datavorous/entrance-exam-dataset huggingface dataset:
The distance of the point Q(0, 2, –2) form the line passing through the point P(5, –4, 3) and perpendicular to the lines r→=−3i^+2k^+λ2i^+3j^+5k^, λ∈ℝ and r→=i^−2j^+k^+μ−i^+3j^+2k^, μ∈ℝ is

Options:
A 86
B 20
C 54
D 74

The correct answer is:
74

Solution:
Plotting the diagram of the given data we get,

Now, let a,b,c be the DR's of the line perpendicular to -3i^+2k^+λ2i^+3j^+5k^ and i^-2j^+k^+μ-i^+3j^+2k^.

⇒2a+3b+5c=0, -a+3b+2c=0

Now, solving 2a+3b+5c=0 & -a+3b+2c=0 we get,

⇒a-9=b-9=c9

⇒a1=b1=c-1

So, the equation will be,

⇒x-51=y+41=z-3-1 ...i

Let R be the foot of perpendicular from Q0,2,-2 to i.

⇒x-51=y+41=z-3-1=k

⇒x=k+5, y=k-4, z=-k+3

⇒R≡k+5, k-4, -k+3

So, DR's of QR are k+5, k-6, 5-k.

Now, using perpendicular condition we get,

⇒1k+5+1k-6+-15-k=0

⇒k=2

⇒R≡7,-2,1

⇒QR=49+16+9

⇒QR=74

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The distance of the Sun from earth is 1.5×1011 m and its angular diameter is 2000'' when observed from the earth. The diameter of the Sun will be

Options:
A 2.45×1010 m
B 1.45×1010 m
C 1.45×109 m
D 0.14×109 m

The correct answer is:
1.45×109 m

Solution:
Diameter of Sun is D=rθ

=1.5×1011×(2000)×13600×π180=1.45×109 m

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The distance of the y-axis from the center of the circle which lies in the first quadrant (see figure) and tangent to the lines y=12x,y=4 and the x-axis is

Options:
A 4+25 units
B 4-855 units
C 2+655 units
D 4-25 units

The correct answer is:
4+25 units

Solution:
Let centre is k,2 whose distance from x-2y=0 is k-45=2
k-4=25
k=4+25 (k=4-25 is rejected)

Topic: Mathematics, Circle, JEE Main
The distance of two points on the axis of a magnet from its centre is 10 cm and 20 cm respectively. The ratio of magnetic intensity at these points is 12.5:1. The length of the magnet will be

Options:
A 5 cm
B 25 cm
C 20 cm
D 10 cm

The correct answer is:
10 cm

Solution:
The magnetic field on the axis of a bar magnet of length 2l, having a pole strength m, at a distance d is

B=μ04πmd-l2-μ04πmd+l2

B=μ0m4πd+l2-d-l2d2-l22=μ0mldπd2-l22

⇒B1B2= d1d2 d22-l2d12-l22
⇒12.51=1020 400-l2100-l22
⇒l= 5cm
Hence, length of magnet =2l= 10cm

Topic: Physics, Magnetic Properties of Matter, JEE Main
The distance through which one has to dig the earth from its surface so as to reach the point where the acceleration due to gravity is reduced by 40% of that at the surface of the earth, is (radius of earth is 6400 km)

Options:
A 2560km
B 3000km
C 3260km
D 1560km

The correct answer is:
2560km

Solution:
Acceleration due to gravity, gd=g1-dR

⇒gdg=1-dR

⇒1-40100=1-dR

⇒dR=40100

⇒d=R40100=2560 km

Topic: Physics, Gravitation, JEE Main
The distance travelled by a body moving along a line in time $t$ is proportional to $t^3$.
The acceleration-time $(a, t)$ graph for the motion of the body will be

Options:
A

B

C

D

The correct answer is:

Solution:
Distance along a line i.e., displacement ( $s$ ) $=t^3\left(\because s \propto t^3\right.$ given $)$
By double differentiation of displacement, we get acceleration.
$V=\frac{d s}{d t}=\frac{d t^3}{d t}=3 t^2$ and
$$
\begin{aligned}
& a=\frac{d v}{d t}=\frac{d 3 t^2}{d t}=6 t \\
& a=6 t \text { or } a \propto t
\end{aligned}
$$
Hence graph (b) is correct.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2012 (12 May Online)
The distance travelled by a particle is related to time t as x = 4t2. The velocity of the particle at t = 5 s is

Options:
A 40 m s–1
B 25 m s–1
C 20 m s–1
D 8 m s–1

The correct answer is:
40 m s–1

Solution:
Given here, x = 4t2

We know that velocity is the rate at which displacement changes with time. So, v=dxdt=d4t2dt=8t

Now, at t=5 s

Velocity of particle is v=8×5=40 m s-1.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The distance travelled by a particle starting from rest and moving with an acceleration $\frac{4}{3} \mathrm{~ms}^{-2}$, in the third second is

Options:
A $\frac{19}{3} m$
B $6 \mathrm{~m}$
C $4 \mathrm{~m}$
D $\frac{10}{3} m$

The correct answer is:
$\frac{10}{3} m$

Solution:
$$
S_{n t h}=u+\frac{a}{2}(2 n-1)
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Mains)
The distance travelled by an object in time t is given by s=2.5t2 . The instantaneous speed of the object at t=5 s will be :

Options:
A 25 m s–1
B 5 m s–1
C 62.5 m s–1
D 12.5 m s–1

The correct answer is:
25 m s–1

Solution:
It is given that

s=2.5t2

Differentiating,

v=dsdt=d2.5t2dt=5t

At t = 5 s, v= 25 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The distance x covered by a paritcle in one dimensional motion varies with time t as x2=at2+2bt+c . If the acceleration of the particle depends on x as x-n , where n is an integer, the value of n is _________

Options:

The correct answer is:
3

Solution:
x2=at2+2bt+c
2xv=2at+2b
xv=at+b
v2+ax=a
ax=a-at+bx2
a=aat2+2bt+c-at+b2x3
a=ac-2x3
a α x-3

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The distillation technique most suited for separating glycerol from spent - lye in the soap industry is:

Options:
A Steam distillation
B Distillation under reduced pressure
C Simple distillation
D Fractional distillation

The correct answer is:
Distillation under reduced pressure

Solution:
Glycerol (B.P. 290oC ) is separated from spent - lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required which might decompose the component.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2016 (03 Apr)
The distillation technique most suited for separating glycerol from spent-lye in the soap industry is

Options:
A Distillation under reduced pressure
B Simple distillation
C Fractional distillation
D Steam distillation

The correct answer is:
Distillation under reduced pressure

Solution:
Glycerol (B.P. 290°C) is separated from spent lye in the soap industry by distillation under reduced pressure, as for simple distillation very high temperature is required which might decompose the component.

Topic: Chemistry, Hydrocarbons, JEE Main
The disturbance $y(x, t)$ of a wave propagating in the positive $x$-direction is given by $y=\frac{1}{1+x^2}$ at time $t=0$ and by $y=\frac{1}{\left[1+\left(x-1^2\right)\right]}$ at $t=2 \mathrm{~s}$, where $x$ and $y$ are in meters. The shape of the wave disturbance does not change during the propagation. The velocity of wave in $\mathrm{m} / \mathrm{s}$ is

Options:
A
$2.0$

B
$4.0$

C
$0.5$

D
$1.0$

The correct answer is:
$0.5$

Solution:
The equation of wave at any time is obtained by putting $\mathrm{X}=x-\mathrm{vt}$
$$
\mathrm{y}=\frac{1}{1+x^2}=\frac{1}{1+(x-v t)^2}
$$
We know at $t=2 \mathrm{sec}$,
$$
y=\frac{1}{1+(x-1)^2}
$$
On comparing (i) and (ii) we get $v t=1$
$$
V=\frac{1}{t}
$$
As $t=2 \mathrm{sec}$
$$
\therefore V=\frac{1}{2}=0.5 \mathrm{~m} / \mathrm{s} \text {. }
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2012 (26 May Online)
The diversity in the type of beaks of finches adapted to different feeding habits on the Galapagos islands, as observed by Darwin, provides evidence for

Options:
A origin of species by natural selection
B intraspecific variations
C intraspecific competition
D interspecific competition

The correct answer is:
origin of species by natural selection

Solution:
Birds of Galapagos islands (Darwin's finches) are believed to have evolved from ancestors on the South American mainland as a result of natural selection due to different feeding niches available to them.

Topic: Biology, Evolution, JIPMER, JIPMER 2011
The domain and range for the function $f(x)=e^{|x| \sin x}$ are domain $=$ IR

Options:
A range $=[0, \infty)$ domain $=$ IR
B range $=[1, \infty)$ domain $=$ IR
C range $=$ IR domain $=I R$
D range $=(0, \infty)$

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

Solution:
$f(x)=e^{|x| \sin x}$ is defined every where
Hence domain of $f(x)$ is $R$
$\begin{aligned}
& \therefore-\infty < |x| \sin x < \infty . \\
& \Rightarrow 0 < e^{|x| \sin x} < \infty \\
& \Rightarrow \text { Range of } f(x) \text { is }(0, \infty)
\end{aligned}$

Topic: Mathematics, Functions, MHT CET, MHT CET 2022 (08 Aug Shift 1)
The domain and range of the function $f$ given by $f(x)=2-|x-5|$ is

Options:
A Domain $=\mathrm{R}^{+}$, Range $=(-\infty, 1)$
B Domain $=\mathrm{R}$, Range $=(-\infty, 2]$
C Domain $=\mathrm{R}$, Range $=(-\infty, 2)$
D Domain $=\mathrm{R}^{+}$, Range $=(-\infty, 2]$

The correct answer is:
Domain $=\mathrm{R}$, Range $=(-\infty, 2]$

Solution:
Given $\mathrm{f}(\mathrm{x})=2-|\mathrm{x}-5|$
Domain of $f(x)$ is defined for all real values of $x$.
Since, $|x-5| \geq 0 \Rightarrow-|x-5| \leq 0$
$\Rightarrow 2-|x-5| \leq 2 \Rightarrow f(x) \leq 2$
Hence, range of $\mathrm{f}(\mathrm{x})$ is $(-\infty, 2]$.

Topic: Mathematics, Functions, VITEEE, VITEEE 2016
The domain and range of the real function $f$ defined by $f(x)=\frac{4-x}{x-4}$ is given by

Options:
A
Domain $=R$, Range $=\{-1,1\}$

B
Domain $=R-\{1\}$, Range $=R$

C
Domain $=R-\{4\}$, Range $=\{-1\}$

D
Domain $=R-\{-1\}$, Range $=\{-1,1\}$

The correct answer is:
Domain $=R-\{4\}$, Range $=\{-1\}$

Solution:
Since, $f(x)=\frac{4-x}{x-4}$ $\Rightarrow x-4 \neq 0 \Rightarrow x \neq 4$ [By definition of function] So, domain of $f=R-\{4\}$
$$
\because f(x)=\frac{4-x}{x-4}=\frac{-(x-4)}{x-4}=-1 \therefore \text { Range }=\{-1\}
$$

Topic: Mathematics, Relations and Functions (Class 11)
The domain and range of the relation R given by $\mathrm{R}=\left\{(x, y) / y=x+\frac{6}{x}, x, y \in \mathrm{N}\right.$ and $\left.x < 6\right\}$ are

Options:
A Domain $=\{2,3\}$, Range $=\{5\} .$
B Domain $=\{1,2\}$, Range $=\{5,7\} .$
C Domain $=\{1,2,3,4,5\}$, Range $=\left\{7,5, \frac{6}{4}, \frac{6}{5}\right\} .$
D Domain $=\{1,2,3\}$, Range $=\{5,7\} .$

The correct answer is:
Domain $=\{1,2,3\}$, Range $=\{5,7\} .$

Solution:
$y=x+\frac{6}{x}, x, y \in N$ and $x < 6$
When $x=1, y=1+6=7$
When $x=2, y=2+3=5$
When $x=3, y=3+2=5$
When $x=4, y=4+1.5=5.5$
When $x=5, y=5+1.2=6.2$
Since $x, y \in N$ and $x < 6$, we write Domain $=\{1,2,3,4,5\}$ and Range $=\{7,5\}$

Topic: Mathematics, Functions, JEE Main
The domain and range of the relation R given by $\mathrm{R}=\left\{(x, y) / y=x+\frac{6}{x}, x, y \in \mathrm{N}\right.$ and $\left.x < 6\right\}$ are

Options:
A Domain $=\{2,3\}$, Range $=\{5\} .$
B Domain $=\{1,2\}$, Range $=\{5,7\} .$
C Domain $=\{1,2,3,4,5\}$, Range $=\left\{7,5, \frac{6}{4}, \frac{6}{5}\right\} .$
D Domain $=\{1,2,3\}$, Range $=\{5,7\} .$

The correct answer is:
Domain $=\{1,2,3\}$, Range $=\{5,7\} .$

Solution:
$y=x+\frac{6}{x}, x, y \in N$ and $x < 6$
When $x=1, y=1+6=7$
When $x=2, y=2+3=5$
When $x=3, y=3+2=5$
When $x=4, y=4+1.5=5.5$
When $x=5, y=5+1.2=6.2$
Since $x, y \in N$ and $x < 6$, we write Domain $=\{1,2,3,4,5\}$ and Range $=\{7,5\}$

Topic: Mathematics, Functions, MHT CET, MHT CET 2020 (20 Oct Shift 1)
The domain and range of y(x)=cosx-3 are respectively

Options:
A R and [-1,1]
B R and [-4,-2]
C R/{0} and [0,1]
D R/(2n+1)π2 and [-1,1]

The correct answer is:
R and [-4,-2]

Solution:
yx=cosx-3

As we know -1≤cosx≤1

⇒-1-3≤cosx-3≤1-3

⇒-4≤cos-3≤-2

⇒yx∈-4,-2

cosx is defined ∀x∈R

⇒ Domain fo yx is R.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
The domain of a function $f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^{2}}}$ is

Options:
A $(4,6]$
B $(-5,5)$
C $[4,5)$
D $(4,5]$

The correct answer is:
$[4,5)$

Solution:
We have $f(y)=\frac{\cos ^{-1}(y-5)}{\sqrt{25-y^{2}}}$
Here $-1 \leq y-5 \leq 1$ and $25-y^{2}>0$ $\therefore \quad 4 \leq y \leq 6 \quad$ and $\quad-5 < y < 5$ Hence domain of $f(y)$ is $[4,5)$

Topic: Mathematics, Functions, MHT CET, MHT CET 2020 (15 Oct Shift 1)
The domain of defined of the function $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is

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

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

Solution:
$f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$
For domain
$$
\frac{1-|x|}{2-|x|} \geq 0
$$
and $2-|x| \neq 0$
$$
\begin{aligned}
|x| & \neq 2 \\
x & = \pm 2
\end{aligned}
$$
Case I When, $x \geq 0$
So, Eq. (i) becomes
$$
\frac{1-x}{2-x} \geq 0 \quad\left\{\because|x|=\left[\begin{array}{cc}
x & x \geq 0 \\
-x & x < 0
\end{array}\right\}\right.
$$
Now, critical points are
$$
x=1,2
$$
Using wavy curve method
But $x \geq 0$
$\therefore$ Solution is $x \in[0,1] \cup(2, \infty)$.
Case II When, $x < 0$
So, Eq. (i) becomes
$$
\frac{1+x}{2+x} \geq 0
$$
Critical points are $x=-1,-2$
Using wavy curve method
But $x < 0$
Solution is $x \in(-\infty,-2) \cup[-1,0)$
$\therefore$ Required solution is union of case I and case II.
$\therefore$ Required solution is
$x \in(-\infty,-2) \cup[-1,1] \cup(2, \infty)$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (24 Aug Shift 1)
The domain of definition of $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$ is Here $(a, b)=\{x: a < x < b\}$ and
$[a, b]=\{x: a \leq x \leq b\}$

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

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

Solution:
We have, $f(x)=\sqrt{\frac{1-|x|}{2-|x|}}$
$f(x)$ is defined for all $x$ satisfying $\frac{1-|x|}{2-|x|} \geq 0 \Rightarrow \frac{|x|-1}{|x|-2} \geq 0$
$\Rightarrow \quad|x| \leq$ lor $|x|>2$
$\Rightarrow \quad x \in[-1,1]$ or $x \in(-\infty,-2) \cup(2, \infty)$
$\Rightarrow \quad x \in(-\infty,-2) \cup[-1,1] \cup(2, \infty)$

Topic: Mathematics, Functions, WBJEE, WBJEE 2018
The domain of definition of the function $f(x)=\sqrt{1+\log _{\mathrm{e}}(1-x)}$ is

Options:
A $-\infty < x \leq 0$
B $-\infty < x \leq \frac{e-1}{e}$
C $-\infty < x \leq 1$
D $x \geq 1-e$

The correct answer is:
$-\infty < x \leq \frac{e-1}{e}$

Solution:
Hints : $1-x>0 \Rightarrow X < 1$
$$
\begin{aligned}
& 1+\log _e(1-x) \geq 0 \\
& \log _e(1-x) \geq-1 \Rightarrow 1-x \geq e^{-1} \\
& x \leq 1-\frac{1}{e} \\
& x \leq \frac{e-1}{e}
\end{aligned}
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2009
The domain of definition of the function fx= logx-1⁡x2+4x+4, is

Options:
A -3,Â -1âª1,Â 2
B -2,Â -1âª2,Â â
C -â, -3âª-2, -1âª2, â
D -2,Â -1âª2,Â â

The correct answer is:
-â, -3âª-2, -1âª2, â

Solution:
log x-1⁡ x2+4x+4≥0
Case 1: 0<x-1<1
i.e., 1<x<2 , then
0<x2+4x+4 ≤1 ⇒ x2+4x+3 ≤0 & x+22>0
⇒ -3 ≤x ≤-1 & x≠-2
So, x∈-2, -1
Case 2: x-1>1
i.e., x>2
x2+4x+4 ≥1⇒x+1x+3≥0⇒x∈ -∞,-3 ∪ -1 ,∞
⇒ x∈-∞, -3∪2, ∞

Hence, domain is -∞,-3 ∪(-2,-1)∪2,∞

Topic: Mathematics, Functions, JEE Main
The domain of definition of the function y=3ex2-1log⁡x-1 is

Options:
A 1, â
B 1,Â â
C R-1
D (-â,Â -1)âª(1,Â â)

The correct answer is:
1, â

Solution:
x2-1â¥0
âx2â¥1
âxâ-ââ, -1âª1,ââ¦â¦.(1)
and x-1>0
âx>1
âxâ1, ââ¦â¦2
From 1 and 2
xâ(1, â)

Topic: Mathematics, Functions, JEE Main
The domain of fx=cos-1x2-5x+6x2-9logx2-3x+2 is

Options:
A x∈-12,1∪2,∞-3
B x∈-12,1∪2,∞-3
C x∈-12,1∪2,∞-3
D x∈-12,1∪2,∞-3

The correct answer is:
x∈-12,1∪2,∞-3

Solution:
We know that, cos-1x is defined when -1⩽x⩽1

So, cos-1x2-5x+6x2-9 is defined when

-1⩽x2-5x+6x2-9⩽1 and x2-9≠0

⇒-1≤x-3x-2x-3x+3≤1 and x≠±3

⇒-1≤x-2x+3≤1 and x≠±3, ⇒-1≤x+3-5x+3≤1 and x≠±3

⇒-1≤1-5x+3≤1 and x≠±3,

⇒-2≤-5x+3⩽0 and x≠±3

⇒0≤5x+3⩽2 and x≠±3, ⇒x+3≥52 and x≠±3

⇒ x≥-12 and x≠±3

∴ x∈-12,∞-3 ⋯1

Now, we know that logx is defined for x>0.

So, logx2-3x+2 is defined for x2-3x+2>0, ⇒x-1x-2>0

⇒x>2 or x<1 ⋯2

from equations 1 & 2

∴ x∈-12,1∪2,∞-3

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The domain of $f(x)=\log \left[(2.5)^{3-x^2}-(0.4)^{x+9}\right]$ is

Options:
A $(-4,3)$
B $(-3,4)$
C $(3,4)$
D $(0, \infty)$

The correct answer is:
$(-3,4)$

Solution:
We have,
$$
f(x)=\log \left[(2.5)^{3-x^2}-(0.4)^{x+9}\right]
$$
$f(x)$ is defined
if $(2.5)^{3-x^2}-(0.4)^{x+9}>0$
$\Rightarrow \quad\left(\frac{25}{10}\right)^{3-x^2}>(0.4)^{x+9}$
$\Rightarrow \quad\left(\frac{10}{4}\right)^{3-x^2}>\left(\frac{4}{10}\right)^{x+9}$
$\Rightarrow \quad\left(\frac{10}{4}\right)^{3-x^2}>\left(\frac{10}{4}\right)^{-x-9}$
$\therefore \quad 3-x^2>-x-9$
$\Rightarrow \quad x^2-x-12 < 0$
$\Rightarrow \quad(x+3)(x-4) < 0 \Rightarrow x \in(-3,4)$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
The domain of fx=logx+1x-2e2logex-2x+3,x∈R is

Options:
A ℝ--1,3
B 2,∞-3
C -1,∞-3
D ℝ-3

The correct answer is:
2,∞-3

Solution:
Given,

Function fx=logx+1x-2e2logex-2x+3, x∈R

To find domain we will take following cases:

Case-1:

x-2>0⇒x>2

x∈2,∞

Case-2:

x+1>0⇒x>-1

And, x+1≠1⇒x≠0

x∈-1,0∪0,∞

Case-3:

x>0

x∈0,∞

Case-4:

e2logex-2x+3⇒x2-2x-3≠0

⇒x-3x+1≠0

x≠-1,3

From case (1) ∩ case (2) ∩ case (3) ∩ case (4) we get,

x∈2,∞-3

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The domain of $f(x)=\sin ^{-1}\left[\log _{2}\left(\frac{x}{2}\right)\right]$ is

Options:
A $0 \leq x \leq 1$
B $0 \leq x \leq 4$
C $1 \leq x \leq 4$
D $4 \leq x \leq 6$

The correct answer is:
$1 \leq x \leq 4$

Solution:
Given function, $f(x)=\sin ^{-1}\left[\log _{2}\left(\frac{x}{2}\right)\right]$
Since, $\quad-1 \leq \sin x \leq-1$
$\Rightarrow \quad-1 \leq \log _{2}\left(\frac{x}{2}\right) \leq 1$
$\Rightarrow \quad 2^{-1} \leq x / 2 \leq 2^{1}$
$\Rightarrow \quad 2^{0} \leq x \leq 2^{2}$
$\Rightarrow \quad 1 \leq x \leq 4$
Required domain is $x \in[1,4]$.

Topic: Mathematics, Inverse Trigonometric Functions, KCET, KCET 2011
The domain of fx=sinlog4-x21-x is

Options:
A 0,5
B 1,5
C -2,1
D 2,3

The correct answer is:
-2,1

Solution:
We have,

fx=sinlog4-x21-x

Domain of sinθ is R.

logax is defined for x,a>0 and a≠1.

For fx to be defined,

4-x21-x>0 and x≠1

⇒11-x>0

Since, 4-x2≥0 ∀x∈R

⇒11-x>0

⇒x∈-∞,1

And, -2,1⊂-∞,1.

Topic: Mathematics, Functions, JEE Main
The domain of fx=sinlog4-x21-x is

Options:
A 0,5
B 1,5
C -2,1
D 2,3

The correct answer is:
-2,1

Solution:
We have,

fx=sinlog4-x21-x

Domain of sinθ is R.

logax is defined for x,a>0 and a≠1.

For fx to be defined,

4-x21-x>0 and x≠1

⇒11-x>0

Since, 4-x2≥0 ∀x∈R

⇒11-x>0

⇒x∈-∞,1

And, -2,1⊂-∞,1.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2022 (04 Jul Shift 2)
The domain of $f(x)=\sqrt{\left(\frac{1}{\sqrt{x}}-\sqrt{x+1}\right)}$ is

Options:
A $x>-1$
B $(-1, \infty)$ K $\{0\}$
C $\left(0, \frac{\sqrt{5}-1}{2}\right]$
D $\left[\frac{1-\sqrt{5}}{2}, 0\right)$

The correct answer is:
$\left(0, \frac{\sqrt{5}-1}{2}\right]$

Solution:
Hint:
$x^{2}+x \leq 1$ and $x>0$
$x \in\left[\frac{-1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right]$
$x \in\left(0, \frac{\sqrt{5}-1}{2}\right]$

Topic: Mathematics, Functions, WBJEE, WBJEE 2020
The domain of $f(x)=\frac{\log _2(x+3)}{x^2+3 x+2}$ is

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

The correct answer is:
$(-3,+\infty)-\{-1,-2\}$

Solution:
Here $x+3\gt0$ and $x^2+3 x+2 \neq 0$
$\therefore x\gt-3$ and $(x+1)(x+2) \neq 0$, i.e. $x \neq-1,-2$
$\therefore$ Domain $=(-3, \infty)-\{-1,-2\}$

Topic: Mathematics, Functions, JEE Main
The domain of fx=x16-x2+log2x3-2x is

Options:
A -2,0∪2,∞
B -2,0∪2,4
C -2,0∪4,∞
D -2,0∪2,4∪4,∞

The correct answer is:
-2,0∪2,4∪4,∞

Solution:
16-x2≠0 and x3-2x>0
x≠4,-4 …..i

and xx2-2>0
xx+2x-2>0
x∈-2,0∪2,∞…ii
Taking intersection of i & ii, we get,

x∈-2,0∪2,4∪4,∞

Topic: Mathematics, Functions, JEE Main
The domain of \(f(x)=\cos ^{-1}\left(\frac{x-3}{2}\right)-\log _{10}(4-x)\) is

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

The correct answer is:
\([1,4)\)

Solution:
Let \(f(x)=\cos ^{-1}\left(\frac{x-3}{2}\right)-\log _{10}(4-x)\)
For domain of \(\log _{10}(4-x) \Rightarrow 4-x > 0 \Rightarrow x < 4\)
For domain of \(\cos ^{-1}\left(\frac{x-3}{2}\right) \Rightarrow-1 \leq \frac{x-3}{2} \leq 1\)
\(\Rightarrow \quad 1 \leq x \leq 5\)
Now domain of \(f(x)\) will be \(x \in[1,4)\)

Topic: Mathematics, Inverse Trigonometric Functions, AP EAMCET, AP EAMCET 2020 (17 Sep Shift 1)
The domain of $\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right]$ is

Options:
A $[1,9]$
B $[-1,9]$
C $[-9,1]$
D $[-9,-1]$

The correct answer is:
$[1,9]$

Solution:
$\begin{aligned} & y=\sin ^{-1}\left[\log _3\left(\frac{x}{3}\right)\right] \Rightarrow-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1 \\ & \Rightarrow \frac{1}{3} \leq \frac{x}{3} \leq 3 \Rightarrow 1 \leq x \leq 9 \Rightarrow x \in[1,9] .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of $\sin ^{-1}\left[\log _3(x / 3)\right]$ is

Options:
A
[1, 9]

B
[-1,9]

C
[-9, 1]

D
[-9, -1]

The correct answer is:
[1, 9]

Solution:
$f(x)=\sin ^{-1}\left(\log _3\left(\frac{x}{3}\right)\right)$ exists if
$-1 \leq \log _3\left(\frac{x}{3}\right) \leq 1 \Leftrightarrow 3^{-1} \leq \frac{x}{3} \leq 3^1 \Leftrightarrow 1 \leq x \leq 9$ or $x \in[1,9]$

Topic: Mathematics, Functions, JEE Main, JEE Main 2002
The domain of $\sin ^{-1} x$ is

Options:
A $(-\pi, \pi)$
B $[-1,1]$
C $(0,2 \pi)$
D $(-\infty, \infty)$

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

Solution:
Given,
$f(x)=\sin ^{-1} x$
Now,
$\sin ^{-1} x$ is defined only for $x \in[-1,1]$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The domain of \(\sqrt{|x|-x}\) is

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

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

Solution:
Given function \(\sqrt{|x|-x}\) is define, if \(|x| \geq x\)
\(\Rightarrow \quad x \in \mathbf{R}\)
Hence, option (c) is correct.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
The domain of the definition of the function fx=14-x2+log10x3-x is:

Options:
A -1, 0∪1, 2∪2, ∞
B (1, 2)∪2, ∞
C -2, -1∪-1, 0∪2, ∞
D -1, 0∪1, 2∪3, ∞

The correct answer is:
-1, 0∪1, 2∪2, ∞

Solution:
We know that log10gx is defined when gx>0 and for a fraction to be defined its denominator must be non-zero.

Thus, the given function is defined only when

x3-x>0 and 4-x2≠0

⇒xx-1x+1>0 and x≠2, -2

The wavy curve for the inequality is
⇒x∈-1, 0∪1, ∞ and x≠2, -2.

Hence, domain is -1, 0∪1, 2∪2, ∞.

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The domain of the definition of the function $f(x)=\frac{1}{4-x^2}+\log _{10}\left(x^3-x\right)$ is

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

The correct answer is:
$(-1,0) \cup(1,2) \cup(2, \infty)$

Solution:
$f(x)=\frac{1}{4-x^2}+\log _{10}\left(x^3-x\right)$
to define $f(x) 4-x^2 \neq 0$ and $x^3-x>0$
$\Rightarrow x^2 \neq 4 \text { and } x(x-1)(x+1)>0$
$\Rightarrow x \neq \pm 2$ and
$\Rightarrow x \in(-1,0) \cup(1,2) \cup(2, \infty)$

Topic: Mathematics, Functions, MHT CET, MHT CET 2022 (11 Aug Shift 1)
The domain of the definition of the function $y(x)$ is given by the equation $2^x+2^y=2$ 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 (13 May Shift 2)
The domain of the derivative of the functions $f(x)\left\{\begin{array}{cc}\tan ^{-1} x, & \text { if }|x| \leq 1 \\ \frac{1}{2}(|x|-1), & \text { if }|x|>1\end{array}\right.$ is given by

Options:
A $R-\{1\}$
B $R-\{0\}$
C $R-\{-1,1\}$
D $R-\{-1\}$

The correct answer is:
$R-\{-1,1\}$

Solution:
$\begin{aligned} & f(x) \begin{cases}\frac{1}{2}(-x-1) & x<-1 \\ \tan ^{-1} x & -1 \leq x \leq 1 \\ \frac{1}{2}(x-1) & x>1\end{cases} \\ & \Rightarrow f^{\prime}(x) \begin{cases}-\frac{1}{2} & x<-1 \\ \frac{1}{1+x^2} & -11\end{cases} \end{aligned}$
$\because f(x)$ discontinuous at $x=-1$ and $x=1$
Hence $f(x)$ is not differentiable at $x=-1$ and $x=1$

Topic: Mathematics, Differentiation, MHT CET, MHT CET 2022 (10 Aug Shift 1)