MCQ
The function $f(x) = |x|$ at $x = 0$ is
- ✓Continuous but non-differentiable
- BDiscontinuous and differentiable
- CDiscontinuous and non-differentiable
- DContinuous and differentiable
Now for differentiability
$f(x) = \,|\,\,x\,\,|\,\, = \,\,|0|\,\, = 0$ and $f(0 + h) = f(h) = \,\,|h|$
$\therefore \,\,\mathop {\lim }\limits_{h \to 0 - } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 - } \,\frac{{|h|}}{h} = - 1$
and $\mathop {\lim }\limits_{h \to 0 + } \,\frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0 + } \,\frac{{|h|}}{h} = 1$.
Therefore it is continuous and non-differentiable.
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and the slope of the curve $y=y(x)$ is never zero, then the value of $10 y(\sqrt{2})$ is . . . .
$STATEMENT -1$ : $\mathrm{P}\left(\mathrm{H}_{\mathrm{i}} \mid \mathrm{E}\right)>\mathrm{P}\left(\mathrm{E} \mid \mathrm{H}_{\mathrm{i}}\right) \cdot \mathrm{P}\left(\mathrm{H}_{\mathrm{i}}\right)$ for $\mathrm{i}=1,2, \ldots, \mathrm{n}$ because
$STATEMENT$ $-2: \sum_{1=1}^{\mathrm{n}} \mathrm{P}\left(\mathrm{H}_{\mathrm{i}}\right)=1$