Question
Function $\text{f}(\text{x})=\log_\text{a}\text{x}$ is increasing on R, if:
- 0 < a < 1
- a > 1
- a < 1
- a > 0
Solution:
$\text{f}(\text{x})=\log_\text{a}\text{x}=\frac{\log\text{x}}{\log\text{a}}$
$\text{f}'(\text{x})=\frac{1}{\text{x}\log\text{a}}$
Given: f(x) is increasing on R.
$\Rightarrow\text{f}'(\text{x})>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\frac{1}{\text{x}\log\text{a}}>0,\forall\ \text{x}\in\text{R}$
$\Rightarrow\text{a}>1$
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$\text{y} = \sin-1\text{x}$
$\text{y} = \text{cosec }\text{x}$
$\text{y} = \sec\text{x}$
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X)$ | $K^2$ | $2K$ | $K$ | $2K$ | $5K^2$ |
Then $\mathrm{P}(\mathrm{X}> 2)$ is equal to