MCQ
If $f(x) = |x - 1|$, then $\int_0^2 {f(x)dx} $ is
- ✓$1$
- B$0$
- C$2$
- D$-2$
$\therefore$ $\int_0^2 {f(x)dx = \int_0^2 {{\rm{ }}|x - 1|dx} } $
$= \int_0^1 {(1 - x)dx + \int_1^2 {(x - 1)dx} } $
$ = \left[ {x - \frac{{{x^2}}}{2}} \right]_0^1 + \left[ {\frac{{{x^2}}}{2} - x} \right]_1^2$
$ = \left( {1 - \frac{1}{2}} \right) + (2 - 2) - \left( {\frac{1}{2} - 1} \right) = 1$.
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$7 x+11 y+\alpha z=13$
$5 x+4 y+7 z=\beta$
$175 x+194 y+57 z=361$
has infinitely many solutions, then $\alpha+\beta+2$ is equal to
| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is: