datavorous/entrance-exam-dataset huggingface dataset:
The temperature of anideal gas is increased from $120 \mathrm{~K}$ to $480 \mathrm{~K}$. If at $120 \mathrm{~K}$, the root mean square speed of gas molecules is $v$, then at $480 \mathrm{~K}$ it will be

Options:
A $4 v$
B $2 v$
C $\frac{v}{2}$
D $\frac{v}{4}$

The correct answer is:
$2 v$

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\
& \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{120}{480}}-=\sqrt{\frac{1}{4}}=\frac{1}{2} \\
& \mathrm{~V}_2=2 v
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2010
The temperature of the water of a pond is $0^{\circ} \mathrm{C}$ while that of the surrounding atmosphere is $-20^{\circ} \mathrm{C}$. If the density of ice is $\rho$ coefficient of thermal conductivity is $k$ and latent heat of melting is $L,$ then the thickness $Z$ of ice layer formed increases as a function of time $t$ is

Options:
A $Z^{2}=\frac{60 k}{\rho L} t$
B $Z=\sqrt{\frac{40 k}{\rho L} t}$
C $Z^{2}=\frac{40 k}{\rho l} \sqrt{t}$
D $Z^{2}=\frac{40 k}{\rho L} t$

The correct answer is:
$Z^{2}=\frac{40 k}{\rho L} t$

Solution:
Temperature of water $=0^{\circ} \mathrm{C}$
Temperature of surrounding atmosphere $=-20^{\circ} \mathrm{C}$
Density of ice $=\rho$
Coefficient of thermal conductivity $=k$
latent heat of melting $=L$

The rate of heat flow $H=\frac{d Q}{d t}=\frac{k A\left(T_{1}-T_{2}\right)}{x}$
$$
\begin{array}{l}
=\frac{k f(0-(-20)]}{x} \\
=\frac{k A \times 20}{x} \\
\frac{d Q}{d t}=\frac{d m}{d t} \cdot L \quad(\because m L=Q) \\
d m=\rho A \cdot d x \\
\frac{d Q}{d t}=\rho A \cdot \frac{d x}{d t} L=\frac{k A \cdot 20}{x} \\
\int_{0}^{Z} x \cdot d x=\frac{20 k}{\rho L} \int_{0}^{t} d t \Rightarrow Z^{2}=\frac{40 k}{\rho L} t
\end{array}
$$
So, thickness $Z$ increases as a function of time $t$ according to the above equation.

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2016
The three sides of a right angled triangle are in GP (geometric progression). If the two acute angles be $\alpha$ and $\beta$, then tan $\alpha$ and tan $\beta$ are

Options:
A $\frac{\sqrt{5}+1}{2}$ and $\frac{\sqrt{5}-1}{2}$
B $\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$
C $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$
D $\frac{\sqrt{5}}{2}$ and $\frac{2}{\sqrt{5}}$

The correct answer is:
$\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$

Solution:
Let $\Delta A B C$ be a right angled triangle at $B$. Let $\angle A$ and $\angle C$ be $\alpha$ and $\beta$

Since, sides are in GP so sides are $a,$ ar, ar $^{2}$
$\begin{array}{lr}\text { Now, } & \begin{array}{r}A C^{2}=A B^{2}+B C^{2} \\ \left(a r^{2}\right)^{2}=a^{2}+(a \cdot r)^{2}\end{array} \\ \Rightarrow \quad a^{2} r^{4}=a^{2}+a^{2} r^{2} \\ \Rightarrow \quad r^{4}-r^{2}-1=0 \\ \Rightarrow \quad r^{2}=\frac{1+\sqrt{5}}{2} \\ \Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}\end{array}$
$\therefore \quad \tan \alpha=\frac{B C}{A B}=\frac{a r}{a}$
$\Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}$
$\begin{aligned} \text { and } \tan \beta &=\frac{A B}{B C}=\frac{a}{a r}=\frac{1}{r} \\ &=\frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}}=\sqrt{\frac{\sqrt{5}-1}{2}} \end{aligned}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2019
The time taken for $10 \%$ completion of a first order reactin is 20 mins. Then, for $19 \%$ completion, the reaction will take

Options:
A $40 \mathrm{mins}$
B 60 mins
C $30 \mathrm{mins}$
D 50 mins

The correct answer is:
$40 \mathrm{mins}$

Solution:
$t=\frac{2.303}{\lambda} \log \frac{\mathrm{N}_0}{\mathrm{~N}}$
$\begin{aligned} & 20=\frac{2.303}{\lambda} \log \frac{100}{90} \\ & t=\frac{2.303}{\lambda} \log \frac{100}{81}\end{aligned}$
equation (i) / (ii)
$\therefore t=40 \mathrm{~min}$.

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2009
The time taken for an electron to complete one revolution in Bohr orbit of hydrogen atom is

Options:
A $\frac{4 m^{2} \pi r^{2}}{n^{2} h^{2}}$
B $\frac{n^{2} h^{2}}{4 m r^{2}}$
C $\frac{4 \pi^{2} m r^{2}}{n h}$
D $\frac{n h}{4 \pi^{2} m r^{2}}$

The correct answer is:
$\frac{4 \pi^{2} m r^{2}}{n h}$

Solution:
Let the distance travelled in $T$ time $=2 \pi r$ (Circumference of orbit)
$\therefore \quad$ Velocity $(v)=\frac{2 \pi r}{T}$
Also, $v=\frac{n \cdot h}{2 \pi m r}$
$\therefore \quad \frac{1}{T \text { (Time period) }}=$ frequency $v=\text{v} \times \frac{1}{2 \pi r}$
From Eq.
$$
\begin{array}{l}
\mathrm{v}=\frac{1}{T}=\frac{n h}{2 \pi \times 2 \pi m r} \\
v=\frac{n h}{4 \pi^{2} r^{2} m}
\end{array}
$$
$T$ (time taken in one revolution)
$$
=\frac{4 \pi^{2} r^{2} m}{n h}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2016
The total number of alkyl bromides (including stereoismers) formed in the reaction $\mathrm{Me}_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{HBr} \rightarrow$ will be

Options:
A 1
B 2
C 3
D No bromide forms

The correct answer is:
3

Solution:
Hint:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2020
The total number of aromatic species generated in the following reactions is

Options:
A zero
B 2
C 3
D 4

The correct answer is:
3

Solution:
Criteria for aromaticity (Huckel rule) The cyclic $\pi$ molecular orbital (electron cloud) formed by overlap of $p$ -orbitals must contalu $(4 n+2) \pi$ -electrons, where $n=$ integer 0,1,2,3 Pte. This is known as Huckel rule.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2015
The total number of injections (one-one into mappings) from $\quad\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right)$ is

Options:
A 400
B 420
C 800
D 840

The correct answer is:
840

Solution:
Let $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$
$\begin{aligned} B &=\left\{b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right\} \\ n(A) &=4, n(B)=7 \end{aligned}$
The total number of infections $={ }^{7} P_{4}$ $=\frac{71}{3 !}=7 \cdot 6 \cdot 5 \cdot 4=840$

Topic: Mathematics, Functions, WBJEE, WBJEE 2012
The total number of isomeric linear dipeptides which can be synthesised from racemic alanine is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Total number of isomeric linear dipeptides which can be synthesised from racemic mixture of alanine are four $(4),$ by $(R)$ type and
(S) type.
($\because$ Has two chiral centres)
i.e.

The possible isomers are
$$
(R R),(S R),(R S) \text { and }(S S)
$$
Hence, total (4) isomeric linear dipeptides for alanine can be synthesised and option
(d) is the correct answer.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2019
The total number of tangents through the point $(3,5)$ that can be drawn to the ellipses $3 x^2+5 y^2=32$ and $25 x^2+9 y^2=450$ is

Options:
A 0
B 2
C 3
D 4

The correct answer is:
3

Solution:
Hints : $(3,5)$ lies outside the ellipse $3 x^2+5 y^2=32$ and on the ellipse $25 x^2+9 y^2=450$. Therefore there will be 2 tangents for the first ellipse and one tangent for the second ellipse.

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2009
The transverse axis of a hyperbola is along the $x$ -axis and its length is $2 a$. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

Options:
A $6 x^{2}-y^{2}=3 a^{2}$
B $x^{2}-3 y^{2}=3 a^{2}$
C $x^{2}-6 y^{2}=3 a^{2}$
D $3 x^{2}-y^{2}=3 a^{2}$

The correct answer is:
$3 x^{2}-y^{2}=3 a^{2}$

Solution:
Let e be the eccentricity of hyperbola and length of conjugate axis be $2 h$. since, vertex $(a, 0)$ bisects the join of centre (0,0) and focus $(a e, 0)$
$$
\begin{array}{l}
a=\frac{a e+0}{2} \Rightarrow e=2 \\
b^{2}=a^{2}\left(c^{2}-1\right)=3 a^{2}
\end{array}
$$
$\therefore$ Required hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{3 a^{2}}=1 \Rightarrow 3 x^{2}-y^{2}=3 a^{2}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2012
The trigonometric equation $\sin ^{-1} x=2 \sin ^{-1} 2 a$ has a real solution, if

Options:
A $|a|>\frac{1}{\sqrt{2}}$
B $\frac{1}{2 \sqrt{2}} < |a| < \frac{1}{\sqrt{2}}$
C $|a|>\frac{1}{2 \sqrt{2}}$
D $|a| \leq \frac{1}{2 \sqrt{2}}$

The correct answer is:
$|a| \leq \frac{1}{2 \sqrt{2}}$

Solution:
We know that, $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{2} \leq 2 \sin ^{-1} 2 a \leq \frac{\pi}{2}$
$\Rightarrow \frac{-\pi}{4} \leq \sin ^{-1} 2 a \leq \frac{\pi}{4}$
$\Rightarrow \sin \left(\frac{-\pi}{4}\right) \leq 2 a \leq \sin \left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{-1}{\sqrt{2}} \leq 2 a \leq \frac{1}{\sqrt{2}}$
$\Rightarrow \frac{-1}{2 \sqrt{2}} \leq a \leq \frac{1}{2 \sqrt{2}} \Rightarrow|a| \leq \frac{1}{2 \sqrt{2}}$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2015
$$
\text { The two structures written below represent }
$$

Options:
A pair of diastereomers
B pair of enantiomers
C same molecule
D both are optically inactive

The correct answer is:
same molecule

Solution:
Hints:
I \& II are same Fischer projection because $180^{\circ}$ rotation doesn't change configuration

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2009
The unit vector in ZOX plane, making angles $45^{\circ}$ and $60^{\circ}$ respectively with $\vec{\alpha}=2 \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{\beta}=j-\hat{k}$ is

Options:
A $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$
B $\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$
C $\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{j}$
D $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$

The correct answer is:
$\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$

Solution:
Hint:
Let the vector be $\vec{r}=x \hat{i}+z \hat{k} \Rightarrow|r|=1$
$\vec{r} \cdot \vec{a}=|r| |\vec{\alpha}| \cos 45^{\circ}$
$\therefore 2 x-z=\frac{3}{\sqrt{2}}$
$\vec{r} \cdot \vec{\beta}=|\vec{r}| |\vec{\beta}| \cos 60^{\circ}$
$z=-\frac{1}{\sqrt{2}}$
$\therefore \mathrm{x}=\frac{1}{\sqrt{2}}$
$\therefore \overrightarrow{\mathrm{r}}=\frac{1}{\sqrt{2}} \hat{i}-\frac{1}{\sqrt{2}} \hat{k}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2020
The units of surface tension and viscosity of liquids respectively are

Options:
A $\mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}, Nm^{-1}$
B $\mathrm{kg} \mathrm{s}^{-2}, \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$
C $\mathrm{Nm}^{-1}, \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-2}$
D $\mathrm{kg} \mathrm{s}^{-1}, \mathrm{kg} \mathrm{m}^{-2} \mathrm{s}^{-1}$

The correct answer is:
$\mathrm{kg} \mathrm{s}^{-2}, \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$

Solution:
From the formula, Surface tension, $(\gamma)$
$$
=\frac{\Delta W}{\Delta A}=\frac{\mathrm{J}}{\mathrm{m}^{2}}=\frac{\mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-2}}{\mathrm{m}^{2}}=\mathrm{kg} \mathrm{s}^{-2}
$$
Form the formula, $F=\eta \cdot A \cdot \frac{d V}{d x}$ or, $\eta$ (coefficient of viscosity)
$$
=\frac{F}{A \cdot \frac{d V}{d x}}=\frac{\mathrm{N}}{\mathrm{m}^{2} \cdot \frac{\mathrm{ms}^{-1}}{\mathrm{m}}}=\mathrm{Nm}^{-2} \cdot \mathrm{s}
$$
or $\mathrm{kg} \mathrm{ms}^{-2} \cdot \mathrm{m}^{-2} \cdot \mathrm{s}=\mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2015
The value of $\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x}} d x$ is

Options:
A $\pi / 4$
B 0
C $\pi / 2$
D 1/2

The correct answer is:
$\pi / 4$

Solution:
$I=\int_0^{\pi / 2} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}+(\sin x)^{\cos x} d x}$
$I=\int_0^{\pi / 2} \frac{(\sin x)^{\cos x} d x}{(\sin x)^{\cos x}+(\cos x)^{\sin x}}$
$\Rightarrow 2 \mathrm{I}=\pi / 2$
$\Rightarrow I=\pi / 4$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
The value of $\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x$ is

Options:
A 72
B 125
C 43
D 69

The correct answer is:
43

Solution:
$\int_{0}^{5} \max \left\{x^{2}, 6 x-8\right\} d x$
Let $f(x)=x^{2}-6 x+8$
$\quad=(x-4)(x-2)$
$\therefore$ For $x \in[0,2] \cup[4,5], f(x)>0$ and in $x \in[2,4], f(x) < 0$
$\int_{0}^{5} \max \left(x^{2}, 6 x-8\right)=\int_{0}^{2} x^{2} d x+\int_{2}^{4}(6 x-8) d x+\int_{4}^{5} x^{2} d x=\frac{8}{3}+20+\frac{61}{3}=43$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
The value of $\int_0^{\infty} \frac{d x}{\left(x^2+4\right)\left(x^2+9\right)}$ is

Options:
A $\frac{\pi}{60}$
B $\frac{\pi}{20}$
C $\frac{\pi}{40}$
D $\frac{\pi}{80}$

The correct answer is:
$\frac{\pi}{60}$

Solution:
$$
\begin{aligned}
& \text { Hints : } \left.\int_0^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^2+4\right)\left(\mathrm{x}^2+9\right)}=\int_0^{\pi / 2} \frac{\sec ^2 \theta}{\left(\tan ^2 \theta+4\right)\left(\tan ^2 \theta+9\right)} \mathrm{d} \theta \text { (putting } \mathrm{x}=\tan \theta\right) \\
& =\frac{1}{5} \int_0^{\pi / 2} \frac{\left\{\left(9+\tan ^2 \theta\right)-\left(4+\tan ^2 \theta\right)\right\} \sec ^2 \theta}{\left(\tan ^2 \theta+4\right)\left(\tan ^2 \theta+9\right)} \mathrm{d} \theta
\end{aligned}
$$
$$
=\frac{1}{5}\left[\int_0^{\pi / 2} \frac{\sec ^2 \theta}{4+\tan ^2 \theta} \mathrm{d} \theta-\int_0^{\pi / 2} \frac{\sec ^2 \theta}{9+\tan ^2 \theta} \mathrm{d} \theta\right]
$$
$$
\begin{aligned}
& =\frac{1}{5}\left[\left.\frac{1}{2} \tan ^{-1}\left(\frac{\tan \theta}{2}\right)\right|_0 ^{\pi / 2}-\left.\frac{1}{3} \tan ^{-1}\left(\frac{\tan \theta}{3}\right)\right|_0 ^{\pi / 2}\right] \\
& =\frac{1}{5}\left[\frac{1}{2} \cdot \frac{\pi}{2}-\frac{1}{3} \cdot \frac{\pi}{2}\right]=\left(\frac{\pi}{2}\right)\left(\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{\pi}{2} \cdot \frac{1}{5} \cdot \frac{1}{6}=\frac{\pi}{60}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2009
The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64}$ is

Options:
A 0
B -1
C 1
D $i$

The correct answer is:
-1

Solution:
We know that, $\omega=\frac{-1+\sqrt{3} i}{2} 1-\sqrt{3} i=-2 \omega$
and $\omega^{2}=\frac{-1-\sqrt{3} i}{2} 1+\sqrt{3} i=-2 \omega^{2}$
Now, $\begin{aligned}\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64} \\ &=\left(\frac{-2 \omega^{2}}{-2 \omega}\right)^{64}+\left(\frac{-2 \omega}{-2 \omega^{2}}\right)^{64} \end{aligned}$
$=\omega^{64}+\frac{1}{\omega^{64}}=\omega+\omega^{2}$
$$
\left[\because \cdot \omega^{3}=1\right]
$$
$$
\left.=-1 \quad \mid \because 1+\omega+\omega^{2}=0\right]
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2015
The value of $(1+\cos \pi / 6)(1+\cos \pi / 3)(1+\cos 2 \pi / 3)(1+\cos 7 \pi / 6)$ is

Options:
A $\frac{3}{16}$
B $\frac{3}{8}$
C $\frac{3}{4}$
D $\frac{1}{2}$

The correct answer is:
$\frac{3}{16}$

Solution:
Hints : $\left(1+\cos \frac{\pi}{6}\right)\left(1+\cos \frac{\pi}{3}\right)\left(1+\cos \frac{2 \pi}{3}\right)\left(1+\cos \frac{7 \pi}{6}\right)$
$$
=\left(1+\frac{\sqrt{3}}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\left(1-\frac{\sqrt{3}}{2}\right)=\left(1-\frac{3}{4}\right)\left(1-\frac{1}{4}\right)=\frac{1}{4} \times \frac{3}{4}=\frac{3}{16}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2009
The value of $2 \cot ^{-1} \frac{1}{2}-\cot ^{-1} \frac{4}{3}$ is

Options:
A $-\frac{\pi}{8}$
B $\frac{3 \pi}{2}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$2 \cot ^{-1} \frac{1}{2}-\cot ^{-1} \frac{4}{3}$
$\begin{array}{l}
=2 \tan ^{-1} 2-\tan ^{-1} \frac{3}{4} \\
=\pi+\tan ^{-1} \frac{4}{-3}-\tan ^{-1} \frac{3}{4} \\
=\pi-\tan ^{-1} \frac{4}{3}-\tan ^{-1} \frac{3}{4}
\end{array}$
$=\pi-\left\{\tan ^{-1} \frac{\frac{4}{3}+\frac{3}{4}}{1-\frac{4}{3} \cdot \frac{3}{4}}\right\}$
$=\pi-\tan ^{-1} \infty=\pi-\frac{\pi}{2}=\frac{\pi}{2}$

Topic: Mathematics, Inverse Trigonometric Functions, WBJEE, WBJEE 2015
The value of $\frac{2}{3 !}+\frac{4}{5 !}+\frac{6}{7 !}+\ldots \ldots$. is

Options:
A $\mathrm{e}^{\frac{1}{2}}$
B $\mathrm{e}^{-1}$
C $\mathrm{e}$
D $\mathrm{e}^{-\frac{1}{3}}$

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

Solution:
$$
\begin{gathered}
\text { Hints: } t_n=\frac{2 n}{(2 n+1) !}=\frac{2 n+1}{(2 n+1) !}-\frac{1}{(2 n+1) !}=\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !} \\
\sum_{n=1}^{\infty} t_n=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \ldots \infty=\mathrm{e}^{-1}
\end{gathered}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2010
The value of $1000\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{999 \times 1000}\right]$ is

Options:
A 1000
B 999
C 1001
D $\frac{1}{999}$

The correct answer is:
999

Solution:
$1000\left[\frac{1}{1 \times 2} \times \frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{999 \times 1000}\right]$
$=1000\left\{\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\right.$
$\left.+\ldots+\left(\frac{1}{999}-\frac{1}{1000}\right)\right]$
$=1000\left(1-\frac{1}{1000}\right)$
$=1000 \times \frac{999}{1000}$
$=999$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
The value of a for which sum of the squares of the roots of the equation $x^2-(a-2) x-a-1=0$ assumes the least value is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
Let $\alpha, \beta$ be the roots, then
$\begin{aligned}
&\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\
&=(a-2)^2+2(a+1) \\
&=a^2-2 a+6 \\
&=(a-1)^2+5 \geq 5 \\
&\therefore \alpha^2+\beta^2 \text { is minimum if } a=1 .
\end{aligned}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2022
The value of ' \(a\) ' for which the scalar triple product formed by the vectors \(\vec{a}=\hat{i}+a \hat{j}+\hat{k}, \vec{\beta}=\hat{j}+a \hat{k}\) and \(\vec{\gamma}=a \hat{i}+\hat{k}\) is maximum, is

Options:
A 3
B -3
C \(-\frac{1}{\sqrt{3}}\)
D \(\frac{1}{\sqrt{3}}\)

The correct answer is:
\(-\frac{1}{\sqrt{3}}\)

Solution:
Hint : \(\Delta=1+\mathrm{a}^3-\mathrm{a}\) For \(\max ; \frac{\mathrm{d} \Delta}{\mathrm{da}}=0\)
\(\frac{\mathrm{d} \Delta}{\mathrm{da}}=3 \mathrm{a}^2-1 \quad \mathrm{a}= \pm \frac{1}{\sqrt{3}}\)
\(\frac{\mathrm{d}^2 \Delta}{\mathrm{da}^2}=6 \mathrm{a} < 0\) for \(\mathrm{a}=-\frac{1}{\sqrt{3}}\)

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2023
The value of $\frac{\cot x-\tan x}{\cot 2 x}$ is

Options:
A 1
B 2
C -1
D 4

The correct answer is:
2

Solution:
Hints: $\frac{\cos ^2 x-\sin ^2 x}{\sin x \cos x} \times \frac{\sin 2 x}{\cos 2 x}=\frac{2 \cos 2 x}{\sin 2 x} \times \frac{\sin 2 x}{\cos 2 x}=2$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2010
The value of $\cos 15^{\circ} \cos 7 \frac{1^0}{2} \sin 7 \frac{1^0}{2}$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{8}$
C $\frac{1}{4}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{8}$

Solution:
$$
\begin{aligned}
& \text { Hints : } \cos 15^{\circ} \cos 7 \frac{1}{2}^0 \sin 7 \frac{1}{2}^0=\frac{1}{2}\left(2 \sin 7 \frac{1}{2}^0 \cos 7 \frac{1}{2}^0\right) \cdot\left(\cos 15^{\circ}\right) \\
& \frac{1}{2}\left(\sin 15^{\circ}\right)\left(\cos 15^{\circ}\right)=\frac{1}{4}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right)=\frac{1}{4} \times \sin 30^0=\frac{1}{8}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2009
The value of $\cos 15^{\circ} \cos 7 \frac{1^{\circ}}{2} \sin 7 \frac{1^{\circ}}{2}$ is

Options:
A $\frac{1}{2}$
B $\frac{1}{8}$
C $\frac{1}{4}$
D $\frac{1}{16}$

The correct answer is:
$\frac{1}{8}$

Solution:
We have, $\cos 15^{\circ} \cdot \cos 7 \frac{1}{2} \cdot \sin 7 \frac{1}{2}$
$$
\begin{array}{l}
=\frac{1}{2} \cos 15\left(2 \cos 7 \frac{1}{2} \cdot \sin 7 \frac{1}{2}\right) \\
=\frac{1}{2} \cos 15\left[\sin 2\left(7 \frac{1}{2}\right)\right] \quad\left[\begin{array}{c}
\because 2 \sin \theta \cdot \cos \theta \\
=\sin 20
\end{array}\right] \\
=\frac{1}{2} \cos 15^{\circ} \sin 15 \\
=\frac{1}{2} \times \frac{1}{2}\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) \\
=\frac{1}{4} \sin 30^{\circ}=\frac{1}{4} \times \frac{1}{2}=\frac{1}{8}
\end{array}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2016
The value of $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}$ $-\cos ^{2} 30^{\circ}-\cos ^{2} 60^{\circ}$ is

Options:
A 0
B 1
C $\frac{1}{2}$
D $\frac{1}{4}$

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

Solution:
Given $\cos ^{2} 75^{\circ}+\cos ^{2} 45^{\circ}+\cos ^{2} 15^{\circ}-\cos ^{2} 30^{\circ}\cos^260^\circ$
$=\left(\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right)^{2}$ $-\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}$
$=\left(\frac{3+1-2 \sqrt{3}}{8}\right)+\frac{1}{2}+\left(\frac{3+1+2 \sqrt{3}}{8}\right)-\frac{3}{4}-\frac{1}{4}$
$=\frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{2}+\frac{1}{2}+\frac{\sqrt{3}}{4}-\frac{3}{4}-\frac{1}{4}=\frac{3}{2}-1=\frac{1}{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2013
The value of \(\cos 15^{\circ}-\sin 15^{\circ}\) is

Options:
A 0
B \(\frac{1}{\sqrt{2}}\)
C \(-\frac{1}{\sqrt{2}}\)
D \(\frac{1}{2 \sqrt{2}}\)

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

Solution:
Hints : \(\cos 15^{\circ}-\sin 15^{\circ}=\sqrt{2} \cos 60^{\circ}=\frac{1}{\sqrt{2}}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2011
The value of det $A$, where $A=\left(\begin{array}{ccc}1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1\end{array}\right)$, lies

Options:
A in the closed interval [1,2]
B in the closed interval [0,1]
C in the open interval (0,1)
D in the open interval (1, 2)

The correct answer is:
in the closed interval [1,2]

Solution:
We have, $|A|=\left|\begin{array}{ccc}1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1\end{array}\right|$
$=1[1-(-\cos \theta)(\cos \theta)]$
$-\cos \theta[-\cos \theta+\cos \theta]+0\left(\cos ^{2} \theta+1\right)$
$=1+\cos ^{2} \theta$
Now, we know that $-1 \leq \cos \theta \leq 1$
$\Rightarrow \quad 0 \leq \cos ^{2} \theta \leq 1$
$\Rightarrow \quad 1 \leq 1+\cos ^{2} \theta \leq 2$
$\Rightarrow \quad 1 \leq|A| \leq 2$
$\therefore \quad|A| \in[1,2]$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2017
The value of determinant $\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$ is

Options:
A 0
B $\left(1+a^{2}+b^{2}\right)$
C $\left(1+a^{2}+b^{2}\right)^{2}$
D $\left(1+a^{2}+b^{2}\right)^{3}$

The correct answer is:
$\left(1+a^{2}+b^{2}\right)^{3}$

Solution:
Let
$\Delta=\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right|$
Apply $\mathrm{C}_{\mathrm{1}} \rightarrow \mathrm{C}_{1}-\mathrm{bC}_{3}$ and $\mathrm{C}_{2} \rightarrow \mathrm{aC}_{3}+\mathrm{C}_{2}$
$=\left|\begin{array}{ccc}1+a^{2}-b^{2}+2 b^{2} & 2 a b-2 a b & -2 b \\ 2 a b-2 a b & 1-a^{2}+b^{2}+2 a^{2} & 2 a \\ 2 b-b+a^{2} b+b^{3} & -2 a+a-a^{3}-a b^{2} & 1-a^{2}-b^{2}\end{array}\right|$
$=\left|\begin{array}{ccc}\left(1+a^{2}+b^{2}\right) & 0 & -2 b \\ 0 & \left(1+a^{2}+b^{2}\right) & 2 a \\ b\left(1+a^{2}+b^{2}\right) & -a\left(1+a^{2}+b^{2}\right) & \left(1-a^{2}-b^{2}\right)\end{array}\right|$
$=\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{ccc}1 & 0 & -2 b \\ 0 & 1 & 2 a \\ b & -a & \left(1-a^{2}-b^{2}\right)\end{array}\right|$
$=\left(1+a^{2}+b^{2}\right)^{2}\left\{\left(1-a^{2}-b^{2}+2 a^{2}\right)+2 b^{2}\right\}$
$=\left(1+a^{2}+b^{2}\right)^{2}\left(1+a^{2}+b^{2}\right)=\left(1+a^{2}+b^{2}\right)^{3}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2013
The value of $f(0)$ so that the function $f(x)=\frac{1-\cos (1-\cos \mathrm{x})}{\mathrm{x}^4}$ is continuous everywhere is

Options:
A $\frac{1}{2}$
B $\frac{1}{4}$
C $\frac{1}{6}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{8}$

Solution:
$$
\text { Hints: } \operatorname{Lim}_{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{x^4}
$$
$$
\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{2 \sin ^2\left(\frac{x}{2}\right)}{2}\right)}{x^4}=2 \operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2\left(\sin ^2\left(\frac{x}{2}\right)\right)\left(\sin ^2\left(\frac{x}{2}\right)\right)^2}{x^4\left(\sin ^2\left(\frac{x}{2}\right)\right)^2}=2 \operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^4\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^4 2^4}=\frac{1}{2^3}=\frac{1}{8}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2010