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JEE Main 2005
Exam
JEE Main
Year
2005
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2005 · Mathematics · Area Under Curves

Let f(x) be a non-negative continuous function such that the area...

Let f(x)beanon−negativecontinuousfunctionsuchthattheareaboundedbythecurvef(x) be a non-negative continuous function such that the area bounded by the curvey=f(x), x−axisandtheordinates-axis and the ordinatesx=\frac{\pi}{4}andandx=\beta>\frac{\pi}{4}is(βsin⁡β+π4cos⁡β+2β)is \left(\beta \sin \beta+\frac{\pi}{4} \cos \beta+\sqrt{2} \beta\right). Then f(π2)f\left(\frac{\pi}{2}\right) is

Options

  1. A(π4+2−1)\left(\frac{\pi}{4}+\sqrt{2}-1\right)
  2. B(π4−2+1)\left(\frac{\pi}{4}-\sqrt{2}+1\right)
  3. C(1−π4−2)\left(1-\frac{\pi}{4}-\sqrt{2}\right)
  4. D(1−π4+2)\left(1-\frac{\pi}{4}+\sqrt{2}\right)

Correct answer

D. (1−π4+2)\left(1-\frac{\pi}{4}+\sqrt{2}\right)

Step-by-step solution

Given that ∫π/4βf(x)dx=βsin⁡β+π4cos⁡β+2β\int_{\pi / 4}^\beta f(x) d x=\beta \sin \beta+\frac{\pi}{4} \cos \beta+2 \beta Differentiating w. r. t β\beta f(β)=βcos⁡β+sin⁡β−π4sin⁡β+2f(\beta)=\beta \cos \beta+\sin \beta-\frac{\pi}{4} \sin \beta+\sqrt{2} f(π2)=(1−π4)sin⁡π2+2=1−π2+2f\left(\frac{\pi}{2}\right)=\left(1-\frac{\pi}{4}\right) \sin \frac{\pi}{2}+\sqrt{2}=1-\frac{\pi}{2}+\sqrt{2}.
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