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JEE Mains · Maths · STD 12 - 8. Application and integration

વિધેય \(f(x)=\left\{\begin{array}{ccc}{x} & {,} & {0 \leq x < \frac{1}{2}} \\ {\frac{1}{2}} & {,} & {x=\frac{1}{2}} \\ {1-x} & {,} & {\frac{1}{2} < x \leq 1}\end{array}\right.\) અને \(g(x)=\left(x-\frac{1}{2}\right)^{2}, x \in R \) આપેલ છે.  તો વક્રો \(y=f(x)\) અને  \(y=g(x)\) દ્વારા રેખાઓ \(2 \mathrm{x}=1\) અને  \(2 \mathrm{x}=\sqrt{3},\) વચ્ચે આવૃત પ્રદેશનું ક્ષેત્રફળ મેળવો.

  1. A \(\frac{1}{3}+\frac{\sqrt{3}}{4}\)
  2. B \(\frac{\sqrt{3}}{4}-\frac{1}{3}\)
  3. C \(\frac{1}{2}+\frac{\sqrt{3}}{4}\)
  4. D \(\frac{1}{2}-\frac{\sqrt{3}}{4}\)
Verified Solution

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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