datavorous/entrance-exam-dataset huggingface dataset:
The smallest ketone and its next homologoue are reacted with $\mathrm{NH}_2 \mathrm{OH}$ to form oxime

Options:
A
two different oximes are formed

B
three different oximes are formed

C
two oximes are optically active

D
all oximes are optically active

The correct answers are:
three different oximes are formed

Solution:
Lowest ketone is $\mathrm{CH}_3 \mathrm{COCH}_3$ (which form an oxime) and its next homologue is $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCH}_3$ which form oxime with $\mathrm{NH}_2 \mathrm{OH}$. Such oxime shows the property of geometrical isomerism, i.e., it shows two isomers, symmetric and antisymmertic. (These are differ w.r.t. arrangement towards the space of $-\mathrm{C}=\mathrm{N}$, their different behaviour, can be confirmed with the help of Beckmann's rearrangement.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2006
The smallest value of $k$, for which both the roots of the equation $x^2-8 k x+16\left(k^2-k+1\right)=0$ are real, distinct and have values atleast 4 , is

Options:

The correct answer is:
2

Solution:
(i) Given, $x^2-8 k x+16\left(k^2-k+1\right)=0$
Now, $\quad D=64\left\{k^2-\left(k^2-k+1\right)\right\}$
$$
\begin{aligned}
& =64(k-1)>0 \\
& \therefore \quad k>1 \\
& 16-32 k+16\left(k^2-k+1\right) \geq 0 \\
& \Rightarrow \quad k^2-3 k+2 \geq 0 \\
& \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1 \text { or } k \geq 2 \\
&
\end{aligned}
$$
Hence, $k=2$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2009 (Paper 2)
The species having bond order different from that in $\mathrm{CO}$ is

Options:
A
$\mathrm{NO}^{-}$

B
$\mathrm{NO}^{+}$

C
$\mathrm{CN}^{-}$

D
$\mathrm{N}_2$

The correct answer is:
$\mathrm{NO}^{-}$

Solution:
$\mathrm{CO}$ : total $e^{-s}=6+8=14$
$$
\begin{gathered}
\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^{1+1}=\pi 2 p_y^{1+1}, \sigma 2 p_z^2 \\
\mathrm{BO}=\frac{10-4}{2}=3 \\
\mathrm{NO}^{-}=\text {Total } e^{-} s=7+8+1=16 \\
\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^{1+1} \\
=\pi 2 p_y^{1+1}, \pi^* 2 p_x^1=\pi^* 2 p_y^1 \\
\mathrm{BO}=\frac{10-6}{2}=\frac{4}{2}=2
\end{gathered}
$$
Total number of $e^{-} s: \mathrm{CN}^{-}=6+7+1=14$
$$
\mathrm{N}_2=7+7=14
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2007 (Paper 1)
The species having pyramidal shape is

Options:
A
$\mathrm{SO}_3$

B
$\mathrm{BrF}_3$

C
$\mathrm{SiO}_3^{2-}$

D
$\mathrm{OSF}_2$

The correct answer is:
$\mathrm{OSF}_2$

Solution:
$\mathrm{OSF}_2$ has pyramidal shape Chemical bonding Basic understanding II

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The species having pyramidal shape is

Options:
A
$\mathrm{SO}_3$

B
$\mathrm{BrF}_3$

C
$\mathrm{SiO}_3^{2-}$

D
$\mathrm{OSF}_2$

The correct answer is:
$\mathrm{OSF}_2$

Solution:
$\mathrm{OSF}_2$ has pyramidal shape Chemical bonding Basic understanding II

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2010 (Paper 2)
The species present in solution when $\mathrm{CO}_2$ is dissolved in water are

Options:
A
$\mathrm{CO}_2, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{HCO}_3^{-}, \mathrm{CO}_3^{2-}$

B
$\mathrm{H}_2 \mathrm{CO}_3, \mathrm{CO}_3^{2-}$

C
$\mathrm{CO}_3^{2-}, \mathrm{HCO}_3^{-}$

D
$\mathrm{CO}_2, \mathrm{H}_2 \mathrm{CO}_3$

The correct answers are:
$\mathrm{CO}_2, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{HCO}_3^{-}, \mathrm{CO}_3^{2-}$

Solution:
When $\mathrm{CO}_2$ is dissolved in water, therefore it forms dibasic carbonic acid, which gives $\mathrm{HCO}_3^{-}$ and $\mathrm{CO}_3^{2-}$ ions on ionisation. Hence in solution $\mathrm{CO}_2, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{HCO}_3^{-}$and $\mathrm{CO}_3^{2-}$ are present.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Advanced, JEE Advanced 2006
The species which by definition has zero standard molar enthalpy of formation at $298 \mathrm{~K}$ is

Options:
A
$\mathrm{Br}_2(g)$

B
$\mathrm{Cl}_2(g)$

C
$\mathrm{H}_2 \mathrm{O}(g)$

D
$\mathrm{CH}_4(g)$

The correct answer is:
$\mathrm{Cl}_2(g)$

Solution:
$\Delta H_f^{\circ}\left(\mathrm{Cl}_2, g\right)=0$
$\Delta H_f^{\circ}$ of elements in their standard state is taken to be zero. $\mathrm{Cl}_2$ is gas, $\mathrm{Br}_2$ is liquid and $\mathrm{I}_2$ is solid at room temperature.
Energetics
Conceptual, implication of rules/laws III

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2010 (Paper 1)
The spin only magnetic moment value (in Bohr magneton units) of \mathrm{Cr}(\mathrm{CO})_6$ is

Options:
A
0

B
$2.84$

C
$4.90$

D
$5.92$

The correct answer is:
0

Solution:
$\mathrm{CO}$ is a strong ligand, all the six electron of the valence shell of $\mathrm{Cr}$ is paired and spin only magnetic moment $=0$.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2009 (Paper 2)
The straight line $2 x-3 y=1$ divides the circular region $x^2+y^2 \leq 6$ into two parts. If $S=\left\{\left(2, \frac{3}{4}\right),\left(\frac{5}{2}, \frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8}, \frac{1}{4}\right)\right\}$, then the number of point(s) in $S$ lying inside the smaller part is

Options:

The correct answer is:
2

Solution:
$x^2+y^2 \leq 6$ and $2 x-3 y=1$ is shown as, For the point to lie in the shaded part, origin and the point lie on opposite side of straight line $L$.
$\therefore$ For any point in shaded part $L>0$ and for any point inside the circle $S < 0$.
Now, for $\left(2, \frac{3}{4}\right) L: 2 x-3 y-1$
$$
L: 4-\frac{9}{4}-1=\frac{3}{4}>0
$$
and $S: x^2+y^2-6$,
$$
S: 4+\frac{9}{16}-6 < 0
$$
$\Rightarrow\left(2 \frac{3}{4}\right)$ lies in shaded part.
For $\left(\frac{5}{2}, \frac{3}{4}\right) L: 5-9-1 < 0 \quad$ [neglect]
For $\left.\left(\frac{1}{4},-\frac{1}{4}\right) L: \frac{1}{2}+\frac{3}{4}-1>0\right\}$
$\therefore\left(\frac{1}{4},-\frac{1}{4}\right)$ lies in the shaded part. For $\left(\frac{1}{8}, \frac{1}{4}\right) L: \frac{1}{4}-\frac{3}{4}-1 < 0$ [neglect] $\Rightarrow$ Only 2 points lie in the shaded part.

Topic: Mathematics, Circle, JEE Main
The straight line $2 x-3 y=1$ divides the circular region $x^2+y^2 \leq 6$ into two parts. If $S=\left\{\left(2, \frac{3}{4}\right),\left(\frac{5}{2}, \frac{3}{4}\right),\left(\frac{1}{4},-\frac{1}{4}\right),\left(\frac{1}{8}, \frac{1}{4}\right)\right\}$, then the number of point(s) in $S$ lying inside the smaller part is

Options:

The correct answer is:
2

Solution:
$x^2+y^2 \leq 6$ and $2 x-3 y=1$ is shown as, For the point to lie in the shaded part, origin and the point lie on opposite side of straight line $L$.
$\therefore$ For any point in shaded part $L>0$ and for any point inside the circle $S < 0$.
Now, for $\left(2, \frac{3}{4}\right) L: 2 x-3 y-1$
$$
L: 4-\frac{9}{4}-1=\frac{3}{4}>0
$$
and $S: x^2+y^2-6$,
$$
S: 4+\frac{9}{16}-6 < 0
$$
$\Rightarrow\left(2 \frac{3}{4}\right)$ lies in shaded part.
For $\left(\frac{5}{2}, \frac{3}{4}\right) L: 5-9-1 < 0 \quad$ [neglect]
For $\left.\left(\frac{1}{4},-\frac{1}{4}\right) L: \frac{1}{2}+\frac{3}{4}-1>0\right\}$
$\therefore\left(\frac{1}{4},-\frac{1}{4}\right)$ lies in the shaded part. For $\left(\frac{1}{8}, \frac{1}{4}\right) L: \frac{1}{4}-\frac{3}{4}-1 < 0$ [neglect] $\Rightarrow$ Only 2 points lie in the shaded part.

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2011 (Paper 2)
The substituents $R_{1}$ and $R_{2}$ for nine peptides are listed in the table given below. How many of these peptides are positively charged at $\mathrm{pH}=7.0$ ?

\begin{array}{|c|c|c|}

\hline Peptide & \mathbf{R}_{\mathbf{1}} & \mathbf{R}_{\mathbf{2}} \\

\hline I & \mathrm{H} & \mathrm{H} \\

\hline II & \mathrm{H} & \mathrm{CH}_{3} \\

\hline III & \mathrm{CH}_{2} \mathrm{COOH} & \mathrm{H} \\

\hline IV & \mathrm{CH}_{2} \mathrm{CONH}_{2} & \left(\mathrm{CH}_{2}\right)_{4} \mathrm{NH}_{2} \\

\hline V & \mathrm{CH}_{2} \mathrm{CONH}_{2} & \mathrm{CH}_{2} \mathrm{CONH}_{2} \\

\hline VI & \left(\mathrm{CH}_{2}\right)_{4} \mathrm{NH}_{2} & \left(\mathrm{CH}_{2}\right)_{4} \mathrm{NH}_{2} \\

\hline VII & \mathrm{CH}_{2} \mathrm{COOH} & \mathrm{CH}_{2} \mathrm{CONH}_{2} \\

\hline VIII & \mathrm{CH}_{2} \mathrm{OH} & \left(\mathrm{CH}_{2}\right)_{4} \mathrm{NH}_{2} \\

\hline IX & \left(\mathrm{CH}_{2}\right)_{4} \mathrm{NH}_{2} & \mathrm{CH}_{3} \\

\hline

\end{array}

Options:

The correct answer is:
4

Solution:
Peptides with isoelectric point (pI) more than 7 would exist as cation in neutral solution $(\mathrm{pH}=7)$ which means the given polypeptide is of basic nature, so it must contain two or more amino groups. Hence IV, VI, VIII and IX are correct options.

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2012 (Paper 1)
The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

Options:
A
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$

B
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$

C
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $$
\mathrm{CH}_3 \mathrm{C} \equiv \equiv \mathrm{CH}
$$

D
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$a)

The correct answer is:
$\mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and
$$
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}
$$a)

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_2$ $-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_3$ There are two (marked) possibility of halides
Thus, in the given context
$$
\begin{array}{r}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CH}+\mathrm{NaNH}_2 \longrightarrow \\
\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{C} \equiv \overline{\mathrm{CNa}^{+}}
\end{array}
$$
$$
\begin{aligned}
\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \overline{\mathrm{CNa}} \\
\longrightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{BrCH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3 \\
\text { 3-octyne } \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_3
\end{aligned}
$$
Hydrocarbon
Conceptual, reaction understanding II

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2010 (Paper 1)
The tangent $P T$ and the normal $P N$ to the parabola $y^2=4 a x$ at a point $P$ on

Options:
A
vertex is $\left(\frac{2 a}{3}, 0\right)$ it meet its axis at points $T$ and $N$,

B
directrix is $x=0$ respectively. The locus of the centroid of $\triangle P T N$ is a parabola whose

C
latusrectum is $\frac{2 a}{3}$

D
focus is $(a, 0)$

The correct answers are:
vertex is $\left(\frac{2 a}{3}, 0\right)$ it meet its axis at points $T$ and $N$,
,
focus is $(a, 0)$

Solution:
Equation of tangent and normal at point $P\left(a t^2, 2 a t\right)$ is $t y=x=a t^2$ and $y=-t x+2 a t+a t^2$ Let centroid of $\triangle P T N$ is $R(h, k)$. $\therefore \quad h=\frac{a t^2+\left(-a t^2\right)+2 a+a t^2}{3}$ and $k=\frac{2 a t}{3}$
$$
\begin{array}{rlrl}
& \Rightarrow & 3 h & =2 a+a \cdot\left(\frac{3 k}{2 a}\right)^2 \\
\Rightarrow & & 3 h & =2 a+\frac{9 k^2}{4 a} \\
\Rightarrow & 9 k^2 & =4 a(3 h-2 a)
\end{array}
$$
$\therefore$ Locus of centroid is
$$
y^2=\frac{4 a}{3}\left(x-\frac{2 a}{3}\right)
$$
$\therefore \operatorname{Vertex}\left(\frac{2 a}{3}, 0\right)$; directrix
$$
x-\frac{2 a}{3}=-\frac{a}{3} \Rightarrow x=\frac{a}{3}
$$
Latusrectum $=\frac{4 a}{3}$
$\therefore$ Focus $\left(\frac{a}{3}+\frac{2 a}{3}, 0\right)$ i.e. $(a, 0)$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2009 (Paper 2)
The tangent to the curve $y=e^x$ drawn at the point $\left(c, e^c\right)$ intersects the line joining the points $\left(c-1, e^{c-1}\right)$ and $\left(c+1, e^{c+1}\right)$

Options:
A
on the left of $x=c$

B
on the right of $x=c$

C
at no point

D
at all points

The correct answer is:
on the left of $x=c$

Solution:
Slope of the line joining the points $\left(c-1, e^{c-1}\right)$ and $\left(c+1, e^{c+1}\right)$ is equal to $\frac{e^{c+1}-e^{c-1}}{2}>e^c$
$$
\text { Tangent to the curve } y=e^x \text { will intersect the given line to the left of the line } x=c \text {. }
$$
ALITER
The equation of the tangent to the curve $y=e^x$ at $\left(c, e^c\right)$ is
$$
y-e^c=e^c(x-c)
$$
Equation of the line joining the given points is
$$
y-e^{c-1}=\frac{e^c\left(e-e^{-1}\right)}{2}[x-(c-1)]
$$
Eliminating $y$ from Eqs. (i) and (ii), we get
$$
\begin{aligned}
{[x-(c-1)]\left[2-\left(e-e^{-1}\right)\right] } & =2 e^{-1} \\
x-c & =\frac{e+e^{-1}-2}{2-\left(e-e^{-1}\right)} < 0 \Rightarrow x < c .
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2007 (Paper 1)
The term that corrects for the attractive forces present in a real gas in the van der Waals' equation is

Options:
A
$n b$

B
$n^2 a / V^2$

C
$-\left(n^2 a / V^2\right)$

D
$-n b$

The correct answer is:
$n^2 a / V^2$

Solution:
In the van der Waals' equation
$$
\left(p+\frac{n^2 a}{V^2}\right)(V-n b)=n R T
$$
The pressure correction factor $\left(n^2 a / V^2\right)$ accounts for intermolecular attraction in real gas.

Topic: Chemistry, States of Matter, JEE Advanced, JEE Advanced 2009 (Paper 1)
The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic $\mathrm{KOH}$ is

Options:

The correct answer is:
5

Solution:
The substrate has three different types of $\mathrm{B}-\mathrm{H}$, therefore, first, three structural isomers of alkenes are expected as
The last two alkenes II and III are also capable of showing geometrical isomerism hence two geometrical isomers for each of them will be counted giving a total of five isomers.

Topic: Chemistry, Hydrocarbons, JEE Main
The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic $\mathrm{KOH}$ is

Options:

The correct answer is:
5

Solution:
The substrate has three different types of $\mathrm{B}-\mathrm{H}$, therefore, first, three structural isomers of alkenes are expected as
The last two alkenes II and III are also capable of showing geometrical isomerism hence two geometrical isomers for each of them will be counted giving a total of five isomers.

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2011 (Paper 1)
The total number of $\alpha$ and $\beta$-particles emitted in the nuclear reaction ${ }_{92} \mathrm{U}^{238} \longrightarrow{ }_{82} \mathrm{~Pb}^{214}$ is

Options:

The correct answer is:
8

Solution:
${ }_{92} \mathrm{U}^{238} \rightarrow_{82} \mathrm{U}^{214}+6_2 \mathrm{He}^{\mathrm{H}}(\alpha)+2_{-1} e^0(\beta)$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2009 (Paper 2)
The total number of basic groups in the following form of lysine is

Options:

The correct answer is:
2

Solution:
Lysine contains two basic groups, e.g., $\mathrm{NH}_2$
Biomolecules
Basic straight
I

Topic: Chemistry, Biomolecules, JEE Advanced, JEE Advanced 2010 (Paper 1)
The total number of contributing structures showing hyperconjugation (involving $\mathrm{C}-\mathrm{H}$ bonds) for the following carbocation is

Options:

The correct answer is:
6

Solution:
$$
\text { These are total } 6 \alpha-\mathrm{H} \text { to } \mathrm{sp}^2 \text { carbon and they all can participate in hyperconjugation. }
$$

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2011 (Paper 2)
The total number of cyclic isomers possible for a hydrocarbon with the molecular formula $\mathrm{C}_4 \mathrm{H}_6$ is

Options:

The correct answer is:
5

Solution:
$\mathrm{C}_4 \mathrm{H}_6$ can have five cyclic isomers
Isomerism (organic)
Structural manipulation III

Topic: Chemistry, General Organic Chemistry, JEE Main
The total number of cyclic isomers possible for a hydrocarbon with the molecular formula $\mathrm{C}_4 \mathrm{H}_6$ is

Options:

The correct answer is:
5

Solution:
$\mathrm{C}_4 \mathrm{H}_6$ can have five cyclic isomers
Isomerism (organic)
Structural manipulation III

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2010 (Paper 1)
The total number of cyclic structural as well as stereoisomers possible for a compound with the molecular formula $=\mathrm{C}_5 \mathrm{H}_{10}$ is

Options:

The correct answer is:
7

Solution:
$$
\text { The total number of cyclic isomers are seven as shown below. }
$$

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2009 (Paper 2)
The total number of diprotic acids among the following is
$\begin{array}{lll}\mathrm{H}_3 \mathrm{PO}_4 & \mathrm{H}_2 \mathrm{SO}_4 & \mathrm{H}_3 \mathrm{PO}_3 \\ \mathrm{H}_2 \mathrm{CO}_3 & \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 & \mathrm{H}_3 \mathrm{BO}_3 \\ \mathrm{H}_3 \mathrm{PO}_2 & \mathrm{H}_2 \mathrm{CrO}_4 & \mathrm{H}_2 \mathrm{SO}_3\end{array}$

Options:

The correct answer is:
6

Solution:
$\mathrm{H}_2 \mathrm{SO}_4, \mathrm{H}_2 \mathrm{CO}_3, \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7, \mathrm{H}_2 \mathrm{CrO}_4, \mathrm{H}_3 \mathrm{PO}_3, \mathrm{H}_2 \mathrm{SO}_3$
Acid-base-salt
Basic structural understanding II

Topic: Chemistry, Ionic Equilibrium, JEE Advanced, JEE Advanced 2010 (Paper 2)
The total number of local maxima and local minima of the function $f(x)=\left\{\begin{array}{lc}(2+x)^3 ; & -3 < x \leq-1 \\ x^{\frac{2}{3}} ; & -1 < x < 2\end{array}\right.$ is

Options:
A
0

B
1

C
2

D
3

The correct answer is:
2

Solution:
Given that, $f(x)=\left\{\begin{array}{l}(2+x)^3 ;-3 < x \leq-1 \\ x^{2 / 3} ; \quad-1 < x < 2\end{array}\right.$
$$
\Rightarrow \quad f^{\prime}(x)=\left\{\begin{array}{l}
3(2+x)^2 ;-3 < x \leq 1 \\
\frac{2}{3} x^{-\frac{1}{3}} ; \quad-1 < x < 2
\end{array}\right.
$$
Clearly, $f^{\prime}(x)$ changes its sign at $x=-1$ from $+$ ve to -ve and so $f(x)$ has local maxima at $x=-1$.
Also, $f^{\prime}(0)$ does not exist but $f^{\prime}\left(0^{-}\right) < 0$ and $f^{\prime}\left(0^{+}\right) < 0$.
It can only be inferred that $f(x)$ has a possibility of a minima at $x=0$.
Hence, the given function has one local maxima at $x=-1$ and one local minima at $x=0$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2008 (Paper 1)
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

Options:
A 75
B 150
C 210
D 243

The correct answer is:
150

Solution:
$\because \quad$ Each person gets at least one ball.

$\therefore \quad 3$ Persons can have 5 balls as follow.

\begin{array}{|c|c|c|}

\hline Person & No. of balls & No. of balls \\

\hline I & 1 & 1 \\

\hline II & 1 & 2 \\

\hline III & 3 & 2 \\

\hline

\end{array}

The number of ways to distribute balls 1,1,3 in first to three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3}$

Also 3 , persons having 1,1 and 3 balls can be arranged in $\frac{3 !}{2 !}$ ways.

$\therefore \quad$ Total no. of ways to distribute $1,1,3$ balls to the three persons

$={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{1} \times{ }^{3} \mathrm{C}_{3} \times \frac{3 !}{2 !}=60$

Similarly, total no. of ways to distribute $1,2,2$ balls to

three persons $={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{2} \times \frac{3 !}{2 !}=90$

$\therefore \quad$ The required number of ways $=60+90=150$

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2012 (Paper 1)
The type of placenta with a unilocular ovary and a single ovule

Options:
A Marginal
B Basal
C Free central
D Axile

The correct answer is:
Basal

Solution:
In basal placentation, the ovary is bicarpellary, syncarpous and unilocular and a single ovule is borne at the base of ovary. Basal placentation is the most advance type of placentation. It is observed in most advanced family like Asteraceae and Poaceae.

Topic: Biology, Morphology of Flowering Plants, NEET
The value of $\frac{(5050) \int_0^1\left(1-x^{50}\right)^{100} d x}{\int_0^1\left(1-x^{50}\right)^{101} d x}$

Options:

The correct answer is:
5051

Solution:
Let $I_2=\int_0^1\left(1-x^{50}\right)^{101} d x$, using integration by parts
$$
\begin{aligned}
& \left.=\left[(1-x)^{50}\right)^{101} \cdot x\right]_0^1+\int_0^1\left(1-x^{50}\right)^{100} 50 \cdot x^{49} \cdot x d x \\
& =0-\int_0^1(50)(101)\left(1-x^{50}\right)^{100}(-x)^{50} d x \\
& =-(50)(101) \int_0^1\left(1-x^{50}\right)^{101} d x+(50)(101) \int_0^1\left(1-x^{50}\right)^{100} d x \\
& =5050 I_2+5050 I_1
\end{aligned}
$$
$$
\begin{aligned}
\therefore & I_2+5050 I_2 & =5050 I_1 \\
\therefore & \frac{(5050) I_1}{I_2} & =5051
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2006
The value of

$6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}} \cdots}}\right) \text { is }$

Options:

The correct answer is:
4

Solution:
Let $x=\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}} \ldots \ldots$

$\Rightarrow \quad x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}$

$\Rightarrow \quad 3 \sqrt{2} x=\sqrt{4-x}$

$\begin{array}{l}

\Rightarrow \quad 18 x^{2}=4-x \\

\Rightarrow \quad 18 x^{2}+x-4=0 \\

\Rightarrow \quad(9 x-4)(2 x+1)=0 \\

\Rightarrow \quad x=\frac{4}{9} \text { or } x=-\frac{1}{2}

\end{array}$

(Not possible because log is not define for -ve value)

$\begin{aligned}

\therefore \quad & 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2} \\

&=6-2=4

\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2012 (Paper 1)
The value of $b$ for which the equations $x^2+b x-1=0, x^2+x+b=0$ have one root in common is

Options:
A
$-\sqrt{2}$

B
$-i \sqrt{3}$

C
$i \sqrt{5}$

D
$\sqrt{2}$

The correct answer is:
$-i \sqrt{3}$

Solution:
If $\quad a_1 x^2+b_1 x+c_1=0$
and $a_2 x^2+b_2 x+c_2=0$
have a common real root, then
$$
\begin{aligned}
\Rightarrow \quad\left(a_1 c_2-a_2 c_1\right)^2=\left(b_1 c_2-\right. & \left.b_2 c_1\right) \\
& \left(a_1 b_2-a_2 b_1\right)
\end{aligned}
$$
$\left.\therefore \begin{array}{l}x^2+b x-1=0 \\ x^2+x+b=0\end{array}\right\}$ have a common root.
$\Rightarrow \quad(1+b)^2=\left(b^2+1\right)(1-b)$
$\Rightarrow b^2+2 b+1=b^2-b^3+1-b$
$\Rightarrow \quad b^3+3 b=0 \Rightarrow b\left(b^2+3\right)=0$
$\Rightarrow \quad b=0, \pm \sqrt{3} i$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2011 (Paper 2)
The value of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log 6-x^2\right)} d x$ is

Options:
A
$\frac{1}{4} \log \frac{3}{2}$

B
$\frac{1}{2} \log \frac{3}{2}$

C
$\log \frac{3}{2}$

D
$\frac{1}{6} \log \frac{3}{2}$

The correct answer is:
$\frac{1}{4} \log \frac{3}{2}$

Solution:
Put $x^2=t \Rightarrow x d x=\frac{d t}{2}$ $\therefore \quad I=\int_{\log 2}^{\log 3} \frac{\sin t \cdot \frac{d t}{2}}{\sin t+\sin (\log 6-t)}$
Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$$
\begin{aligned}
& I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2+\log 3-t)}{\sin (\log 2+\log 3-t)+\sin } d t \\
& I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin (t)} d t \\
& \therefore I=\int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t \quad \ldots(\mathrm{li})
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t+\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t
$$
$$
\begin{gathered}
\therefore \quad 2 I=\frac{1}{2}(t)_{\log 2}^{\log 3}=\frac{1}{2}(\log 3-\log 2) \\
\Rightarrow I=\frac{1}{4} \log \left(\frac{3}{2}\right)
\end{gathered}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2011 (Paper 1)