datavorous/entrance-exam-dataset huggingface dataset:
The set $\mathrm{A}=\{\mathrm{x}: \mathrm{x}+4=4\}$ can also be represented by

Options:
A 0
B $\varphi$
C $\{\varphi\}$
D $\{0\}$

The correct answer is:
$\{0\}$

Solution:
Let $\begin{aligned} \mathrm{A} &=\{\mathrm{x}: \mathrm{x}+4=4\}=\{\mathrm{x}: \mathrm{x}=4-4\} \\ &=\{\mathrm{x}: \mathrm{x}=0\}=\{0\} \end{aligned}$
Let the people who read all three papers $\mathrm{A}, \mathrm{B}, \mathrm{C}=\mathrm{x} \%$ So, people who read only A and $\mathrm{B}$ not $\mathrm{C}=(30-\mathrm{x}) \%$ People who read only $\mathrm{B}$ and $\mathrm{C}$, not $\mathrm{A}=(28-\mathrm{x}) \%$ People who read only A and $\mathrm{C}$, not $\mathrm{B}=(36-\mathrm{x}) \%$
Venn diagram representing these is shown below
Remaining numbers in circles are filled asshown below.
People who read only $\mathrm{A}+30-\mathrm{x}+\mathrm{x}+36-\mathrm{x}=42$
People reading only $\mathrm{A}=42-30-36+\mathrm{x}=(\mathrm{x}-24) \%$
Similarly, people who read only $\mathrm{B}=51-(30-\mathrm{x}+\mathrm{x}+28-\mathrm{x})$
$=51-(58-x)=51-58+x=(x-7) \%$
People who read only $\mathrm{C}=68-(36-\mathrm{x}+\mathrm{x}+28-\mathrm{x})$
$68-(64-x)=(x+4) \%$
Let $x \%$ people read all the three newspap
$8 \%$ people do not read any newspapers. $(x-24)+(x-7)+(x+4)+(30-x)+(36-x)+(28-x)+x=92$
$x+98-31=92$
$x=92-67=25$

Topic: Mathematics, Sets and Relations, NDA, NDA 2012 (Phase 1)
The set containing only transition metals is

Options:
A Ti, Nb, Ra
B Pd, Ir, Ta
C Ag, Au, In
D W, Pt, Po

The correct answer is:
Pd, Ir, Ta

Solution:
Transition metals = Ti, Nb, Pd, Ir, Ta, Ag, Au, W, Pt
Representative elements = Ra, In, Po

Topic: Chemistry, d and f Block Elements, JEE Main
The set having ions which are coloured and paramagnetic both is -

Options:
A Cu2+, Cr3+, Sc+
B Cu2+, Zn2+, Mn4+
C Sc3+, V5+, Ti4+
D Ni2+, Mn7+, Hg2+

The correct answer is:
Cu2+, Cr3+, Sc+

Solution:
Cu2+:[Ar]3 d94 s0
Cr3+:[Ar]3 d34 s0
Sc+:[Ar]3 d14 s1

All are coloured and paramagnetic due to presence of unpaired electrons.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (22 Jul Shift 1)
The set of all $\alpha$, for which the vectors $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in \mathbb{R}$, is

Options:
A $\left(-\frac{4}{3}, 1\right)$
B $[0,1)$
C $\left(-\frac{4}{3}, 0\right]$
D $(-2,0]$

The correct answer is:
$\left(-\frac{4}{3}, 0\right]$

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}=\mathrm{t} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \alpha \hat{\mathrm{k}} \\ & \text { so } \overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} < 0, \forall \mathrm{t} \in \mathrm{R} \\ & \alpha \mathrm{t}^2-12+6 \alpha \mathrm{t} < 0 \\ & \alpha \mathrm{t}^2+6 \alpha \mathrm{t}-12 < 0, \forall \mathrm{t} \in \mathrm{R} \\ & \alpha < 0 \text {, and } \mathrm{D} < 0 \\ & 36 \alpha^2+48 \alpha < 0 \\ & 12 \alpha(3 \alpha+4) < 0 \\ & \frac{-4}{3} < \alpha < 0\end{aligned}$
also for $\mathrm{a}=0, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}} < 0$
hence a $\alpha \in\left(\frac{-4}{3}, 0\right]$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The set of all possible values of θ in the interval 0,π for which the points1, 2 and sinθ, cosθ lie on the same side of the line x+y=1 is?

Options:
A 0,π2
B π4,3π4
C 0,3π4
D 0,π4

The correct answer is:
0,π2

Solution:
We know that if Ax1,y1 & Bx2,y2 lies on the same side to line L=0, then LAx1,y1 LBx2,y2>0.

Given, the points A1,2 & Bsinθ,cosθ lies on the same side of the line x+y-1=0.

∴1+2-1sinθ+cosθ-1>0

⇒sinθ+cosθ>1

⇒sinθ×12+cosθ×12>1×12

⇒sinθ.cosπ4+cosθ.sinπ4>12

⇒sinθ+π4>12

⇒π4<θ+π4<3π4

⇒0<θ<π2

⇒θ∈0,π2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The set of all α∈R, for which w=1+1-8αz1-z is a purely imaginary number, for all z∈C satisfying z=1 and Rez≠1, is :

Options:
A 0
B 0,14,-14
C equal to R
D an empty set

The correct answer is:
0

Solution:
For w to be purely imaginary, 1+1-8αz1-z+1+1-8αz¯1-z¯=0

⇒1-z¯+1-8αz-1-8α+1-z+1-8αz¯-1-8α=0

⇒2-z+z¯+1-8αz+z¯-2+16α=0

⇒8αz+z¯=16α

⇒z+z¯=2 or α=0

For all z∈C, we have α=0.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2018 (15 Apr)
The set of all $\alpha \in R$, for which $w=\frac{1+(1-8 \alpha) z}{1-z}$ is a purely imaginary number, for all $z \in C$ satisfying $|z|=1$ and $\operatorname{Re} z \neq 1$, is

Options:
A
$\{0\}$

B
an empty set

C
$\left\{0, \frac{1}{4},-\frac{1}{4}\right\}$

D
equal to $R$

The correct answer is:
$\{0\}$

Solution:
$$
\begin{aligned}
&\because|z|=1 \text { \& } \operatorname{Re} z \neq 1 \\
&\text { Suppose } z=x+i y \Rightarrow x^2+y^2=1 \ldots \ldots(\mathrm{i}) \\
&\text { Now, } w=\frac{1+(1-8 \alpha) z}{1-z} \\
&\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y)}{1-(x+i y)} \\
&\Rightarrow w=\frac{1+(1-8 \alpha)(x+i y))((1-\mathrm{x})+\mathrm{iy})}{1-(x+i y))((1-\mathrm{x})+\mathrm{iy})} \\
&\Rightarrow \quad w=\frac{\left[\left(1+x(1-8 \alpha)(1-x)-(1-8) y^2\right]\right.}{(1-x)^2+y^2} \\
&+i \frac{[(1+x(1-8 \alpha)) y-(1-8 \alpha) y(1-x)]}{(1-x)^2+y^2}
\end{aligned}
$$
If, $w$ is purely imaginary. So,
$$
\begin{aligned}
&\operatorname{Re} w=\frac{\left[(1+x(1-8 \alpha))(1-\alpha)-(1-8 \alpha) y^2\right]}{(1-x)^2+y^2} \\
&=0 \\
&\Rightarrow(1-x)+x(1-8 \alpha)(1-x)=(1-8) y^2 \\
&\Rightarrow(1-x)+x(1-8 \alpha)-x^2(1-8 \alpha)=(1-8 x) y^2 \\
&\Rightarrow(1-x)+x(1-8 \alpha)=1-8 \alpha \\
&{\left[\text { From }(i), x^2+y^2=1\right]}
\end{aligned}
$$
$\Rightarrow 1-8 \alpha=1$
$\Rightarrow \alpha=0$
$\therefore \alpha \in\{0\}$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles $x^2+y^2-4 x-4 y+6=0$ and $\mathrm{x}^2+\mathrm{y}^2-10 \mathrm{x}-10 \mathrm{y}+\lambda=0$ is the interval:

Options:
A
$(12,32)$

B
$(18,42)$

C
$(12,24)$

D
$(18,48)$

The correct answer is:
$(18,42)$

Solution:
The equations of the circles are
$$
x^2+y^2-10 x-10 y+\lambda=0
$$
and $x^2+y^2-4 x-4 y+6=0$
$$
\begin{aligned}
&C_1=\text { centre of }(1)=(5,5) \\
&\mathrm{C}_2=\text { centre of }(2)=(2,2) \\
&d=\text { distance between centres } \\
&=\mathrm{C}_1 \mathrm{C}_2=\sqrt{9+9}=\sqrt{18} \\
&r_1=\sqrt{50-\lambda}, r_2=\sqrt{2} \\
&\text { For exactly two common tangents we } \\
&\text { have } \\
&r_1-r_2 < C_1 C_2 < r_1+r_2 \\
&\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\
&\Rightarrow \sqrt{50-\lambda}-\sqrt{2} < 3 \sqrt{2} \text { or } 3 \sqrt{2} < \sqrt{50-\lambda}+\sqrt{2} \\
&\Rightarrow \sqrt{50-\lambda} < 4 \sqrt{2} \text { or } 2 \sqrt{2} < \sqrt{50-\lambda} \\
&\Rightarrow 50-\lambda < 32 \text { or } 8 < 50-\lambda \\
&\Rightarrow \lambda>18 \text { or } \lambda < 42 \\
&\text { Required interval is }(18,42) \\
&
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2014 (11 Apr Online)
The set of all real values of λ for which the quadratic equation λ2+1x2-4λx+2=0 always have exactly one root in the interval (0, 1) is :

Options:
A −3, −1
B 0, 2
C (1, 3]
D (2, 4]

The correct answer is:
(1, 3]

Solution:
f0·f1≤0

⇒2λ2+1-4λ+2≤0⇒2λ2-4λ+3≤0

⇒λ-1λ-3≤0

⇒λ∈1, 3

But at λ=1, both roots are 1.

So, λ≠1

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The set of all the possible real values of a such that the inequality x-a-1x-a2+2<0 holds for all x∈-1,3, is

Options:
A 1,∞
B -∞,-1
C -∞,1
D 0,1

The correct answer is:
-∞,-1

Solution:
a-1<-1 and a2+2>3
a<0 and a2>1
a<0 and a>1 or a<-1
⇒a<-1

Topic: Mathematics, Quadratic Equation, JEE Main
The set of all the values of \( x \) satisfying the inequality \( \left(\cot ^{-1} x\right)^{2}-7\left(\cot ^{-1} x\right)+10>0 \) is

Options:
A \( (-\infty, \cot 5) \cup(\cot 4, \cot 2) \)
B \( (\cot 5, \cot 4) \)
C \( (-\infty, \cot 5) \cup(\cot 2, \infty) \)
D \( (\cot 2, \infty) \)

The correct answer is:
\( (\cot 2, \infty) \)

Solution:
We have, cot-1x2-7cot-1x+10>0

⇒cot-1x-5cot-1x-2>0
⇒cot-1x∈-∞,2∪5,∞ ...1

But, cot-1x∈0, π ...2

Taking intersection of 1 and 2 we get,

⇒cot-1x∈0,2

⇒x∈cot2,∞

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The set of all values of a for which the equation cos⁡2x+asin⁡x=2a-7 has a solution is

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

The correct answer is:
[2,6]

Solution:
cos⁡2x+asin⁡x=2a-7
i.e. 2sin2⁡x-asin⁡x+2a-8=0
sin⁡x=a±a2-82a-84=a±a-84
sin⁡x=a-42 or 2
Hence, -1≤a-42≤1

The range of a is [2,6]

Topic: Mathematics, Trigonometric Equations, JEE Main
The set of all values of λ for which the equation cos22x-2sin4x-2cos2x=λ

Options:
A -2,-1
B -2,-32
C -1,-12
D -32,-1

The correct answer is:
-32,-1

Solution:
Given:

λ=cos22x-2sin4x-2cos2x

⇒λ=2cos2x-12-21-cos2x2-2cos2x

⇒λ=4cos4x-4cos2x+1-21-2cos2x+cos4x-2cos2x

⇒λ=2cos4x-2cos2x+1-2

⇒λ=2cos4x-2cos2x-1

⇒λ=2cos4x-cos2x-12

⇒λ=2cos2x-122-34

So,

λmax=214-34=-1 (max Value)

λmin=20-34=-32 (Minimum Value)

So, range of λ is-32,-1

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The set of all values of λ for which the system of linear equations:
2 x 1 - 2 x 2 + x 3 = λ x 1
2 x 1 - 3 x 2 + 2 x 3 = λ x 2
- x 1 + 2 x 2 = λ x 3
has a non-trivial solution,

Options:
A Contains more than two elements.
B Is an empty set.
C Is a singleton.
D Contains two elements.

The correct answer is:
Contains two elements.

Solution:
2 - Î» x 1 - 2 x 2 + x 3 = 0
2 x 1 - Î» + 3 x 2 + 2 x 3 = 0
- x 1 + 2 x 2 - Î» x 3 = 0
The systems of linear equations will have a non-trivial solution
â 2 - Î» - 2 1 2 - Î» - 3 2 - 1 2 - Î» = 0
â 2 - Î» Î» 2 + 3 Î» - 4 + 2 - 2 Î» + 2 + 1 4 - Î» - 3 = 0
2 Î» 2 + 6 Î» - 8 - Î» 3 - 3 Î» 2 + 4 Î» - 4 Î» + 4 + 1 - Î» = 0
- Î» 3 - Î» 2 + 5 Î» - 3 = 0
Î» 3 + Î» 2 - 5 Î» + 3 = 0
Î» - 1 Î» 2 + 2 Î» - 3 = 0
Î» - 1 Î» + 3 Î» - 1 = 0
â Î» = -3 , 1 , 1

Topic: Mathematics, Determinants, JEE Main, JEE Main 2015 (04 Apr)
The set of all values of λ for which the system of linear equations
x-2y-2z=λx
x+2y+z=λy
-x-y=λz
has a non-trivial solution :

Options:
A is an empty set
B contains more than two elements
C is a singleton
D contains exactly two elements

The correct answer is:
is a singleton

Solution:
For non-trivial solution, ∆=0.

⇒1-λ-2-212-λ1-1-1-λ=0

⇒1-λ-2λ+λ2+1+1-λ+1-1 -1+2-λ=0

⇒λ2-3λ2+3λ-1=0

⇒λ-13=0

⇒λ=1

Singleton set

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The set of meta directing functional groups from the following sets is:

Options:
A -CN,-NH2,-NHR,-OCH3
B -NO2,-NH2,-COOH,-COOR
C -NO2,-CHO,-SO3H,-COR
D -CN,-CHO,-NHCOCH3,-COOR

The correct answer is:
-NO2,-CHO,-SO3H,-COR

Solution:
Due to the presence of some groups, the electron density decreases on ortho and para positions. So, the electron density on meta position is relatively higher than ortho and para positions. Thus, these groups directs the electrophilic substitution on meta position.

Due to the presence of some groups, the electron density decreases on ortho and para positions. So, the electron density on meta position is relatively higher than ortho and para positions. Thus, these groups directs the electrophilic substitution on meta position. If the atom of group which is attached to benzene ring has pi bond then that group act as meta directing group.

All are –M, Hence meta directing groups.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The set of points on the axis of the parabola y2-2y-4x+5=0 from which all the three normals drawn to the parabola are real and distinct, is

Options:
A {x,1:x>3}
B {x,-1:x≥1}
C {x,3:x≥1}
D {x,-3:x≥3}

The correct answer is:
{x,1:x>3}

Solution:
Given equation of the parabola can be rewritten as
(y-1)2=4(x-1)
Axis of the parabola is y=1,
equation of normal is
y-1=mx-1-2m-m3
Let (h, 1) is a point on its axis, then
0=mh-1-2m-m3
âm2=h-3
âh>3 for real & distinct values of m

Topic: Mathematics, Parabola, JEE Main
The set of points where $f(x)=\frac{x}{1+|x|}$ is differentiable is

Options:
A
$(-\infty, 0) \cup(0, \infty)$

B
$(-\infty,-1) \cup(-1, \infty)$

C
$(-\infty, \infty)$

D
$(0, \infty)$

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

Solution:
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{1-x}, & x < 0 \\
\frac{x}{1+x}, & x \geq 0
\end{array} \quad \Rightarrow f^{\prime}(x)= \begin{cases}\frac{1}{(1-x)^2}, & x < 0 \\
\frac{1}{(1+x)^2}, & x \geq 0\end{cases}\right.
$$
$\therefore f^{\prime}(x)$ exist at everywhere.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2006
The set of real values of $\alpha$ for which the system of linear equations
$$
\begin{aligned}
& x+(\sin \alpha) y+(\cos \alpha) z=0 \\
& x+(\cos \alpha) y+(\sin \alpha) z=0 \\
& -x+(\sin \alpha) y-(\cos \alpha) z=0
\end{aligned}
$$
has a non-trivial solution is

Options:
A $\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}+\frac{\pi}{8}$ ( $n$ is an integer)
B $\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}$ ( $n$ is an integer)
C $\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}$ ( $n$ is an integer)
D $\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}-\frac{\pi}{8}$ ( $n$ is an integer)

The correct answer is:
$\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}$ ( $n$ is an integer)

Solution:
$$
\begin{aligned}
x+\sin \alpha y+\cos \alpha z & =0, \quad x+\cos \alpha y+\sin \alpha z=0 \\
-x+\sin \alpha y-\cos \alpha z & =0
\end{aligned}
$$
$$
\begin{aligned}
& \text { Non-trivial sol, so }\left|\begin{array}{ccc}
1 & \sin \alpha & \cos \alpha \\
1 & \cos \alpha & \sin \alpha \\
-1 & \sin \alpha & -\cos \alpha
\end{array}\right|=0 \\
& \Rightarrow 1\left(-\sin ^2 \alpha-\cos ^2 \alpha\right)-\sin \alpha(-\cos \alpha+\sin \alpha) \\
& +\cos \alpha(\sin \alpha+\cos \alpha) \mid=0 \\
& -1-\sin \alpha(-\cos \alpha+\sin \alpha]+\cos \alpha[\sin \alpha+\cos \alpha]=0 \\
& -1+\sin \alpha \cos \alpha-\sin ^2 \alpha+\sin \alpha \cos \alpha+\cos ^2 \alpha=0 \\
& -1+\sin 2 \alpha+\cos 2 \alpha=0 \\
& \sin 2 \alpha+\cos 2 \alpha=1 \\
& \Rightarrow \quad \frac{1}{\sqrt{2}} \sin 2 \alpha+\frac{1}{\sqrt{2}} \cos 2 \alpha=\frac{1}{\sqrt{2}} \\
& \Rightarrow \quad \sin \left(2 \alpha+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad 2 \alpha+\frac{\pi}{4} & =n \pi+(-1)^n \frac{\pi}{4} \\
2 \alpha & =n \pi+(-1)^n \frac{\pi}{4}-\frac{\pi}{4} \\
\Rightarrow \quad \alpha & =\frac{n \pi}{2}+(-1)^n \frac{\pi}{8}-\frac{\pi}{8}
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
The set of values of a for which limx→ax-5-2x+2=0, where, ζ denotes the greatest integer less than or equal to ζ is equal to

Options:
A -7.5,-6.5
B (-7.5,-6.5]
C -7.5,-6.5
D [-7.5,-6.5)

The correct answer is:
-7.5,-6.5

Solution:
We have,

limx→ax-5-2x+2=0

⇒limx→ax-5-2x-2=0

⇒limx→ax-2x-7=0

⇒limx→ax-2x=7

Now when a∈I (integer) then a-2a=7⇒a=-7

Now when a∈I+f where f is fraction,

So, I+f-2I+2f=7

⇒I+f-2I-2f=7

⇒I+0-2I-2f=7

⇒-I-2f=7

Now taking case 1 when f∈0,12

Then ⇒-I-0=7⇒I=-7⇒a∈-7,-6.5

Now taking case 2 when f∈12,1

So, -I-1=7⇒I=-8⇒a∈-7.5,7

Hence, from both the cases we can say that a∈-7.5,-6.5.

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The set of value(s) of \( a \) for which the function \( f(x)=\frac{a x^{3}}{3}+(a+2) x^{2}+(a-1) x+2 \) possess a negative point of
inflection, is

Options:
A empty set.
B \( \left\{-\frac{4}{5}\right\} \)
C \( (-2,0) \)
D \( (-\infty,-2) \cup(0, \infty) \)

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

Solution:
Given fx=ax33+a+2x2+a-1x+2

Differentiating both sides with respect to x, we get

f'x=3ax23+2a+2x+a-1=ax2+2a+2x+a-1

Again, differentiating both sides with respect to x, we get

f''x=2ax+2a+2

Since, fx has a point of inflection.

So, f''x=0

⇒2ax+2a+2=0

⇒x=-a+2a

Since, fx has a negative point of inflection.

So, x=-a+2a<0

⇒a+2a>0

⇒a∈-∞,-2∪0,∞

Topic: Mathematics, Application of Derivatives, JEE Main
The set of values of b for which local extremum values of the function f(x) are positive where fx=23 a2 x3-5a2 x2+3x+b and maxima occurs at x=13 is

Options:
A -4,Â â
B -38, â
C -10,38
D None of these

The correct answer is:
-38, â

Solution:
f′x=2a2 x2-5ax+3=ax-1 2ax-3=0
x=1a, 32a
If a>0, then local maxima occurs at x=1a and minima at x=32a
∵ maxima occurs at x=1a=13 ⇒ a=3
Minima occurs at x=32a=12
f12>0⇒ 38+b>0⇒ b>-38
If a<0, then maxima shall occur at x=32a and minima at x=12a
i.e. 32a=13⇒ a=92>0 (not acceptable)
Hence, b>-38

Topic: Mathematics, Application of Derivatives, JEE Main
The set of values of k for which the circle C:4x2+4y2-12x+8y+k=0 lies inside the fourth quadrant and the point 1,-13 lies on or inside the circle C is

Options:
A An empty set
B 6,959
C 809,10
D 9,929

The correct answer is:
9,929

Solution:
Given equation of circle C:4x2+4y2-12x+8y+k=0

⇒x2+y2-3x+2y+k4=0

Centre 32,-1;r=13-k2⇒k≤13 …1

case (i) Point 1,-13 lies on or inside circle C

⇒S1≤0⇒k≤929 ⋯2

case (ii) C lies in 4th quadrant

r<1

⇒13-k2<1

⇒k<9 ⋯3

Hence from equation 1∩2∩3⇒k∈(9,929]

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The set of values of $\alpha$ such that $f: \mathbf{R} \rightarrow\left[0, \frac{\pi}{2}\right)$ defined by $f(x)=\tan ^{-1}\left(x^2+x+\alpha^2\right)$ is onto is

Options:
A $\left(\frac{-1}{2}, \frac{1}{2}\right)$
B $\left(\frac{-1}{4}, \frac{1}{4}\right)$
C $\left(-\infty, \frac{-1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$
D $\left(-\infty, \frac{-1}{4}\right) \cup\left(\frac{1}{4}, \infty\right)$

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

Solution:
Let $A=\left\{x: 0 \leq x < \frac{\pi}{2}\right\}$
Since $f: \mathbf{R} \rightarrow A$ is an onto function, therefore,
Range of $f=A$
$\Rightarrow 0 \leq f(x) \leq \frac{\pi}{2}$
for all $x \in \mathbf{R}$
$\Rightarrow 0 \leq \tan ^{-1}\left(x^2+x+\alpha^2\right) \leq \frac{\pi}{2}$
for all $x \in \mathbf{R}$
$\Rightarrow \quad 0 \leq x^2+x+\alpha^2 \leq \infty \quad$ for all $x \in \mathbf{R}$
$\Rightarrow \quad x^2+x+\alpha^2 \geq 0 \quad$ for all $x \in \mathbf{R}$
$\Rightarrow \quad 1-4 \alpha^2 \leq 0 \Rightarrow \alpha^2 \geq \frac{1}{4}$
$\therefore \quad\left(-\infty,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The set of values of $x$ such that $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ is

Options:
A $\phi$
B $\left\{\frac{1}{2}\right\}$
C $\left\{\frac{1}{3}, 2\right\}$
D $\left\{\frac{1}{3}, 4\right\}$

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

Solution:
We have,
$\begin{aligned}
& \tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right) \\
& \tan ^{-1} \frac{\left(\frac{x}{x-2}-\frac{x}{2 x-1}\right)}{1+\left(\frac{x}{x-2}\right)\left(\frac{x}{2 x-1}\right)}=\tan ^{-1} \frac{2}{3} \\
& \Rightarrow \frac{2 x^2-x-x^2+2 x}{2 x^2-4 x-x+2+x^2}=\frac{2}{3} \Rightarrow \frac{x^2+x}{3 x^2-5 x+2}=\frac{2}{3} \\
& \Rightarrow \quad 3 x^2+3 x=6 x^2-10 x+4 \\
& \Rightarrow \quad 3 x^2-13 x+4=0 \Rightarrow(3 x-1)(x-4)=0 \\
& \Rightarrow \quad x=\frac{1}{3}, 4, \quad x \in\left\{\frac{1}{3}, 4\right\}
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The set representing the correct order of ionic radii is

Options:
A Li +  >  Be 2+  >  Na +  >  Mg 2+
B Li +  >  Na +  >  Mg 2+  >  Be 2+
C Mg 2+  >  Be 2+  >  Li +  >  Na +
D Na + â>â Li + â>â Mg 2+ â>â Be 2+

The correct answer is:
Na +  >  Li +  >  Mg 2+  >  Be 2+

Solution:
2 + charge cation is smaller in size than 1 + charge cation due to more effective nuclear charge.
Na +  >  Li + (down the group)
Mg 2+  >  Be 2+ (down the group)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The set representing the correct order of ionic radius is

Options:
A Li+>Be2+>Na+>Mg2+
B Na+>Li+>Mg2+>Be2+
C Li+>Na+>Mg2+>Be2+
D Mg2+>Be2+>Li+>Na+

The correct answer is:
Na+>Li+>Mg2+>Be2+

Solution:
In general the ionic radius increases on moving from top to bottom in group and decreases on moving from left to right in period. So, Na+ > Li+ and Mg2+ > Be2+.
The size of Mg2+ is however smaller than Li+. Despite the Mg atom having a slightly larger atomic size, the Mg2+ ion is marginally smaller in size than the Li+ ion because of its higher nuclear charge
Hence, the correct trend is
Na0.98â«+>Li0.68â«+>Mg0.65â«2+>Be0.62â«2+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The set representing the correct order of ionic radius is :

Options:
A
$\mathrm{Li}^{+}>\mathrm{Be}^{2+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}$

B
$\mathrm{Na}^{+}>\mathrm{Li}^{+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}$

C
$\mathrm{Li}^{+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}$

D
$\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}>\mathrm{Li}^{+}>\mathrm{Na}^{+}$

The correct answer is:
$\mathrm{Na}^{+}>\mathrm{Li}^{+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}$

Solution:
Follow the periodic trends

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2009
The set $S=\{1,2,3, \ldots, 12)$ is to be partitioned into three sets $A, B, C$ of equal size. Thus, $A \cup B \cup C=S, A \cap B=B \cap C=A \cap C=\phi$. The number of ways to partition $S$ is

Options:
A
$\frac{12 !}{3 !(4 !)^3}$

B
$\frac{12 !}{3 !(3 !)^4}$

C
$\frac{12 !}{(4 !)^3}$

D
$\frac{12 !}{(3 !)^4}$

The correct answer is:
$\frac{12 !}{(4 !)^3}$

Solution:
Number of ways is ${ }^{12} \mathrm{C}_4 \times{ }^8 \mathrm{C}_4 \times{ }^4 \mathrm{C}_4$
$=\frac{12 !}{(4 !)^3}$.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2007
The set that contains atomic numbers of only transition elements, is :

Options:
A 37,42,50,64
B 21,25,42,72
C 9,17,34,38
D 21,32,53,64

The correct answer is:
21,25,42,72

Solution:
A transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of the oxidation state. Zn(z = 30), Cd(z = 48), Hg(z = 80) are not transition element.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The set that represents the pair of neutral oxides of nitrogen is:

Options:
A NO and N2O
B N2O and N2O3
C N2O and NO2
D NO and NO2

The correct answer is:
NO and N2O

Solution:
N2O and NO are neutral oxides of nitrogen NO2 and N2O3 are acidic oxides.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (17 Mar Shift 2)
The set which does not have ambidentate ligand(s) is

Options:
A

C2O42-, ethylene diammine, H2O

B

EDTA ,4-NCS-,C2O42-

C

NO2-,C2O42-, EDTA 4-

D

C2O42-,NO2-,NCS-

The correct answer is:

C2O42-, ethylene diammine, H2O

Solution:
Ligand which has two different donor atoms and either of the two donor atom is attached to the metal during complex formation is called ambidentate ligand. Among the given ligands NCS-SCN- and NO2-ONO- are ambidentate ligands

C2O42- bidentate,ethylenediamine(bidentate) and H2O(monodentate) ligands are not ambidentate.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The set which was all the species planar is: -

Options:
A I3–,CO2,XeF4,I2Cl6
B CO2,SF4,ClF3,BrF5
C XeF4,H2O,XeO4,PF3
D ICl2+,ICl2–,CO2,XeO3

The correct answer is:
I3–,CO2,XeF4,I2Cl6

Solution:
CO2→O=C=O

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The set {x∈R:cos⁡2x+2cos2⁡x=2} is equal to

Options:
A 2nπ+π3:n∈Z
B nπ±π6:n∈Z
C nπ+π3:n∈Z
D 2nπ-π3:n∈Z

The correct answer is:
nπ±π6:n∈Z

Solution:
Given, 2cos2â¡x-1+2cos2â¡x=2
â cos 2 x= 3 4
cos 2 x= cos 2 Ï 6
â´x=nÏÂ±Ï6 :nâZ

Topic: Mathematics, Trigonometric Equations, JEE Main
The setting time of Cement is increased by adding

Options:
A Clay
B Silica
C Gypsum
D Limestone

The correct answer is:
Gypsum

Solution:
Cement or Portland cement mainly contains oxides of calcium, silicon, aluminium, and iron. When water is mixed with cement, the oxides get hydrated, and cement sets extremely fast. To control the setting time of cement, gypsum is added to it, which increases the setting time of cement.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (06 Apr Shift 1)