BITSAT 2010
- Exam
- BITSAT
- Year
- 2010
- Subject
- Mathematics
- Chapter
- Trigonometric Equations
BITSAT 2010 · Mathematics · Trigonometric Equations
If (f(x)= x 1+x^2 ), then (fof of )(x) is
If (f(x)=\frac{x}{\sqrt{1+x^2}}), then (fof of is
Options
- A(\frac{3 x}{\sqrt{1+x^2}})
- B(\frac{\mathrm{x}}{\sqrt{1+3 \mathrm{x}^2}})
- C(\frac{3 x}{\sqrt{1-x^2}})
- DNone of these
Correct answer
B. (\frac{\mathrm{x}}{\sqrt{1+3 \mathrm{x}^2}})
Step-by-step solution
(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}})
((f o f)(x)=\frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\frac{x^2}{1+x^2}}}=\frac{x}{\sqrt{2 x^2+1}})
((fofof)(x)=\frac{\frac{x}{\sqrt{2 x^2+1}}}{\sqrt{1+\frac{x^2}{2 x^2+1}}}=\frac{x}{\sqrt{1+3 x^2}})