datavorous/entrance-exam-dataset huggingface dataset:
Consider the following linear equations
$$
a x+b y+c z=0, b x+c y+a z=0, c x+a y+b z=0
$$
Match the conditions/expressions in Column I with statements in Column II.

Options:
A
A-p; B-p, s; C-p; D-q

B
A-r, q; B-r, s; C-r, s; D-r, s

C
A-s; B-p; C-q; D-s

D
A-r; B-q; C-p; D-s

The correct answer is:
A-r; B-q; C-p; D-s

Solution:
Let $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=-\frac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
(A) If $a+b+c \neq 0$ and $a^2+b^2+c^2=a b+b c+c a$
$$
\Rightarrow \quad \Delta=0 \text { and } a=b=c \neq 0
$$
The equations represents identical planes.
(B) $a+b+c=0$ and $a^2+b^2+c^2 \neq a b+b c+c a$
$$
\Rightarrow \quad \Delta=0
$$
The equations have infinitely many solutions.
$$
\begin{array}{rlrl}
& & a x+b y & =(a+b) z \\
\Rightarrow & b x+c y & =(b+c) z \\
\Rightarrow & & \left(b^2-a c\right) y & =\left(b^2-a c\right) z \Rightarrow y=z \\
\Rightarrow & a x+b y+c y & =0 \\
& a x & =a y \Rightarrow x=y=z
\end{array}
$$
(C) $a+b+c \neq 0$ and $a^2+b^2+c^2 \neq a b+b c+c a$
$$
\Rightarrow \quad \Delta \neq 0
$$
The equation represent planes meeting at only one point.
(D) $a+b+c=0$ and $a^2+b^2+c^2=a b+b c+c a$
$$
\Rightarrow \quad a=b=c=0
$$
The equations represent whole of the three dimensional space.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2007 (Paper 1)
Consider the lines given by
$$
L_1: x+3 y-5=0, L_2: 3 x-k y-1=0, L_3: 5 x+2 y-12=0
$$
Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

Options:
A
(A) r, (B) p,s, (C) s, (D) p,s

B
(A) s, (B) p,q,r, (C) r, (D) p,q

C
(A) s, (B) p,q, (C) r, (D) p,q,s

D
(A) r, (B) p,q, (C) s, (D) p,q

The correct answer is:
(A) s, (B) p,q, (C) r, (D) p,q,s

Solution:
(A) Solving equations $L_1$ and $L_3$,
$$
\begin{aligned}
\frac{x}{-36+10} & =\frac{y}{+12-25}=\frac{1}{2-15} \\
x & =2, y=1
\end{aligned}
$$
$L_1, L_2, L_3$ are concurrent, if point $(2,1)$ lies on $L_2$.
$$
\therefore \quad 6-k-1=0 \Rightarrow k=5
$$
(B) Either $L_1$ is parallel to $L_2$, or $L_3$ is parallel to $L_2$, then
$$
\begin{aligned}
& \frac{1}{3}=\frac{3}{-k} \text { or } \frac{3}{5}=\frac{-k}{2} \\
& k=-9 \quad \text { or } \quad k=-\frac{6}{5}
\end{aligned}
$$
$$
\Rightarrow \quad k=-9 \text { or } k=-\frac{6}{5}
$$
(C) $L_1, L_2, L_3$ form a triangle, if they are not concurrent, or not parallel.
$$
\therefore \quad k \neq 5,-9,-\frac{6}{5} \Rightarrow k=\frac{5}{6}
$$
(D) $L_1, L_2, L_3$ do not form a triangle, if
$$
k=5,-9,-\frac{6}{5}
$$

Topic: Mathematics, Determinants, JEE Main
Consider the lines given by
$$
L_1: x+3 y-5=0, L_2: 3 x-k y-1=0, L_3: 5 x+2 y-12=0
$$
Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

Options:
A
(A) r, (B) p,s, (C) s, (D) p,s

B
(A) s, (B) p,q,r, (C) r, (D) p,q

C
(A) s, (B) p,q, (C) r, (D) p,q,s

D
(A) r, (B) p,q, (C) s, (D) p,q

The correct answer is:
(A) s, (B) p,q, (C) r, (D) p,q,s

Solution:
(A) Solving equations $L_1$ and $L_3$,
$$
\begin{aligned}
\frac{x}{-36+10} & =\frac{y}{+12-25}=\frac{1}{2-15} \\
x & =2, y=1
\end{aligned}
$$
$L_1, L_2, L_3$ are concurrent, if point $(2,1)$ lies on $L_2$.
$$
\therefore \quad 6-k-1=0 \Rightarrow k=5
$$
(B) Either $L_1$ is parallel to $L_2$, or $L_3$ is parallel to $L_2$, then
$$
\begin{aligned}
& \frac{1}{3}=\frac{3}{-k} \text { or } \frac{3}{5}=\frac{-k}{2} \\
& k=-9 \quad \text { or } \quad k=-\frac{6}{5}
\end{aligned}
$$
$$
\Rightarrow \quad k=-9 \text { or } k=-\frac{6}{5}
$$
(C) $L_1, L_2, L_3$ form a triangle, if they are not concurrent, or not parallel.
$$
\therefore \quad k \neq 5,-9,-\frac{6}{5} \Rightarrow k=\frac{5}{6}
$$
(D) $L_1, L_2, L_3$ do not form a triangle, if
$$
k=5,-9,-\frac{6}{5}
$$

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2008 (Paper 2)
Consider the motion of a positive point charge in a region where there are simultaneous uniform electric and magnetic fields $\vec{E}=E_{0} \hat{j}$ and $\vec{B}=B_{0} \hat{j}$. At time $t=0$, this charge has velocity $\vec{v}$ in the in the $x$-y plane, making an angle $\theta$ with the $x$-axis. Which of the following option(s) is (are) correct for time $t>0$ ?

Options:
A If $\theta=0^{\circ}$, the charge moves in a circular path in the $x$ - plane.
B If $\theta=0^{\circ}$, the charge undergoes helical motion with constant pitch along the $y$-axis.
C If $\theta=10^{\circ}$, the charge undergoes helical motion with its pitch increasing with time, along the $y$-axis.
D If $\theta=90^{\circ}$, the charge undergoes linear but accelerated motion along the $y$-axis.

The correct answers are:
If $\theta=10^{\circ}$, the charge undergoes helical motion with its pitch increasing with time, along the $y$-axis., If $\theta=90^{\circ}$, the charge undergoes linear but accelerated motion along the $y$-axis.

Solution:
If $\theta=0^{\circ}$, the charged particle is projected along $x$-axis, due to magnetic field, $\vec{B}$ the charged particle will tend to move in a circular path in $y-z$ plane but due to force of electric field $\vec{E}$, the particle will move in a helical path with increasing pitch. Hence options (A) and (B) are wrong. If $\theta=10^{\circ}$, we can resolve velocity into two rectangular components. One along $x$-axis $\left(v \cos 10^{\circ}\right)$ and one along $y$-axis $\left(v \sin 10^{\circ}\right)$. Due to $v \cos 10^{\circ}$, the particle will move in circular path and due to $v \sin 10^{\circ}$ plus the force due to electric field, the particle will undergo helical motion with its pitch increasing.

If $\theta=90^{\circ}$, the charge is moving along the magnetic field. Therefore the force due to magnetic field is zero. But the force due to electric field will accelerate the particle along $y$-axis.

Topic: Physics, Magnetic Effects of Current, JEE Advanced, JEE Advanced 2012 (Paper 1)
Consider the parabola $y^2=8 x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latusrectum and the point $P\left(\frac{1}{2}, 2\right)$ on the parabola and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latusrectum. Then, $\frac{\Delta_1}{\Delta_2}$ is

Options:

The correct answer is:
2

Solution:
As, we know area of triangle formed by three points on parabola is twice the area of triangle formed by corresponding tangents, i.e. area of $\triangle P Q R=2$ area of $\Delta T_1 T_2 T_3$.
$\therefore \quad \Delta_1=2 \Delta_2$ or $\frac{\Delta_1}{\Delta_2}=2$

Topic: Mathematics, Parabola, JEE Main
Consider the parabola $y^2=8 x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latusrectum and the point $P\left(\frac{1}{2}, 2\right)$ on the parabola and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latusrectum. Then, $\frac{\Delta_1}{\Delta_2}$ is

Options:

The correct answer is:
2

Solution:
As, we know area of triangle formed by three points on parabola is twice the area of triangle formed by corresponding tangents, i.e. area of $\triangle P Q R=2$ area of $\Delta T_1 T_2 T_3$.
$\therefore \quad \Delta_1=2 \Delta_2$ or $\frac{\Delta_1}{\Delta_2}=2$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2011 (Paper 1)
Consider the planes $3 x-6 y-2 z=15$ and $2 x+y-2 z=5$.
Statement I The parametric equations of the line of intersection of the given planes are $x=3+14 t, y=1+2 t$ and $z=15 t$.
Statement II The vectors $14 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}$ is parallel to the line of intersection of the given planes.

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
Given planes are $3 x-6 y-2 z=15$ and $2 x+y-2 z=5$
For $z=0$, we get $x=3, y=-1$
Direction ratios of planes are
$$
< 3-6-2>\text { and } < 2 \quad 1-2>
$$
then the dr's of line of intersection of planes is $\langle 14215\rangle$ and line is
$$
\begin{aligned}
\frac{x-3}{14} & =\frac{y+1}{2}=\frac{z-0}{15}=\lambda \text { (say) } \\
\Rightarrow \quad x & =14 \lambda+3, y=2 \lambda-1, z=15 \lambda
\end{aligned}
$$
Hence, Statement I is false.
But Statement II is true.

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2007 (Paper 2)
Consider the system of equations
where $\quad a, b, c, d \in\{0,1\}$
Statement 1 The probability that the system of equations has a unique solution, is $3 / 8$.
Statement 2 The probability that the system of equations has a solution, is 1 .

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Solution:
The number of all possible deteminants of the form
Out of which only 10 determinants given by
vanish and remaining six determinants have non-zero values.
Hence, the required probability $=\frac{6}{16}=\frac{3}{8}$
$\Rightarrow$ Statement 1 is correct.
Statement 2 is also correct as the homogeneous equations have always a solution and Statement 2 is not a correct explanation of Statement 1.

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2008 (Paper 1)
Consider the system of equations $x-2 y+3 z=-1, x-3 y+4 z=1$ and $-x+y-2 z=k$
Statement 1 The system of equations has no solution for $k \neq 3$.
Statement 2 The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $k \neq 3$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
$$
\text { The given system of equations can be expressed as }
$$
When $k \neq 3$, the given system of equations has no solution.
$\Rightarrow$ Statement 1 is true.Clearly, Statement 2 is also true as it is rearrangement of rows and columns of
Hence, option (a) is correct.

Topic: Mathematics, Determinants, JEE Main
Consider the system of equations $x-2 y+3 z=-1, x-3 y+4 z=1$ and $-x+y-2 z=k$
Statement 1 The system of equations has no solution for $k \neq 3$.
Statement 2 The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $k \neq 3$.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

Solution:
$$
\text { The given system of equations can be expressed as }
$$
When $k \neq 3$, the given system of equations has no solution.
$\Rightarrow$ Statement 1 is true.Clearly, Statement 2 is also true as it is rearrangement of rows and columns of
Hence, option (a) is correct.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2008 (Paper 1)
Consider the two curves
$C_1: y^2=4 x$
$C_2: x^2+y^2-6 x+1=0$, then

Options:
A
$C_1$ and $C_2$ touch each other only at one point

B
$C_1$ and $C_2$ touch each other exactly at two points

C
$C_1$ and $C_2$ intersect (but do not touch) at exactly two points

D
$C_1$ and $C_2$ neither intersect nor touch each other

The correct answer is:
$C_1$ and $C_2$ touch each other exactly at two points

Solution:
For the points of intersection of the two given curves
$$
C_1: y^2=4 x \text { and } C_2: x^2+y^2-6 x+1=0 \text {, }
$$
we have
$$
\begin{aligned}
& x^2+4 x-6 x+1=0 \\
& \Rightarrow \quad x^2-2 x+1=0 \\
& \Rightarrow \quad(x-1)^2=0 \\
& \Rightarrow \quad x=1,1 \\
& \Rightarrow \quad y=2,-2 \\
&
\end{aligned}
$$
Thus, the given curves touch each other exactly at two points $(1,2)$ and $(1,-2)$.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2008 (Paper 1)
Consider three planes
$$
P_1: x-y+z=1, P_2: x+y-z=-1
$$
and $P_3: x-3 y+3 z=2$
Let $L_1, L_2$ and $L_3$ be the lines of intersection of the planes $P_2$ and $P_3, P_3$ and $P_1, P_1$ and $P_2$, respectively.
Statement 1 Atleast two of the lines $L_1, L_2$ and $L_3$ are non-parallel.
Statement 2 The three planes do not have a common point.

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is false, Statement 2 is true

Solution:
The given equations are
$$
\begin{aligned}
& x-y+z=1, \\
& x+y-z=-1 \text { and } x-3 y+3 z=2
\end{aligned}
$$
The system of equations can be put in matrix form as $A X=B$
which is inconsistent as $\rho(A: B) \neq \rho(A)$.
$\Rightarrow$ The three planes do not have a common point.
$\Rightarrow$ Statement 2 is true.
Since, planes $P_1, P_2$ and $P_3$ are pairwise intersection, their lines of intersection are parallel.
Statement 1 is false.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2008 (Paper 1)
Consider three points $P=(-\sin (\beta-\alpha),-\cos \beta), Q=(\cos (\beta-\alpha), \sin \beta)$ and $R=(\cos (\beta-\alpha+\theta), \sin (\beta-\theta)$, where $0 < \alpha, \beta, \theta < \frac{\pi}{4}$. Then,

Options:
A
$P$ lies on the line segment $R Q$

B
$Q$ lies on the line segment $P R$

C
$R$ lies on the line segment $Q P$

D
$P, Q, R$ are non- collinear

The correct answer is:
$P, Q, R$ are non- collinear

Solution:
For collinear points
$$
\Delta=\left|\begin{array}{ccc}
-\sin (\beta-\alpha) & -\cos \beta & 1 \\
\cos (\beta-\alpha) & \sin \beta & 1 \\
\cos (\beta-\alpha+\theta) & \sin (\beta-\theta) & 1
\end{array}\right|
$$
Clearly, $\Delta \neq 0$ for any value of $\alpha, \beta, \theta$, hence points are non-collinear.

Topic: Mathematics, Straight Lines, JEE Advanced, JEE Advanced 2008 (Paper 2)
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as:

Options:
A In-situ conservation of biodiversity
B Advanced Ex-situ conservation of biodiversity
C In-situ conservation by sacred groves
D In-situ cryoconservation of biodiversity

The correct answer is:
Advanced Ex-situ conservation of biodiversity

Solution:
Cryopreservation of gametes of threatened species in viable and fertile conditions can be referred to as advanced ex-situ conservation of biodiversity because these gametes are stored in liquid nitrogen at a temperature of about - 196°C.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2015 (Phase 1)
Cryopreservation of gametes of threatened species in viable and fertile condition can be referred to as:

Options:
A in situ conservation of biodiversity
B Advanced ex-situ conservation of biodiversity
C in situ conservation by sacred groves
D in situ cryo-conservation of biodiversity

The correct answer is:
Advanced ex-situ conservation of biodiversity

Solution:
Cryopreservation of gametes of threatened species in viable and fertile conditions can be referred to as advanced ex-situ conservation of biodiversity because these gametes are stored in liquid nitrogen at a temperature of about - 196°C.

Topic: Biology, Biodiversity and its Conservation, NEET
$\mathrm{CuSO}_4$ decolourises on addition of $\mathrm{KCN}$, the product is

Options:
A
$\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$

B
$\mathrm{Cu}^{2+}$ get reduced to form $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$

C
$\mathrm{Cu}(\mathrm{CN})_2$

D
$\mathrm{CuCN}$

The correct answer is:
$\mathrm{CuCN}$

Solution:
$\mathrm{CuSO}_4$ decolourise on addition of $\mathrm{KCN}$, the product is $\mathrm{CuCN}$ along with $(\mathrm{CN})_2$ but in excess of KCN is form colourless potassium tetra cyano cuperate (I)

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2006
Cyclohexen on ozonolysis followed by reaction with zinc dust and water gives compound $E$. Compound $E$ on further treatment with aqueous $\mathrm{KOH}$ yields compound $F$. Compound $F$ is :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2007 (Paper 2)
$\frac{d^2 x}{d y^2}$ equals

Options:
A
$\left(\frac{d^2 y}{d x^2}\right)^{-1}$

B
$-\left(\frac{d^2 y}{d x^2}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}$

C
$\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}$

D
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

The correct answer is:
$-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$

Solution:
Since,
$$
\begin{aligned}
\Rightarrow \quad \frac{d}{d y}\left(\frac{d x}{d y}\right) & =\frac{d}{d x}\left(\frac{d y}{d x}\right)^{-1} \frac{d x}{d y} \\
\frac{d^2 x}{d y^2} & =-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}\left(\frac{d x}{d y}\right) \\
& =-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2007 (Paper 2)
Dissolving $120 \mathrm{~g}$ of urea (molecular weight 60) in $1000 \mathrm{~g}$ of water gave a solution of density $1.15 \mathrm{~g} / \mathrm{mL}$. The molarity of the solution is

Options:
A
$1.78 \mathrm{M}$

B
$2.00 \mathrm{M}$

C
$2.05 \mathrm{M}$

D
$2.22 \mathrm{M}$

The correct answer is:
$2.05 \mathrm{M}$

Solution:

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Advanced, JEE Advanced 2011 (Paper 1)
Electrolysis of dilute aq. $\mathrm{NaCl}$ solution was carried out by passing $10 \mathrm{~mA}$ current. The time required to liberate $0.01$ mole of $\mathrm{H}_2$ gas at the cathode is (1 Faraday $=96500 \mathrm{C} \mathrm{mol}^{-1}$ )

Options:
A
$9.65 \times 10^4 \mathrm{~s}$

B
$19.3 \times 10^4 \mathrm{~s}$

C
$28.95 \times 10^4 \mathrm{~s}$

D
$38.6 \times 10^4 \mathrm{~s}$

The correct answer is:
$19.3 \times 10^4 \mathrm{~s}$

Solution:
$2 \mathrm{H}_2 \mathrm{O}+2 e^{-} \longrightarrow \mathrm{H}_2+2 \mathrm{OH}^{-}$
For $0.01$ mole of $\mathrm{H}_2, 0.02$ mole of electrons are consumed Charge required $=0.02 \times 96500 \mathrm{C}=i \times t$
$$
\text { Time required }=\frac{0.02 \times 96500}{10 \times 10^{-3}}=19.3 \times 10^4 \mathrm{~s}
$$

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2008 (Paper 2)
Electrons with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-rays is

Options:
A
$\lambda_0=\frac{2 m c \lambda^2}{h}$

B
$\lambda_0=\frac{2 h}{m c}$

C
$\lambda_0=\frac{2 m^2 c^2 \lambda^3}{h^2}$

D
$\lambda_0=\lambda$

The correct answer is:
$\lambda_0=\frac{2 m c \lambda^2}{h}$

Solution:
Momentum of striking electrons, $p=\frac{h}{\lambda}$
$\therefore$ Kinetic energy of striking electrons, $K=\frac{p^2}{2 m}=\frac{h^2}{2 m \lambda^2}$
This is also, maximum energy of $X$-ray photons.
Therefore, $\quad \frac{h c}{\lambda_0}=\frac{h^2}{2 m \lambda^2}$ or $\lambda_0=\frac{2 m \lambda^2 c}{h}$
$\therefore$ Correct option is (a).

Topic: Physics, Dual Nature of Matter, JEE Advanced, JEE Advanced 2007 (Paper 2)
Equation of common tangent of $y=x^2, y=-x^2+4 x-4$ is

Options:
A
$y=4(x-1)$

B
$y=0$

C
$y=-4(x-1)$

D
$y=-30 x-50$

The correct answers are:
$y=4(x-1)$
,
$y=0$

Solution:
The equation of tangent to $y=x^2$, be $y=m x-\frac{m^2}{4}$, putting in $y=-x^2+4 x-4$, we should only get one value of $x$ i.e. Discriminate must be zero.
$$
\begin{array}{rrr}
\therefore & m x-\frac{m^2}{4} & =-x^2+4 x-4 \\
\Rightarrow & x^2+x(m-4)+4-\frac{m^2}{4} & =0 \\
\Rightarrow & m(m-4) & =0 \\
& \therefore y=0 & \text { or } m=0,4
\end{array}
$$
$$
\therefore y=0 \text { and } y=4(x-1) \text { are tangents. }
$$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2006
Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

Options:
A
$x+2 y-2 z=0$

B
$3 x+2 y-2 z=0$

C
$x-2 y+z=0$

D
$5 x+2 y-4 z=0$

The correct answer is:
$x-2 y+z=0$

Solution:
The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. $\Rightarrow \quad \overrightarrow{\mathbf{n}}_1=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 2 \\ 4 & 2 & 3\end{array}\right|=(8 \hat{\mathbf{i}}-\hat{\mathbf{j}}-10 \hat{\mathbf{k}})$
Also, equation of plane containing $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of normal to be $\overrightarrow{\mathbf{n}}_2=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$
$$
\begin{array}{ll}
\Rightarrow & a x+b y+c z=0 \\
\text { where, } \overrightarrow{\mathbf{n}}_1 \cdot \overrightarrow{\mathbf{n}}_2=0 \\
\Rightarrow & 8 a-b-10 c=0 \\
\text { and } & \overrightarrow{\mathbf{n}}_2 \perp(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\
\Rightarrow & 2 a+3 b+4 c=0
\end{array}
$$
From Eqs. (ii) and (iii), we get
$$
\begin{aligned}
& \frac{a}{-1 \quad-10}=\frac{b}{8}=\frac{c}{-1} \\
\Rightarrow \quad \frac{a}{-4+30} & =\frac{b}{-20-32}=\frac{c}{24+2} \\
\Rightarrow \quad \quad \quad \frac{a}{26} & =\frac{b}{-52}=\frac{c}{26} \\
\Rightarrow \quad \quad \quad \quad \frac{a}{1} & =\frac{b}{-2}=\frac{c}{1}
\end{aligned}
$$
From Eqs. (i) and (iv), required equation of plane, is $x-2 y+z=0$.

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2010 (Paper 1)
Extra pure $\mathrm{N}_2$ can be obtained by heating

Options:
A
$\mathrm{NH}_3$ with $\mathrm{CuO}$

B
$\mathrm{NH}_4 \mathrm{NO}_4$

C
$\left(\mathrm{NH}_4\right)_2 \mathrm{Cr}_2 \mathrm{O}_7$

D
$\mathrm{Ba}\left(\mathrm{N}_3\right)_2$

The correct answer is:
$\mathrm{Ba}\left(\mathrm{N}_3\right)_2$

Solution:
$\mathrm{Ba}\left(\mathrm{N}_3\right)_2 \stackrel{\text { Heat }}{\longrightarrow} \mathrm{Ba}(\mathrm{s})+3 \mathrm{~N}_2(\mathrm{~g})$
Azide salt of barium can be obtained in purest form as well as the decomposition product contain solid $\mathrm{Ba}$ as byproduct along with gaseous nitrogen, hence no additional step of separation is required.
Other reactions are:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2011 (Paper 1)
Extraction for zinc from zinc blende is achieved by

Options:
A
electrolytic reduction

B
roasting followed by reduction with carbon

C
roasting followed by reduction with another metal

D
roasting followed by self-reduction

The correct answer is:
roasting followed by reduction with carbon

Solution:
Zinc blende is roasted and then treated with coke for the reduction.
$$
\begin{gathered}
2 \mathrm{ZnS}+3 \mathrm{O}_2 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{ZnO}+2 \mathrm{SO}_2 \uparrow \\
\mathrm{ZnO}+\mathrm{C} \stackrel{\Delta}{\longrightarrow} \mathrm{Zn}+\mathrm{CO} \uparrow
\end{gathered}
$$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2007 (Paper 1)