JEE Main 2023 (Online) 1st February Morning Shift | Functions Question 30 | Mathematics

Let $$f(x) = \left| {\matrix{ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

$${\alpha ^2} - {\beta ^2} = 4\sqrt 3 $$

$${\beta ^2} - 2\sqrt \alpha = {{19} \over 4}$$

$${\beta ^2} + 2\sqrt \alpha = {{19} \over 4}$$

$${\alpha ^2} + {\beta ^2} = {9 \over 2}$$

JEE Main 2023 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function

defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.

Then range of $f$ is

$ \left(-\infty,-\frac{21}{4}\right] \cup[1, \infty) $

$\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty) $

$\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty) $

$\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$

JEE Main 2023 (Online) 31st January Evening Shift

MCQ (Single Correct Answer)

-1

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

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

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

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

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