Q17 / 50 · Manipal (MET) 2022 · Mathematics
Let $f(x)=\sqrt{1+x^2}$, then
- A$f(x y)=f(x) \cdot f(y)$
- B$f(x y) \geq f(x) \cdot f(y)$
- Correct answerC$f(x, y) \leq f(x) \cdot f(y)$
- D$f(x y)=f(x)-f(y)$
Solution
Given that, $f(x)=\sqrt{1+x^2}$
$\begin{array}{ll}
\therefore & f(y)=\sqrt{1+y^2} \\
\text { and } & f(x y)=\sqrt{1+x^2 y^2}
\end{array}$
Now, $\quad \| f(x y)=f(x) f(y)$ $\begin{aligned} \Rightarrow \quad \sqrt{1+x^2 y^2} & =\sqrt{1+x^2} \sqrt{1+y^2} \\ & =\sqrt{\left(1+x^2\right)\left(1+y^2\right)} \end{aligned}$
On squaring both sides, we get $\begin{aligned} & 1+x^2 y^2=\left(1+x^2\right)\left(1+y^2\right) \\ \Rightarrow \quad & 1+x^2 y^2=1+y^2+x^2+x^2 y^2 \end{aligned}$ $\begin{aligned} & \therefore \quad 1+x^2 y^2 \leq 1+x^2+y^2+x^2 y^2 \\ & \text { Hence, } \quad f(x y) \leq f(x) \cdot f(y)\end{aligned}$
Now, $\quad \| f(x y)=f(x) f(y)$ $\begin{aligned} \Rightarrow \quad \sqrt{1+x^2 y^2} & =\sqrt{1+x^2} \sqrt{1+y^2} \\ & =\sqrt{\left(1+x^2\right)\left(1+y^2\right)} \end{aligned}$
On squaring both sides, we get $\begin{aligned} & 1+x^2 y^2=\left(1+x^2\right)\left(1+y^2\right) \\ \Rightarrow \quad & 1+x^2 y^2=1+y^2+x^2+x^2 y^2 \end{aligned}$ $\begin{aligned} & \therefore \quad 1+x^2 y^2 \leq 1+x^2+y^2+x^2 y^2 \\ & \text { Hence, } \quad f(x y) \leq f(x) \cdot f(y)\end{aligned}$
Hence, the correct answer is option: 'C'.