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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Functions

BITSAT 2023 · Paper 1 · Mathematics · Functions

If f(x)= x 1+x^2 , then (fof of) (x) is

If f(x)=x1+x2f(x)=\frac{x}{\sqrt{1+x^2}}, then (fof of) (x)(x) is

Options

  1. A3x1+x2\frac{3 x}{\sqrt{1+x^2}}
  2. Bx1+3x2\frac{x}{\sqrt{1+3 x^2}}
  3. C3x1−x2\frac{3 x}{\sqrt{1-x^2}}
  4. DNone of these

Correct answer

B. x1+3x2\frac{x}{\sqrt{1+3 x^2}}

Step-by-step solution

f(x)=x1+x2f(x)=\frac{x}{\sqrt{1+x^2}}
fof=x1+x21+x21+x2=x2x2+1=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}}
fofof =x2x2+11+x22x2+1=x1+3x2=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}}
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