JEE Main 2023 (Online) 31st January Evening Shift | Functions Question 35 | Mathematics

The absolute minimum value, of the function

$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,

where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :

$\frac{3}{4}$

$\frac{3}{2}$

$\frac{1}{4}$

$\frac{5}{4}$

JEE Main 2023 (Online) 31st January Morning Shift

MCQ (Single Correct Answer)

-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

$$\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$

$$\left(\frac{5}{37}, \frac{2}{5}\right]$$

$$\left(\frac{5}{26}, \frac{2}{5}\right]$$

$$\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$

JEE Main 2023 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

-1

The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is :

$[2 \sqrt{2}, \sqrt{11}]$

$[\sqrt{5}, \sqrt{13}]$

$[\sqrt{2}, \sqrt{7}]$

$[\sqrt{5}, \sqrt{10}]$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-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

8000

8400

8100

8200

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