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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Match the Statements / Expressions in Column $I$ with the Statements / Expressions in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS.$
| Column $I$ | Column $II$ |
| $(A)$ $ \mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ are concurrent, if | $(p)$ $\mathrm{k}=-9$ |
| $(B)$ One of $\mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ is parallel to at least one of the other two, if | $(q)$ $\mathrm{k}=-\frac{6}{5}$ |
| $(C)$ $\mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ form a triangle, if | $(r)$ $\mathrm{k}=\frac{5}{6}$ |
| $(D)$ $ \mathrm{L}_1, \mathrm{~L}_2, \mathrm{~L}_3$ do not form a triangle, if | $(s)$ $\mathrm{k}=5$ |
| Column $I$ | Column $II$ |
| $(A)$ Circle | $(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$ |
| $(B)$ Parabola | $(q)$ Points $z$ in the complex plane satisfying $|z+2|-|z-2|= \pm 3$ |
| $(C)$ Ellipse | $(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$ |
| $(D)$ Hyperbola | $(s)$ The eccentricity of the conic lies in the interval $1 \leq x<\infty$ |
| $(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=|z|^2+1$ |