MCQ
Let $\text{f(x)}=\sqrt{\text{x}^2+1}$ Then which of the following is correct?
- A$\text{f(xy)}=\text{f(x)}\text{f(y)}$
- B$\text{f(xy)}\geq\text{f(x)}\text{f(y)}$
- C$\text{f(xy)}\leq\text{f(x)}\text{f(y)}$
- DNone os these.
Solution:
Given, $\text{f(x)}=\sqrt{\text{x}^2+1}\ ...(\text{i})$
Replacing x by y in (i), we get
$\text{f(y)}=\sqrt{\text{y}^2+1}$
$\therefore\ \text{f(x)}\text{f(y)}=\sqrt{\text{x}^2+1}\sqrt{\text{y}^2+1}$
$=\sqrt{(\text{x}^2+1)(\text{y}^2+1)}$
$=\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
Also, replacing x by xy in (i), we get
$\text{f(xy)}=\sqrt{\text{x}^2\text{y}^2+1}$
Now,
$\text{x}^2\text{y}^2+1\leq\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1$
$\Rightarrow\sqrt{\text{x}^2\text{y}^2+1}\leq\sqrt{\text{x}^2\text{y}^2+\text{x}^2+\text{y}^2+1}$
$\Rightarrow\text{f}(\text{xy})\leq\text{f(x)}\text{f(y)}$
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The events A, B, C are mutually exclusive events such that $\text{P(A)}=\frac{(3\text{x}+1)}{3,\text{P(B)}}=\frac{(\text{x}-1)}{4\text{ and}\text{ P}}\text{(C)}=\frac{(1-2\text{x})}{4}.$The set of possible values of x are in the interval: