datavorous/entrance-exam-dataset huggingface dataset:
The most contributing tautomeric enol form of $\mathrm{MeCOCH}_{2} \mathrm{CO}_{2} \mathrm{Et}$ is

Options:
A $\mathrm{CH}_{2}=\mathrm{C}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{Et}$
B $\mathrm{MeC}(\mathrm{OH})=\mathrm{CHCO}_{2} \mathrm{Et}$
C MeCOCH = C(OH)OEt
D $\mathrm{CH}_{2}=\mathrm{C}(\mathrm{OH}) \mathrm{CH}=\mathrm{C}(\mathrm{OH}) \mathrm{OEt}$

The correct answer is:
$\mathrm{MeC}(\mathrm{OH})=\mathrm{CHCO}_{2} \mathrm{Et}$

Solution:

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
The most likely protonation site in the following molecule is

Options:
A $\mathrm{C}-1$
B $\mathrm{C}-2$
C $\mathrm{C}-3$
D $C-6$

The correct answer is:
$\mathrm{C}-1$

Solution:
Addition of a proton at $\mathrm{C}^{-1}$ results in the formation of tropylium carbocation (an aromatic species with more stability due to delocalisation of $\pi$ electrons). thus, it is the most reactive site towards protonation. Addition of proton at any other carbon atom interupt in the delocalisation of it electrons by disturbing planarity of molecules and herce makes it less stable. Thus, these all are lest reactive towards protonation.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The normal to a curve at $P(x, y)$ meets the $x$-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$ then the curve is

Options:
A a parabola
B a circle
C a hyperbola
D an ellipse

The correct answer is:
a hyperbola

Solution:
equation of normal
$Y-y=\left(\frac{\frac{-1}{d y}}{d x}\right) \cdot(X-x) \quad Y=0 \quad \therefore \quad y \frac{d y}{d x}+x=X \quad \therefore \quad a / c \text { to } Q \quad y \cdot \frac{d y}{d x}+x=2 x$
$\mathrm{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x} \Rightarrow \int \mathrm{y} \mathrm{dy}=\int \mathrm{xdx} \quad \frac{\mathrm{y}^{2}}{2}=\frac{\mathrm{x}^{2}}{2}+\mathrm{c} \quad \frac{\mathrm{y}^{2}-\mathrm{x}^{2}}{2}=\mathrm{c}$ a hyperbola

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2021
The normal to the curve $y=x^{2}-x+1$, drawn at the points with the abscissa $x_{1}=0, x_{2}=-1$ and $x_{3}=5 / 2$

Options:
A are paraliel to each other
B are pairwise perpendicular
C are concurrent
D are not concurrent

The correct answer is:
are concurrent

Solution:
Given equation of curve.
$$
\begin{aligned}
y &=x^{2}-x+1 \\
\Rightarrow \quad & \frac{d y}{d x}=2 x-1
\end{aligned}
$$
Slope of normal $=\frac{1}{1-2 x}$
Now, at $x_{1}=0, y_{1}=1$
$\therefore$ Slope of normal at $(0,1)=1$
$\therefore$ Equation of normal, $y-1=1(x-0)$ $\Rightarrow \quad x-y+1=0$
At $x_{2}=-1, y_{2}=3$
Slope of normal at $(-1,3)=\frac{1}{3}$ Equation of normal, $y-3=\frac{1}{3}(x+1)$
$\Rightarrow \quad 3 y-9=x+1$
$x-3 y+10=0$
At $x_{3}=\frac{5}{2}, y_{3}=\frac{19}{4}$
Slope of normal at $\left(\frac{2}{5}, \frac{19}{4}\right)=-\frac{1}{4}$
Equation of normal, $y-\frac{19}{4}=-\frac{1}{4}\left(x-\frac{5}{2}\right)$
$\begin{array}{lr}\Rightarrow & x+4 y=\frac{43}{2} \\ \Rightarrow & 2 x+8 y=43\end{array}$
Here, intersecting point of Eqs.(i) and (ii) is
(ii) is $\left(\frac{7}{2}, \frac{9}{2}\right)$ and normal (iii) passes through it. Hence, normals are concurrent.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
The normality of 30 volume \(\mathrm{H}_2 \mathrm{O}_2\) is

Options:
A \(2.678 \mathrm{~N}\)
B \(5.336 \mathrm{~N}\)
C \(8.034 \mathrm{~N}\)
D \(6.685 \mathrm{~N}\)

The correct answer is:
\(5.336 \mathrm{~N}\)

Solution:
Hints: Volume strength \(=5.6 \times\) normality
\(\begin{aligned}
& 30=5.6 \times \mathrm{N} \\
& \Rightarrow \mathrm{N}=\frac{30}{5.6}=5.3
\end{aligned}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2011
The nucleus $^{64}_{29}Cu$ accepts an orbital clectron to yield.

Options:
A $\frac{65}{28} \mathrm{Ni}$
B $\frac{64}{30} \mathrm{Zn}$
C $\frac{64}{28} \mathrm{Ni}$
D $\frac{65}{30} \mathrm{zn}$

The correct answer is:
$\frac{64}{28} \mathrm{Ni}$

Solution:
The nucleus $^{64}_{29}Cu$ accepts an orbital electron to yield $^{64}_{28}Ni$. The atomic number of $\mathrm{Cu}$ is 29 , which is equal to the number of electrons and also equal to the number of protons. When $^{64}_{29}Cu$ accepts an orbital electron then electrons subtract from the atomic number, i.e.
$29-1=28$
$$
^{64}_{29}Cu+{ }_{-1} e^{0} \longrightarrow{ }_{28}^{64} \mathrm{Ni}
$$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2018
The number $(101)^{100}-1$ is divisible by

Options:
A $10^{4}$
B $10^{6}$
C $10^{8}$
D $10^{12}$

The correct answer is:
$10^{4}$

Solution:
$(101)^{100}-1=(1+100)^{100}-1$
$=\left(1+{ }^{100} C_{1} \cdot 100+{ }^{100} C_{2} 100^{2}+\ldots \ldots\right)-1$
$={ }^{100} C_{1} 100+{ }^{100} C_{2}(100)^{2}+$
$\quad{ }^{100} C_{3}(100)^{3}+\ldots \ldots+{ }^{100} C_{100}(100)^{100}$
$=10^{4}\left(1+{ }^{100} C_{2}+{ }^{100} C_{3} 10^{2}+\ldots .\right.$
$\quad+{ }^{100} C_{100}(100)^{98}$
$=10^{4}(1+$ an integer multiple of 10$)$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2018
The number \((101)^{100}-1\) is divisible by

Options:
A \(10^4\)
B \(10^6\)
C \(10^8\)
D \(10^{12}\)

The correct answer is:
\(10^4\)

Solution:
\(\begin{aligned}\text {Hints: }
& (101)^{100}-1={ }^{100} \mathrm{C}_1 100+{ }^{100} \mathrm{C}_2 100^2+{ }^{100} \mathrm{C}_3 100^3+\ldots \ldots +{ }^{100} \mathrm{C}_{100} 100^{100} \\
& =100^2\left[1+{ }^{100} \mathrm{C}_2+{ }^{100} \mathrm{C}_3 100+\ldots \ldots .\right] \\
& =\left(10^4\right)
\end{aligned}\)

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2011
The number of acidic protons in $\mathrm{H}_{3} \mathrm{PO}_{3}$ are

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
There are two acidic protons in $\mathrm{H}_{3} \mathrm{PO}_{3}$.

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2012
The number of alkene (s) which can produce 2 -butanol by the successive treatment of
(i) $\mathrm{B}_{2} \mathrm{H}_{6}$ in tetrahydrofuran solvent and
(ii) alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$ solution is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
The number of alkene(s) that can produce 2-butanol by the successive treatment of $\mathrm{B}_{2} \mathrm{H}_{6}$ in tetrahydrofuran solvent and alkaline $\mathrm{H}_{2} \mathrm{O}_{2}$ solution is $2 .$ The alkenes are $c i s$ -but -2 - ene and trans-but-2-ene. The hydration product is obtained in accordance with opposite Markownikoff's rule. In this reaction, addition takes place through the initial formation of $\pi$ -complex which changes into a cyclic four centre transition state with the addition of boron atom to the less hindered carbon atom.
$\mathrm{H}_{3} \mathrm{C}-\mathrm{HC}=\mathrm{CH}-\mathrm{CH}_{3} \frac{(\mathrm{i}) \mathrm{B}_{2} \mathrm{H}_{6}}{(\mathrm{ii}) \mathrm{H}_{2} \mathrm{O}_{2} \mathrm{OH}} \mathrm{CH}_{3}-\mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{3}$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2018
The number of all numbers having 5 digits, with distinct digits is

Options:
A 99999
B $9 \times{ }^{9} P_{4}$
C ${ }^{10} P_{5}$
D ${ }^{9} P_{4}$

The correct answer is:
$9 \times{ }^{9} P_{4}$

Solution:
Total number of 5-digit numbers having all the digits distinct $={ }^{10} \mathrm{P}_{5}-{ }^{9} \mathrm{P}_{4}$
$$
\begin{array}{l}
=\frac{10 !}{5 !}-\frac{91}{5 !}=\frac{10 \times 9 !}{5 !}-\frac{91}{51} \\
=\frac{91}{51}(10-1)=9 \times \frac{9!}{(9-4) !} \\
=9 \times 9 P_{4}
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2017
The number of amino acids and number of peptide bonds in a linear tetrapeptide (made of different amino acids) are respectively

Options:
A 4 and 4
B 5 and 5
C 5 and 4
D 4 and 3

The correct answer is:
4 and 3

Solution:
Combination of four amino acid units gives linear tetrapeptide, in which three peptide bonds are present.

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2014
The number of $\sigma$ and $\pi$ bonds between two carbon atoms in calcium carbide are

Options:
A one $\sigma,$ one $\pi$
B one $\sigma$, two $\pi$
C two $\sigma$, one $\pi$
D one $\sigma, 1 \frac{1}{2} \pi$

The correct answer is:
one $\sigma$, two $\pi$

Solution:
Calcium carbide is calcium acetylide, having structure

Thus, it has one sigma ($\sigma$) and two pi ($\pi$) bonds.

Topic: Chemistry, s Block Elements, WBJEE, WBJEE 2016
The number of atoms in body centred and face centred cubic unit cell respectively are

Options:
A 2 and 4
B 4 and 3
C 1 and 2
D 4 and 6

The correct answer is:
2 and 4

Solution:
$Z_{\mathrm{BCC}}=\frac{1}{8} \times 8+1=2 \quad Z_{\mathrm{FCC}}=\frac{1}{8} \times 8+\frac{1}{2} \times 6=1+3=4$

Topic: Chemistry, Solid State, WBJEE, WBJEE 2022
The number of atoms of a radioactive substance of half-life $T$ is $N_{0}$ at $t=0 .$ The time necessary to decay from $N_{0} / 2$ atoms to $N_{0} / 10$ atoms will be

Options:
A $\frac{5}{2} T$
B $\operatorname{tlog} 5$
C $T \log \left[\frac{5}{2}\right]$
D $\frac{T}{2} \log 5$

The correct answer is:
$T \log \left[\frac{5}{2}\right]$

Solution:
We have, $\frac{N(t)}{N_{0}}=e^{-\lambda t}$ or $N(t)=N_{0} e^{-\lambda t}$
$\therefore$ For the given condition. $\frac{N_{0}}{2}=N_{0} e^{-\lambda t_{1}}$ and $\frac{N_{0}}{10}=N_{0} e^{-\lambda t_{2}}$
$\Rightarrow \quad \frac{1}{2}=e^{-\lambda t_{1}}$ and $\frac{1}{10}=e^{-\lambda t_{2}}$
or $\quad e^{\lambda t_{1}}=2$ and $e^{\lambda t_{2}}=10$
Taking log on both sides, $\lambda t_{1}=\log 2 \Rightarrow t_{1}=\frac{\log 2}{\lambda}$
or $\quad t_{1}=\frac{\log 2 \times \mathrm{T}}{\log 2}$
and $\quad \lambda t_{2}=\log 10$
$\Rightarrow \quad t_{2}=\frac{\log 10}{\lambda}$
of $\quad t_{2}=\frac{\log 10 \times \mathrm{T}}{\log 2}$
$\therefore \quad\left(t_{2}-t_{1}\right)=T\left[\frac{\log 10}{\log 2}-1\right]=T\left[\frac{\log 10-\log 2}{\log 2}\right]$
$=T\left[\frac{\log 5}{\log 2}\right]$
$\Rightarrow\left(t_{2}-t_{1}\right)=T \left[\frac{\log 5}{\log 2}\right]$

Topic: Physics, Nuclear Physics, WBJEE, WBJEE 2013
The number of complex numbers $\mathrm{p}$ such that $|\mathrm{p}|=1$ and imaginary part of $\mathrm{p}^{4}$ is 0, is

Options:
A 4
B 2
C 8
D infinitely many

The correct answer is:
4

Solution:
Hint:
Let $p=x+i y, p^{2}=\left(x^{2}-y^{2}\right)+2 i x y, p^{4}=\left(x^{2}-y^{2}\right)^{2}-4 x^{2} y^{2}+4 i x y\left(x^{2}-y^{2}\right)$
Now, $x y\left(x^{2}-y^{2}\right)=0, \quad x=\pm y \Rightarrow y^{2}=\frac{1}{2} \Rightarrow y=\pm \frac{1}{\sqrt{2}}$
Four complex numbers.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2020
The number of de-Broglie wavelengths contained in the second Bohr orbit of hydrogen atom is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
The number of de-Broglie wavelength contained in the second Bohr orbit of Hydrogen atom is 2

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2016
The number of diagonals in a polygon is 20 . The number of sides of the polygon is

Options:
A 5
B 6
C 8
D 10

The correct answer is:
8

Solution:
Hints: \({ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=20\)
\(\mathrm{n}=8\)

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2011
The number of diagonals in a regular polygon of 100 sides is

Options:
A 4950
B 4850
C 4750
D 4650

The correct answer is:
4850

Solution:
The number of diagonals in a regular polygon of $n$ sides is given by the following formula ${ }^{n} C_{2}-n$
$\therefore$ The number of diagonals in a regular polygon of 100 sides
$$
\begin{array}{l}
={ }^{100} C_{2}-100=\frac{100 \cdot 99}{2}-100 \\
=\frac{9900-200}{2}=\frac{9700}{2}=4850
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2012
The number of digits in $20^{301}$ (given, $\log _{10} 2=0.3010$ is

Options:
A 602
B 301
C 392
D 391

The correct answer is:
392

Solution:
Let $y=20^{301}$
Number of digits $=$ Integral part of
$\left(301 \log _{10} 20\right)+1$
Integral part of $\left.\left[301 \log _{10} 10+\log _{10} 2\right)\right]+1$
$=$ Integral part of $[301(1+0.3010)]+1$
$=$ Integral part of $[301 \times 1.3010]+1$
Integral part of $[391.601]+1$
$=391+1=392$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2014
The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

Options:
A 0
B 2
C 1
D >2

The correct answer is:
1

Solution:
Given, $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$
$\sin x\left(\sin ^{2} x-\cos ^{2} x\right)-\cos x\left(\sin x \cos x-\cos ^{2} x \right ) + \cos x \left(\cos ^{2} x-\sin x \cos x\right)=0$
$\Rightarrow \sin x(\sin x-\cos x)(\sin x+\cos x)$ $-\cos ^{2} x(\sin x-\cos x)-\cos ^{2} x$ $(\sin x-\cos x)=0$
$\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+\sin x \cos x-2 \cos ^{2} x\right)=0$
$\Rightarrow(\sin x-\cos x)\left(\sin ^{2} x+2 \sin x\right.\left.\cos x-\sin x \cos x-2 \cos ^{2} x\right)=0$
$\Rightarrow(\sin x-\cos x)^{2}(\sin x+2 \cos x)=0$
$\Rightarrow \sin x-\cos x=0$ or $\sin x+2 \cos x=0$
$\Rightarrow \quad \tan x=1$ or $\tan x=-2$
But $\quad x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$
$\Rightarrow$ Only one solution i.e. $x=\frac{\pi}{4}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015
The number of electron in 2 coulomb of charge is

Options:
A $5 \times 10^{29}$
B $12.5 \times 10^{18}$
C $1.6 \times 10^{19}$
D $9 \times 10^{11}$

The correct answer is:
$12.5 \times 10^{18}$

Solution:
$n=\frac{2}{1.6 \times 10^{-19}}=12.5 \times 10^{18}$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2009
The number of integer values of $m$, for which the $x$ -coordinate of the point of intersection of the lines $3 x+4 y=9$ and $y=m x+1$ is also an integer, is

Options:
A 0
B 2
C 4
D 1

The correct answer is:
2

Solution:
Given lines are, $3 x+4 y=9$
$$
y=m x+1
$$
Solving Eqs. (i) and (ii), we get
$$
x=\frac{5}{3+4 m}
$$
If $m=-1,$ then $x=-5$ and $m=-2,$ then $x=-1$ are the only
possibilities.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2012
The number of irrational terms in the binomial expansion of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is

Options:
A 90
B 88
C 93
D 94

The correct answer is:
94

Solution:
General term of $\left(3^{1 / 5}+7^{1 / 3}\right)^{100}$ is given by $T_{\gamma+1}={ }^{100} C_{r}\left(3^{1 / 5}\right)^{100-r}\left(7^{1 / 3}\right)^{r}$
$={ }^{100} C_{1} \cdot 3^{\frac{100-r}{5}} \cdot 7^{\frac{r}{3}}$
For a rational term. $\frac{100-r}{5}$ and $\frac{r}{3}$ must be integer.
Hence, $r=0,15,30,45,60,75,90$
$\therefore$ There are seven rational terms. Hence, number of irrational terms
$$
=101-7=94
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2015
The number of irrational terms in the expansion of $\left(3^{\frac{1}{8}}+5^{\frac{1}{4}}\right)^{84}$ is

Options:
A 73
B 74
C 75
D 76

The correct answer is:
74

Solution:
Given, $\left(3^{\frac{1}{8}}+5^{\frac{1}{4}}\right)^{84}$
$\begin{aligned} \text { here, } T_{r+1} &={ }^{84} C_{r}\left(3^{\frac{1}{8}}\right)^{84-r}\left(5^{\frac{1}{4}}\right)^{r} \\ &={ }^{84} C_{r}\left(3^{\frac{84-r}{8}}\right) \cdot 5^{\frac{r}{4}} \end{aligned}$
since, $T_{r}=1$ is rational for $r=4,12,20,28,36,44$ 52,60,68,76,84
no. of rational terms $=11$
Here, total number of terms $=85$ Number of irrational terms $=85-11$
$$
=74
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2019
The number of lines which pass through the point (2,-3) and are at a distance 8 from the point (-1,2) is

Options:
A infinite
B 4
C 2
D 0

The correct answer is:
0

Solution:
The distance from the point (-1,2) of the line which passes through (2,-3)
$$
\begin{array}{l}
=\sqrt{(2+1)^{2}+(-3-2)^{2}} \\
=\sqrt{9+25}=\sqrt{34} < 8
\end{array}
$$
Given that, the distance between the point (-1,2) and the line is 8 But the maximum distance of the line passing through (2,-3) from (-1,2) is $\sqrt{34}$. So, there is no such line possible.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2013
The number of lone pairs of electrons on the central atoms of $\mathrm{H}_{2} \mathrm{O}, \mathrm{SnCl}_{2}, \mathrm{PCl}_{3}$ an $\mathrm{XeF}_{2}$
respectively, are

Options:
A 2,1,1,3
B 2,2,1,3
C 3.1,1,2
D 2,1,2,3

The correct answer is:
2,1,1,3

Solution:
The number of lone pairs of electrons on the central atoms of $\mathrm{H}_{2} \mathrm{O}, \mathrm{SnCl}_{2}, \mathrm{PCl}_{3}$ and $\mathrm{XeF}_{2}$ are 2 .
1,1 and 3 respectively.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2015
The number of onto functions from the set $\{1,2, \ldots, 11\}$ to the set $\{1,2, \ldots, 10\}$ is

Options:
A $5 \times 11 !$
B $10 !$
C $\frac{11 !}{2}$
D $10 \times 11 !$

The correct answer is:
$10 \times 11 !$

Solution:
Let $\quad A=\{1,2, \ldots, 11\}$
$\begin{array}{ll}\therefore & n(A)=11 \text { and } B=\{1,2, \ldots, 10\} \\ \therefore & n(B)=10\end{array}$
$\therefore \text{Hence number of onto function}$ $=^{n(A)} C_{n(B)} \times n(B) ! \times n(B)$
$={ }^{11} C_{10} \times 10 ! \times 10$
$=(11 \times 10 !) \times 10=11 ! \times 10$

Topic: Mathematics, Functions, WBJEE, WBJEE 2013
The number of permutations by taking all letters and keeping the vowels of the word COMBINE in the odd places is

Options:
A 96
B 144
C 512
D 576

The correct answer is:
576

Solution:
Hints: Vowels : O, I, E
No. of Odd place : 4
No of ways $={ }^4 \mathrm{P}_3 \times 4 !=576$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2010
The number of points at which the function $f(x)=\max \{a-x, a+x, b\},-\infty < x < \infty$ $0 < a < b$ cannot be differentiable, is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
Graph of $f(x)$

There are two sharp turns. Hence, $f(x)$ cannot be differentiable at two points.

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2016
$$
\text { The number of points of intersection of } 2 y=1 \text { and } y=\sin x \text {, in }-2 \pi \leq x \leq 2 \pi \text { is }
$$

Options:
A 1
B 2
$(8)^{1+|\cos x|+\left|\cos { }^2\right|+}$
C $\infty^3=4^3$
D 4

The correct answer is:
4

Solution:
Hints : $y=\frac{1}{2}=\sin \mathrm{x}$
$-2 \pi \leq x \leq 2 \pi$
$$
x=\frac{\pi}{6}, \frac{5 \pi}{6},-\frac{7 \pi}{6},-\frac{11 \pi}{6}
$$
No. of sol $^{\mathrm{n}} 4$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2010
The number of points on the line $x+y=4$ which are unit distance apart from the line $2 x+2 y=5$ is

Options:
A 0
B 1
C 2
D Infinity

The correct answer is:
0

Solution:
Hints : $x+y=4$
$$
\begin{aligned}
& x+y=\frac{5}{2} \\
& \mathrm{PQ}=\frac{4-5 / 2}{\sqrt{2}}=\frac{3}{2 \sqrt{2}}=\frac{3 \sqrt{2}}{4}
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2009
The number of possible organobromine compounds which can be obtained in the allylic bromination of 1 -butene with N-bromosuccinimide is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
NBS (N-Bromosuccinimide) does allylic bromination through free radical substitution.

\\(	herefore\) Total 4 isomers are possible.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2018
The number of real roots of equation $log_ex+e x=0$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
We have, $\log _{e} x+e x=0$
$\log _{e} x=-e x$
Since, both the graphs intersect at only one since.

Hence, the number of real roots of equation is one

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2015
The number of real solutions of the equation $(\sin x-x)\left(\cos x-x^{2}\right)=0$ is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given, $(\sin x-x)\left(\cos x-x^{2}\right)=0$
$\Rightarrow \quad \sin x=x$ or $\cos x=x^{2}$
Now, if $\sin x=x,$ then only one solution ie $x=0$ is possible. Also, if $\cos x=x^{2}$, then two solutions are possible. Hence, there are total 3 solutions.

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2015
The number of real values of $\alpha$ for which the system of equations
$
\begin{array}{l}
x+3 y+5 z=\alpha x \\
5 x+y+3 z=\alpha y
\end{array}
$
$3 x+5 y+z=\alpha z$
has infinite number of solutions is

Options:
A 1
B 2
C 4
D 6

The correct answer is:
1

Solution:
The system of equations are
$$
\begin{array}{l}
x+3 y+5 z=\alpha x \\
5 x+y+3 z=\alpha y \\
3 x+5 y+z=\alpha z \\
(1-\alpha) x+3 y+5 z=0 \\
5 x+(1-\alpha) y+3 z=0 \\
3 x+5 y+(1-\alpha) z=0
\end{array}
$$
For infinite number of solutions, we must have $\left|\begin{array}{ccc}1-\alpha & 3 & 5 \\ 5 & 1-\alpha & 3 \\ 3 & 5 & 1-\alpha\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{ccc}9-\alpha & 3 & 5 \\ 9-\alpha & 1-\alpha & 3 \\ 9-\alpha & 5 & 1-\alpha\end{array}\right|=0,$
$C_{1} \rightarrow C_{2}+C_{2}+C_{3}$
Taking $(9-\alpha)$ common from the first column, we get $\begin{array}{l}\Rightarrow \quad(9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & 1-\alpha & 3 \\ 1 & 5 & 1-\alpha\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 0 & -\alpha-2 & -2 \\ 0 & 2 & -\alpha-4\end{array} \mid=0\right. \\ \Rightarrow & (9-\alpha)\left|\begin{array}{cc}\alpha+2 & 2 \\ -2 & \alpha+4\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left(\alpha^{2}+6 \alpha+8+4\right)=0 \\ (\alpha-9)\left(\alpha^{2}+6 \alpha+12\right)=0\end{array}$
$\Rightarrow \alpha=9$ is the only real value of $\alpha$.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2012
The number of selection of $n$ objects from $2 n$ objects of which $n$ are identical and the rest are different, is

Options:
A $2^{n}$
B $2^{n-1}$
C $2^{n}-1$
D $2^{n-1}+1$

The correct answer is:
$2^{n}$

Solution:
Number of ways of selection of $n$ objects from $2 n$ objects, where as $n$ objects are identical in out of $2 n$ objects. $n$ identical and no different object $=1$ ways
$$
={ }^{n} C_{0}
$$
$n-1$ identical and 1 different object $=1 \times{ }^{n} C_{1}$
$n-2$ identical and 2 different object $=1 \times{ }^{n} C_{2}$
0 identical and $n$ different objects $=1 \times{ }^{n} C_{n}$
$$
={ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots \ldots+{ }^{n} C_{n}=2^{n}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2018
The number of solutions of \(2 \sin x+\cos x=3\) is

Options:
A 1
B 2
C infinite
D No solution

The correct answer is:
No solution

Solution:
Hints: \(\sqrt{5} < 3\) No solution

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2011
The number of solutions of the equation $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$ is

Options:
A 0
B 1
C 2
D infinite

The correct answer is:
2

Solution:
Given equation is $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$
$\Rightarrow \quad \frac{1}{2} \cdot\left(\frac{1}{1 / 2}\right) \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \log _{3}(x+5)^{2}=1$
$\left(\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right)$
$\Rightarrow \quad \frac{2}{2} \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \cdot 2 \log _{3}(x+5)=\log _{3} 3$
$\Rightarrow \quad \log _{3}\left\{\left(\frac{x+1}{x+5} \cdot(x+5)\right)\right\}=\log _{3} 3$
$\left(\because \log m+\log n=\log m n\right.$ and $\log _{n} n=1$
$\Rightarrow \quad(x+1)=3 \Rightarrow x=2$
So, only one solution is possible.

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2013
The number of solutions of the equation $\log _{2}\left(x^{2}+2 x-1\right)=1$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
Given $\log _{2}\left(x^{2}+2 x-1\right)=1$
$\Rightarrow \quad \log _{2}\left(x^{2}+2 x-1\right)=\log _{2} 2$
$\Rightarrow$ $x^{2}+2 x-1=2$
$\Rightarrow \quad x^{2}+2 x-3=0$
$\Rightarrow \quad x^{2}+3 x-x-3=0$
$\Rightarrow \quad x(x+3)-1(x+3)=0$
$=(x+3)(x-1)=0$
$x=1,-3$
Since, $x=1$ and $x=-3$ are satisfy the given equation therefore the number of solutions of the equation are two.

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2012