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}} $.
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$x+3 y+k^{2} z=0$
$3 x+y+3 z=0$
has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to