JEE MAIN Function Previous Year Questions (PYQs) – Page 3 of 16

Let $f\left( x \right) = \left\{ {\matrix{ { - 1} & { - 2 \le x < 0} \cr {{x^2} - 1,} & {0 \le x \le 2} \cr } } \right.$ and

$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$

Then, in the interval (–2, 2), g is :

Let $f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$, 0 < x < 1. Then :

Let f : R $\to$ R be defined as$f(x) = \left\{ {\matrix{ {{{\lambda \left| {{x^2} - 5x + 6} \right|} \over {\mu (5x - {x^2} - 6)}},} & {x < 2} \cr {{e^{{{\tan (x - 2)} \over {x - [x]}}}},} & {x > 2} \cr {\mu ,} & {x = 2} \cr } } \right.$ where [x] is the greatest integer is than or equal to x. If f is continuous at x = 2, then $\lambda$ + $\mu$ is equal to :

Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $f(x)=\dfrac{x}{1+x^{2}},\ x\in\mathbb{R}$. Then the range of $f$ is :

Let $f, g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :

$f(x)=|x-1| \text { and } g(x)= \begin{cases}\mathrm{e}^x, & x \geq 0 \\ x+1, & x \leq 0 .\end{cases}$

Then the function $f(g(x))$ is

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

Let $f\colon A\to B$ be a function defined as $f(x)=\dfrac{x-1}{x-2}$, where $A=\mathbb{R}-{2}$ and $B=\mathbb{R}-{1}$. Then $f$ is :

Let [t] denote the greatest integer less than or equal to t. Let f(x) = x $-$ [x], g(x) = 1 $-$ x + [x], and h(x) = min{f(x), g(x)}, x $\in$ [$-$2, 2]. Then h is :

The domain of the function $f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$ is :

If the range of the function $f(x)=\dfrac{5-x}{x^2-3x+2}$, $x\ne1,2$, is $(-\infty,\alpha]\cup[\beta,\infty)$, then $\alpha^2+\beta^2$ is equal to:f