MCQ
The area between the curve $y = 4 + 3x - {x^2}$ and $x -$ axis is
- ✓$125/6$
- B$125/3$
- C$125/2$
- DNone of these
we get $x = - 1,\,4$.
Curve does not intersect $x -$ axis between $x = - 1$ and $x = 4$.
$\therefore$ Area $ = \int_{ - 1}^4 {(4 + 3x - {x^2})dx = \frac{{125}}{6}} $.
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$\left( {1 + \alpha } \right)x + \beta y + z = 2$ ; $\alpha x + \left( {1 + \beta } \right)y + z = 3$ ; $\alpha x + \beta y + 2z = 2$ has a unique solution, is