- A$1$
- B$0$
- C$\infty $
- ✓Does not exist
$f'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - 1}}{{ - h}} = 0$
Hence, $f'(0)$ does not exist.
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$L _1: \overrightarrow{ r }=\lambda \hat{ i }, \lambda \in R ,$
$L _2: \overrightarrow{ r }=\hat{ k }+\mu \hat{ j }, \mu \in R \text { and }$
$L _3: \overrightarrow{ r }=\hat{ i }+\hat{ j }+ vk , v \in R$
are given. For which point(s) $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ so that $P$, $Q$ and $R$ are collinear?
$(1)$ $\hat{k}+\hat{j}$ $(2)$ $\hat{ k }$ $(3)$ $\hat{ k }+\frac{1}{2} \hat{ j }$ $(4)$ $\hat{k}-\frac{1}{2} \hat{j}$
$2 x+y-z=3$
$x-y-z=\alpha$
$3 x+3 y+\beta z=3$
has infinitely many solution, then $\alpha+\beta-\alpha \beta$ is equal to .... .