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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Area Under Curves

BITSAT 2014 · Mathematics · Area Under Curves

The area bounded by the x -axis, the curve y=f(x) and the lines...

The area bounded by the xx -axis, the curve y=f(x)y=f(x) and the lines x=1,\mathrm{x}=1,\mathrm{x}=\mathrm{b},, is equal to \mathrmb2+1−2\sqrt{\mathrmb^{2}+1}-\sqrt{2} for all b>1,b > 1, then f(x)f(x) is

Options

  1. Ax−1\sqrt{x-1}
  2. Bx+1\sqrt{x+1}
  3. Cx2+1\sqrt{x^{2}+1}
  4. Dx1+x2\frac{x}{\sqrt{1+x^{2}}}

Correct answer

D. x1+x2\frac{x}{\sqrt{1+x^{2}}}

Step-by-step solution

Given ∫1bf(x)dx=\mathrmb2+1−2\int_{1}^{\mathrm{b}} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\sqrt{\mathrmb^{2}+1}-\sqrt{2} Differentiate with respect to b\mathrm{b} f(b)=bb2+1⇒\frac{b}{\sqrt{b^{2}+1}} \Rightarrow f(x)=xx2+1\frac{x}{\sqrt{x^{2}+1}}
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