- A${\tan ^{ - 1}}({x^2} + 1)$
- B${\tan ^{ - 1}}({x^2} + x)$
- C${\tan ^{ - 1}}(x + 1)$
- ✓${\tan ^{ - 1}}({x^2} + x + 1)$
$ = {\tan ^{ - 1}}\,\left[ {\frac{{x + \frac{1}{{x + 1}}}}{{1 - \frac{x}{{x + 1}}}}} \right] = {\tan ^{ - 1}}\,({x^2} + x + 1)$.
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| X: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| P(X): | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
Find the events E = {X : X is a prime number}, F{X : X < 4}, the probability $\text{P}(\text{E}\cup\text{F})$ is:
If $\text{A}=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\cot^{-1}(\pi\text{x})\end{bmatrix}$ and $\text{B}=\frac{1}{\pi}\begin{bmatrix}-\cos^{-1}(\text{x}\pi)&\tan^{-1}\Big(\frac{\text{x}}{\pi}\Big)\\\sin^{-1}\Big(\frac{\text{x}}{\pi}\Big)&\tan^{-1}(\pi\text{x})\end{bmatrix}$ then A - B is:
$\text{I}$
$0$
$2\text{I}$
$\frac{1}{2}\text{I}$