MCQ
The equation $2{\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{{11\pi }}{6}$ has
- ✓No solution
- BOnly one solution
- CTwo solutions
- DThree solutions
==> ${\cos ^{ - 1}}x + ({\cos ^{ - 1}}x + {\sin ^{ - 1}}x) = \frac{{11\pi }}{6}$
==> ${\cos ^{ - 1}}x + \frac{\pi }{2} = \frac{{11\pi }}{6}$
$ \Rightarrow {\cos ^{ - 1}}x = 4\pi /3$
which is not possible as ${\cos ^{ - 1}}x \in [0,\,\pi ]$.
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Minimize $z=2 x+3 y$ the coordinates of the corner points of the bounded feasible region are $A\,(3,3), B\,(20,3),$ $\mathrm{C}\,(20,10), \mathrm{D}\,(18,12)$ and $\mathrm{E}\,(12,12) .$ The minimum value of $z$ is $\ldots \ldots$