datavorous/entrance-exam-dataset huggingface dataset:
The solution of the differential equation $y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y \quad$ satisfying $y\left(\frac{\pi}{4}\right)=1 \mathrm{is}$

Options:
A $\cos \frac{x}{y}=-\log _{e} y+\frac{1}{\sqrt{2}}$
B $\sin \frac{x}{y}=\log _{e} y+\frac{1}{\sqrt{2}}$
C $\sin \frac{x}{y}=\log _{e} x-\frac{1}{\sqrt{2}}$
D None of the above

The correct answer is:
None of the above

Solution:
Given differential equation is
$$
\begin{array}{l}
y \sin \left(\frac{x}{y}\right) d x=\left\{x \sin \left(\frac{x}{y}\right)-y\right\} d y \\
\Rightarrow \quad \frac{d x}{d y}=\frac{x \sin \left(\frac{x}{y}\right)-y}{y \sin \left(\frac{x}{y}\right)}=\frac{x}{y}-\frac{1}{\sin \left(\frac{x}{y}\right)}
\end{array}
$$
On putting $v=\frac{x}{y} \Rightarrow x=y$
$\Rightarrow \quad \frac{d x}{d y}=v \cdot 1+y \frac{d v}{d y}$ in Eq. $0 .$ we get
$v+y \frac{d v}{d y}=v-\frac{1}{\sin v}$
$\Rightarrow \quad y \frac{d v}{d y}=-\frac{1}{\sin v}$
$\Rightarrow \quad-\int \sin v d v=\int \frac{d y}{y}$
(on integrating)
$\Rightarrow \quad \cos v=\log y+C$
$\Rightarrow \quad \cos \left(\frac{x}{y}\right)=\log y+C$
Given at $x=\frac{\pi}{4}, y=1$ then from $\mathrm{Eq}$. (i) $\Rightarrow \quad \cos \left(\frac{\pi}{4}\right)=\log (1)+C$
$\Rightarrow \quad\left(C=\frac{1}{\sqrt{2}}\right)$
On putting the value of $C$ in Eq. (0), we get $\cos \left(\frac{x}{y}\right)=\log _{e} y+\frac{1}{\sqrt{2}}$
Which is the required solution. So option D is correct.

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2013
The solution of the equation $\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$ is

Options:
A 3
B 7
C 9
D 49

The correct answer is:
9

Solution:
Given, $\log _{101} \log _{7}(\sqrt{x+7}+\sqrt{x})=0$
$\therefore \quad \log _{7}(\sqrt{x+7}+\sqrt{x})=(101)^{0}$
$\Rightarrow \quad \log _{7}(\sqrt{x+7}+\sqrt{x})=1$
$\Rightarrow \quad(\sqrt{x+7}+\sqrt{x})=7^{1}$
$\Rightarrow \quad \sqrt{x+7}+\sqrt{x}=7$
On squaring both sides, we get
$(x+7)+x+2 \sqrt{x^{2}+7 x}=49$
$\Rightarrow \quad 2 x-42=-2 \sqrt{x^{2}+7 x}$
$\Rightarrow \quad x-21=-\sqrt{x^{2}+7 x}$
Again, squaring both sides, we get
$x^{2}+441-42 x=x^{2}+7 x$
$\Rightarrow \quad 49 x=441$
$\Rightarrow \quad x=\frac{441}{49} \Rightarrow x=9$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2014
The solutions set of inequation \(\cos ^{-1} x < \sin ^{-1} x\) is

Options:
A \([-1,1]\)
B \(\left[\frac{1}{\sqrt{2}}, 1\right]\)
C \([0,1]\)
D \(\left(\frac{1}{\sqrt{2}}, 1\right)\)

The correct answer is:
\(\left(\frac{1}{\sqrt{2}}, 1\right)\)

Solution:
Hints: \(\cos ^{-1} x < \sin ^{-1} x\)
\(x \in\left(\frac{1}{\sqrt{2}}, 1\right), \quad \cos ^{-1} x < \sin ^{-1} x\)

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2011
The species in which nitrogen atom is in a state of sp hybridisation is

Options:
A \(\mathrm{NO}_3^{-}\)
B \(\mathrm{NO}_2\)
C \(\mathrm{NO}_2^{+}\)
D \(\mathrm{NO}_2^{-}\)

The correct answer is:
\(\mathrm{NO}_2^{+}\)

Solution:
Hint : Steric Number in \(\mathrm{NO}_2^{+}=\frac{5+0-1}{2}=2\)
\(\therefore\) sp hybridization

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2023
The specific heat $C$ of a solid at low temperature shows temperature dependence according to the relation $C=D T^{3}$, where $D$ is a constant and $T$ is the temperature is kelvin. A piece of this solid of mass $m \mathrm{kg}$ is taken and its temperature is raised from $20 \mathrm{K}$ to $30 \mathrm{K}$. The amount of heat required in the process in energy units is

Options:
A $5 \times 10^{4} \mathrm{Dm}$
B $(33 / 4) \times 10^{4} D m$
C $(65 / 4) \times 10^{4} D m$
D $(5 / 4) \times 10^{4} D m$

The correct answer is:
$(65 / 4) \times 10^{4} D m$

Solution:
Amount of heat required, $Q=\int d Q$ $=\int_{T_{1}=20}^{T_{2}=30} \mathrm{mcdT}$
Given, $C=D T^{3}$
$\therefore \quad Q=\int_{20}^{30} m D T^{3} d T=m D \int_{20}^{30} T^{3} d T$
$$
=m D \frac{1}{4}\left[(30)^{2}-(20)^{2}\right]=\frac{65}{4} \times 10^{4} m D
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2013
The stability of Me $\mathcal{C}=\mathrm{CH}_{2}$ is more than that of $\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$ due to

Options:
A inductive effect of the Me groups
B resonance effect of the Me groups
C hyperconjugative effect of the Me groups
D resonance as well as inductive effect of the Me groups

The correct answer is:
hyperconjugative effect of the Me groups

Solution:
2-hyperconltjugate forms. Therefore, $\mathrm{Me} \underline{\mathcal{C}}=\mathrm{CH}_{2}$ is more stable than
$\mathrm{MeCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}$

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2012
The stable bivalency of $\mathrm{Pb}$ and trivalency of $\mathrm{Bi}$ is

Options:
A due to $d$ contraction in $\mathrm{Pb}$ and $\mathrm{Bi}$
B due to relativistic contraction of the $6 s$ orbitals of $\mathrm{Pb}$ and $\mathrm{Bi}$, leading to inert pair effect
C due to screening effect
D due to attainment of noble liquid configuration

The correct answer is:
due to relativistic contraction of the $6 s$ orbitals of $\mathrm{Pb}$ and $\mathrm{Bi}$, leading to inert pair effect

Solution:
$_{82} \mathrm{Pb}=6 \mathrm{s}^{2} 6 p^{2}$
$_{83}{\mathrm{Bi}}=6 s^{2} 6 p^{3}$
Due to inertness of $6 s^{2}$ electrons in $\mathrm{Pb}$ and $\mathrm{Bi}$, they show bivalency and trivalency respectively instead of tetra and pentavalency.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2012
The state of hybridization of the central atom and the number of lone pairs over the central atom in $\mathrm{POCl}_{3}$ are

Options:
A $s p, 0$
B $s p^{2}, 0$
C $s p^{3}, 0$
D $d s p^{2}, 1$

The correct answer is:
$s p^{3}, 0$

Solution:
$4 \sigma$ bonds $\Rightarrow s p^{3}$ hybridisation without lone pair of electrons.

Topic: Chemistry, Chemical Bonding and Molecular Structure, WBJEE, WBJEE 2012
The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_1$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ be the Planck's constant and $e$ be the charge of an electron, then the frequency of light in the second case is

Options:
A $v_{1}-\frac{e}{h}\left(V_{2}+V_{1}\right)$
B $v_{1}+v \frac{e}{h}\left(V_{2}+V_{1}\right)$
C $v_{1}-\frac{e}{h}\left(V_{2}-V_{1}\right)$
D $v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)$

The correct answer is:
$v_{1}+\frac{e}{h}\left(V_{2}-V_{1}\right)$

Solution:
Here, $h v_{1}=\phi_{0}=e V_{1}$ ...(i)
and $h v_{2}=\phi_{0}+e V_{2}$...(ii)
From Eqs. (i) and (ii), we have
$\begin{aligned}
& h\left(v_{2}-v_{1}\right)=e\left(V_{2}-v_{1}\right) \\
\Rightarrow \quad & v_{2}=\frac{e}{h}\left(V_{2}-v_{1}\right)+v_{1}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2013
The straight line $3 x+y=9$ divides the line segment joining the points $(1,3)$ and $(2,7)$ in the ratio

Options:
A $3: 4$ externally
B $3: 4$ internally
C $4: 5$ internally
D $5: 6$ externally

The correct answer is:
$3: 4$ internally

Solution:
Hints : Ratio $=-\frac{3+3-9}{6+7-9}=\frac{3}{4}$ internally

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
The straight line through the origin which divides the area formed by the curves $y=2 x-x^{2}, y=0$ and $x=1$ into two equal halves is

Options:
A $y=x$
B $y=2 x$
C $y=\frac{3}{2} x$
D $y=\frac{2}{3} x$

The correct answer is:
$y=\frac{2}{3} x$

Solution:
$y=2 x-x^{2}, y=0, x=1$
Area $=A=\int_{0}^{1}\left(2 x-x^{2}\right) d x$ or, $A=\frac{2}{3}$ sq units
Let $y=m x$ line divides area in two equal halves
$\therefore \frac{1}{2} \times 1 \times \mathrm{m}=\frac{1}{3}$ or, $\mathrm{m}=\frac{2}{3} \therefore$ equation of line is $\mathrm{y}=\frac{2}{3} \mathrm{x}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2021
The straight line $x+y-1=0$ meets the circle $x^2+y^2-6 x-8 y=0$ at $A$ and $B$. Then the equation of the circle of which $A B$ is a diameter is

Options:
A $x^2+y^2-2 y-6=0$
B $x^2+y^2+2 y-6=0$
C $2\left(x^2+y^2\right)+2 y-6=0$
D $3\left(x^2+y^2\right)+2 y-6=0$

The correct answer is:
$x^2+y^2-2 y-6=0$

Solution:
Hints: $x^2+y^2-6 x-8 y+\lambda(x+y-1)=0$
Centre $=\left(3-\frac{\lambda}{2} \cdot 4-\frac{\lambda}{2}\right)$ Lie on $x+y-1=0$
$$
\begin{aligned}
& 3-\frac{\lambda}{2}+4-\frac{\lambda}{2}-1=0, \lambda=6 \\
& x^2+y^2-6 x-8 y+6 x+6 y-6=0 ; \quad x^2+y^2-2 y-6=0
\end{aligned}
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2010
The straight lines $x+y=0,5 x+y=4$ and $x+5 y=4$ form

Options:
A an isosceles triangle
B an equilateral triangle
C a scalene triangle
D a right angled triangle

The correct answer is:
an isosceles triangle

Solution:
Given lines are
$$
\begin{array}{r}
x+y=0 \\
5 x+y=4
\end{array}
$$
and
$$
x+5 y=4
$$
On solving above equations pairwise, we get
$$
\begin{array}{l}
A(1,-1), B\left(\frac{2}{3}, \frac{2}{3}\right), C(-1,1) \\
Now, A B=\sqrt{\left(\frac{2}{3}-1\right)^{2}+\left(\frac{2}{3}+1\right)^{2}}
\end{array}
$$

$$
\begin{array}{l}
=\sqrt{\frac{1}{9}+\frac{25}{9}}=\sqrt{\frac{26}{9}}=\frac{\sqrt{26}}{3} \\
\begin{aligned}
B C &=\sqrt{\left(-1-\frac{2}{3}\right)^{2}+\left(1-\frac{2}{3}\right)^{2}} \\
&=\sqrt{\frac{25}{9}+\frac{1}{9}}=\sqrt{\frac{26}{9}}=\frac{\sqrt{26}}{3} \\
C A=\sqrt{(1+1)^{2}+(-1-1)^{2}}
\end{aligned}
\end{array}
$$
$$
=\sqrt{4+4}=2 \sqrt{2}
$$
Here, we see that $A B=B C$
Hence, given lines form an isosceles triangle.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2014
The straight lines \(x+2 y-9=0,3 x+5 y-5=0\) and \(a x+\) by \(-1=0\) are concurrent if the straight line \(35 x-22 y+\) \(1=0\) passes through the point

Options:
A \((-a,-b)\)
B \((a,-b)\)
C \((-a, b)\)
D \((a, b)\)

The correct answer is:
\((a, b)\)

Solution:
Hint : \(\left|\begin{array}{lll}a & b & -1 \\ 1 & 2 & -9 \\ 3 & 5 & -5\end{array}\right|=0 \Rightarrow 35 a-22 b+1=0\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2023
The stress along the length of a rod (with rectangular cross-section) is $1 \%$ of the Young's modulus of its material. What is the approximate percentage of change of its volume? (Poisson's ratio of the material of the rod is $0.3 .$ )

Options:
A $3 \%$
B $1 \%$
C $0.7 \%$
D $0.4 \%$

The correct answer is:
$0.4 \%$

Solution:
$\because$ Stress, $\frac{F}{\Delta A}=1 \%$ of $Y=\frac{Y}{100}$
But Young's modulus, $Y=\frac{\text { stress }}{\text { strain }}=\frac{\frac{F}{\Delta A}}{\frac{\Delta l}{l}}$
$\therefore \quad Y=\frac{\frac{Y}{100}}{\frac{\Delta}{l}} \quad\left(\right.$ putting $\left.\frac{F}{\Delta A}=\frac{Y}{100}\right)$
$\therefore \quad \frac{\Delta l}{l}=\frac{1}{100}$
Poisson's ratio, $\sigma=\frac{-\frac{\Delta r}{r}}{\frac{\Delta l}{l}}$
$\therefore \quad \frac{\Delta r}{r}=-\sigma \cdot \frac{\Delta l}{l}=\frac{-0.3}{100}$
$$
\frac{\Delta r}{r}=\frac{-0.3}{100}
$$
$\therefore$ Change in volume, $\frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta l}{l}$
$$
\begin{array}{l}
=\frac{2 \times(-0.3)}{100}+\frac{1}{100} \\
=\frac{1-0.6}{100}=\frac{0.4}{100}
\end{array}
$$
$\Delta V \%=0.4 \%$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2018
The structure of the product $P$ of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
($\because$ Acetic anhydride is limiting reagent) Thus, we get one mole of acetanilide and molar mass of
(ii)

$\because O^{-}$ is more reactive than $-$ OMe and p-position is occupied by - OMe, the substitution occurs at ortho-position
w.r.t - O group.

Topic: Chemistry, Alcohols Phenols and Ethers, WBJEE, WBJEE 2017
The sum $1 \times 1!+2 \times 2!+\ldots+50 \times 50$ ! equals

Options:
A $51 !$
B $51 !-1$
C $51 !+1$
D $2 \times 51 !$

The correct answer is:
$51 !-1$

Solution:
$1 \times 1!+2 \times 2!+\ldots+50 \times 50!$
$=(2-1) 1!+(3-1) 2!+(4-1) 3 !$
$+\ldots+(51-1) 50!$
$=(2 !-1!)+(3 !-2!)+(4 !-3 !)+(51!-50!)$
$+\ldots+(511-501)$
$=511-11=511-1$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
The sum of all real roots of the equation $|x-2|^2+|x-2|-2=0$

Options:
A 7
B 4
C 1
D 5

The correct answer is:
4

Solution:
Hints: Put $1 x-21=y$
$$
\begin{aligned}
& y^2+y-2=0 \\
& (y-1)(y+2)=0 \\
& y=1 \\
& |x-2|=1 \\
& x-2= \pm 1 \\
& x=2 \pm 1 \\
& x=3,1
\end{aligned}
$$
$$
y=-2
$$
(Not possible)
Sum $=4$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2009
The sum of $n$ terms of the following series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ is

Options:
A $n^{2}\left(2 n^{2}-1\right)$
B $n^{3}(n-1)$
C $n^{3}+8 n+4$
D $2 n^{4}+3 n^{2}$

The correct answer is:
$n^{2}\left(2 n^{2}-1\right)$

Solution:
We have. $\mathrm{1}^{3}+3^{3}+5^{3}+7^{3}+\ldots$
$\begin{aligned} \text { Now, } T_{n} &=(2 n-1)^{3} \\ &=8 n^{3}-3(2 n)^{2}(1)+3(2 n)\left(0^{2}-(\pi)^{3}\right.\\ &=8 n^{3}-12 n^{2}+6 n-1 \end{aligned}$
$\therefore \quad S_{n}=\Sigma T_{n}$
$=\frac{8 n^{2}(n+1)^{2}}{4}-\frac{12 n(n+0) 2 n+1}{6}$$+\frac{6 n(n+1)}{2}-n$
$=2 n^{2}(n+1)^{2}-2 n(n+1)(2 n+1)+3 n(n+1)-n$
$=n\left[2 n(n+1)^{2}-2(n+1)(2 n+1)+3(n+1)-1\right]$
$=n\left[2 n\left(n^{2}+1+2 n\right)-2\left(2 n^{2}+3 n+1\right)+3n+3-1\right].$
$=n\left[2 n^{3}+2 n+4 n^{2}-4 n^{2}-6 n-2+3 n+2\right]$
$=n\left[2 n^{3}-n\right]=n^{2}\left(2 n^{2}-1\right).$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2016
The sum of the infinite series $1+\frac{1}{2 !}+\frac{1.3}{4 !}+\frac{1.3 .5}{6 !}+\ldots \ldots$. is

Options:
A $e$
B $e^2$
C $\sqrt{e}$
D $\frac{1}{e}$

The correct answer is:
$\sqrt{e}$

Solution:
Hints : $\mathrm{T}_{\mathrm{n}}=\frac{1.3 .5 \ldots(2 n-1)}{\mid 2 n}$
$$
\begin{aligned}
& =\frac{\mid 2 n}{\lfloor 2 n(2.4 \ldots 2 n)} \\
& =\frac{\lfloor 2 n}{2^{\mathrm{n}}\lfloor n \mid 2 n} \\
& =\frac{x^{\mathrm{n}}}{\lfloor n} \\
& \therefore \frac{x}{\lfloor 1}+\frac{x^2}{\lfloor 2}+\ldots=e^x-1 \quad \frac{1}{2}=x \\
& \exp =1+e^x-1=e^x=e^{1 / 2}
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2009
The sum of the infinite series $1+\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots$ is equal to

Options:
A $\sqrt{2}$
B $\sqrt{3}$
C $\sqrt{\frac{3}{2}}$
D $\sqrt{\frac{1}{3}}$

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

Solution:
Given series, $1+\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots \infty$
$=1+\frac{1}{3}+\frac{\frac{1}{2} \cdot \frac{3}{2}}{1 \cdot 2}\left(\frac{2}{3}\right)^{2}+\frac{1}{2} \cdot \frac{3}{1 \cdot 2 \cdot 3} \cdot \frac{5}{3}\left(\frac{2}{3}\right)^{3} \ldots \infty$
$=1+\frac{1}{2} \cdot \frac{2}{3}+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2 !}\left(\frac{2}{3}\right)^{2}+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)}{3 !}$
$=\left(1-\frac{2}{3}\right)^{-1 / 2}=3^{1 / 2}=\sqrt{3}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
The sum of the series
$\frac{1}{1 \times 2}^{25} C_{0}+\frac{1}{2 \times 3}{ }^{25} C_{1}+\frac{1}{3 \times 4}{ }^{25} C_{2}+\ldots \\
+\frac{1}{26 \times 27}{ }^{25} C_{25} \text { is }$

Options:
A $\frac{2^{27}-1}{26 \times 27}$
B $\frac{2^{27}-28}{26 \times 27}$
C $\frac{1}{2}\left(\frac{2^{26}+1}{26 \times 27}\right)$
D $\left(\frac{2^{26}-1}{52}\right)$

The correct answer is:
$\frac{2^{27}-28}{26 \times 27}$

Solution:
Given series is, $\frac{1}{1 \times 2} \stackrel{25}{C_{0}}+\frac{1}{2 \times 3}{ }^{25} C_{1}+\frac{1}{3 \times 4}{ }^{25} C_{2}$
$+\ldots+\frac{1}{26 \times 27}^{25} \mathrm{C}_{25}$
$\begin{aligned} \int_{0}^{x}(1+x)^{25} d x &=\int_{0}^{x}\left[1^{25} C_{0}+{ }^{25} C_{1} x\right.\\ &\left.+{ }^{25} C_{2} x^{2}+\ldots+{ }^{25} C_{25} x^{25}\right] d x \end{aligned}$
On integrating w. I.t. $x$, taking limits 0 to $x$, we get $\left[\frac{(1+x)^{26}}{26}\right]_{0}^{x}$
$=\left[2^{2} \mathrm{C}_{0} x+{ }^{25} \mathrm{C}_{1} \cdot \frac{x^{2}}{2}+{ }^{2} \mathrm{C}_{2} \frac{x^{3}}{3}+\ldots+{ }^{2} \mathrm{C}_{25} \cdot \frac{x^{26}}{26}\right]_{0}^{x}$
$\begin{aligned} \Rightarrow \quad &\left\{\frac{1}{26}(1+x)^{26}-\frac{1}{26}\right\} \\ &={ }^{25} \mathrm{C}_{0} x+{ }^{25} \mathrm{C}_{1} \cdot \frac{x^{2}}{2}+\ldots+{ }^{25} \mathrm{C}_{25} \cdot \frac{x^{26}}{26} \end{aligned}$
Again, integrating w.r.t. $x$, taking limits 0 to $1,$ we get
$\left.\frac{1}{26} \int_{0}^{1}[1+x)^{26}-1\right] d x$
$\begin{aligned} &\left.\frac{1}{26} \int_{0}^{1}[1+x)^{26}-1\right] d x \\=& \int_{0}^{1}\left[2^{25} C_{0} x+{ }^{25} C_{1} \cdot \frac{x^{2}}{2}+\ldots+{ }^{25} C_{25} \frac{x^{26}}{26}\right] d x \\ \Rightarrow \quad & \frac{1}{26}\left[\frac{(1+x)^{27}}{27}-x\right]_{0}^{1} \end{aligned}$
$=\left[{ }^{25} \mathrm{C}_{0} \frac{x^{2}}{2}+{ }^{25} \mathrm{C}_{1} \cdot \frac{x^{3}}{2 \times 3}+\ldots+{ }^{25} \mathrm{C}_{25} \frac{x^{27}}{26 \times 27}\right]_{0}^{1}$
$\begin{aligned} \Rightarrow \quad \frac{1}{26}\left\{\frac{2^{27}}{27}-1-\frac{1}{27}\right\}=\frac{1}{2} & C_{0}+\frac{1}{2 \times 3} \cdot{ }^{25} C_{1} \\ &+\ldots+\frac{1}{26 \times 27} \cdot{ }^{25} C_{25} \end{aligned}$
$\begin{aligned} \therefore \quad \frac{1}{1 \times 2} \cdot^{25} \mathrm{C}_{0} &+\frac{1}{2 \times 3}{ }^{25} \mathrm{C}_{1}+\frac{1}{3 \times 4}{ }^{25} \mathrm{C}_{2} \\ \quad &+\ldots+\frac{1}{26 \times 27} \cdot{ }^{25} \mathrm{C}_{25}=\frac{2^{27}-28}{26 \times 27} \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
The sum of the series \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\ldots . \infty\) is

Options:
A \(2 \log _{\mathrm{e}} 2+1\)
B \(2 \log _{\mathrm{e}} 2\)
C \(2 \log _{\mathrm{e}} 2-1\)
D \(\log _e 2-1\)

The correct answer is:
\(2 \log _{\mathrm{e}} 2-1\)

Solution:
\(\begin{aligned}\text {Hints: }
& s=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4} \ldots \ldots \infty \\
= & \left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right) \ldots . \\
= & \frac{1}{1}-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5} \ldots \\
= & \frac{1}{1}-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}+\frac{2}{5}-\ldots \\
= & 2\left[\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} \ldots .\right]-1=2 \log 2-1
\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2011
The sum of the series $\sum_{n=1}^{\infty} \sin \left(\frac{n ! \pi}{720}\right)$ is

Options:
A $\sin \left(\frac{\pi}{180}\right)+\sin \left(\frac{\pi}{360}\right)+\sin \left(\frac{\pi}{540}\right)$
B $\sin \left(\frac{\pi}{6}\right)+\sin \left(\frac{\pi}{30}\right)+\sin \left(\frac{\pi}{120}\right)+\sin \left(\frac{\pi}{360}\right)$
C $\sin \left(\frac{\pi}{6}\right)+\sin \left(\frac{\pi}{30}\right)+\sin \left(\frac{\pi}{120}\right)$ $+\sin \left(\frac{\pi}{360}\right)+\sin \left(\frac{\pi}{720}\right)$
D $\sin \left(\frac{\pi}{180}\right)+\sin \left(\frac{\pi}{360}\right)$

The correct answer is:
$\sin \left(\frac{\pi}{6}\right)+\sin \left(\frac{\pi}{30}\right)+\sin \left(\frac{\pi}{120}\right)$ $+\sin \left(\frac{\pi}{360}\right)+\sin \left(\frac{\pi}{720}\right)$

Solution:
We know that
$$
\begin{array}{c}
\sin n \pi=0, \forall n \in N \\
\therefore E=\sum_{n=1}^{\infty} \sin \left(\frac{n ! \pi}{720}\right)=\sin \left(\frac{\pi}{720}\right)+\sin \left(\frac{2 ! \pi}{720}\right) \\
+\sin \left(\frac{3 ! \pi}{720}\right)+\sin \left(\frac{4 ! \pi}{720}\right)+\sin \left(\frac{5 ! \pi}{720}\right) \\
+\sin \left(\frac{6 ! \pi}{720}\right)+\ldots+\sin \left(\frac{720 ! \pi}{720}\right) \\
+\sin \left(\frac{\pi}{720}\right)+\sin \left(\frac{\pi}{320}\right)+\sin \left(\frac{\pi}{120}\right)+\sin \left(\frac{\pi}{30}\right) \\
\end{array}
$$
Here, we see that after five terms of the above series all angles are multiple of $\pi,$ i.e., $\sin n \pi$, so all further values are zero.
$$
\begin{aligned}
\therefore \quad E=\sin \left(\frac{\pi}{720}\right)+\sin \left(\frac{\pi}{320}\right)+\sin \left(\frac{\pi}{120}\right) \\
&+\sin \left(\frac{\pi}{30}\right)+\sin \left(\frac{\pi}{6}\right)
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2014
The system of equations $$ \begin{array}{l} \lambda x+y+3 z=0 \\ 2 x+\mu y-z=0 \\ 5 x+7 y+z=0 \end{array} $$ has infinitely many solutions in $R$. Then,

Options:
A $\lambda=2, \mu=3$
B $\lambda=1, \mu=2$
C $\lambda=1, \mu=3$
D $\lambda=3, \mu=1$

The correct answer is:
$\lambda=1, \mu=3$

Solution:
Given, system of equations
$$
\begin{array}{l}
\lambda x+y+3 z=0 \\
2 x+\mu y-z=0 \\
5 x+7 y+z=0
\end{array}
$$
System has infinitely many solutions in $R$, if
$$
\left|\begin{array}{ccc}
\lambda & 1 & 3 \\
2 & \mu & -1 \\
5 & 7 & 1
\end{array}\right|=0
$$
$\Rightarrow \quad \lambda(\mu+7)-1(2+5)+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda(\mu+7)-7+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-7+42-15 \mu=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-15 \mu+35=0$
By checking the options, we get option (c) $(\lambda=1, \mu=3)$ satisfies the given equation.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2019
The system of linear equations $\lambda x+y+z=3$
$x-y-2 z=6$
$-x+y+z=\mu$ has

Options:
A infinite number of solutions for $\lambda \neq-1$ and all $\mu$
B infinite number of solutions for $\lambda=-1$ and $\mu=3$
C no solution for $\lambda, x-1$
D unique solution for $\lambda=-1$ and $\mu=3$

The correct answer is:
infinite number of solutions for $\lambda=-1$ and $\mu=3$

Solution:
Augmented matrix [A:B] = $\left[\begin{array}{rrrr}1 & -1 & -2 & 6 \\ -1 & 1 & 1 & \mu \\ \lambda & 1 & 1 & 3\end{array}\right]$
$\sim\left[\begin{array}{cccc}1 & -1 & -2 & 6 \\ 0 & 0 & -1 & \mu+6 \\ \lambda+1 & 0 & -1 & 9\end{array}\right]$
Applying
$$
-\left[\begin{array}{cccc}1 & -1 & -2 & R_{2}+R_{1}, R_{3} \rightarrow R_{3}+R_{1} \\ 0 & 0 & -1 & \mu+6 \\ \lambda+1 & 0 & 0 & 3-\mu\end{array}\right]
$$
Infinite number of solutions for $\lambda=-1$ and $\mu=3$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2012
The system that contains the maximum number of atoms is

Options:
A $4.25 \mathrm{g}$ of $\mathrm{NH}_{3}$
B $8 \mathrm{g}$ of $\mathrm{O}_{2}$
C $2 \mathrm{g}$ of $\mathrm{H}_{2}$
D $4 \mathrm{g}$ of He

The correct answer is:
$2 \mathrm{g}$ of $\mathrm{H}_{2}$

Solution:
Number of atoms $=\frac{\text { mass }}{\text { molar mass }} \times N_{A}$ x Number of atoms in 1 mole
$\therefore$ Number of atoms in $4.25 \mathrm{g} \mathrm{NH}_{3}$
$$
\begin{array}{l}
=\frac{4.25}{17} \times N_{A} \times 4 \\
=N_{A}
\end{array}
$$
Number of atoms in $8 \mathrm{g}$
$$
\mathrm{O}_{2}=\frac{8}{32} \times \mathrm{N}_{A} \times 2=\frac{N_{\mathrm{A}}}{2}
$$
Number of atoms in $2 \mathrm{g}$
$$
\mathrm{H}_{2}=\frac{2}{2} \times \mathrm{N}_{A} \times 2=2 \mathrm{N}_{\mathrm{A}}
$$
Number of atoms in $4 \mathrm{g}$
$$
\mathrm{He}=\frac{4}{4} \times N_{A} \times 1=N_{A}
$$
Thus, $2 \mathrm{g} \mathrm{H}_{2}$ contains the maximum number of atoms among the given.

Topic: Chemistry, Some Basic Concepts of Chemistry, WBJEE, WBJEE 2014
The tangent at point \((a \cos \theta, b \sin \theta), 0 < \theta < \frac{\pi}{2}\), to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the \(x\)-axis at \(T\) and \(y\)-axis at \(T_1\). Then the value of \(\min _{0 < \theta < \frac{\pi}{2}}(O T)\left(O T_1\right)\) is

Options:
A \(a b\)
B \(2 \mathrm{ab}\)
C 0
D 1

The correct answer is:
\(2 \mathrm{ab}\)

Solution:
Hint: Tangent : \(x(b \cos \theta)+y(a \sin \theta)=a b\)
\(\therefore \mathrm{OT} \cdot \mathrm{OT}_1=\frac{\mathrm{ab}}{\cos \theta \sin \theta}=\frac{2 \mathrm{ab}}{\sin 2 \theta}\)

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2023
The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} \mathrm{C}$ to $1000^{\circ} \mathrm{C}$. The heat required in the process is

Options:
A $4.79 \times 10^{-4}$ cal
B $9.21 \times 10^{-5}$ cal
C $2.17 \times 10^{-4} \mathrm{cal}$
D Information not sufficient

The correct answer is:
Information not sufficient

Solution:
Information in the statement of the questions is insufficient to calculate heat required to raise the temperature from $100^{\circ} \mathrm{C}$ to $1000^{-\mathrm{C}}$.

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2016