1
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the function $$f$$ is given by $$f(x)=x^3-3(a-2) x^2+3 a x+7$$, for some $$\mathrm{a} \in \mathbb{R}$$, is increasing in $$(0,1]$$ and decreasing in $$[1,5)$$, then a root of the equation $$\frac{\mathrm{f}(x)-14}{(x-1)^2}=0(x \neq 1)$$ is

A
$$-$$7
B
6
C
7
D
5
2
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The unit vector perpendicular to each of the vectors $$\bar{a}+\bar{b}$$ and $$\bar{a}-\bar{b}$$, where $$\bar{a}=\hat{i}+\hat{j}+\hat{k}$$ and $$\overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$$ is

A
$$\frac{-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$$
B
$$\frac{14 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$$
C
$$\frac{14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$$
D
$$\frac{-14 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$$
3
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The logical statement $$(\sim(\sim \mathrm{p} \vee \mathrm{q}) \vee(\mathrm{p} \wedge \mathrm{r})) \wedge(\sim \mathrm{q} \wedge \mathrm{r})$$ is equivalent to

A
$$\sim p \vee r$$
B
$$(p \wedge \sim q) \vee r$$
C
$$(\mathrm{p} \wedge \mathrm{r}) \wedge \sim \mathrm{q}$$
D
$$(\sim p \wedge \sim q) \wedge r$$
4
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $$\bar{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \bar{b}=\hat{i}+\hat{j}$$ and $$\bar{c}$$ be a vector such that $$|\bar{c}-\bar{a}|=4,|(\bar{a} \times \bar{b}) \times \bar{c}|=3$$ and the angle between $$\overline{\mathrm{c}}$$ and $$\overline{\mathrm{a}} \times \overline{\mathrm{b}}$$ is $$\frac{\pi}{6}$$, then $$\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}$$ is equal to

A
$$-3$$
B
$$\frac{3}{2}$$
C
3
D
$$\frac{-3}{2}$$
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