1
JEE Main 2023 (Online) 11th April Evening Shift
+4
-1

The domain of the function $$f(x)=\frac{1}{\sqrt{[x]^{2}-3[x]-10}}$$ is : ( where $$[\mathrm{x}]$$ denotes the greatest integer less than or equal to $$x$$ )

A
$$(-\infty,-2) \cup[6, \infty)$$
B
$$(-\infty,-3] \cup[6, \infty)$$
C
$$(-\infty,-2) \cup(5, \infty)$$
D
$$(-\infty,-3] \cup(5, \infty)$$
2
JEE Main 2023 (Online) 10th April Morning Shift
+4
-1

If $$f(x) = {{(\tan 1^\circ )x + {{\log }_e}(123)} \over {x{{\log }_e}(1234) - (\tan 1^\circ )}},x > 0$$, then the least value of $$f(f(x)) + f\left( {f\left( {{4 \over x}} \right)} \right)$$ is :

A
2
B
4
C
0
D
8
3
JEE Main 2023 (Online) 6th April Evening Shift
+4
-1

Let the sets A and B denote the domain and range respectively of the function $$f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$$, where $$\lceil x\rceil$$ denotes the smallest integer greater than or equal to $$x$$. Then among the statements

(S1) : $$A \cap B=(1, \infty)-\mathbb{N}$$ and

(S2) : $$A \cup B=(1, \infty)$$

A
only $$(\mathrm{S} 2)$$ is true
B
only (S1) is true
C
neither (S1) nor (S2) is true
D
both (S1) and (S2) are true
4
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1

Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to

A
$$\frac{9}{4}$$
B
$$\frac{7}{4}$$
C
$$\frac{7}{3}$$
D
$$\frac{9}{2}$$
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