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

A fair die is tossed twice in succession. If $$\mathrm{X}$$ denotes the number of fours in two tosses, then the probability distribution of $$\mathrm{X}$$ is given by

A
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{1}{36}$$ $$\frac{25}{36}$$ $$\frac{5}{18}$$
B
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{25}{36}$$ $$\frac{1}{36}$$ $$\frac{5}{18}$$
C
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{25}{36}$$ $$\frac{5}{18}$$ $$\frac{1}{36}$$
D
$$\mathrm{X}=x_i$$ 0 1 2
$$\mathrm{P_i}$$ $$\frac{5}{18}$$ $$\frac{1}{36}$$ $$\frac{25}{36}$$
2
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}(x)=\frac{3 x+4}{5 x-7}$$ and $$\mathrm{g}(x)=\frac{7 x+4}{5 x-3}$$, then $$\mathrm{f}(\mathrm{g}(x))=$$

A
$$\frac{x^3+1}{x^2+2}$$
B
$$41 x$$
C
$$\mathrm{g}(\mathrm{f}(x))$$
D
$$\frac{5 x-7}{41}$$
3
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
4
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
$$\frac{-14 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$$
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