Q17 / 50 · Manipal (MET) 2023 · Mathematics
Let $f(x)=\frac{1}{\sqrt{1+x^2}}$, then
- A$f(x, y)=f(x) \cdot f(y)$
- Correct answerB$f(x, y) \geq f(x) \cdot f(y)$
- C$f(x, y) \leq f(x) \cdot f(y)$
- D$f(x, y)=f(x)-f(y)$
Solution
Given that
$f(x)=\frac{1}{\sqrt{1+x^2}}$
$\therefore \quad f(y)=\frac{1}{\sqrt{1+y^2}}$
and $f(x, y)=\frac{1}{\sqrt{1+x^2 y^2}}$
Now, $f(x, y)=f(x) \cdot f(y)$
$\frac{1}{\sqrt{1+x^2 y^2}}=\frac{1}{\sqrt{1+x^2}} \times \frac{1}{\sqrt{1+y^2}}$
$\frac{1}{\sqrt{1+x^2 y^2}}=\frac{1}{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}}$
On squaring both sides, we get
$\frac{1}{1+x^2 y^2}=\frac{1}{\left(1+x^2\right)\left(1+y^2\right)}$
$\frac{1}{1+x^2 y^2} \geq \frac{1}{\left(1+x^2\right)\left(1+y^2\right)}$
$\left[\because 1+x^2 y^2 \leq 1+x^2+y^2+x^2 y^2\right]$
Hence, $f(x, y) \geq f(x) \cdot f(y)$
Hence, the correct answer is option: 'B'.