqrt{3}}{4}\)
- B \(\frac{\sqrt{3}}{4}-\frac{1}{3}\)
- C \(\frac{1}{2}+\frac{\sqrt{3}}{4}\)
- D \(\frac{1}{2}-\frac{\sqrt{3}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\sqrt{3}}{4}-\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Required area \(=\) Area of trepezium ABCD - Area of parabola between \(x=\frac{1}{2}\) and \( x=\frac{\sqrt{3}}{2}\)…
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