datavorous/entrance-exam-dataset huggingface dataset:
Paragraph: If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z =xy. If the errors in x, y and z are ∆x, ∆y and ∆z , respectively, then
z±∆z=x±∆xy±∆y=xy1±∆xx1±∆yy-1
The series expansion for 1±∆yy-1, to first power in ∆y/y, is 1∓∆y/y. The relative errors in independent variables are always added. So the error in z will be
Δz=z( Δx x + Δy y ).
The above derivation makes the assumption that \(\Delta r / x \ll 1, \Delta y / y < 1\). Therefore, the higher powers of these quantities are neglected.
Question : Consider the ratio \(r=\frac{(1-a)}{(1+a)}\) to be determined by measuring a dimensionless quantity a If the error in the measurement of a is \(\Delta a(\frac{\Delta a}{a}<<1)\), then what is the error \(\Delta r\) in determining \(r\) ?

Options:
A ∆a1+a2
B 2∆a1+a2
C 2∆a1-a2
D 2a∆a1-a2

The correct answer is:
2∆a1+a2

Solution:
r=1-a1+a
∆rr=∆1-a 1-a+∆1+a1+a
= ∆a1-a+∆a1+a
= ∆a1+a+1-a1-a1+a
∴ ∆r=2∆a1-a1+a1-a1+a=2∆a1+a2

Topic: Physics, Mathematics in Physics, JEE Advanced, JEE Advanced 2018 (Paper 1)
Paragraph: If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z =xy. If the errors in x, y and z are ∆x, ∆y and ∆z , respectively, then
z±∆z=x±∆xy±∆y=xy1±∆xx1±∆yy-1
The series expansion for 1±∆yy-1, to first power in ∆y/y, is 1∓∆y/y. The relative errors in independent variables are always added. So the error in z will be
Δz=z( Δx x + Δy y ).
Question : The above derivation makes the assumption that Δr x ≪1, Δy y ≪1 . Therefore, the higher powers of these quantities are neglected.

Options:
A 0.04
B 0.03
C 0.02
D 0.01

The correct answer is:
0.02

Solution:
N=N0e-λt
ln⁡N=ln⁡N0-λt
dNN=-dλt
Converting to error,
∆NN=∆λt
∴ ∆λ=402000×L=0.02 ( N is number of nuclei left undecayed)

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2018 (Paper 1)
Paragraph:
In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised $\mathrm{He}$ atom), $\mathrm{H}$ atoms and $\mathrm{He}^{+}$ions are excited to their respective first excited states. Subsequently, $\mathrm{H}$ atoms transfer their total excitation energy to $\mathrm{He}^{+}$ions (by collisions). Assume that the Bohr model of atom is exactly valid.Question:
The quantum number $n$ of the state finally populated in $\mathrm{He}^{+}$ions is

Options:
A
2

B
3

C
4

D
5

The correct answer is:
4

Solution:
Energy given by $\mathrm{H}$-atom in transition from $n=2$ to $n=1$ is equal to energy taken by $\mathrm{He}^{+}$ atom in transition from $n=2$ to $n=4$.
$\therefore$ correct option is (c).

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised $\mathrm{He}$ atom), $\mathrm{H}$ atoms and $\mathrm{He}^{+}$ions are excited to their respective first excited states. Subsequently, $\mathrm{H}$ atoms transfer their total excitation energy to $\mathrm{He}^{+}$ions (by collisions). Assume that the Bohr model of atom is exactly valid.Question:
The ratio of the kinetic energy of the $n=2$ electron for the $\mathrm{H}$ atom to that of $\mathrm{He}^{+}$ ion is

Options:
A
$\frac{1}{4}$

B
$\frac{1}{2}$

C
1

D
2

The correct answer is:
$\frac{1}{4}$

Solution:
Kinetic energy $K \propto Z^2$
$$
\frac{K_H}{K_{\mathrm{He}^{+}}}=\left(\frac{1}{2}\right)^2=\frac{1}{4}
$$
$\therefore$ correct option is (a).

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised $\mathrm{He}$ atom), $\mathrm{H}$ atoms and $\mathrm{He}^{+}$ions are excited to their respective first excited states. Subsequently, $\mathrm{H}$ atoms transfer their total excitation energy to $\mathrm{He}^{+}$ions (by collisions). Assume that the Bohr model of atom is exactly valid.Question:
The wavelength of light emitted in the visible region by $\mathrm{He}^{+}$ions after collisions with $\mathrm{H}$ atoms is

Options:
A
$6.5 \times 10^{-7} \mathrm{~m}$

B
$5.6 \times 10^{-7} \mathrm{~m}$

C
$4.8 \times 10^{-7} \mathrm{~m}$

D
$4.0 \times 10^{-7} \mathrm{~m}$

The correct answer is:
$4.8 \times 10^{-7} \mathrm{~m}$

Solution:
Visible light lies in the range, $\lambda_1=4000 Å$ to $\lambda_2=7000 Å$.
Energy of photons corresponding to these wavelengths (in $\mathrm{eV}$ ) would be;
$$
\begin{aligned}
& E_1=\frac{12375}{4000}=3.09 \mathrm{eV} \\
& E_2=\frac{12375}{7000}=1.77 \mathrm{eV}
\end{aligned}
$$
From energy level diagram of $\mathrm{He}^{+}$atom we can see that in transition from $n=4$ to $n=3$, energy of photon released will lie between $E_1$ and $E_2$.
$$
\Delta E_{43}=-3.4-(-6.04)=2.64 \mathrm{eV}
$$
Wavelength of photon corresponding to this energy.
$$
\lambda=\frac{12375}{2.64} Å=4687.5 Å=4.68 \times 10^{-7} \mathrm{~m}
$$
Therefore, (c) is the most correct option.

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ' $r$ '.Question:
The empty space in this $\mathrm{HCP}$ unit cell is

Options:
A
$74 \%$

B
$47.6 \%$

C
$32 \%$

D
$26 \%$

The correct answer is:
$26 \%$

Solution:
Packing fraction $=\frac{\text { Volume of the atoms in one unit cell }}{\text { Volume of one unit cell }}=74 \%$ Empty space $=26 \%$

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ' $r$ '.Question:
The number of atoms in this HCP unit cell is

Options:
A
4

B
6

C
12

D
17

The correct answer is:
6

Solution:
The effective number of atoms in the hcp is $=12 \times \frac{1}{6}+2 \times \frac{1}{2}+3=6$

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
In hexagonal system of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of the three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ' $r$ '.Question:
The volume of this $\mathrm{HCP}$ unit cell is

Options:
A
$24 \sqrt{2} r^3$

B
$16 \sqrt{2} r^3$

C
$12 \sqrt{2} r^3$

D
$\frac{64}{3 \sqrt{3}} r^3$

The correct answer is:
$24 \sqrt{2} r^3$

Solution:
Height of unit cell $=\sqrt{\frac{2}{4}} 4 r$
$$
\begin{aligned}
& \text { Base area }=6 \times \frac{\sqrt{3}}{4}(2 r)^2 \\
& \text { Volume }=\frac{6 \sqrt{3}}{4}(2 r)^2 \cdot \sqrt{\frac{2}{3}} 4 r=24 \sqrt{2} r^3
\end{aligned}
$$

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
In the following reaction sequence, product $I . J$ and $L$ are formed. K represent a reagent
Question:
$$
\text { The structure of product } L \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{H}_2, \mathrm{Pd} / \mathrm{BaSO}_4$, quinoline is well known reagent.
Rosenmund's reaction (conversion of acyl halide to aldehyde) $\mathrm{H}_2, \mathrm{Pd}, \mathrm{BaSO}_4$ is known as Lindlar's reagent and reduces triple bond to double bond (syn addition) giving cis product. Reagent has simultaneous action because usually reaction is not selective-follows radical mechanism.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Paragraph:
In the following reaction sequence, product $I . J$ and $L$ are formed. K represent a reagent
Question:
$$
\text { The structure of product } L \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{H}_2, \mathrm{Pd} / \mathrm{BaSO}_4$, quinoline is well known reagent.
Rosenmund's reaction (conversion of acyl halide to aldehyde) $\mathrm{H}_2, \mathrm{Pd}, \mathrm{BaSO}_4$ is known as Lindlar's reagent and reduces triple bond to double bond (syn addition) giving cis product. Reagent has simultaneous action because usually reaction is not selective-follows radical mechanism.

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
In the following reaction sequence, product $I . J$ and $L$ are formed. K represent a reagent
Question:
The structure of the product $I$ is

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{NaBH}_4$ reduces $-\mathrm{CHO}$ selectively into $-\mathrm{CH}_2 \mathrm{OH}$, which is further converted into corresponding bromide by $\mathrm{PBr}_3$ without affecting the triple bond.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
In the following reaction sequence, product $I . J$ and $L$ are formed. K represent a reagent
Question:
$$
\text { The structures of compounds } J \text { and } K \text {, respectively, are }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Bromide converted into Grignard reagent and further carbonation (I. $\mathrm{CO}_2, \mathrm{II} \mathrm{H}_3 \mathrm{O}^{+}$) produces carboxylic acid $(J)$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Let $A_1, G_1, H_1$ denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For $n \geq 2$, let $A_{n-1}$ and $H_{n-1}$ has arithmetic, geometric and harmonic means as $A_n, G_n, H_n$, respectively.Question:
Which of the following statements is correct?

Options:
A
$\mathrm{H}_1>\mathrm{H}_2>\mathrm{H}_3>\ldots$

B
$H_1 < H_2 < H_3 < \ldots$
C
$H_1>H_3>H_5>\ldots$ and $H_2 < H_4 < H_6 < \ldots$
D
$\mathrm{H}_1 < \mathrm{H}_3 < \mathrm{H}_5 < \ldots$ and $\mathrm{H}_2>\mathrm{H}_4>\mathrm{H}_6>\ldots$

The correct answer is:
$H_1 < H_2 < H_3 < \ldots$

Solution:
As above $A_1>H_2>H_1, A_2>H_3>H_2$ $\therefore \quad H_1 < H_2 < H_3 < \ldots$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Let $A_1, G_1, H_1$ denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For $n \geq 2$, let $A_{n-1}$ and $H_{n-1}$ has arithmetic, geometric and harmonic means as $A_n, G_n, H_n$, respectively.Question:
Which of the following statements is correct?

Options:
A
$A_1>A_2>\ldots$

B
$A_1 < A_2 < A_3 < \ldots$
C
$A_1>A_3>A_5>\ldots$ and $A_2 < A_4 < A_6 < \ldots$
D
$A_1 < A_3 < A_5 < \ldots$ and $A_2>A_4>A_6>\ldots$

The correct answer is:
$A_1>A_2>\ldots$

Solution:
Let $f(x)=k e^x-x$
$\begin{aligned} & & f^{\prime}(x) & =k e^x-1=0 \\ & & x & =-\ln k \\ & \therefore & f^{\prime \prime}(x) & =k e^x \\ & \text { Hence, } & {\left[f^{\prime \prime}(x)\right]_{x=-\ln k} } & =1>0 \\ & & f(-\ln k) & =1+\ln k\end{aligned}$
For one root of given equation
$$
\begin{aligned}
1+\ln k & =0 \\
Hence,
k & =\frac{1}{e} .
\end{aligned}
$$
$A_2$ is $A M$ of $A_1$ and $H_1$ and $A_1>H_1$
$$
\Rightarrow \quad A_1>A_2>H_1
$$
$A_3$ is $\mathrm{AM}$ of $\mathrm{A}_2$ and $\mathrm{H}_2$ and $\mathrm{A}_2>\mathrm{H}_2$
$$
\begin{array}{ll}
\Rightarrow & A_2>A_3>H_2 \\
\therefore & A_1>A_2>A_3>\ldots
\end{array}
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Let $A_1, G_1, H_1$ denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For $n \geq 2$, let $A_{n-1}$ and $H_{n-1}$ has arithmetic, geometric and harmonic means as $A_n, G_n, H_n$, respectively.Question:
Which one of the following statements is correct?

Options:
A
$G_1>G_2>G_3>\ldots$

B
$G_1 < G_2 < G_3 < \ldots$
C
$G_1=G_2=G_3=\ldots$

D
$G_1 < G_3 < G_5 < \ldots$ and $G_2>G_4>G_6>\ldots$

The correct answer is:
$G_1=G_2=G_3=\ldots$

Solution:
$A_1=\frac{a+b}{2}, G_1=\sqrt{a b}$ and $H_1=\frac{2 a b}{a+b}$ $\begin{aligned} A_n=\frac{A_{n-1}+H_{n-1}}{2}, G_n & =\sqrt{A_{n-1} H_{n-1}}, \\ H_n & =\frac{2 A_{n-1} H_{n-1}}{A_{n-1}+H_{n-1}}\end{aligned}$
Clearly, $\quad G_1=G_2=G_3=\ldots=\sqrt{a b}$.

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2007 (Paper 2)
Paragraph:
Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$Question:
If the point $P(a, b, c)$, with reference to Eq. (E), lies on the plane $2 x+y+z=1$, then the value of $7 a+b+c$ is

Options:
A
0

B
12

C
7

D
6

The correct answer is:
6

Solution:
$$
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
$$
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
$$
7 a+c=0
$$
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
$$
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
$$
As $(a, b, c)$ lies on $2 x+y+z=1$
$$
\begin{array}{lc}
\Rightarrow & 2 a+b+c=1 \\
\Rightarrow & 2 a+6 a-7 a=1 \\
\Rightarrow & a=1, b=6 \text { and } c=-7 \\
\therefore & 7 a+b+c=7+6-7=6
\end{array}
$$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$Question:
Let $b=6$, with $a$ and $c$ satisfying Eq. (E). If $\alpha$ and $\beta$ are the roots of the quadratic equation $a x^2+b x+c=0$, then $\sum_{n=0}^{\infty}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^n$ is

Options:
A
6

B
7

C
$\frac{6}{7}$

D
$\infty$

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
$$
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
$$
7 a+c=0
$$
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
$$
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
$$
If $b=6, a=1$ and $c=-7$
$$
\begin{array}{ll}
\therefore & a x^2+b x+c=0 \\
\Rightarrow & x^2+6 x-7=0 \\
\Rightarrow & (x+7)(x-1)=0 \\
\therefore & x=1,-7 \\
\Rightarrow & \quad \sum_{n=0}^{\infty}\left(\frac{1}{1}-\frac{1}{7}\right)^n \Rightarrow \sum_{n=0}^{\infty}\left(\frac{6}{7}\right)^n \\
\Rightarrow & 1+\frac{6}{7}+\left(\frac{6}{7}\right)^n+\ldots \infty \\
& =\frac{1}{1-\frac{6}{7}}=\frac{1}{1 / 7}=7
\end{array}
$$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Let $a$, b and $c$ be three real numbers satisfying $\left[\begin{array}{lll}a & b & c\end{array}\right]\left[\begin{array}{lll}1 & 9 & 7 \\ 8 & 2 & 7 \\ 7 & 3 & 7\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$Question:
Let $\omega$ be a solution of $x^3-1=0$ with $\operatorname{Im}(\omega)>0$. If $a=2$, with $b$ and $c$ satisfying Eq. (E), then the value of $\frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c}$

Options:
A
$-2$

B
2

C
3

D
$-3$

The correct answer is:
$-2$

Solution:
$$
\begin{aligned}
& \text { Given, }[a b c]_{1 \times 3}\left[\begin{array}{lll}
1 & 9 & 7 \\
8 & 2 & 7 \\
7 & 3 & 7
\end{array}\right]_{3 \times 3}=\left[\begin{array}{lll}
0 & 0 & 0
\end{array}\right] \\
& \Rightarrow \quad\left[\begin{array}{c}
a+8 b+7 c \\
9 a+2 b+3 c \\
7 a+7 b+7 c
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right] \\
& \Rightarrow \quad a+8 b+7 c=0 \\
& \Rightarrow 9 a+2 b+3 c=0 \\
& \Rightarrow \quad a+b+c=0 \\
&
\end{aligned}
$$
On multiplying Eq. (iii) by 2 , then subtract from Eq. (ii), we get
$$
7 a+c=0
$$
Again multiplying Eq. (iii) by 3 , then subtract from Eq. (ii), we get
$$
\begin{array}{ll}
& 6 a-b=0 \\
\therefore \quad & b=6 a \text { and } c=-7 a
\end{array}
$$
If $a=2, b=12$ and $c=-14$
$$
\begin{aligned}
\therefore & \frac{3}{\omega^a}+\frac{1}{\omega^b}+\frac{3}{\omega^c} \\
\Rightarrow & \frac{3}{\omega^2}+\frac{1}{\omega^{12}}+\frac{3}{\omega^{-14}}=\frac{3}{\omega^2}+1+3 \omega^2 \\
& =3 \omega+1+3 \omega^2 \\
& =1+3\left(\omega+\omega^2\right) \\
& =1-3=-2
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Let $A, B, C$ be three sets of complex numbers as defined below.
$A=\{z ; \operatorname{lm} z \geq 1\} ; B=\{z:|z-2-i|=3\} ; C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\}$.Question:
Let $z$ be any point in $A \cap B \cap C$ and let $w$ be any point satisfying $|w-2-i| < 3$. Then, $|z|-|w|+3$ lies between

Options:
A
$-6$ and 3

B
$-3$ and 6

C
$-6$ and 6

D
$-3$ and 9

The correct answer is:
$-3$ and 9

Solution:
$$
\begin{array}{ll}
& |w-(2+i)| < 3 \\
\Rightarrow & \quad|| w|-| 2+i|| < 3 \\
\Rightarrow & -3+\sqrt{5} < |w| < 3+\sqrt{5} \\
\Rightarrow & -3-\sqrt{5} < -|w| < 3-\sqrt{5}
\end{array}
$$
Also, $|z-(2+i)|=3$
$$
\begin{aligned}
\Rightarrow \quad-3+\sqrt{5} & \leq|z| \leq 3+\sqrt{5} \\
-3 & < |z|-|w|+3 < 9
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Let $A, B, C$ be three sets of complex numbers as defined below.
$A=\{z ; \operatorname{lm} z \geq 1\} ; B=\{z:|z-2-i|=3\} ; C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\}$.Question:
Let $z$ be any point in $A \cap B \cap C$. The $|z+1-i|^2+|z-5-i|^2$ lies between

Options:
A
25 and 29

B
30 and 34

C
35 and 39

D
40 and 44

The correct answer is:
35 and 39

Solution:
$|z+1-i|^2+|z-5-i|^2=(x+1)^2+(y-1)^2+(x-5)^2+(y-1)^2$
$$
\begin{aligned}
& =2\left(x^2+y^2-4 x-2 y\right)+28 \\
& =2(4)+28 \\
& {\left[\because x^2+y^2-4 x-2 y=4\right]} \\
& =36 \\
&
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Let $A, B, C$ be three sets of complex numbers as defined below.
$A=\{z ; \operatorname{lm} z \geq 1\} ; B=\{z:|z-2-i|=3\} ; C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\}$.Question:
The number of elements in the set $A \cap B \cap C$ is

Options:
A
0

B
1

C
2

D
$\infty$

The correct answer is:
1

Solution:
Let $z=x+i y$
Set $A$ corresponds to the region $y \geq 1$.
Set $B$ consists of points lying on the circle, centred at $(2,1)$ and radius 3 , i.e., $x^2+y^2-4 x-2 y=4$
$$
\text { Set } C \text { consists of points lying on the } x+y=\sqrt{2}
$$
Clearly, there is only one point of intersection of the line
$$
x+y=\sqrt{2} \text { and circle } x^2+y^2-4 x-2 y=4 .
$$

Topic: Mathematics, Complex Number, JEE Main
Paragraph:
Let $A, B, C$ be three sets of complex numbers as defined below.
$A=\{z ; \operatorname{lm} z \geq 1\} ; B=\{z:|z-2-i|=3\} ; C=\{z: \operatorname{Re}((1-i) z)=\sqrt{2}\}$.Question:
The number of elements in the set $A \cap B \cap C$ is

Options:
A
0

B
1

C
2

D
$\infty$

The correct answer is:
1

Solution:
Let $z=x+i y$
Set $A$ corresponds to the region $y \geq 1$.
Set $B$ consists of points lying on the circle, centred at $(2,1)$ and radius 3 , i.e., $x^2+y^2-4 x-2 y=4$
$$
\text { Set } C \text { consists of points lying on the } x+y=\sqrt{2}
$$
Clearly, there is only one point of intersection of the line
$$
x+y=\sqrt{2} \text { and circle } x^2+y^2-4 x-2 y=4 .
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Let $A$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices in $A$ is

Options:
A
12

B
6

C
9

D
3

The correct answer is:
12

Solution:
Case $I$ When all three diagonal elements are 1 , then
Number of matrices $={ }^3 C_1=3$
Case II When two diagonal elements are zero and one element is one no, then Number of matrices $={ }^3 C_1 \cdot{ }^3 C_1=9$
$\therefore$ Total matrices $=3+9=12$

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Let $A$ be the set of all $3 \times 3$ symmetric matrices all of whose entries are either 0 or 1 . Five of these entries are 1 and four of them are 0 .Question:
The number of matrices $A$ in $A$ for which the system of linear equations $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ has a unique solution, is

Options:
A
less than 4

B
atleast 4 but less than 7

C
atleast 7 but less than 10

D
at least 10

The correct answer is:
atleast 4 but less than 7

Solution:
Given, $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$
For unique solution, $\operatorname{det}(A) \neq 0$
$$
\begin{gathered}
\text { Case I } \operatorname{det}(A)=\left|\begin{array}{lll}
1 & a & b \\
a & 1 & c \\
b & c & 1
\end{array}\right| \\
=1-a^2-b^2-c^2+2 a b c \neq 0
\end{gathered}
$$
Here $a, b, c$ is selected from $1,0,0$. (No case is possible)
Case II
(i) $\operatorname{det}(A)=\left|\begin{array}{lll}1 & a & b \\ a & 0 & c \\ b & c & 0\end{array}\right|=2 a b c-c^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
(ii) $\operatorname{det}(A)=\left|\begin{array}{lll}0 & a & b \\ a & 1 & c \\ b & c & 0\end{array}\right|=2 a b c-b^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
(iii) $\operatorname{det}(A)=\left|\begin{array}{lll}0 & a & b \\ a & 0 & c \\ b & c & 1\end{array}\right|=2 a b c-a^2 \neq 0$
Here $a, b, c$ are selected from $1,1,0$. (2 cases are possible)
Hence, there are exactly 6 matrices for unique solution. Hence, option (b) is correct.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2009 (Paper 1)