1
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The volume of the tetrahedron whose coterminous edges are represented by

$$ \bar{a}=-12 \hat{i}+p \hat{k}, \bar{b}=3 \hat{j},-\hat{k}, \bar{c}=2 \hat{i}+\hat{j}-15 \hat{k} $$

570 cu. units, then $\mathrm{p}=$

A
7
B
-12
C
-482
D
482
2
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

With usual notations, the perimeter of a triangle ABC is 6 times the arithmetic mean of sine of its angles. If $\mathrm{a}=1$, then $\angle \mathrm{A}=$

A
$\frac{\pi}{6}$
B
$\frac{\pi}{3}$
C
$\frac{\pi}{2}$
D
$\frac{2 \pi}{3}$
3
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The angle between the lines whose direction cosines are $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}$ and $\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}$ is

A
$90^{\circ}$
B
$120^{\circ}$
C
$45^{\circ}$
D
$30^{\circ}$
4
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\sin \left(\frac{\pi}{4} \cot \theta\right)=\cos \left(\frac{\pi}{4} \tan \theta\right)$, then the general solution of $\theta$ is

A
$n \pi+\frac{\pi}{4}, n \in \mathbb{Z}$
B
$\quad n \pi+(-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$
C
$2 \mathrm{n} \pi \pm \frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
D
$\quad 2 \mathrm{n} \pi \pm 3 \frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
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