1
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Let $$a, b, c$$ be the lengths of sides of triangle $$A B C$$ such that $$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$$. Then $$\frac{(\mathrm{A}(\triangle \mathrm{ABC}))^2}{\mathrm{k}^4}=$$

A
36
B
32
C
38
D
40
2
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the mean and S.D. of the data $$3,5,7, a, b$$ are 5 and 2 respectively, then $$a$$ and $$b$$ are the roots of the equation

A
$$x^2-10 x+18=0$$
B
$$2 x^2-20 x+19=0$$
C
$$x^2-10 x+19=0$$
D
$$x^2-20 x+18=0$$
3
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A man takes a step forward with probability 0.4 and backwards with probability 0.6 . The probability that at the end of eleven steps, he is one step away from the starting point is

A
$${ }^{11} \mathrm{C}_6(0.24)^6$$
B
$${ }^{11} \mathrm{C}_6(0.24)^5$$
C
$${ }^{11} \mathrm{C}_6(0.4)^6(0.6)^5$$
D
$${ }^{11} \mathrm{C}_6(0.4)^5(0.6)^6$$
4
MHT CET 2023 9th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}$$ and $$\mathrm{f}(1)=4$$, then $$\mathrm{f}(x)$$ is

A
$$\frac{x^2}{2}+\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}$$
B
$$\frac{x^2}{2}-\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}$$
C
$$\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}$$
D
$$\frac{x^2}{2}-\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}$$
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