MCQ
If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $'a'$ is equal to
- A$2$
- ✓$\frac {3}{2}$
- C$\frac {1}{2}$
- D$\frac {2}{3}$
$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$
$\, = e\left[ {\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}} - 1} \right)2x} \right]$
$\therefore 2a = 3 \Rightarrow a = 3/2$
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$\mathrm{x}_{\mathrm{i}}$ $\ \ 3\ \ 8\ \ 11\ \ 10\ \ 5\ \ 4$
$\mathrm{f}_{\mathrm{i}}$ $\ \ 5 \ \ 2 \ \ 3 \ \ 2 \ \ 4 \ \ 4$
Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) The mean of the above data is | $(1) 2.5$ |
| ($Q$) The median of the above data is | $(2) 5$ |
| ($R$) The mean deviation about the mean of the above data is | $(3) 6$ |
| ($S$) The mean deviation about the median of the above data is | $(4) 2.7$ |
| $(5) 2.4$ |
The correct option is :