1
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The equation $\mathrm{e}^{\sin x}-\mathrm{e}^{-\sin x}=4$ has ̱_________ solutions.

A
2
B
4
C
3
D
no
2
MHT CET 2024 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $\alpha, \beta$ be the roots of the equation $x^2-\mathrm{p} x+\mathrm{r}=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^2-q x+r=0$. Then the value of r is

A
$\frac{2}{9}(\mathrm{p}-\mathrm{q})(2 \mathrm{q}-\mathrm{p})$
B
$\frac{2}{9}(\mathrm{q}-\mathrm{p})(2 \mathrm{p}-\mathrm{q})$
C
$\frac{2}{9}(q-2 p)(2 q-p)$
D
$\frac{2}{9}(2 p-q)(2 q-p)$
3
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation $(\operatorname{cosp}-1) x^2+(\operatorname{cosp}) x+\sin p=0$ in the variable $x$, has real roots. Then p can take any value in the interval

A
$(0,2 \pi)$
B
$(-\pi, 0)$
C
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
D
$(0, \pi)$
4
MHT CET 2024 3rd May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation $(\operatorname{cosp}-1) x^2+(\cos p) x+\operatorname{sinp}=0$ in the variable $x$, has real roots. Then p can take any value in the interval

A
$(0,2 \pi)$
B
$(-\pi, 0)$
C
$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
D
$(0, \pi)$
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