datavorous/entrance-exam-dataset huggingface dataset:
In the given circuit, the $\mathrm{AC}$ source has $\omega=100 \mathrm{rad} / \mathrm{s}$. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)

Options:
A The current through the circuit, $I$ is $0.3 \mathrm{~A}$.
B The current through the circuit, $I$ is $0.3 \sqrt{2} A$
C The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{~V}$
D The voltage across $50 \Omega$ resistor $=10 \mathrm{~V}$

The correct answers are:
The current through the circuit, $I$ is $0.3 \mathrm{~A}$., The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{~V}$

Solution:
Impedance across $A B, R C$ part of the circuit

$\begin{aligned} \mathrm{Z}_{1}=& \sqrt{X_{c}^{2}+R_{1}^{2}}=\sqrt{\left(\frac{1}{\omega C}\right)^{2}+R_{1}^{2}} \\=& \sqrt{(100)^{2}+(100)^{2}}=100 \sqrt{2} \end{aligned}$

$\therefore \quad I_{1}=\frac{V}{Z_{1}}=\frac{20}{100 \sqrt{2}}$

$\quad\left[\right.$ leads emf by $\left.\phi_{1}\right]$

where $\cos \phi_{1}=\frac{R}{Z_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$

Impedance across $C D, L R$ part of the circuit.

$\mathrm{Z}_{2}=\sqrt{X_{L}^{2}+R_{2}^{2}}=\sqrt{(\omega L)^{2}+R_{2}^{2}}$

$=\sqrt{(0.5 \times 100)^{2}+(50)^{2}}=50 \sqrt{2} \Omega$

$\therefore \quad I_{2}=\frac{V}{Z_{2}}=\frac{20}{50 \sqrt{2}}$

where $\cos \phi_{2}=\frac{R}{Z_{2}}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{2}=45^{\circ}$ $\therefore \quad$ Current $I$ from the circuit

$I=\frac{20}{100 \sqrt{2}}+\frac{20}{50 \sqrt{2}}=\mathrm{I}_{1}+\mathrm{I}_{2} \simeq 0.3 \mathrm{~A}$

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2012 (Paper 2)
In the Newman projection for 2,2-dimethylbutane
$X$ and $Y$ can respectively be

Options:
A
$\mathrm{H}$ and $\mathrm{H}$

B
$\mathrm{H}$ and $\mathrm{C}_2 \mathrm{H}_5$

C
$\mathrm{C}_2 \mathrm{H}_5$ and $\mathrm{H}$

D
$\mathrm{CH}_3$ and $\mathrm{CH}_3$

The correct answers are:
$\mathrm{H}$ and $\mathrm{C}_2 \mathrm{H}_5$
,
$\mathrm{CH}_3$ and $\mathrm{CH}_3$

Solution:
Conformation projection along $\mathrm{C}_1-\mathrm{C}_2$
$\mathrm{C}_1$ contains all three $\mathrm{Hs}$
So, $\quad X=\mathrm{H}$
$\mathrm{C}_2$ contains two methyl and one ethyl group
So, $\quad Y=\mathrm{C}_2 \mathrm{H}_5$
Conformational projection along
$$
\mathrm{C}_2-\mathrm{C}_3
$$
$\mathrm{C}_2$ contains three methyl groups
$\left(\mathrm{C}_2\right.$ form back carbon in the given structure)
So, $\quad Y=\mathrm{CH}_3$
$\mathrm{C}_3$ contains two Hs and one methyl group
$\left(\mathrm{C}_3\right.$ form front carbon in the given structure)
So, $\quad X=\mathrm{CH}_3$
$\mathrm{C}_3$ contains two Hs and one methyl group
$\left(\mathrm{C}_3\right.$ form front carbon in the given structure)
So, $\quad X=\mathrm{CH}_3$
Isomerism (Stereochemistry)
Conceptual (Structural visualisation) III

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
In the options given below, let $E$ denote the rest mass energy of a nucleus and $n$ a neutron. The correct option is

Options:
A
$E\left({ }_{92}^{236} \mathrm{U}\right)>E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{36}^{97} Y\right)+2 E(n)$

B
$\left.E{ }_{92}^{236} \mathrm{U}\right) < E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{39}^{97} Y\right)+2 E(n)$
C
$E\left({ }_{92}^{236} \mathrm{U}\right) < E\left(5_{56}^{140} \mathrm{Ba}\right)+E\left({ }_{36}^{94} \mathrm{Kr}\right)+2 E(n)$
D
$E\left(9_{92}^{236} \mathrm{U}\right) < E\left(5_{56}^{140} \mathrm{Ba}\right)+E\left({ }_{36}^{94} \mathrm{Kr}\right)+2 E(n)$

The correct answer is:
$E\left({ }_{92}^{236} \mathrm{U}\right)>E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{36}^{97} Y\right)+2 E(n)$

Solution:
Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei. Therefore, option (a) will be correct.

Topic: Physics, Nuclear Physics, JEE Advanced, JEE Advanced 2007 (Paper 1)
In the reaction,
$$
2 \mathrm{X}+\mathrm{B}_2 \mathrm{H}_6 \longrightarrow\left[\mathrm{BH}_2(\mathrm{X})_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}
$$
The amine (s) $X$, is (are)

Options:
A
$\mathrm{NH}_3$

B
$\mathrm{CH}_3 \mathrm{NH}_2$

C
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$

D
$\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$

The correct answers are:
$\mathrm{NH}_3$
,
$\mathrm{CH}_3 \mathrm{NH}_2$
,
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$

Solution:
Small amines such as $\mathrm{NH}_3 \cdot \mathrm{CH}_3 \mathrm{NH}_2$ and $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$ give unsymmetrical cleavage of diborane according to following reaction,
$$
\begin{aligned}
& \mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NH}_3 \longrightarrow \\
& {\left[\mathrm{H}_2 \mathrm{~B}\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}} \\
&
\end{aligned}
$$
Large amines, such as $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$ give symmetrical cleavage of diborane according to following reaction,

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2009 (Paper 2)
In the reaction,
$\stackrel{\mathrm{HBr}}{\longrightarrow}$
the products are,

Options:
A

B

C

D

The correct answer is:

Solution:
Alcohol-phenol-ether
Conceptual, understanding of mechanism
II (Given options raised the difficulty level)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 1)
In the reaction, the intermediate(s) is (are)

Options:
A

B

C

D

The correct answers are:
,

Solution:
- $\mathrm{OH}$ and in aqueous in alkaline medium $-\overline{\mathrm{O}}$ are ortho and para directing. We expect the main intermediate products as mono ortho/para substituted product. Remember! if we consider (b), which could be there in small amount then we need to consider even (d) because $-\overline{\mathrm{O}}$ is strongest activating group that facilitates electrophilic substitution reaction overall on the ring. $-\overline{\mathrm{O}}$
activates ortho and para position more but doesn't deactivate the meta position. We get a small amount of even the meta substituted product. But, as the question appears to be directly concerned answer should be only (a) and (c).
Reaction mechanism (Electrophilic substitution)
Conceptual
II

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 1)
In the reaction the structure of the product $T$ is

Options:
A

B

C

D

The correct answer is:

Solution:
Acids and their derivatives Conceptual (Reaction understanding) II

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the reaction the structure of the product $T$ is

Options:
A

B

C

D

The correct answer is:

Solution:
Acids and their derivatives Conceptual (Reaction understanding) II

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 2)
In the scheme given below, the total number of intramolecular aldol condensation products formed from ' $Y$ ' is

Options:

The correct answer is:
1

Solution:
For aldol condensation $\mathrm{C}-5$ and $\mathrm{C}-7$ can attack to C-1 similarly C-2 and C-10 can attack to C-6 but all give same product.
Carbonyl compounds
Conceptual
III

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Advanced, JEE Advanced 2010 (Paper 1)
Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100 \mathrm{~W}, 60 \mathrm{~W}$ and $40 \mathrm{~W}$ bulbs have filament resistances $R_{100}, R_{60}$ and $R_{40}$, respectively, the relation between these resistances is

Options:
A
$\frac{1}{R_{100}}=\frac{1}{R_{40}}+\frac{1}{R_{60}}$

B
$R_{100}=R_{40}+R_{60}$

C
$R_{100}>R_{60}>R_{40}$

D
$\frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$

The correct answer is:
$\frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$

Solution:
$R=\frac{V^2}{P}$ or $R \propto \frac{1}{P}$ $\therefore \quad \frac{1}{R_{100}}>\frac{1}{R_{60}}>\frac{1}{R_{40}}$
Hence, the correct option is (d).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2010 (Paper 1)
Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is $2 \pi / 3$, is $\sqrt{3}$, then area of triangle is

Options:
A
$4 \sqrt{3}$

B
$12-7 \sqrt{3}$

C
$12+7 \sqrt{3}$

D
None of these

The correct answer is:
$12+7 \sqrt{3}$

Solution:
Let $A B=A C$ and $\angle A=120^{\circ}$
$\therefore$ Area of $\Delta=\frac{1}{2} a^2 \sin 120^{\circ}$
where, $a=A D+B D$
$=\sqrt{3} \tan 30^{\circ}+\sqrt{3} \cot 15^{\circ}$
$=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ} \tan 30^{\circ}}{\tan 45^{\circ}-\tan 30^{\circ}}\right)$
$=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$
$\therefore \quad a=4+2 \sqrt{3}$
$\therefore$ Area of $\Delta=\frac{1}{2}(4+2 \sqrt{3})^2\left(\frac{\sqrt{3}}{2}\right)=12+7 \sqrt{3}$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2006
Internal bisector of $\angle A$ of $\triangle A B C$ meets side $B C$ at $D$. A line drawn through $D$ perpendicular to $A D$ intersects the side $A C$ at $E$ and side $A B$ at $F$. If $a, b$ and $c$ represent sides of $\triangle A B C$, then

Options:
A
$A E$ is $\mathrm{HM}$ of $b$ and $c$

B
$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$

C
$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$

D
the $\triangle A E F$ is isosceles

The correct answers are:
$A E$ is $\mathrm{HM}$ of $b$ and $c$
,
$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$
,
$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$
,
the $\triangle A E F$ is isosceles

Solution:
We have, $\triangle A B C=\triangle A B D+\triangle A C D$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2} \\
& \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2}
\end{aligned}
$$
Again, $\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}$
$\therefore A E$ is $\mathrm{HM}$ of $b$ and $c$.
$$
\begin{aligned}
E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\
& =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}
\end{aligned}
$$
As $A D \perp E F$ and $D E=D F$ and $A D$ is bisector.
$\Rightarrow \triangle A E F$ is isosceles.

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2006
Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \times$ $\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)>0$ and

$-1 < \sin \theta < -\frac{\sqrt{3}}{2}$, then $\varphi$ cannot satisfy

Options:
A $0 < \varphi < \frac{\pi}{2}$
B $\frac{\pi}{2} < \varphi < \frac{4 \pi}{3}$
C $\frac{4 \pi}{3} < \varphi < \frac{3 \pi}{2}$
D $\frac{3 \pi}{2} < \varphi < 2 \pi$

The correct answers are:
$0 < \varphi < \frac{\pi}{2}$, $\frac{4 \pi}{3} < \varphi < \frac{3 \pi}{2}$, $\frac{3 \pi}{2} < \varphi < 2 \pi$

Solution:
As $\tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]$

Hence $\frac{3 \pi}{2} < \theta < \frac{5 \pi}{3}$

Now $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1$

$\Rightarrow 2 \cos \theta(1-\sin \varphi)=2 \sin \theta \cos \varphi-1$

$\Rightarrow 2 \cos \theta+1=2 \sin (\theta+\varphi)$

As $\quad \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right), 1 < 2 \sin (\theta+\varphi) < 2$

As $\theta+\varphi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ or $(\theta+\varphi) \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right)$

We have $\varphi \in\left(-\frac{3 \pi}{2},-\frac{2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right)$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $\theta \in\left(0, \frac{\pi}{4}\right)$ and $t_1=(\tan \theta)^{\tan \theta}, t_2=(\tan \theta)^{\cot \theta}$ and $t_4=(\cot \theta)^{\tan \theta}$, then

Options:
A
$t_1>t_2>t_3>t_4$

B
$t_4>t_3>t_1>t_2$

C
$t_3>t_1>t_2>t_4$

D
$t_2>t_3>t_1>t_4$

The correct answer is:
$t_4>t_3>t_1>t_2$

Solution:
As when $\theta \in\left(0, \frac{\pi}{4}\right)$
$\tan \theta < \cot \theta$
Since, $\tan \theta < 1$ and $\cot \theta>1$
$\therefore(\tan \theta)^{\cot \theta} < 1$ and $(\cot \theta)^{\tan \theta}>1$
$\therefore t_4>t_1$ which only holds in option (b).

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Advanced, JEE Advanced 2006
Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\left[\begin{array}{ccc}1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1\end{array}\right]$, where each of $a, b$ and $c$ is either $\omega$ or $\omega^2$. Then, the number of distinct matrices in the set $S$ is

Options:
A
2

B
6

C
4

D
8

The correct answer is:
2

Solution:
$|A| \neq 0$, as non-singular.
$$
\begin{aligned}
& \therefore \quad\left|\begin{array}{ccc}
1 & a & b \\
\omega & 1 & c \\
\omega^2 & \omega & 1
\end{array}\right| \neq 0 \\
& \Rightarrow \quad 1(1-c \omega)-a\left(\omega-c \omega^2\right) \\
& +b\left(\omega^2-\omega^2\right) \neq 0 \\
& \Rightarrow \quad 1-c \omega-a \omega+a c \omega^2 \neq 0 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad(1-c \omega)(1-a \omega) \neq 0 \\
& \Rightarrow \quad a \neq \frac{1}{\omega}, c \neq \frac{1}{\omega} \Rightarrow a=\omega, c=\omega
\end{aligned}
$$
and $b \in\left\{\omega, \omega^2\right\} \Rightarrow 2$ solutions

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15,27-2 a_2>0$ and $a_k=2 a_{k-1}-a_{k-2}$ for $k=3,4, \ldots, 11$.
If $\frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11}=90$, then the value of $\frac{a_1+a_2+\ldots+a_{11}}{11}$ is equal to

Options:

The correct answer is:
0

Solution:
$$
\text { } \begin{aligned}
& a_k=2 a_{k-1}-a_{k-2} \\
& \Rightarrow \quad a_1, a_2, \ldots, a_{11} \text { are in AP } \\
& \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \\
&=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \\
& \Rightarrow \quad 225+35 d^2+150 d=90 \\
& 35 d^2+150 d+135=0 \\
& \Rightarrow \quad d=-3,-\frac{9}{7}
\end{aligned}
$$
Given, $a_2 < \frac{27}{2} \therefore d=-3$ and $d \neq-\frac{9}{7}$
$$
\Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11}
$$
$$
=\frac{11}{2}[30-10 \times 3]=0
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2010 (Paper 2)
Let $a_1, a_2, a_3, \ldots ., a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i$, $1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_m}{S_n}$ does not depend on $n$, then $a_2$ is

Options:

The correct answer is:
3

Solution:
Given, $a_1=3, m=5 n$ and $a_1, a_2, \ldots$ are in AP. $\therefore \frac{S_m}{S_n}=\frac{S_{5 n}}{S_n}$ is independent of $n$. Now, $\quad \frac{\frac{5 n}{2}[2 \times 3+(5 n-1) d]}{\frac{n}{2}[2 \times 3+(n-1) d]}$ $\Rightarrow \quad \frac{f\{(6-d)+5 n\}}{(6-d)+n}$ independent of $n$, if $\begin{aligned} \quad 6-d= & 0 \Rightarrow d=6 \\ \therefore \quad a_2= & a_1+d=3+6=9 \\ & \text { Or }\end{aligned}$
If $d=0$ $\frac{S_m}{S_n}$ is independent of $n$.
$$
\therefore \quad a_2=3
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $a_{1}, a_{2}, a_{3}, \ldots . .$ be in harmonic progression with $a_{1}=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_{n} < 0$ is

Options:
A 22
B 23
C 24
D 25

The correct answer is:
25

Solution:
$\because a_{1}, a_{2}, a_{3}, \ldots \ldots$ are in H.P.

$\begin{array}{l}

\therefore \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \ldots \ldots \text { are in A.P. } \\

\therefore \quad \frac{1}{a_{1}}=\frac{1}{5} \text { and } \frac{1}{a_{20}}=\frac{1}{25} \\

\frac{1}{a_{1}}+19 d=\frac{1}{a_{20}} \Rightarrow \frac{1}{5}+19 d=\frac{1}{25} \Rightarrow d=\frac{-4}{475} \\

\text { Now } \frac{1}{a_{n}}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right) \\

\text { Clearly } a_{n} < 0, \text { if } \frac{1}{a_{n}} < 0 \Rightarrow \frac{1}{5}-\frac{4 n}{475}+\frac{4}{475} < 0 \\

\Rightarrow \quad-4 n < -99 \text { or } n>\frac{99}{4}=24 \frac{3}{4} \quad \therefore n \geq 25 \\

\therefore \quad \text { Least value of } n \text { is } 25 .

\end{array}$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a$ $>-1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{x \rightarrow 0^{+}} \beta(a)$ are

Options:
A $-\frac{5}{2}$ and 1
B $-\frac{1}{2}$ and $-1$
C $-\frac{7}{2}$ and 2
D $-\frac{9}{2}$ and 3

The correct answer is:
$-\frac{1}{2}$ and $-1$

Solution:
$(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$

Let $a+1=y$, then equation reduces to $\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0$

On dividing both sides by $y-1$, we get $\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0$

On taking limit as $y \rightarrow 1$ i.e. $a \rightarrow 0$ on both sides, we get $\frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^{2}+3 x+1=0$

$\Rightarrow x=-1,-\frac{1}{2}$ (roots of the equation)

$\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $a$ and $b$ be non-zero and real numbers. Then, the equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ represents

Options:
A
Four straight lines, when $c=0$ and $a, b$ are of the same sign

B
Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$

C
Two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$

D
A circle and an ellipse, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$

The correct answer is:
Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$

Solution:
Let $a$ and $b$ be non- zero real numbers.
Therefore, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ implies either
$$
\begin{array}{rc}
& x^2-5 x y+6 y^2=0 \\
\Rightarrow & (x-2 y)(x-3 y)=0 \\
\Rightarrow & x=2 y \text { and } x=3 y
\end{array}
$$
represent two straight lines passing through origin.
Or $\quad a x^2+b y^2+c=0$
When $c=0$ and $a$ and $b$ are of same signs, then
$$
\begin{aligned}
& a x^2+b y^2+c=0 \\
\Rightarrow & x=0 \text { and } y=0
\end{aligned}
$$
which is a point specified as the origin.
When $a=b$ and $c$ is of sign opposite to that of $a, a x^2+b y^2+c=0$ represents a circle.
Hence, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$
may represent two straight lines and a circle.

Topic: Mathematics, Pair of Lines, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let $A$ and $B$ be two distinct points on the parabola $y^2=4 x$. If the axis of the parabola touches a circle of radius $r$ having $A B$ as its diameter, then the slope of the line joining $A$ and $B$ can be

Options:
A
$-\frac{1}{r}$

B
$\frac{1}{r}$

C
$\frac{2}{r}$

D
$-\frac{2}{r}$

The correct answers are:
$\frac{2}{r}$
,
$-\frac{2}{r}$

Solution:
Here, coordinate $M=\left(\frac{t_1^2+t_2^2}{2}, t_1+t_2\right) i e$, mid point of chord $A B$.
$M P=t_1+t_2=r$ Also, $\quad m_{A B}=\frac{2 t_2-2 t_1}{t_2^2-t_1^2}=\frac{2^{\ldots(i)}}{t_2+t_1}$ (when $A B$ is chord)
$$
\Rightarrow \quad m_{A B}=\frac{2}{r}
$$
[from Eq. (i)]
Also, $\quad m_{A^{\prime} B^{\prime}}=-\frac{2}{r}$ (when $A^{\prime} B^{\prime}$ is chord)
Hence, (c, d) is the correct option.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be unit vectors such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$. Which one of the following is correct?

Options:
A
$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a}=\mathbf{0}$

B
$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0}$

C
$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{a} \times \mathbf{c}=\mathbf{0}$

D
$\mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ are mutually perpendicular

The correct answer is:
$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0}$

Solution:
Since $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors and $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ represent an equilateral triangle.
$$
\therefore \quad \mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0} .
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $a, b, c, p, q$ be real numbers.
Suppose, $\alpha, \beta$ are the roots of the equation $x^2+2 p x+q=0$ and $\alpha, \frac{1}{\beta}$ are the roots of the equation $a x^2+2 b x+c=0$, where $\beta^2 \notin\{-1,0,1\}$.
Statement $1\left(p^2-q\right)\left(b^2-a c\right) \geq 0$.
Statement $2 b \neq p a$ or $c \neq q a$.

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:
Given, $\alpha$ and $\beta$ are the roots of $x^2+2 p x+q=0$.
$$
\begin{array}{rlrl}
& \therefore & \alpha+\beta & =-2 p \\
\text { and } & \alpha \beta & =q
\end{array}
$$
$\alpha$ and $\frac{1}{\beta}$ are the roots of $a x^2+2 b x+c=0$
and
$$
\alpha+\frac{1}{\beta}=\frac{-2 b}{a}
$$
and
$$
\frac{\alpha}{\beta}=\frac{c}{a}
$$
$$
\text { Now, } \begin{aligned}
\left(p^2-q\right)\left(b^2-a c\right) & =\left[\left(\frac{\alpha+\beta}{-2}\right)^2-\alpha \beta\right]\left[\left(\frac{\alpha+\frac{1}{\beta}}{2}\right)^2-\frac{\alpha}{\beta}\right] a^2 \\
& =\frac{(\alpha-\beta)^2}{16}\left(\alpha-\frac{1}{\beta}\right)^2 \cdot a^2 \geq 0
\end{aligned}
$$
Statement 1 is true.
$$
\begin{array}{ll}
\text { Again now, } & p a=-\left(\frac{\alpha+\beta}{2}\right) a=-\frac{a}{2}(\alpha+\beta) \\
\text { and } & b=-\frac{a}{2}\left(\alpha+\frac{1}{\beta}\right) \\
& p a \neq b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta \\
\Rightarrow & \beta \neq 1 \\
\because & \beta^2 \neq\{-1,0,1\}, \text { correct. } \\
\text { Similarly, if } & c \neq q x \\
\Rightarrow & a \frac{\alpha}{\beta} \neq a \alpha \beta \Rightarrow \alpha\left(\beta-\frac{1}{\beta}\right) \neq 0
\end{array}
$$
$$
\begin{array}{ll}
\Rightarrow & \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \\
\Rightarrow & \beta \neq\{-1,0,1\}
\end{array}
$$
Statement 2 is true.
Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2008 (Paper 2)