datavorous/entrance-exam-dataset huggingface dataset:
The domain of the function fx=x2-8x+12.ln2⁡x-3 is

Options:
A 3,∞
B 4,∞
C 6,∞∪2,4
D 6,∞∪4

The correct answer is:
6,∞∪4

Solution:
x2-8x+12ln2x-3â¥0 and x-3>0

âx2-8x+12â¥0âxâ(-â,2]âª[6,â)

Also, lnx-3=0âx=4

Hence, the solution is xâ4âª[6,â)

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\frac{1}{\sqrt{[x]^2-[x]-2}}$ is
Here $[x]$ denotes the greatest integer not exceeding the value of $[x]$

Options:
A $(-\infty,-2) \cup(1, \infty)$
B $(-\infty,-2) \cup(0, \infty)$
C $(-\infty,-2) \cup(2, \infty)$
D $(-\infty,-1) \cup(3, \infty)$

The correct answer is:
$(-\infty,-1) \cup(3, \infty)$

Solution:
We have,
$$
f(x)=\frac{1}{\sqrt{[x]^2-[x]-2}}
$$
$f(x)$ is defined when
$$
\begin{aligned}
& {[x]^2-[x]-2>0} \\
& ([x]-2)([x]+1)>0 \\
& {[x]>2 \text { and }[x] < -1} \\
& x \geq 3 \text { and } x < -1
\end{aligned}
$$
$\therefore$ Domain of $f(x)$ is $x \in(-\infty,-1) \cup(3, \infty)$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
The domain of the function $f(x)=$ $\sqrt{\frac{1}{|x-2|-(x-2)}}$ is :

Options:
A $\quad(-\infty, 2]$
B $(2, \infty)$
C $(-\infty, 2)$
D $[2, \infty)$

The correct answer is:
$(-\infty, 2)$

Solution:
$$
\begin{array}{l}
\quad|x-2|=\left\{\begin{array}{l}
x-2, x \geq 2 \\
2-x, x < 2
\end{array}\right. \\
\Rightarrow \quad|x-2|-(x-2)=\left\{\begin{array}{l}
0, x \geq 2 \\
4-2 x, x < 2
\end{array}\right. \\
\Rightarrow \quad \text { given expression is defined for }(-\infty, 2)
\end{array}
$$

Topic: Mathematics, Functions, VITEEE, VITEEE 2019
The domain of the function $f(x)=\sqrt{x}$ is

Options:
A $R-\{0\}$
B $R^{+}$
C $\mathrm{R}^{+} \cup\{0\}$
D R

The correct answer is:
$\mathrm{R}^{+} \cup\{0\}$

Solution:
(B)
For $\mathrm{f}(\mathrm{x})$ to be defined, the term under the square root should be greater than or equal to zero.
$x \geq 0$
So domain is $[0, \infty]$ i.e. $\mathrm{R}^{+} \cup\{0\}$.

Topic: Mathematics, Functions, MHT CET, MHT CET 2020 (16 Oct Shift 2)
The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$ is

Options:
A
$(0, \infty)$

B
$(-\infty, 0)$

C
$(-\infty, \infty)-\{0\}$

D
$(-\infty, \infty)$

The correct answer is:
$(-\infty, 0)$

Solution:
$\frac{1}{\sqrt{|x|-x}} \Rightarrow|x|-x>0 \Rightarrow|x|>x \Rightarrow x$ is negative
$x \in(-\infty, 0)$

Topic: Mathematics, Functions, JEE Main, JEE Main 2011
The domain of the function $f(x)=\frac{1}{\sqrt{x+|x|}}$ is

Options:
A $(-\infty, 0)$
B $(2,5)$
C $(0, \infty)$
D $(-\infty, \infty)$

The correct answer is:
$(0, \infty)$

Solution:
$$
f(x)=\frac{1}{\sqrt{x+|x|}}
$$
Here $x+|x| \geq 0$ and $\sqrt{x+|x|} \neq 0$
$$
\therefore \mathrm{x}+|\mathrm{x}|>0
$$
Now when $x>0, x+|x|=x+x \Rightarrow 2 x>0$.
When $\mathrm{x} < 0, \mathrm{x}+|\mathrm{x}|=\mathrm{x}-\mathrm{x}=0$
$\therefore \mathrm{x}>0$ is the required domain.

Topic: Mathematics, Functions, MHT CET, MHT CET 2021 (20 Sep Shift 1)
The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$ is

Options:
A R
B $(-\infty, 0)$
C $(0, \infty)$
D $(-\infty, 1)$

The correct answer is:
$(-\infty, 0)$

Solution:
Given, $f(x)=\frac{1}{\sqrt{|x|-x}}$
For $f(x)$ to be defined, $|x|-x>0 \Rightarrow x < |x|$ $\Rightarrow x \in(-\infty, 0)$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
The domain of the function $f(x)=\frac{1}{\sqrt{|x|-x}}$ is

Options:
A $(0, \infty)$
B $(-\infty, 0)$
C $\frac{(-\infty, \infty)}{\{0\}}$
D $(-\infty, \infty)$

The correct answer is:
$(-\infty, 0)$

Solution:
Given, $f(x)=\frac{1}{\sqrt{|x|-x}}$
$f(x)$ is defined if $|x|-x>0$
$|x|>x$
It is possible only $x \in(-\infty, 0)$
$\therefore$ The domain of $f(x)$ is $(-\infty, 0)$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 2)
The domain of the function $\mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$ is

Options:
A $[0, \infty)$
B $(-\infty, 0)$
C $[1, \infty)$
D $(-\infty, 0]$

The correct answer is:
$(-\infty, 0)$

Solution:
$\mathrm{f}(\mathrm{x})=\frac{1}{\sqrt{|\mathrm{x}|-\mathrm{x}}}$
$|x|-x \neq 0$
So $\mathrm{x} < 0$
$\mathrm{x}=(-\infty, 0)$
again $|x|-x>0$ $|x|>x$
it is possible only when $\mathrm{x}$ is negative.
$\mathrm{x}=(-\infty, 0)$
So from eq. (1) \& eq. (2)
domain $\mathrm{x}=(-\infty, 0)$

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 2)
The domain of the function $f(x)=\sqrt{x-x^2}+\sqrt{4+x}+\sqrt{4-x}$ is

Options:
A $[-4, \infty)$
B $[-4,4]$
C $[0,4]$
D $[0,1]$

The correct answer is:
$[0,1]$

Solution:
$f(x)=\sqrt{x-x^2}+\sqrt{4+x}+\sqrt{4-x}$
Clearly $f(x)$ is defined, if $4+x \geq 0 \Rightarrow x \geq-4$
$\begin{aligned}
& 4-x \geq 0 \Rightarrow x \leq 4 \\
& x(1-x) \geq 0 \quad \Rightarrow x \geq 0 \text { and } x \leq 1
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function \( f(x)=e^{\sin (x-[x])}+[x] \cos \left(\frac{\pi}{[x+1]}\right),( \) where \( [\cdot] \) represents greatest integer function ), is:

Options:
A \( R \)
B \( R-[-1,0] \)
C \( R-[0,1] \)
D \( R-[-1,0) \)

The correct answer is:
\( R-[-1,0) \)

Solution:
We have,

fx=esin (x-x)+xcosπx+1

For fx to be defined,

x+1≠0

⇒x+1≠0

⇒x≠–1

⇒x∉–1, 0

Hence, domain of fx is x∈R – [–1, 0).

Topic: Mathematics, Functions, JEE Main
The domain of the function \( f(x)=\sqrt{\cos x} \) is

Options:
A \( \left[0, \frac{\Pi}{2}\right] \)
B \( \left[0, \frac{\Pi}{2}\right] \cup\left[\frac{3 \Pi}{2}, 2 \Pi\right] \)
C \( \left[\frac{3 \Pi}{2}, 2 \Pi\right] \)
D \( \left[-\frac{\Pi}{2}, \frac{\Pi}{2}\right] \)

The correct answer is:
\( \left[0, \frac{\Pi}{2}\right] \cup\left[\frac{3 \Pi}{2}, 2 \Pi\right] \)

Solution:
We know that, square root is defined only for non negative number
Hence for \( f(x)=\sqrt{\cos x} \) to be defined
\[
\Rightarrow \cos \geq 0
\]
\( \Rightarrow x \) should lies in ist and \( 4 \) th quadrant including ends of the quadrant
\( \Rightarrow x \in\left[2 n \Pi+0,2 n \Pi+\frac{\Pi}{2}\right] \cup\left[2 n \Pi+\frac{3 \pi}{2}, 2 n \Pi+2 \Pi\right] \)
, where \( n \in Z \), for \( n=0 \)

Topic: Mathematics, Mathematical Reasoning, KCET, KCET 2014
The domain of the function given by $2^x+2^y=2$ is

Options:
A $0 < x \leq 1$
B $0 \leq x \leq 1$
C $-\infty < x \leq 0$
D $-\infty < x < 1$

The correct answer is:
$-\infty < x < 1$

Solution:
$\begin{aligned} & 2^x+2^y=2 \\ & \Rightarrow 2^y=2-2^x \\ & \Rightarrow y=\log _2\left(2-2^x\right) \text { is defined, if } 2-2^x>0 \\ & \Rightarrow 2^x < 2 \\ & \Rightarrow 2^{x-1} < 1 \\ & \Rightarrow x-1 < 0 \\ & \Rightarrow-\infty < x < 1\end{aligned}$

Topic: Mathematics, Functions, MHT CET, MHT CET 2023 (11 May Shift 1)
The domain of the function $\log _{10}\left(x^2-5 x+6\right)$ is

Options:
A $(-\infty, \infty)$
B $(-\infty, 2) \cup(3, \infty)$
C $(2,3)$
D None of these

The correct answer is:
$(-\infty, 2) \cup(3, \infty)$

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\log _{10}\left(\mathrm{x}^2-5 \mathrm{x}+6\right) \\ & \therefore \mathrm{x}^2-5 \mathrm{x}+6>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}-3)>0 \\ & \Rightarrow \mathrm{x}>3 \text { or } \mathrm{x} < 2 \\ & \therefore \mathrm{x} \in(-\infty, 2) \cup(3, \infty)\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function $\log _{10}\left(x^2-5 x+6\right)$ is

Options:
A $(-\infty, \infty)$
B $(-\infty, 2) \cup(3, \infty)$
C $(2,3)$
D None of these

The correct answer is:
$(-\infty, 2) \cup(3, \infty)$

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\log _{10}\left(\mathrm{x}^2-5 \mathrm{x}+6\right) \\ & \therefore \mathrm{x}^2-5 \mathrm{x}+6>0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}-3)>0 \\ & \Rightarrow \mathrm{x}>3 \text { or } \mathrm{x} < 2 \\ & \therefore \mathrm{x} \in(-\infty, 2) \cup(3, \infty)\end{aligned}$

Topic: Mathematics, Functions, MHT CET, MHT CET 2021 (24 Sep Shift 1)
The domain of the function $\sqrt{\log \left(x^2-6 x+6\right)}$ is

Options:
A $(-\infty, \infty)$
B $(-\infty, 3-\sqrt{3}) \cup(3+\sqrt{3}, \infty)$
C $(-\infty, 1] \cup[5, \infty)$
D $[0, \infty)$

The correct answer is:
$(-\infty, 1] \cup[5, \infty)$

Solution:
The function $f(x)=\sqrt{\log \left(x^2-6 x+6\right)}$ is defined when $\log \left(x^2-6 x+6\right) \geq 0$
$\Rightarrow x^2-6 x+6 \geq 1 \Rightarrow(x-5)(x-1) \geq 0$
This inequality holds if $x \leq 1$ or $x \geq 5$. Hence, the domain of the function will be $(-\infty, 1] \cup[5, \infty)$

Topic: Mathematics, Functions, JEE Main
The domain of the function
$\sqrt{x^2-5 x+6}+\sqrt{2 x+8-x^2}$ is

Options:
A $[2,3]$
B $[-2,4]$
C $[-2,2] \cup[3,4]$
D $[-2,1] \cup[2,4]$

The correct answer is:
$[-2,2] \cup[3,4]$

Solution:
$f(x)=\sqrt{(x-2)(x-3)}+\sqrt{-(x-4)(x+2)}$
The first part is real outside $(2,3)$ and the second is real in $[-2,4]$ so that the domain is $[-2,2] \cup[3,4]$.

Topic: Mathematics, Functions, VITEEE, VITEEE 2021
The domain of the function $y=f(x)$, where $x$ and $y$ are related by $2^x+2^y=2$ is

Options:
A $(-\infty, \infty)$
B $(-\infty, 1)$
C $(-1, \infty)$
D $(1, \infty)$

The correct answer is:
$(-\infty, 1)$

Solution:
Given that $2^x+2^y=2 \Rightarrow 2^y=2-2^x$
$$
\begin{aligned}
& \because 2^y>0 \Rightarrow 2-2^x>0 \\
& \Rightarrow 2^x < 2 \Rightarrow x < 1 \Rightarrow x \in(-\infty, 1)
\end{aligned}
$$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
The domain of the function $y=\sqrt{x-2}+\sqrt{1-x}$ is

Options:
A $x \leq 2$
B $x \geq 2$
C Null set
D The set of all numbers

The correct answer is:
Null set

Solution:
Given, function is
$$
y=\sqrt{x-2}+\sqrt{1-x}
$$
Since, $x-2 \geq 0$ and $1-x \geq 0$
$\begin{array}{ll}\Rightarrow & x \geq 2 \quad \text{...(i)}\\ \Rightarrow & x \leq 1 \quad \text{...(ii)} \\ \therefore \text { From Eqs. (i) and (ii), } x=\phi\end{array}$

Topic: Mathematics, Functions, COMEDK, COMEDK 2014
The domain of the function $y=\frac{1}{\sqrt{|x|-x}}$ is

Options:
A $(-\infty, 0)$
B $(-\infty, 0]$
C $(-\infty,-1)$
D $(-\infty, \infty)$

The correct answer is:
$(-\infty, 0)$

Solution:
For it must $|x|-x \gt 0$
$|x| \gt x$ but $|x|=x$ for $x$ positive and $|x| \gt x$ for $x$ negative. So, domain will be $(-\infty, 0)$

Topic: Mathematics, Functions, JEE Main
The domain of the real valued function $f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}}$ is

Options:
A $\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)$
B $(-\infty, 1) \cup(2, \infty)$
C $\left(-\frac{1}{3}, \frac{5}{2}\right]$
D $\left(-\infty, \frac{-1}{3}\right] \cup\left[\frac{5}{2}, \infty\right)$

The correct answer is:
$\left(-\infty,-\frac{1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)$

Solution:
Given function $f(x)=\sqrt{\frac{2 x^2-7 x+5}{3 x^2-5 x-2}}$
Here, $\mathrm{f}(\mathrm{x})$ should be greater than or equal to 0 .
$$
\begin{aligned}
& \text { So, } \\
& 2 x^2-7 x+5=0 \\
& 2 x^2-5 x-2 x+5=0 \\
& x(2 x-5)-1(2 x-5)=0 \\
& (x-1)(2 x-5) \\
& x=1, \frac{5}{2} \\
& 3 x^2-5 x-2=0 \\
& 3 x^2-6 x+x-2=0 \\
& 3 x(x-2)+1(x-2)=0 \\
& (x-2)(3 x+1)=0 \\
& x=1, \frac{-1}{3}
\end{aligned}
$$
Now,
When we include $\mathrm{x}=\frac{-1}{3}, 2$ then $\mathrm{f}(\mathrm{x})$ would give not define value so, we will exclude these values from the domain. When we take values between $\left(-\frac{1}{3}, 1\right)$ and $\left(2, \frac{5}{2}\right)$ then it would give negative values so, these interval will not include in domain.
So, domain is $\left(-\infty, \frac{-1}{3}\right) \cup[1,2) \cup\left[\frac{5}{2}, \infty\right)$
Therefore, option (a) is correct.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2022 (20 Jul Shift 1)
The domain of the real valued function
$f(x)=\sqrt{\frac{2-|x|}{3-|x|}}$ is

Options:
A $(-\infty, \infty)$
B $(-\infty,-3) \cup(2, \infty)$
C $(-\infty,-3] \cup(-2,2) \cup[3, \infty)$
D $(-\infty,-3) \cup[-2,2] \cup(3, \infty)$

The correct answer is:
$(-\infty,-3) \cup[-2,2] \cup(3, \infty)$

Solution:
Given, $f(x)=\sqrt{\frac{2-|x|}{3-|x|}}$
For $f(x)$ to be defined, $\frac{2-|x|}{3-|x|} \geq 0$
Case I If $x \geq 0$, then $\frac{2-x}{3-x} \geq 0 \Rightarrow \frac{x-2}{x-3} \geq 0$
Thus, $x \in(-\infty, 2] \cup(3, \infty)$...(i)
Case II If $x < 0$, then $\frac{2+x}{3+x} \geq 0$
Thus, $x \in(-\infty,-3) \cup[-2, \infty)$...(ii)
Case III $\because 3-|x| \neq 0 \Rightarrow|x| \neq 3$
$\Rightarrow \quad x \neq 3,-3$...(iii)
From Eqs. (i), (ii) and (iii), we get
$x \in(-\infty,-3) \cup[-2,2] \cup(3, \infty)$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 2)
The domain of the real valued function fx=6x2+5x-64-x-x+4 is

Options:
A -4,-32∪23,4
B -∞,-32∪23,∞
C -4,4
D -32,23

The correct answer is:
-4,-32∪23,4

Solution:
Given:

fx=6x2+5x-64-x-x+4

For fx to be defined

1 6x2+5x-6≥0

⇒3x-22x+3≥0

⇒x∈(-∞,-32]∪[23,∞) ....i

2 4-x-x+4≠0

⇒4-x≠x+4

⇒4-x≠x+4

⇒x≠0 .....ii

3 4-x≥0⇒x∈(-∞,4] .....iii

4 4+x≥0⇒x∈[-4,∞) .....iv

So, i∩ii∩iii∩iv is

-4,-32∪23,4

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
The domain of the real valued function $f(x)=\frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$ is

Options:
A $(3,4]$
B $[4, \infty)$
C $(1, \infty)$
D $(1,3)$

The correct answer is:
$(3,4]$

Solution:
Given, $f(x)=\frac{\sqrt{\log _{0.5}(x-3)}}{\sqrt{x-1}}$
$\log _{0.5}(x-3) \geq 0$
$\Rightarrow \quad x-3 \leq(0.5)^0$
$\Rightarrow \quad x \leq 4$ ...(A)
$x-3>0$
$x>3$ ...(B)
$x-1>0$
$x>1$ ...(C)
$A \cap B \cap C \Rightarrow 3 < x \leq 4$
i.e. $x \in(3,4]$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 1)
The domain of the real valued function $f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}}$ is

Options:
A $(-3, \infty)$
B $(-3,-1) \cup(-1, \infty)$
C $(-3,-2) \cup(-2,-1) \cup(-1, \infty)$
D $(-3,-2) \cup(-1, \infty)$

The correct answer is:
$(-3,-2) \cup(-1, \infty)$

Solution:
$$
\begin{aligned}
& \text { Given, } f(x)=\frac{\log _2(x+3)}{\sqrt{x^2+3 x+2}} \\
& \\
& \quad x^2+3 x+2>0 \\
& \Rightarrow \quad x^2+2 x+x+2>0 \\
& \Rightarrow \quad x(x+2)+1(x+2)>0 \\
& \Rightarrow \quad(x+2)(x+1)>0 \\
& \Rightarrow \quad x \in(-\infty,-2) \cup(-1, \infty) \\
& \text { But } x+3>0 \\
& \quad x>-3
\end{aligned}
$$
Hence, $x \in(-3,-2) \cup(-1, \infty)$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 2)
The domain of the real valued function fx=x-23-x is…

Options:
A 2,3
B 2,3
C 2,3
D 2,3

The correct answer is:
2,3

Solution:
We have,
fx=x-23-x
Clearly, fx will be defined when
3-x>0⇒x<3
And x-2â¥0⇒xâ¥2
â´ Domain of fx is 2,3 .

Topic: Mathematics, Functions, JEE Main
The domain of the real valued function fx=x-23-x is…

Options:
A 2,3
B 2,3
C 2,3
D 2,3

The correct answer is:
2,3

Solution:
We have,
fx=x-23-x
Clearly, fx will be defined when
3-x>0⇒x<3
And x-2â¥0⇒xâ¥2
â´ Domain of fx is 2,3 .

Topic: Mathematics, Functions, MHT CET, MHT CET 2019 (Shift 1)
The domain of the real valued function
$$
f(x)=\frac{\sqrt{\log _{10}\left(\frac{x}{x-2}\right)}}{\sqrt{[x]^2-5[x]+6}} \text { is }
$$
(Here $[\mathrm{x}]$ denotes the greatest integer function)

Options:
A $(-\infty, 0) \cup(2, \infty)$
B $(2, \infty)$
C $(-\infty, 2) \cup(4, \infty)$
D $[4, \infty)$

The correct answer is:
$[4, \infty)$

Solution:
$f(x)=\frac{\sqrt{\log _{10} \frac{x}{x-2}}}{\sqrt{[x]^2-5[x]+6}}$ (given)
$$
\begin{aligned}
& {[x]^2-5[x]+6>0} \\
& \Rightarrow([x]-3)([x]-2)>0 \\
& \Rightarrow[x] < 2 \text { or }[x]>3 \Rightarrow x \in(-\infty, 2) \cup[4, \infty) \\
& \text { and } \log _{10} \frac{x}{x-2} \geq 0 \Rightarrow \frac{x}{x-2} \geq 1 \Rightarrow \frac{x}{x-2}-1 \\
& \Rightarrow \frac{2}{x-2} \geq 0 \quad \therefore x \in(2, \infty)
\end{aligned}
$$
Taking common solution $\therefore x \in[4, \infty)$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2023 (13 May Shift 1)
The domain of the real valued function $f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}}$ is

Options:
A $[-1,2]$
B $(-1,2)$
C $[-1, \infty)$
D $[2, \infty)$

The correct answer is:
$[-1,2]$

Solution:
Given, function is
$$
\begin{aligned}
& f(x)=\frac{\sqrt{2-x}+\sqrt{1+x}}{\sqrt{x+3}} \\
& x+3>0 \Rightarrow x>-3 ...(i)\\
& 2-x \geq 0 \Rightarrow x \leq 2 ...(ii)\\
& 1+x \geq 0 \Rightarrow x \geq-1...(iii)
\end{aligned}
$$
From the above three inequalities, it can be concluded that $x \in[-1,2]$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 2)
The domain of the real valued function $f(x)=\frac{\sqrt{|x|-x}}{\sqrt{x-[x]}}$ is

Options:
A $\mathbb{Z}$
B $\phi$
C $\mathbb{R}-\mathbb{Z}$
D $\mathbb{R}$

The correct answer is:
$\mathbb{R}-\mathbb{Z}$

Solution:
$\begin{aligned} & \text f(x)=\sqrt{\frac{|x|-x}{x-[x]}} \\ & x-[x]=\{x\} \\ & \text { for } x \in Z,\{x\}=0 \\ & \therefore \quad \text { Domain : } R-Z\end{aligned}$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2023 (13 May Shift 2)
The domain of trigonometric functions can be restricted to any one of their branch (not necessarily principal value) in order to obtain their inverse functions.

Options:
A
True

B
False

The correct answer is:
True

Solution:
No Solution Available

Topic: Mathematics, Inverse Trigonometric Functions
The domain set of the function f( x )= tan −1 x− cot −1 x+ cos −1 ( 2−x ) is

Options:
A 0, 1
B [-1, 1]
C 1, 3
D None of these

The correct answer is:
1, 3

Solution:
Since, tan-1â¡x and cot-1â¡x exists for all xâ R and cos-1â¡2-x exist if -1â¤2-xâ¤1
â1â¤xâ¤3
Thus, the domain of the given function is [1, 3].

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The domestic sewage in large cities

Options:
A has a high BOD as it contains both aerobic and anaerobic bacteria
B is processed by aerobic and then anaerobic bacteria in the secondary treatment in Sewage Treatment Plants (STPs)
C when treated in STPs does not really require the aeration step as the sewage contains adequate oxygen
D has very high amounts of suspended solids and dissolved salts

The correct answer is:
is processed by aerobic and then anaerobic bacteria in the secondary treatment in Sewage Treatment Plants (STPs)

Solution:
Sewage is waste water having food residue, animal and human excreta, detergents, discharges from commercial and industrial establishments. The domestic sewage is processed first by aerobic and then by anaerobic bacteria in secondary treatment in Sewage Treatment Plant (STPs).

Topic: Biology, Microbes in Human Welfare, NEET, NEET 2012 (Mains)
The door of a working refrigerator is left open in a well insulated room. The temperature of air in the room will

Options:
A
decrease

B
increase in winters and decrease in summers

C
remain the same

D
increase

The correct answer is:
increase

Solution:
In a refrigerator, the heat dissipated in the atmosphere is more than that taken from the cooling chamber, therefore the room is heated. If the door of a refrigerator is kept open.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2012 (26 May Online)
The door of an operating refrigerator is kept open. As a result, the temperature of the room will

Options:
A remain unchanged
B increases
C decreases
D depends on the contents inside the refrigerator

The correct answer is:
increases

Solution:
A refrigerator is a heat engine which extracts heat from low temperature reservoir and transfer it to high temperature. If a refrigerator's door is kept open, then room will become hot, because in this situation, refrigerator exhaust more heat into the room than earlier. Hence, temperature of the room increases and room becomes hot.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
The Doppler's effect is applicable for

Options:
A Light waves
B Sound waves
C Space waves
D Both (1) and (2)

The correct answer is:
Both (1) and (2)

Solution:
Doppler's effect is applicable for both light and sound waves.

Topic: Physics, Waves and Sound, JEE Main
The d.r. of normal to the plane through $(1,0,0),(0,1,0)$ which makes an angle $\pi / 4$ with plane $x+y=3$ are

Options:
A
$1, \sqrt{2}, 1$

B
$1,1, \sqrt{2}$

C
$1,1,2$

D
$\sqrt{2}, 1,1$

The correct answer is:
$1,1, \sqrt{2}$

Solution:

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2002
The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen in mL at 20°C and one bar will be released when 0.15 g of aluminium reacts? (Take 1 bar = 0.987 atm )

Options:

The correct answer is:
203

Solution:
The reaction between aluminium and caustic soda is
2Al2×27=54g+2NaOH+2H2O→2NaAlO2+3H23×22.4LatSTP
So 54 g of Al produces H2 at STP =3×22.4 L
0.15 g of Al will produce H2 at STP
=3×22.454×0.15
=0.186L

At STP	Given Conditions

P1=1 atm	P2=1 bar =0.987 atm

V1=0.186 L	V2=?

T1=273 K	T2=273+20=293K

Applying ideal gas equation, P1V1T1=P2V2T2
1×0.1867273=0.987×V2293
V2=2930.987×1×0.1867273
=0.2030 L =203 mL

Topic: Chemistry, States of Matter, JEE Main
The drawing is of a magnified human liver cell.

Which of the following is true about structure labeled X?

Options:
A It is mainly found in cells associated with the synthesis of lipids and steroid hormones.
B It is mainly found in cells away from the nucleus and transport its products to the chloroplast.
C It is mainly associated with nuclear membrane reformation in late anaphase and packaging of protein and synthesis of glycolipids.
D Is mainly associated with nuclear membrane reformation during telophase stage and transport its synthetic products to dictyosomes in plants.

The correct answer is:
Is mainly associated with nuclear membrane reformation during telophase stage and transport its synthetic products to dictyosomes in plants.

Solution:
The structure labelled X is rough endoplasmic reticulum. Eukaryotic cells have a network or reticulum of tiny tubular structures scattered in the cytoplasm that is called the endoplasmic reticulum (ER). The endoplasmic reticulum bearing ribosomes on their surface is called the rough endoplasmic reticulum (RER). RER is frequently observed in the cells actively involved in protein synthesis and secretion. The golgi apparatus principally performs the function of packaging materials to be delivered either to the intra-cellular targets or secreted outside the cell. Energy production is the function of mitochondria

Topic: Biology, Cell - The Unit of Life, NEET
The drift velocity of electrons for a conductor connected in an electrical circuit is Vd. The conductor is now replaced by another conductor with same material and same length but double the area of cross-section. The applied voltage remains same. The new drift velocity of electrons will be

Options:
A Vd
B Vd2
C Vd4
D 2Vd

The correct answer is:
Vd

Solution:
Drift velocity of electrons through a conductor kept in an external electric field is given by Vd=-eEmτ=-eVmlτ

Where, e=magnitude of charge on electron, m=mass of electron, E=external electric field and τ=average relaxation time.

As, drift velocity is independent of area. So, the new drift velocity of electrons will be Vd.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The drift velocity of electrons in silver wire with cross-sectional area $3.14 \times 10^{-6} \mathrm{~m}^{2}$ carrying a current of $20 \mathrm{~A}$ is. Given atomic weight of $\mathrm{Ag}=$ $108,$ density of silver $=10.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Options:
A $2.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
B $67.98 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
C $0.67 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$
D $6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$

Solution:
$\quad$ Number of electrons per kg of silver

$$

=\frac{6.023 \times 10^{26}}{108}

$$

Number of electrons per unit volume of silver

$$

\begin{array}{c}

n=\frac{6.023 \times 10^{26}}{108} \times 10.5 \times 10^{3} \\

\qquad \begin{array}{c}

v_{d}=\frac{I}{n e A} \\

=\frac{20}{6.023 \times 10^{26} \times 10.5 \times 10^{3} \times 1.6 \times 10^{-19} \times 3.14 \times 10^{-6}} \times 108 \\

=6.798 \times 10^{-4} \mathrm{~m} / \mathrm{sec} .

\end{array}

\end{array}

$$

Topic: Physics, Current Electricity, JEE Main
The drift velocity of the electrons in a copper wire of length 2 m under the application of a potential difference of 220 V is 0.5 m s-1. Their mobility (in m2 V-1 s-1)

Options:
A 2.5×10-3
B 2.5×10-2
C 5×102
D 4.5×10-3

The correct answer is:
4.5×10-3

Solution:
Drift velocity=mobility× intensity of the electric field
or vd=μE or μ=vdE
or μ=vdV/l ∵E=Vl
∴μ=0.5×2220=4.5×10−3 m2 V−1 s−1

Topic: Physics, Current Electricity, JEE Main