MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin (2 + x) - \sin (2 - x)}}{x} = $
- A$\sin 2$
- B$2\sin 2$
- ✓$2\cos 2$
- D$2$
$i.e.,$ $\mathop {\lim }\limits_{x \to 0} \frac{{\sin (2 + x) - \sin (2 - x)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2\cos 2.\sin x}}{x}$
$ = 2\cos 2.\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 2\cos 2$
You may also apply $L-$ Hospital rule.
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$ A=\{z: \operatorname{Im} z \geq 1\} $
$ B=\{z:|z-2-i|=3\} $
$ C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\} .$
$1.$ The number of elements in the set $\mathrm{A} \cap \mathrm{B} \cap \mathrm{C}$ is
$(A)$ $0$ $(B)$ $1$ $(C)$ $2$ $(D)$ $\infty$
$2.$ Let $z$ be any point in $A \cap B \cap C$. Then, $|z+1-i|^2+|z-5-i|^2$ lies between
$(A)$ $25$ and $29$ $(B)$ $30$ and $34$ $(C)$ $35$ and $39$ $(D)$ $40$ and $44$
$3.$ Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w-2-i|<3$. Then, $|z|-|w|+3$ lies between
$(A)$ $-6$ and $3$ $(B)$ $-3$ and $6$
$(C)$ $-6$ and $6$ $(D)$ $-3$ and $9$
Give the answer question $1,2$ and $3.$