1
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$\int \sqrt{x^2-4 x+2} d x=$$

A
$$ \frac{1}{2}(x-2) \sqrt{x^2-4 x+2}+\log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C $$
B
$$ (x-2) \sqrt{x^2-4 x+2}+\frac{1}{2} \log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C $$
C
$$ \frac{1}{2}(x-2) \sqrt{x^2-4 x+2}-\sin ^{-1} \frac{x-2}{\sqrt{2}}+C $$
D
$$ \frac{1}{2}(x-2) \sqrt{x^2-4 x+2}-\log \left|(x-2)+\sqrt{x^2-4 x+2}\right|+C $$
2
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

$$4\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \text { is equal to }$$

A
$$ \frac{1}{8} $$
B
$$ \frac{1}{4} $$
C
$$ \frac{1}{2} $$
D
1
3
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

The minimum value of $$Z=150 x+200 y$$ for the given constraints

$$\begin{aligned} & 3 x+5 y \geq 30 \\ & x+y \geq 8 ; x \geq 0, y \geq 0 \text { is } \end{aligned}$$

A
0
B
1600
C
1350
D
1200
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

The vector equation of two lines are

$$\begin{aligned} & \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\ & \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k} \end{aligned}$$

Then the shortest distance between them is

A
$$\frac{4}{29}$$
B
$$\frac{4}{\sqrt{29}}$$
C
$$\frac{8}{29}$$
D
$$\frac{8}{\sqrt{29}}$$
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