datavorous/entrance-exam-dataset huggingface dataset:
The domain of the function cos-12sin-114x2-1π is

Options:
A -∞,-12∪12,∞∪0
B -∞,-12∪12,∞
C -∞,-12∪12,∞∪0
D R--12,12

The correct answer is:
-∞,-12∪12,∞∪0

Solution:
fx=cos-12sin-114x2-1π

Case I: -1≤14x2-1≤1

⇒4x2-1≥1 or 4x2-1≤-1 and 4x2-1≠0

∴x2≥12 or x2≤0⇒x=0 and x≠±12

∴x∈-∞,-12∪12,∞∪0

Case II: -1≤2sin-114x2-1π≤1

⇒-π2≤sin-114x2-1≤π2 which is always true.

Hence, x∈-∞,-12∪12,∞∪0

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The domain of the function cosec-11+xx is :

Options:
A -12,∞-{0}
B -1,-12∪(0,∞)
C -12,0∪[1,∞)
D -12,∞-{0}

The correct answer is:
-12,∞-{0}

Solution:
Domain of cosec-1x is x≤-1 or x≥1

1+xx≤-1 or 1+xx≥1

1+x+xx≤0 or 1+x-xx≥0

2x+1x≤0 or 1x≥0

x∈-12,0 or x∈(0,∞)

x∈-12,∞-{0}

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The domain of the function defined by
$f(x)=\frac{-5}{4 x^2+1}+\sqrt{x^2-4}$ is

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

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

Solution:
$f(x)=\frac{-5}{4 x^2+1}+\sqrt{x^2-4}$
$\because f(x)$ will be undefined if $4 x^2+1=0$ or $x^2-4 < 0$
$\because 4 x^2+1 \neq 0$ for $x \in R \quad\left\{x^2\right.$ is always positive $\}$
Now, let $x^2-4 < 0$
$\begin{array}{ll}\Rightarrow & x^2 < 4 \\ \Rightarrow & x \in(-2,2)\end{array}$
$\Rightarrow$ Function will be undefined when $x \in(-2,2)$
$\therefore$ Domain of $f(x)=R-(-2,2)$
$=(-\infty,-2] \cup[2, \infty)$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
The domain of the function defined by $f(x)=\cos ^{-1} \sqrt{x-1}$ is

Options:
A $[1,2]$
B $[0,2]$
C $[-1,1]$
D $[0,1]$

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

Solution:
We have, $f(x)=\cos ^{-1} \sqrt{x-1}$ $f(x)$ is defined when $\sqrt{x-1} \in[0,1]$
$\begin{array}{ll}
\therefore \quad & 0 \leq \sqrt{x-1} \leq 1 \\
& 0 \leq x-1 \leq 1 \\
& 1 \leq x \leq 2
\end{array}$
$\therefore$ Domain of $f(x)$ is $[1,2]$.

Topic: Mathematics, Functions, KCET, KCET 2020
The domain of the function defined by $f(x)$ $=\sin ^{-1} \sqrt{x-1}$ is

Options:
A
$[1,2]$

B
$[-1,1]$

C
$[0,1]$

D
none of these

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

Solution:
No Solution Available

Topic: Mathematics, Inverse Trigonometric Functions
The domain of the function $\bar{f}$ defined by $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$ is equal to

Options:
A
$(-\infty,-1) \cup(1,4]$

B
$(-\infty,-1] \cup(1,4]$

C
$(-\infty,-1) \cup[1,4]$

D
$(-\infty,-1) \cup[1,4)$

The correct answer is:
$(-\infty,-1) \cup(1,4]$

Solution:
Since, $f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^2-1}}$ $\Rightarrow 4-x \geq 0$ and $x^2-1>0$ [By definition of function] $\Rightarrow x-4 \leq 0$ and $(x+1)(x-1)>0$
$\Rightarrow x \leq 4$ and $x < -1$ and $x>1$
So, domain of $f=(-\infty,-1) \cup(1,4]$

Topic: Mathematics, Relations and Functions (Class 11)
The domain of the function $\mathrm{f}$ defined by $\mathrm{f}(\mathrm{x})=\log _{\mathrm{x}} 10$ is

Options:
A $x>10$
B $x>0$ excluding $x=10$
C $x \geq 10$
D $x>0$ excluding $x=1$

The correct answer is:
$x>0$ excluding $x=1$

Solution:
$\mathrm{f}(\mathrm{x})=\log _{\mathrm{x}} 10$
Domain of logarithmic function is $\mathrm{x}>0$ excluding $\mathrm{x}=1$.

Topic: Mathematics, Functions, NDA, NDA 2019 (Phase 1)
The domain of the function f given by fx=1x2-x-6 is

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

The correct answer is:
-∞,-2∪4,∞

Solution:
Given that fx=1x2-x-6>0 ,
f is defined if x2-x-6>0 or x-3x+2>0,
⇒x<-2 or x>3
⇒x<-2 or x≥4
Hence, domain =-∞,-2∪4,∞

Topic: Mathematics, Functions, JEE Main
The domain of the function \( f: R \rightarrow R \) defined by \( f(x)=\sqrt{x^{2}-7 x+12} \) is

Options:
A \((-\infty, 3] \cup(4, \infty) \)
B \( (3,4) \)
C \((-\infty, 3] \cup[4, \infty) \)
D \((-\infty, 3] \cap[4, \infty) \)

The correct answer is:
\((-\infty, 3] \cup[4, \infty) \)

Solution:
$(C)$
$f(x)=\sqrt{x^{2}-7 x+12}$
$x^{2}-7 x+12 \geq 0$
$(x-4)(x-3) \geq 0$
$\Rightarrow x \in(-\infty, 3] \cup[4, \infty)$

Topic: Mathematics, Functions, KCET, KCET 2019
The domain of the function fx=19-x2+log20x3-3x is

Options:
A -3,0∪3,∞
B -3,0∪3,3
C -3,0∪3,∞
D -3,0∪3,3∪3,∞

The correct answer is:
-3,0∪3,3∪3,∞

Solution:
9-x2≠0 and x3-3x>0
x≠-3,3 ....(1)

and xx2-3>0
⇒xx+3x-3>0
x∈-3,0∪3,∞ …. 2
From 1 ∩ 2, we get,

x∈-3,0∪3,3∪3,∞

Topic: Mathematics, Functions, JEE Main
The domain of the function f(x)=1[x]-1, where [x] is greatest integer function of x is

Options:
A R−(1,2)
B R−{1}
C R−{0,1}
D R−[1,2)

The correct answer is:
R−[1,2)

Solution:
Given that, f(x)=1[x]-1

Domain =R-points at which f(x) is not defined

Now, f(x) will be undefined, when x-1=0⇒x=1

Now, all the values for which the greatest integer function is 1.

⇒x∈[1, 2)

Therefore, Domain of f(x) is R-[1, 2)

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
The domain of the function fx=1x2-3x-10 is (where [x] denotes the greatest integer less than or equal to x)

Options:
A (-∞,-3]∪(5,∞)
B (-∞,-2)∪[6,∞)
C (-∞,-2)∪(5,∞)
D (-∞,-3]∪[6,∞)

The correct answer is:
(-∞,-2)∪[6,∞)

Solution:
Given:

fx=1x2-3x-10

For fx to be defined,

x2-3x-10>0

⇒x-5x+2>0

⇒x<-2 & x>5

⇒x<-2 & x≥6

So, domain is -∞,-2∪[6,∞).

Topic: Mathematics, Functions, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The domain of the function fx=1x2-x-20 is (where, . represents the greatest integer function)

Options:
A -â,-4âª[6,â)
B (-â,4]âª[6,â)
C -∞,-4∪(6,∞)
D None of these

The correct answer is:
-∞,-4∪[6,∞)

Solution:
fx is defined for
x2-x-20>0⇒x-5x+4>0
⇒x<-4 or x>5
But x<-4⇒x=-5,-6,-7…
∴x<-4
Also, x>5⇒x=6,7,8…
∴x≥6
∴ Domain of f=-∞,-4∪[6,∞)

Topic: Mathematics, Functions, JEE Main
The domain of the function fx=2sin-1⁡log2⁡12x2 is

Options:
A -2,-1∪1,2
B -2,-1∪1,2
C -2,-1∪1,2
D -2,-1∪1,2

The correct answer is:
-2,-1∪1,2

Solution:
For fx to be defined, we must have
-1≤log212x2≤1⇒2-1≤12x2≤21 as the base=2>1
⇒1≤x2≤4….i
Now, 1≤x2⇒x2-1≥0 i.e. x-1x+1≥0
⇒x≤-1 or x≥1….ii
Also, x2≤4⇒x2-4≤0 i.e. x-2x+2≤0
⇒-2≤x≤2…..iii
From Equation ii and iii, we get,

the domain of f=-∞,-1∪1,∞∩-2,2
=-2,-1∪1,2.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The domain of the function $\mathrm{f}(\mathrm{x})=\sqrt{(2-x)(x-3)}$ is

Options:
A $(0, \infty)$
B $[0, \infty]$
C $[2,3]$
D $(2,3)$

The correct answer is:
$[2,3]$

Solution:
$\begin{aligned} & f(x)=\sqrt{(2-x)(x-3)} \\ & \text { Here }(2-\mathrm{x})(\mathrm{x}-3) \geq 0 \\ & \Rightarrow-(\mathrm{x}-2)(\mathrm{x}-3) \geq 0 \\ & \Rightarrow(\mathrm{x}-2)(\mathrm{x}-3) \leq 0 \\ & \Rightarrow \mathrm{x} \in[2,3] \end{aligned}$

Topic: Mathematics, Functions, NDA, NDA 2019 (Phase 1)
The domain of the function fx=4cos-1⁡1-x2 is

Options:
A -∞,-3∪3,∞
B -3,3
C -∞,-3∪3,∞
D ϕ

The correct answer is:
-3,3

Solution:
cos-1⁡1-x2 is defined if

-1≤1-x2≤1
⇒-2≤1-x≤2⇒-3≤-x≤1
⇒-1≤x≤3
x≥-1 is true for all real values of x .
x≤3⇒-3≤x≤3
Also, 4cos-1⁡1-x2 is defined if

cos-1⁡1-x2≥0
⇒1-x2≥cos⁡0=1
⇒1-x≥2
⇒x≤-1 (not possible)
∴ Domain of fx=-3,3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The domain of the function fx=4x12-x9+x4-x+1 is

Options:
A -∞,-1
B 1,∞
C -1,1
D -∞,∞

The correct answer is:
-∞,∞

Solution:
fx is defined for x12-x9+x4-x+1>0
âx4x8+1-xx8+1+1>0
âx8+1xx3-1+1>0
If xâ¥1 or xâ¤-1, then the above expression is positive.
If -1<xâ¤0, the above inequality still holds.
If 0<x<1, then x12-xx8+1+x4+1>0
as x4+1>x8+1 so x4+1>xx8+1
Hence, the domain of f=-â,â.

Topic: Mathematics, Functions, JEE Main
The domain of the function

$f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^{2}}}$

Options:
A [1,2]
B [2,3)
C [1,2]
D [2,3]

The correct answer is:
[2,3)

Solution:
$f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^{2}}}$ is defined

if (i) $-1 \leq x-3 \leq 1 \Rightarrow 2 \leq x \leq 4$ and

(ii) $9-x^{2}>0 \Rightarrow-3
Taking common solution of (i) and (ii), we get $2 \leq x<3$

$\therefore$ Domain $=[2,3)$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}$ is

Options:
A $[1,2)$
B $[2,3)$
C $[1,2]$
D $[2,3]$

The correct answer is:
$[2,3)$

Solution:
To define $f(x), 9-x^2\gt3 \Rightarrow-3 \lt x \lt 3$ ...(i)
$-1 \leq(x-3) \leq 1 \Rightarrow 2 \leq x \leq 4$ ....(ii)
From (i) and (ii), $2 \leq x \lt 3$ i.e., $[2,3$ ).

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\frac{\sin ^{-1}(x-3)}{\sqrt{9-x^2}}$ is

Options:
A
$[2,3]$

B
$[2,3)$

C
$[1,2]$

D
$[1,2)$

The correct answer is:
$[2,3)$

Solution:
$9-x^2>0$ and $-1 \leq x-3 \leq 1 \Rightarrow x \in[2,3)$

Topic: Mathematics, Functions, JEE Main, JEE Main 2004
The domain of the function $f(x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$ is

Options:
A $(-3,3)$
B $[-3,3]$
C $(-\infty,-3) \mathrm{U}(3, \infty)$
D $(-\infty,-3] \mathrm{U}[3, \infty)$

The correct answer is:
$[-3,3]$

Solution:
Hints: $f(x)=\sqrt{\cos ^{-1}\left(\frac{1-|x|}{2}\right)}$
$$
-1 \leq \frac{1-|\mathrm{x}|}{2} \leq 1 \Rightarrow-2-1 \leq-|\mathrm{x}| \leq 2-1 \Rightarrow-3 \leq-|\mathrm{x}| \leq 1 \Rightarrow-1 \leq \mathrm{x} \mid \leq 3 \Rightarrow \mathrm{x} \in[-3,3]
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2010
The domain of the function $f(x)=\exp \left(\sqrt{5 x-3-2 x^2}\right)$ is

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

The correct answer is:
$\left[1, \frac{3}{2}\right]$

Solution:
$f(x)=e^{\sqrt{5 x-3-2 x^2}} \Rightarrow 5 x-3+ 2 x^2\gt0 \text { or } \quad(x-1)\left(x-\frac{3}{2}\right) \geq 0$
$\therefore D \in[1,3 / 2]$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\frac{\sin ^{-1}(3-x)}{\ln (|x|-2)}$

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

The correct answer is:
$(2,3) \cup(3,4]$

Solution:
$f(x)=\frac{\sin ^{-1}(3-x)}{\log [x \mid-2]}$
Let $g(x)=\sin ^{-1}(3-x) \Rightarrow-1 \leq 3-x \leq 1$
Domain of $g(x)$ is $[2,4]$
and let $h(x)=\log [x \mid-2] \Rightarrow|x|-2\gt0$
$\Rightarrow|x|\gt2 \Rightarrow x \lt -2 \text { or } x\gt2 \Rightarrow(-\infty,-2) \cup(2, \infty)$
we know that
$\begin{aligned}
&(f / g)(x)=\frac{f(x)}{g(x)} \forall x \in D_1 \cap D_2-\{x \in R: g(x)=0\} \\
& \therefore \text { Domain of } f(x)=(2,4]-\{3\} \quad=(2,3) \cup(3,4]
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function fx=log21-log12x2-5x+16 is

Options:
A 1,4
B -∞,4
C 1,∞
D 1,4

The correct answer is:
1,4

Solution:
The function fx is defined if
1-log12x2-5x+16>0 and x2-5x+16>0
⇒log12x2-5x+16<1 and x-522+394>0
⇒x2-5x+16<12 ∵x-522+394>0, ∀x∈R
⇒x2-5x+4<0
⇒x-1x-4<0⇒1<x<4
∴ Domain of f=1,4

Topic: Mathematics, Functions, JEE Main
The domain of the function
$f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ is

Options:
A $]-3,-2.5[\cap]-2.5,-2[$
B $[-2,0[\cup] 0,1]$
C $] 0,1[$
D None of the above

The correct answer is:
$[-2,0[\cup] 0,1]$

Solution:
$x+2 \geq 0$, i.e., $x \geq-2$ or $-2 \leq x$ $\log _{10}(1-x) \neq 0$ $\Rightarrow 1-x \neq 1 \Rightarrow x \neq 0$ Again, $1-x>0$ $\Rightarrow 1>x \Rightarrow x < 1$
Combining all the results for values of $x$, we get
$-2 \leq x < 0 \text { and } 0 < x < 1$

Topic: Mathematics, Functions, VITEEE, VITEEE 2013
The domain of the function $f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}$ is

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

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

Solution:
Given, function
$$
f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}
$$
$f(x)$ will be defined if,
$$
1-x>0, \neq 1 \text { and } x+2 \geq 0
$$
or $\quad 1>x, 1-x \neq 1$ and $x \geq-2$
or $x < 1, x \neq 0$ and $x \geq-2$
$$
\Rightarrow \quad x \in[-2,0) \cup(0,1)
$$

Topic: Mathematics, Functions, KCET, KCET 2022
The domain of the function
$f(x)=\log (1-x)+\sqrt{x^{2}-1}$

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

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

Solution:
We have, $f(x)=\log (1-x)+\sqrt{x^{2}-1}$ $\log (1-x)$ is defined if $1-x>0 \Rightarrow x < 1$
$\Rightarrow \quad x \in(-\infty,-1) \quad \text{...(i)}$
and $\sqrt{x^{2}-1}$ is defined if $x^{2}-1 \geq 0$
$$
\begin{array}{ll}
\Rightarrow & x^{2} \geq 1 \\
\Rightarrow & x \geq 1 \text { or } x \leq-1 \quad \text{...(ii)}
\end{array}
$$
By Eqs. (i) and (ii), required domain is $(-\infty,-1]$.

Topic: Mathematics, Functions, COMEDK, COMEDK 2012
The domain of the function $f(x)=\log _{3+x}\left(x^2-1\right)$ is

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

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

Solution:
$\quad f(x)$ is to be defined when $x^2-1\gt0$
$\begin{aligned}
& \Rightarrow x^2\gt1, \Rightarrow x \lt -1 \text { or } x\gt1 \text { and } 3+x\gt0 \\
& \therefore x\gt-3 \text { and } x \neq-2 \\
& \therefore D_f=(-3,-2) \cup(-2,-1) \cup(1, \infty)
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\sqrt{\log _{0.5} x !}$ is

Options:
A $\{0,1,2,3, \ldots\}$
B $\{1,2,3, \ldots\}$
C $(0, \infty)$
D $\{0,1\}$

The correct answer is:
$\{1,2,3, \ldots\}$

Solution:
We have, $f(x)=\sqrt{\log _{0.5} x !}$
So, $f(x)$ will be defined when,
$\begin{aligned}
& \log _{0.5} x ! \geq 0 \\
& x ! \leq(0.5)^0 \\
& x ! \leq 1 \quad \therefore x \in\{0,1\}
\end{aligned}$
Hence, the domain of the function is $\{0,1\}$.

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2016
The domain of the function, $f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}$

Options:
A $[0,1]$
B $[1,4]$
C $[4,5]$
D $(-\infty, \infty)$

The correct answer is:
$[1,4]$

Solution:
We have,
$f(x)=\sqrt{\log _{10}\left(\frac{5 x-x^2}{4}\right)}$
For $f(x)$ to be defined
$\begin{aligned}
& \therefore \quad x \in[1,4] \\
&
\end{aligned}$
So, domain of $f(x)$ is $[1,4]$.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 1)
The domain of the function $f(x)=\sqrt{\log \frac{1}{|\sin x|}}$ is

Options:
A $R-\{2 n \pi, n \in I\}$
B $R-\{n \pi, n \in I\}$
C $R-\{-\pi, \pi\}$
D $(-\infty, \infty)$

The correct answer is:
$R-\{n \pi, n \in I\}$

Solution:
$\begin{aligned} f(x) & =\sqrt{\log \frac{1}{|\sin x|}} \Rightarrow \sin x \neq 0 \Rightarrow x \neq n \pi+(-1)^n 0 \\ & \Rightarrow x \neq n \pi \text {. Domain of } f(x)=R-\{n \pi, n \in l\} .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\log (\sqrt{x-4}+\sqrt{6-x})$ is

Options:
A $[4, \infty)$
B $(-\infty, 6]$
C $[4,6]$
D None of these

The correct answer is:
$[4,6]$

Solution:
$\begin{aligned} & f(x)=\log (\sqrt{x-4}+\sqrt{6-x}) \\ & \Rightarrow x-4 \geq 0 \text { and } 6-x \geq 0 \Rightarrow x \geq 4 \text { and } x \leq 6 \\ & \therefore \text { Domain of } f(x)=[4,6] .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function fx=log3⁡1-log6⁡x2-7x+16 is

Options:
A 2,5
B ∞,5
C 2,∞
D 2,5

The correct answer is:
2,5

Solution:
fx is defined if
1-log6⁡x2-7x+16>0 and x2-7x+16>0
⇒log6⁡x2-7x+16<1 and x-722+154>0
⇒x2-7x+16<6 ∵x-722+154>0 for all real x
⇒x2-7x+10<0
⇒x-2x-5<0⇒2<x<5
∴ Domain of f=2,5

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\frac{\sqrt{4-x^{2}}}{\sin ^{-1}(2-x)}$ is

Options:
A $[0,2]$
B $[0,2)$
C $[1,2)$
D $[1,2]$

The correct answer is:
$[1,2)$

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\sqrt{4-\mathrm{x}^{2}}}{\sin ^{-1}(2-\mathrm{x})} \\ & \sqrt{4-\mathrm{x}^{2}} \text { is defined for } 4-\mathrm{x}^{2} \geq 0 \text {. } \\ & \Rightarrow \mathrm{x}^{2} \leq 4 \\ & \Rightarrow-2 \leq \mathrm{x} \leq 2 \\ & \text { and } \sin ^{-1}(2-\mathrm{x}) \text { is defined for }-1 \leq 2-\mathrm{x} \leq 1 \\ & \Rightarrow-3 \leq-\mathrm{x} \leq-1 \\ & \Rightarrow 1 \leq \mathrm{x} \leq 3 \end{aligned}$
Also, $\sin ^{-1}(2-x)=0$ for $x=2$
$\begin{aligned} \therefore \text { Domain off }(\mathrm{x}) &=[-2,2] \cap[1.31-\{2\\ &=[1,2) \end{aligned}$

Topic: Mathematics, Functions, VITEEE, VITEEE 2012
The domain of the function fx=sec-13x-4+tanh-1x+35 is

Options:
A -8,1∪53,2
B 1,53
C -8,1∪53,2
D (-8,1]∪53,2

The correct answer is:
(-8,1]∪53,2

Solution:
Given that

fx=sec-13x-4+tanh-1x+35

Domain of sec-1x

x ∈ (-∞,-1] ∪[1,∞)

sec-13x-4 is

⇒3x-4 ∈(-∞,-1] ∪[1,∞)

⇒3x ∈(-∞,3]∪[5,∞)

⇒x ∈(-∞,1]∪53,∞ ...(I)

We know that domain of tanh-1x is -1,1

⇒x+35 ∈-1,1

⇒x+3 ∈ -5,5

⇒x ∈-8,2 ...(II)

From I & (II)

(-8,1]∪53,2.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
The domain of the function $f(x)=\sin ^{-1}\left\{\left(1+e^x\right)^{-1}\right\}$ is

Options:
A $\left(\frac{1}{4}, \frac{1}{3}\right)$
B $[-1,0]$
C $[0,1]$
D $[-1,1]$

The correct answer is:
$\left(\frac{1}{4}, \frac{1}{3}\right)$

Solution:
Clearly
$-1 \leq \frac{1}{1+e^x} \leq 1$
But $2 \lt e^x \lt 3 \Rightarrow 3 \lt \left(e^x+1\right) \lt 4 \Rightarrow \frac{1}{4} \lt \frac{1}{1+e^x} \lt \frac{1}{3}$
$\therefore$ Domain of
$f(x)=\left(\frac{1}{4}, \frac{1}{3}\right)$

Topic: Mathematics, Functions, JEE Main
The domain of the function, fx=sin-13x2+x-1(x-1)2+cos-1x-1x+1 is:

Options:
A 0, 12
B 0, 14
C 14, 12∪0
D -2, 0∪14, 12

The correct answer is:
14, 12∪0

Solution:
The domain of the function

fx=sin-13x2+x-1(x-1)2+cos-1x-1x+1

For, cos-1x-1x+1

-1≤x-1x+1≤1

⇒-1≤1-2x+1≤1

⇒-2≤-2x+1≤0

⇒0≤1x+1≤1

⇒x+1∈[1,∞)

⇒x∈[0,∞) ....i

For, sin-13x2+x-1(x-1)2

-1≤3x2+x-1x-12≤1

⇒-x-12≤3x2+x-1≤x-12, x≠1

Now, -x-12≤3x2+x-1, x≠1

⇒4x2-x≥0 , x≠1

⇒x4x-1≥0, x≠1

⇒x∈(-∞, 0]∪14, ∞ -1 ....ii

And 3x2+x-1≤x-12, x≠1

⇒2x2+3x-2≤0, x≠1

⇒x+22x-1≤0, x≠1

⇒x∈-2, 12 ....iii

Domain of the function sin-13x2+x-1(x-1)2 from the equations ii & iii is

⇒x∈-2, 0∪14, 12 ....iv

Now the domain of the given function will be the intersection of the equation i & iv

Hence, domain is x∈14, 12∪{0}

Topic: Mathematics, Functions, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The domain of the function $f(x)=\sin ^{-1}\left(\log _2\left(\frac{x^2}{2}\right)\right)$ is

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

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

Solution:
$\begin{aligned} & f(x)=\sin ^{-1}\left(\log _2\left(\frac{x^2}{2}\right)\right) \\ \Rightarrow & -1 \leq \log _2\left(\frac{x^2}{2}\right) \leq 1 \\ \Rightarrow & \frac{1}{2} \leq \frac{x^2}{2} \leq 2 \Rightarrow 1 \leq x^2 \leq 4 \\ \therefore & x \in[-2,-1] \cup[1,2] .\end{aligned}$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2023 (12 May Shift 2)
The domain of the function $f(x)=\sin ^{-1}\left[\log _2(x / 2)\right]$ is

Options:
A $[1,4]$
B $[-4,1]$
C $[-1,4]$
D None of these

The correct answer is:
$[1,4]$

Solution:
$\quad f(x)=\sin ^{-1}\left[\log _2(x / 2)\right]$, Domain of $\sin ^{-1} x$ is $x \in[-1,1]$
$\begin{aligned}
& \Rightarrow{ }^{-1 \leq \log _2(x / 2) \leq 1} \Rightarrow{ }^{\frac{1}{2} \leq \frac{x}{2} \leq 2} \Rightarrow 1 \leq x \leq 4 \\
& \therefore x \in[1,4] .
\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The domain of the function
$f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}$ is

Options:
A $[-1,1]$
B $[1,4]$
C $(0,16]$
D $[1,16]$

The correct answer is:
$[1,16]$

Solution:
We have,
$$
f(x)=\sin ^{-1}\left[\log _4\left(\frac{x}{4}\right)\right]+\sqrt{17 x-x^2-16}
$$
$f(x)$ is defined of
$$
\begin{aligned}
& \log _4\left(\frac{x}{4}\right) \in[-1,1] \text { and } 17 x-x^2-16 \geq 0 \\
& \therefore 4^{-1} \leq \frac{x}{4} \leq 4 \text { and } x^2-17 x+16 \leq 0 \\
& \Rightarrow 1 \leq x \leq 16 \text { and }(x-16)(x-1) \leq 0 \\
& \Rightarrow 1 \leq x \leq 16 \text { and } 1 \leq x \leq 16 \\
& \therefore \text { Domain of } f(x) \text { is }[1,16] .
\end{aligned}
$$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2019 (04 May Shift 1)
The domain of the function $\mathrm{f}(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)$ is $(-\infty,-a] \cup[a, \infty)$. Then $a$ is equal to

Options:
A $\frac{\sqrt{17}}{2}+1$
B $\frac{\sqrt{17}-1}{2}$
C $\frac{1+\sqrt{17}}{2}$
D $\frac{\sqrt{17}}{2}-1$

The correct answer is:
$\frac{1+\sqrt{17}}{2}$

Solution:
$f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^2+1}\right)$
$\mathrm{f}(x)$ is defined, if
$\begin{aligned}
& -1 \leq \frac{|x|+5}{x^2+1} \leq 1 \\
& \Rightarrow 0 \leq \frac{|x|+5}{x^2+1} \leq 1 \\
& \Rightarrow x^2-|x|-4 \geq 0 \\
& \Rightarrow\left(|x|-\frac{1-\sqrt{17}}{2}\right)\left(|x|-\frac{1+\sqrt{17}}{2}\right) \geq 0 \\
& \Rightarrow|x| \geq \frac{1+\sqrt{17}}{2} \\
& \text { or }|x| \leq \frac{1-\sqrt{17}}{2}, \text { which is not possible } \\
& \Rightarrow x \in\left(-\infty,-\frac{1+\sqrt{17}}{2}\right] \cup\left[\frac{1+\sqrt{17}}{2}, \infty\right) \\
& \Rightarrow \mathrm{a}=\frac{1+\sqrt{17}}{2}
\end{aligned}$

Topic: Mathematics, Functions, MHT CET, MHT CET 2023 (10 May Shift 2)
The domain of the function fx=sin-1x2-3x+2x2+2x+7 is

Options:
A [1,∞)
B (-1,2]
C [-1,∞)
D (-∞,2]

The correct answer is:
[-1,∞)

Solution:
Given,

fx=sin-1x2-3x+2x2+2x+7

We know that domain of sin-1fx is -1≤fx≤1

So, x2-3x+2x2+2x+7≥-1 and x2-3x+2x2+2x+7≤1

Now solving x2-3x+2x2+2x+7≥-1

Since denominator x2+2x+7 is always positive,

So, x2-3x+2≥-x2+2x+7

⇒2x2-x+9≥0

⇒x∈R .....1

Now solving x2-3x+2x2+2x+7≤1

⇒x2-3x+2≤x2+2x+7

⇒5x≥-5

⇒x≥-1 .....2

So from equation 1 & 2 we get domain x∈[-1,∞)

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The domain of the function f(x)=sin−1x+5x2+1 is −∞, −a∪a, ∞, then a is equal to

Options:
A 172
B 17−12
C 1+172
D 172+1

The correct answer is:
1+172

Solution:
fx=sin−1x+5x2+1

⇒-1≤x+5x2+1≤1, ∵x∈-1,1 for sin-1x to exist.

Case 1

-x2-1≤x+5

⇒x2+1+x+5≥0, true for all x∈R.

Case 2

x+5≤x2+1

⇒x2-x-4≥0

Let x=t⇒t2-t-4≥0

⇒x+17-12x-17+12≥0

⇒x∈-∞,-17+12∪17+12,∞

∴a=1+172

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The domain of the function $f(x)=\sqrt{x-1}+\sqrt{6-x}$ is

Options:
A $[1, \infty)$
B $[1,6]$
C $(-\infty, 6]$
D $(-\infty, 6)$

The correct answer is:
$[1,6]$

Solution:
$$
f(x)=\sqrt{x-1}+\sqrt{6-x}
$$
Here $\mathrm{f}(\mathrm{x})$ is defined when
$x-1 \geq 0$ and $6-x \geq 0 \quad$ i.e.
$x \geq 1$ and $x-6 \leq 0 \quad$ i.e. $x \geq 1$ and $x \leq 6$
$\therefore$ Domain of $\mathrm{f}(\mathrm{x})$ is $[1,6]$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\sqrt{x-1}+\sqrt{6-x}$ is

Options:
A $[1, \infty)$
B $[1,6]$
C $(-\infty, 6]$
D $(-\infty, 6)$

The correct answer is:
$[1,6]$

Solution:
$$
f(x)=\sqrt{x-1}+\sqrt{6-x}
$$
Here $\mathrm{f}(\mathrm{x})$ is defined when
$x-1 \geq 0$ and $6-x \geq 0 \quad$ i.e.
$x \geq 1$ and $x-6 \leq 0 \quad$ i.e. $x \geq 1$ and $x \leq 6$
$\therefore$ Domain of $\mathrm{f}(\mathrm{x})$ is $[1,6]$

Topic: Mathematics, Functions, MHT CET, MHT CET 2021 (22 Sep Shift 1)
The domain of the function fx=x-1-x24 is

Options:
A -1,-12∪12,1
B -1,1
C -∞,-12∪12,∞
D 12,1

The correct answer is:
12,1

Solution:
fx=x-1-x24
For fx to be real
x-1-x2≥0 and 1-x2≥0
⇒2x2-1≥0 and x2-1≤0
⇒x+12x-12≥0 and x+1x-1≤0
⇒x≤-12 or x≥12 and x∈-1,1
When x∈-1,-12, then x-1-x2 is negative
Hence, the required domain is 12,1

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, where $[x]$ denotes the greatest integer not more than $x$, is

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

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

Solution:
Given function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, is define if
$$
\frac{4-x^2}{[x]+2} \geq 0 \Rightarrow \frac{x^2-4}{[x]+2} \leq 0
$$
So, either $x^2-4 \leq 0$
From intervals Eqs. (iii) and (iv),
$$
x \in(-\infty,-2) \cup[-1,2]
$$

Topic: Mathematics, Functions, JEE Main
The domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, where $[x]$ denotes the greatest integer not more than $x$, is

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

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

Solution:
Given function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$, is define if
$$
\frac{4-x^2}{[x]+2} \geq 0 \Rightarrow \frac{x^2-4}{[x]+2} \leq 0
$$
So, either $x^2-4 \leq 0$
From intervals Eqs. (iii) and (iv),
$$
x \in(-\infty,-2) \cup[-1,2]
$$

Topic: Mathematics, Functions, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 2)