1
JEE Main 2023 (Online) 31st January Morning Shift
+4
-1
If the domain of the function $$f(x)=\frac{[x]}{1+x^{2}}$$, where $$[x]$$ is greatest integer $$\leq x$$, is $$[2,6)$$, then its range is
A
$$\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
B
$$\left(\frac{5}{37}, \frac{2}{5}\right]$$
C
$$\left(\frac{5}{26}, \frac{2}{5}\right]$$
D
$$\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
2
JEE Main 2023 (Online) 30th January Evening Shift
+4
-1
The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is :
A
$[2 \sqrt{2}, \sqrt{11}]$
B
$[\sqrt{5}, \sqrt{13}]$
C
$[\sqrt{2}, \sqrt{7}]$
D
$[\sqrt{5}, \sqrt{10}]$
3
JEE Main 2023 (Online) 29th January Evening Shift
+4
-1

Consider a function $$f:\mathbb{N}\to\mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$$ with $$f(1)=1$$. Then $$\frac{1}{f(2022)}+\frac{1}{f(2028)}$$ is equal to

A
8000
B
8400
C
8100
D
8200
4
JEE Main 2023 (Online) 29th January Morning Shift
+4
-1

The domain of $$f(x) = {{{{\log }_{(x + 1)}}(x - 2)} \over {{e^{2{{\log }_e}x}} - (2x + 3)}},x \in \mathbb{R}$$ is

A
$$( - 1,\infty ) - \{ 3\}$$
B
$$\mathbb{R} - \{ - 1,3)$$
C
$$(2,\infty ) - \{ 3\}$$
D
$$\mathbb{R} - \{ 3\}$$
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