1
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$\mathrm{A}=\{1,3,4,6,9\}$$ and $$\mathrm{B}=\{2,4,5,8,10\}$$. Let $$\mathrm{R}$$ be a relation defined on $$\mathrm{A} \times \mathrm{B}$$ such that $$\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.$$ and $$\left.b_{1} \leq a_{2}\right\}$$. Then the number of elements in the set R is :

A
180
B
26
C
52
D
160
2
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the $$1011^{\text {th }}$$ term from the end in the binominal expansion of $$\left(\frac{4 x}{5}-\frac{5}{2 x}\right)^{2022}$$ is 1024 times $$1011^{\text {th }}$$R term from the beginning, then $$|x|$$ is equal to

A
$$ \frac{5}{16} $$
B
8
C
12
D
15
3
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)$$, then $$\lambda, \frac{\lambda}{3}$$ are the roots of the equation :

A
$$4 x^{2}+24 x-27=0$$
B
$$4 x^{2}-24 x+27=0$$
C
$$4 x^{2}-24 x-27=0$$
D
$$4 x^{2}+24 x+27=0$$
4
JEE Main 2023 (Online) 11th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let the mean of 6 observations $$1,2,4,5, \mathrm{x}$$ and $$\mathrm{y}$$ be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :

A
$$\frac{10}{3}$$
B
$$\frac{8}{3}$$
C
$$\frac{7}{3}$$
D
3
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