1
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The unit vectors perpendicular to the plane determined by the points $\mathrm{A}(1,-1,2), \mathrm{B}(2,0,-1)$, $\mathrm{C}(0,2,1)$ is

A

$\pm\left(\frac{3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{11}}\right)$

B

$\pm\left(\frac{-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)$

C

$\pm\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$

D

$\pm\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right)$

2
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2 y-z+1=0$ and $3 x-y-4 z+3=0$ is

A
$\quad 4 x-3 y-2 z+1=0$
B
$3 x-2 y+2 z-3=0$
C
$8 x-5 y-11 \mathrm{z}+8=0$
D
$5 x-4 y+2 z-3=0$
3
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

A manufacturing company produces two items, A and B. Each toy should be processed by two machines, I and II. Machine I can be operated for maximum 10 hours 40 minutes. It takes 20 minutes for an item of A and 15 minutes for B. Machine II can be operated for a total time at 8 hours 20 minutes. It takes 5 minutes for an item A and 8 minutes for B . The profit per item of $A$ is $Rs 25$ and per item of $B$ is ₹ 18 . The formulation of an L.P.P. to maximize the profit (where $x$ is number of items A and $y$ is the number of item $B$ ) is

A

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 20 x+15 y \leq 640 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 5 x+8 y \geq 500 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x, y \geq 0 \end{aligned} $$

B

Maximize $z=25 x+18 y$

$$ \begin{aligned} \text { subject to } 20 x+15 y & \leq 640 \\ 5 x+8 y & \leq 500 \\ x, y & \geq 0 \end{aligned} $$

C

$$ \begin{array}{r} \text { Maximize } z=25 x+18 y \\ \text { subject to } 20 x+5 y \leq 8 \\ 5 x+8 y \leq 10 \\ x, y \geq 0 \end{array} $$

D

$$ \begin{aligned} & \text { Maximize } \mathrm{z}=25 x+18 y \\ & \text { subject to } 4 x+3 y \leq 128 \\ & \qquad \begin{array}{r} 5 x+8 y \geq 500 \\ x, y \geq 0 \end{array} \end{aligned} $$

4
MHT CET 2025 26th April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \begin{array}{r}If\,\,\,\, y=\tan ^{-1}\left(\frac{1}{1+x+x^2}\right)+\tan ^{-1}\left(\frac{1}{x^2+3 x+3}\right) +\tan ^{-1}\left(\frac{1}{x^2+5 x+7}\right) \end{array} $$

then the value of $y^{\prime}(0)$ is

A

$\frac{9}{10}$

B

$\frac{1}{10}$

C

$-\frac{9}{10}$

D

$-\frac{1}{10}$

MHT CET Papers
EXAM MAP