datavorous/entrance-exam-dataset huggingface dataset:
Silver chloride dissolves in excess of ammonium hydroxide solution. The cation present in the resulting solution is

Options:
A $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+}$
B $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{4}\right]^{+}$
C $\mathrm{Ag}^{+}$
D $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$

The correct answer is:
$\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}$

Solution:
$$
\begin{aligned}
&\begin{array}{l}
\text { Silver chloride dissolves in excess of } \\
\text { ammonium hydroxide solution. The cation } \\
\text { present in the resulting solution is }\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \\
\text {In aqueous solution, silver chloride exist as } \mathrm{Ag}^{+} \\
\text {=and } \mathrm{Cl}^{-} \text {. } \mathrm{Ag}^{+} \text {present in solution reacts with } \mathrm{NH}_{3}
\end{array}\\
&\begin{array}{l}
\text { form }\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \\
\mathrm{Ag}^{+} \mathrm{Cl}^{-}+2 \mathrm{NH}_{3} \longrightarrow\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+} \mathrm{Cl}^{-}
\end{array}
\end{aligned}
$$

Topic: Chemistry, Coordination Compounds, WBJEE, WBJEE 2018
Simplest form of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 x}}}}$ is

Options:
A $\sec \frac{x}{2}$
B $\sec x$
C $\operatorname{cosec} x$
D 1

The correct answer is:
$\sec \frac{x}{2}$

Solution:
$$
\text { Hints : } \begin{aligned}
& \frac{2}{\sqrt{2+\sqrt{2+\sqrt{2.2 \cos ^2 2 x}}}}=\frac{2}{\sqrt{2+\sqrt{2+2 \cos 2 x}}}=\frac{2}{\sqrt{2+\sqrt{2.2 \cos ^2 x}}} \\
& =\frac{2}{\sqrt{2+2 \cos x}}=\frac{2}{2 \cos \frac{x}{2}}=\sec \frac{x}{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2009
$\mathrm{SiO}_{2}$ is attacked by which one / ones of the following?

Options:
A HF
B conc. HCl
C hot $\mathrm{NaOH}$
D Fluorine

The correct answers are:
HF, hot $\mathrm{NaOH}$, Fluorine

Solution:
Hint:
$\mathrm{SiO}_{2}(\mathrm{~s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{SiF}_{4}(\mathrm{aq} .)+\mathrm{O}_{2}(\mathrm{~g})$
$\mathrm{SiO}_{2}(\mathrm{~s})+6 \mathrm{HF}$ (aq.) $\rightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}$ (aq.) $+2 \mathrm{H}_{2} \mathrm{O}(\ell)$
$\mathrm{SiO}_{2}(\mathrm{~s})+2 \mathrm{NaOH}($ aq. $) \rightarrow \mathrm{Na}_{2} \mathrm{SiO}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\ell)$

Topic: Chemistry, p Block Elements (Group 13 & 14), WBJEE, WBJEE 2020
Six molecules of an ideal gas have velocities \(1,3,5,5,6\) and \(5 \mathrm{~m} / \mathrm{s}\) respectively. At any given temperature, if \(\overline{V}\) and \(V_{\mathrm{rms}}\) represent average and rms speed of the molecules, then

Options:
A \(\overline{V}=5 \mathrm{~m} / \mathrm{s}\)
B \(V_{r m s} > \overline{V}\)
C \(V_{\text {rms }}^2 < \overline{V}^2\)
D \(V_{r m s}=\overline{V}\)

The correct answer is:
\(V_{r m s} > \overline{V}\)

Solution:
Hint : \(\overline{V}=\frac{1+3+5+5+6+5}{6}=\frac{25}{6}=4.16\)
\(\begin{aligned}
& \overline{V}_{\mathrm{ms}}=\sqrt{\frac{1+9+25 \times 3+36}{6}}=\sqrt{\frac{121}{6}}=\frac{11}{\sqrt{6}}=4.48 \\
& \mathrm{~V}_{\mathrm{rms}} > \overline{V}
\end{aligned}\)

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2023
Six numbers are in AP such that their sum is 3 . The first term is 4 times the third term. Then, the fifth term is

Options:
A -15
B -3
C 9
D -4

The correct answer is:
-4

Solution:
Let the numbers be $a-5 d t, a-3 d, a-d, a+d, a+3 d, a+5 d$
Also, given $T_{1}=4 T_{3},$ where $T_{1}, T_{3}$ are respectively, first and third terms of AP. $\Rightarrow \quad a-5 d=4(a-d)$
$\Rightarrow \quad d=-3 a=-\frac{3}{2}$
$\therefore$ The fifth term $a+3 d=\frac{1}{2}+3\left(-\frac{3}{2}\right)=\frac{1}{2}-\frac{9}{2}=-4$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2012
Six positive numbers are in GP, such that their product is 1000 . If the fourth term is 1 , then the last term is

Options:
A 1000
B 100
C $\frac{1}{100}$
D $\frac{1}{1000}$

The correct answer is:
$\frac{1}{100}$

Solution:
Let the six terms of GP are
$$
\frac{a}{r^{5}} \cdot \frac{a}{r^{3}}, \frac{a}{r}, a r, a r^{3}, a r^{5}
$$
Now, according to the question $\frac{a}{r^{5}} \cdot \frac{a}{r^{3}} \cdot \frac{a}{r} \cdot a r \cdot a r^{3} \cdot a r^{5}=1000$
$\Rightarrow \quad a^{6}=(10)^{3} \Rightarrow a^{2}=10$
Also, given fourth term $=a r=1$
$\Rightarrow \quad a^{2} r^{2}=1$
$\Rightarrow \quad r^{2}=\frac{1}{10} \quad \ldots$ (ii)
$\therefore$ Last term $=a r^{5}=\left[a^{2}\left(r^{2}\right)^{5}\right]^{1 / 2}$
$=\left[10 \times \frac{1}{10^{5}}\right]^{1 / 2}[$ from Eqs. (1) and $(i i)]$
$=\left(\frac{1}{10^{4}}\right)^{1 / 2}=\frac{1}{100}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
Six wires, each of resistance $r$, are connected so as to form a tetrahedron. The equivalent resistance of the combination when current enters through one corner and leaves through some other corner is

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

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

Solution:
Six wires each of resistance $r$ form a tetrahedron as shown in the following figure

It is circuit of Wheatstone the equivalent resistance of upper circuit $\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}=\frac{2}{2 r}=\frac{1}{r}$
$$
R=r
$$
It will in parallel with outer resistance $\frac{1}{R_{e q}}=\frac{1}{r}+\frac{1}{r}=\frac{2}{r}$
$R_{e q}=\frac{r}{2}$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2017
Sodium nitroprusside is

Options:
A $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}_2\right]$
B $\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$
C $\mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$
D $\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}_3\right]$

The correct answer is:
$\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}\right]$

Solution:
Fact

Topic: Chemistry, d and f Block Elements, WBJEE, WBJEE 2022
Solubility products $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of the salts of types $\mathrm{MX}, \mathrm{MX}_{2}$ and $\mathrm{M}_{3} \mathrm{X}$ at temperature $\mathrm{T}$ are $4.0 \times 10^{-8}, 3.2 \times 10^{-14}$ and $2.7 \times$ $10^{-15}$ respectively. Solubilities (in $\mathrm{mol} \mathrm{dm}^{-3}$ ) of the salts at temperature $\mathrm{T}$ are in the order

Options:
A $\quad M X>M X_{2}>M_{3} X$
B $M_{3} X>M X_{2}>M X$
C $\mathrm{MX}_{2}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}$
D $\mathrm{MX}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}$

The correct answer is:
$\mathrm{MX}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}$

Solution:
$\begin{array}{ll}
\mathrm{MX} & \mathrm{K}_{\mathrm{sp}}=4 \times 10^{-8}=\mathrm{S}^{2} \therefore \mathrm{S}=2 \times 10^{-4} \\
\mathrm{MX}_{2} & \mathrm{~K}_{\mathrm{sp}}=3.2 \times 10^{-14}=4 \mathrm{~S}^{3}
\end{array}$
$\therefore \mathrm{S}^{3}=\frac{3.2 \times 10^{-14}}{4}$
$\mathrm{~S}^{3}=8 \times 10^{-15} \quad \mathrm{~S}=2 \times 10^{-5}$
$\mathrm{M}_{3} \mathrm{X} \quad \mathrm{K}_{\mathrm{sp}}=2.7 \times 10^{-15}=27 \mathrm{~S}^{4} \quad \therefore \mathrm{S}=10^{-4}$
$\therefore \mathrm{MX}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}$

Topic: Chemistry, Ionic Equilibrium, WBJEE, WBJEE 2021
Solution of the differential equation $x d y-y d x=0$ represents a

Options:
A parabola
B circle
C hyperbola
D straight line

The correct answer is:
straight line

Solution:
Hints: $x . d y-y . d x=0 \Rightarrow x d y=y d x$
$$
\begin{aligned}
& \frac{d y}{y}=\frac{d x}{x} \Rightarrow \log y=\log x+\log c \\
& y=x c
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2010
Solution of $(x+y)^{2} \frac{d y}{d x}=a^{2}$ ('a' being a constant) is

Options:
A $\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C$ is an arbitrary constant
B $x y=a \tan C x, C$ is an arbitrary constant
C $\frac{x}{a}=\tan \frac{y}{C}, C$ is an arbitrary constant
D $x y=\tan (x+C), C$ is an arbitrary constant

The correct answer is:
$\frac{(x+y)}{a}=\tan \frac{y+C}{a}, C$ is an arbitrary constant

Solution:
We have, $(x+y)^{2} \frac{d y}{d x}=a^{2}$
Let $x+y=v$
$\therefore$
$1+\frac{d y}{d x}=\frac{d v}{d x}$
$\frac{d y}{d x}=\frac{d y}{d x}-1$
$v^{2}\left(\frac{d v}{d x}-1\right)=a^{2}$
$\Rightarrow$
$v^{2} \frac{d v}{d x}=v^{2}+a^{2}$
$\quad \frac{d v}{d x}=\frac{v^{2}+a^{2}}{v^{2}}$
$\Rightarrow \quad \frac{v^{2}}{v^{2}+d^{2}} d v=d x$
On integrating both sides, we get $\int \frac{v^{2}}{v^{2}+a^{2}} d v=\int d x$
$\int\left(1-\frac{a^{2}}{v^{2}+a^{2}}\right) \mathrm{d} v=x+C^{\prime}$
$\Rightarrow \quad v-\frac{a^{2}}{a} \tan ^{-1} \frac{v}{a}=x+C^{\prime}$
$\Rightarrow \quad x+y-a \tan ^{-1} \frac{x+y}{a}=x+C^{\prime}$
$\Rightarrow \quad y=a \tan ^{-1} \frac{x+y}{a}+C^{\prime}$
$=\frac{y-C^{\prime}}{a}=\tan ^{-1} \frac{x+y}{a}$
$\Rightarrow \quad\left(\frac{x+y}{a}\right)=\tan \left(\frac{y+C}{a}\right)$
where $-C^{\prime}=C$.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2017
Sound waves are passing through two routes-one in atraight path and the other along a semicireular path of radius $r$ and are again combined into one pipe and superposed as shown in the figure. If the velocity of sound waves in the pipe is $v$, then frequencies of resultant waves of maximum amplitude will be integral multiples of

Options:
A $\frac{v}{r(\pi-2)}$
B $\frac{v}{r(\pi-1)}$
C $\frac{2 v}{r(\pi-1)}$
D $\frac{v}{r(\pi+1)}$

The correct answer is:
$\frac{v}{r(\pi-2)}$

Solution:
Path difference $=(\pi r-2 r)$

$=(\pi-2) r=n \lambda$ (wheref $=$ frequency)
$$
\begin{aligned}
v &=f \times \lambda \\
\Rightarrow \quad \frac{v}{\lambda} &=f \Rightarrow f=n \left[\frac{v}{(\pi-2) r}\right]^{n}
\end{aligned}
$$
So, multiples integral $=\left[\frac{v}{(\pi-2) r}\right]$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2014
\(\stackrel{\mathrm{L}}{=}~\xrightarrow [\text { (ii) } \mathrm{H}_3 \mathrm{O}^{\oplus}]{\text { (i) } \mathrm{PhMgBr}}\stackrel{\mathrm{M}}{=}~\xrightarrow []{\mathrm{CrO}_3 / \mathrm{H}^{\oplus}}\stackrel{\mathrm{N}}{=}~\xrightarrow []{\mathrm{Ph}_3 \mathrm{P}=\mathrm{CH}_2}\mathrm{Ph}_2 \mathrm{C}= \mathrm{CH}_2\)
"L" in the above sequence of reaction is/are (where \(L \neq M \neq N\))

Options:
A Benzaldehyde
B Methyl benzoate
C Benzoyl chloride
D Benzonitrile

The correct answer is:
Benzaldehyde

Solution:
Hint :

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2023
Straight lines $x-y=7$ and $x+4 y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $A B=A C$. The equation of line $A C$ passing through (2,-7) is

Options:
A $x-y-9=0$
B $23 x+7 y+3=0$
C $2 x-y-11=0$
D $7 x-6 y-56=0$

The correct answers are:
$x-y-9=0$, $23 x+7 y+3=0$

Solution:
Given equations of lines are
$x-y=7$ ....(i)
and $x+4y=2$ ...(ii)
By solving Eqs.(i) and (ii), we get the point $B(6,-1)$

Let the slope of $A C$ be $m$ then, $A B=A C$
$\therefore \quad\left|\frac{m+\frac{1}{4}}{1-\frac{m}{4}}\right|=\left|\frac{-\frac{1}{4}-1}{1-\frac{1}{4}}\right|$
$\Rightarrow \quad m=\frac{-23}{7}, 1$
When $m=-\frac{23}{7},$ then equation of line $y+7=\frac{-23}{7}(x-2)$
$\Rightarrow \quad 7 y+49=-23 x+46$
$\Rightarrow \quad 23 x+7 y+3=0$
When $m=1$, then equation of line
$y+7=(x-2)$
$\Rightarrow \quad x-y-9=0$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
Structure of \(M\) is,

Options:
A \(\mathrm{Ph}-\mathrm{C} \equiv \mathrm{CH}\)
B \(\mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_3\)
C \(\mathrm{H}_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}\)
D \(\mathrm{H}_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_3\)

The correct answer is:
\(\mathrm{Ph}-\mathrm{C} \equiv \mathrm{CH}\)

Solution:
Hint :
Benzaldehyde \((\mathrm{O})\) gives Cinnamic acid \((\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{COOH})\) on reaction with \(\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}\) and \(\mathrm{CH}_3 \mathrm{COONa}\) known as Perkin's condensation

Topic: Chemistry, Aldehydes and Ketones, WBJEE, WBJEE 2023
Sulphan' is

Options:
A a mixture of $\mathrm{SO}_{3}$ and $\mathrm{H}_{2} \mathrm{SO}_{3}$
B $100 \%$ conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
C a mixture of gypsum and conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
D $100 \%$ oleum (a mixture of $100 \% \mathrm{SO}_{3}$ in $100 \%$ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$

The correct answer is:
$100 \%$ oleum (a mixture of $100 \% \mathrm{SO}_{3}$ in $100 \%$ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)$

Solution:
$\mathrm{H}_{2} \mathrm{SO}_{4}$ saturated with $\mathrm{SO}_{3}$ is called oleum of sulphan.
$$
\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{SO}_{3} \longrightarrow \mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{7}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2013
Sulphuryl chloride $\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right.$ ) reacts with white phosphorus $\left(\mathrm{P}_{4}\right)$ to give

Options:
A $P C l_{5}, S O_{2}$
B $\mathrm{OPCl}_{3}, \mathrm{SOCl}_{2}$
C $P C l_{5}$. $S O_{2}, S_{2} C_{2}$
D $\mathrm{OPCl}_{3}, \mathrm{SO}_{2}, \mathrm{S}_{2} \mathrm{Cl}_{2}$

The correct answer is:
$P C l_{5}, S O_{2}$

Solution:
When sulphuryl chloride $\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)$, reacts with white phosphorus $\left(\mathrm{P}_{4}\right),$ it give phosphorus pentachloride $\left(\mathrm{PCl}_{5}\right)$ and sulphur dioxide $\left(\mathrm{SO}_{2}\right)$. The reaction is as follows:
$$
\mathrm{P}_{4}+10 \mathrm{SO}_{2} \mathrm{Cl}_{2} \longrightarrow 4 \mathrm{PCl}_{5}+10 \mathrm{SO}_{2}
$$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2015
Sum of $n$ terms of the following series $1^3+3^3+5^3+7^3+\ldots \ldots \ldots$ is

Options:
A $n^2\left(2 n^2-1\right)$
B $\mathrm{n}^3(\mathrm{n}-1)$
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:
Hints : $\sum(2 n-1)^3$
$$
\begin{aligned}
& \sum\left\{\left(8 n^3-3.4 n^2+3.2 n-1\right)\right\} \\
= & 2 n^2(n+1)^2-2 n(n+1)(2 n+1)+3 n(n+1)-n \\
= & 2 n^4+4 n^3+2 n^2-2 n\left[2 n^2+3 n+1\right]+3 n^2+3 n-n \\
= & 2 n^4+4 n^3+2 n^2-4 n^3-6 n^2-2 n+3 n^2+3 n-n \\
= & 2 n^4-n^2 \\
= & n^2\left(2 n^2-1\right)
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2010
Sum of the last 30 coeffivients in the expansion of $(1+x)^{59}$, when expanded in ascending powers of $x$ is

Options:
A $2^{59}$
B $2^{58}$
C $2^{30}$
D $2^{29}$

The correct answer is:
$2^{58}$

Solution:
Hints: Total terms $=60$
Sum of first 30 terms $=\frac{\text { Sum of all the terms }}{2}=\frac{2^{59}}{2}=2^{58}$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2010
Supose in a hypothetical world the angular momentum is quantized to be even integral multiples of $\frac{\mathrm{h}}{2 \pi}$. The largest possible wavelength emitted by hydrogen atoms in visible range in a world according to Bohr's model will be, (Consider hc $=1242 \mathrm{Mev}-\mathrm{fm}$ )

Options:
A $153 \mathrm{~nm}$
B $409 \mathrm{~nm}$
C $121 \mathrm{~nm}$
D $487 \mathrm{~nm}$

The correct answer is:
$487 \mathrm{~nm}$

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{(2)^2}-\frac{1}{(4)^2}\right]$
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{3}{16}\right], \quad \lambda=\frac{912 \times 16}{3}=4864 \mathrm{~A}^0$
$\lambda \approx 487 \mathrm{~nm}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2022
Suppose a gaseous mixture of \(\mathrm{He}, \mathrm{Ne}, \mathrm{Ar}\) and \(\mathrm{Kr}\) is treated with photons of the frequency appropriate to ionize Ar. What ion(s) will be present in the mixture?

Options:
A \(\mathrm{Ar}^{+}\)
B \(\mathrm{Ar}^{+}+\mathrm{Kr}\)
C \(\mathrm{Ar}^{+}+\mathrm{He}^{+}+\mathrm{Ne}^{+}\)
D \(\mathrm{He}^{+}+\mathrm{Ar}^{+}+\mathrm{Kr}^{+}\)

The correct answer is:
\(\mathrm{Ar}^{+}+\mathrm{Kr}\)

Solution:
Hint: \(\mathrm{He} > \mathrm{Ne} > \mathrm{Ar} > \mathrm{Kr} > \mathrm{Xe} > \mathrm{Rn}\) (Order of lonization energy)
Energy of photon is sufficient to ionize \(\mathrm{Ar}\), hence \(\mathrm{Kr}\) will also ionize.
Therefore mixture contains \(\mathrm{Ar}^{+}\)and \(\mathrm{Kr}^{+}\)

Topic: Chemistry, Classification of Elements and Periodicity in Properties, WBJEE, WBJEE 2023
Suppose a machine produces metal parts that contain some defective parts witl probability $0.05 .$ How many parts should be produced in order that the probability 0 atleast one part being defective is $1 / 20$ more? (Given that, $\log _{10} 95=1.977$ an $\log _{10} 2=0.3$ )

Options:
A 11
B 12
C 15
D 14

The correct answers are:
15, 14

Solution:
Given probability of delective pat $=0.05=\frac{1}{20}$
Probability of non-defective part $=1-0.05=0.95=\frac{19}{20}$
We know that, $P(X=r)={ }^{n} Cr p^{r} q^{n-r}$
where, $\quad p=\frac{1}{20}, q=\frac{19}{20}$
$r \geq 1$ and $n=?$
Also, $P(X \geq 1) \geq \frac{1}{2}$
$\Rightarrow 1-P(X=0) \geq \frac{1}{2}$
$\Rightarrow 1-{ }^{n} C_{0}\left(\frac{1}{20}\right)^{0}\left(\frac{19}{20}\right)^{n-0} \geq \frac{1}{2}$
$\Rightarrow \quad 1-\frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$
$\Rightarrow$
$\frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$
$\Rightarrow$
$\frac{1}{2} \geq\left(\frac{95}{100}\right)^{n}$
$\Rightarrow \quad \log 2^{-1} \geq n[\log 95-\log 100 \mid$
$\Rightarrow \quad-\log 2 \geq n[\log 95-2]$
$\Rightarrow \quad-(0.3) \geq n[1977-2]$
$\Rightarrow \quad n \geq \frac{03}{0.023}$
$\Rightarrow \quad n \geq \frac{300}{23}$
$\Rightarrow \quad n \geq 1304$
$\therefore \quad n=14,15$

Topic: Mathematics, Probability, WBJEE, WBJEE 2015
Suppose $M=\int_{0}^{\pi / 2} \frac{\cos x}{x+2} d x$ $N=\int_{0}^{\pi / 4} \frac{\sin x \cos x}{(x+1)^{2}} d x$ Then, the value of $(M-N)$ equals

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

The correct answer is:
$\frac{2}{\pi+4}$

Solution:
Given, $M=\int_{0}^{\pi / 2} \frac{\cos x}{(x+2)} d x$
and $\quad N=\int_{0}^{\sqrt{4}} \frac{\sin x \cos x}{(x+1)^{2}} d x$
$\Rightarrow \quad N=\int_{0}^{\pi / 4} \frac{1}{2} \frac{\sin 2 x}{(x+1)^{2}} d x$
Put $2 x=t \Rightarrow d x=\frac{d t}{2}$
$\begin{aligned} N &=\int_{0}^{\pi / 2} \frac{\sin t}{4\left(\frac{t}{2}+1\right)^{2}} d t \\ &=\int_{0}^{\pi / 2} \frac{\sin t}{(t+2)^{2}} d t \\ &=\int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x \end{aligned}$
$M-N=\int_{0}^{\pi / 2} \cos x \times \frac{1}{(x+2)} d x$
$=\left[\frac{\sin x}{x+2}\right]_{0}^{\pi / 2}-\int_{0}^{\pi / 2}-\frac{\sin x}{(x+2)^{2}} d x$
$-\int_{0}^{\pi / 2} \frac{\sin x}{(x+2)^{2}} d x$
$=\frac{\sin \frac{\pi}{2}}{\frac{\pi}{2}+2}=\frac{1}{\frac{\pi+4}{2}}=\frac{2}{\pi+4}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2014
Suppose that $f(x)$ is a differentiable function such that $f^{\prime}(x)$ is continuous, $f^{\prime}(0)=1$ and $f^{\prime \prime}(0)$ does not exist. Let $g(x)=x f^{\prime}(x) .$ Then,

Options:
A $g^{\prime}(0)$ does not exist
B $g^{\prime}(0)=0$
C $g^{\prime}(0)=1$
D $g^{\prime}(0)=2$

The correct answer is:
$g^{\prime}(0)=1$

Solution:
Given, $f^{\prime}(0)=1$ and $f^{\prime \prime}$ " (0) does not exist.
Also, given $g(x)=x f^{\prime}(x)$
$\therefore \quad g^{\prime}(x)=x f^{\prime \prime}(x)+f^{\prime}(x)$
Put $x=0,$ we get
$$
\begin{aligned}
g^{\prime}(0) &=0 f^{\prime \prime}(0)+f^{\prime}(0) \\
&=0+1=1
\end{aligned}
$$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2014
Suppose that the equation $f(x)=x^{2}+b x+c=0$ has two distinct real roots $\alpha$ and $\beta$. The angle between the tangent to the curve $y=f(x)$ at the point $\left(\frac{\alpha+\beta}{2}, f\left(\frac{\alpha+\beta}{2}\right)\right)$ and the positive direction of the $x$ -axis is

Options:
A $0^{\circ}$
B $30^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$0^{\circ}$

Solution:
Since, $\alpha$ and $\beta$ are the roots of
$$
\begin{aligned}
f(x) &=x^{2}+b x+c \\
\therefore \quad \alpha+\beta &=-b \text { and } \alpha \beta=c \\
\therefore \quad f\left(\frac{\alpha+\beta}{2}\right) &=\left(\frac{\alpha+\beta}{2}\right)^{2}+b\left(\frac{\alpha+\beta}{2}\right)+c \\
&=\left(-\frac{b}{2}\right)^{2}+b\left(-\frac{b}{2}\right)+c \\
&=\frac{b^{2}}{4}-\frac{b^{2}}{2}+c=-\frac{b^{2}}{4}+c
\end{aligned}
$$
Now, $\quad \frac{d y}{d x}=f^{\prime}(x)=2 x+b$
At point $\left(\frac{\alpha+\beta}{2}, f\left(\frac{\alpha+\beta}{2}\right)\right)$
i.e., $\begin{aligned}\left(-\frac{b}{2},-\frac{b^{2}}{4}+c\right) \\ \frac{d y}{d x} &=2\left(-\frac{b}{2}\right)+b=0 \end{aligned}$
Hence, the slope of the tangent to the curve and the positive direction of $x$ -axis is $0^{\circ}$.

Topic: Mathematics, Parabola, WBJEE, WBJEE 2014
Suppose that $z_{1}, z_{2}, z_{3}$ are three vertices of an equilateral triangle in the angand plane. Let $\alpha=\frac{1}{2}(\sqrt{3}+i)$ and $\beta$ be a non-zero complex number. The points $a z_{1}+\beta, \alpha z_{2}+\beta, \alpha z_{3}+\beta$ will be

Options:
A the vertices of an equilateral triangle
B the vertices of an isosceles triangle
C collinear
D the vertices of a scalene triangle

The correct answer is:
the vertices of an equilateral triangle

Solution:
Since, $z_{1}, z_{2}$ and $z_{3}$ are the vertices of an
equilateral triangle, therefore
$$
\begin{aligned}
\left|z_{1}-z_{2}\right| &=\left|z_{2}-z_{3}\right| \\
&=\left|z_{3}-z_{1}\right|=k \\
\alpha &=\frac{1}{2}(\sqrt{3}+i)
\end{aligned}
$$
Also,
$\Rightarrow \quad|\alpha|=\frac{1}{2} \sqrt{3+1}=\frac{1}{2} \times 2=1$
Let $A=\alpha z_{1}+\beta, B=\alpha z_{2}+\beta$
and
$$
C=\alpha z_{3}+\beta
$$
Now. $|A B|=\left|\alpha z_{2}+\beta-\left(\alpha z_{1}+\beta\right)\right|$
$$
\begin{aligned}
&=\left|\alpha\left(z_{2}-z_{1}\right)\right| \\
&=|\alpha|\left|z_{2}-z_{1}\right| \\
&=|1|\left|z_{2}-z_{1}\right| \\
&=1\left|z_{2}-z_{1}\right| \\
&=\left|z_{2}-z_{1}\right|=k \\
\text { Similarly, } & B C=C A=k
\end{aligned}
$$
Hence, the points $\alpha z_{1}+\beta, \alpha z_{2}+\beta$ and $\alpha z_{3}+\beta$
are the vertices of an equilateral triangle.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
Suppose the mass of a single Ag-atom is $m$. Ag metal crystallises in fcc lattice with unit cell of length $a$. The density of Ag metal in terms of $a$ and $m$ is

Options:
A $\frac{4 m}{a^{3}}$
B $\frac{2 m}{a^{3}}$
C $\frac{m}{a^{3}}$
D $\frac{m}{4 a^{3}}$

The correct answer is:
$\frac{4 m}{a^{3}}$

Solution:
Given that,
Mass of single Ag-atom $=m$
Unit cell length $=a$
Total atoms present per unit cell of fcc lattice $=4$
Therefore, mass of unit cell becomes $4 \mathrm{m}$.
Volume of unit cell $=a^{3}$
Density $(\rho)=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}=\frac{4 \mathrm{m}}{\mathrm{a}^{3}}$

Topic: Chemistry, Solid State, WBJEE, WBJEE 2015
$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

Options:
A $\frac{2 a}{b}$
B $\frac{2 b}{a}$
C $\frac{a}{b}$
D $\frac{b}{a}$

The correct answer is:
$\frac{2 b}{a}$

Solution:
Hints : Let $\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)=\theta$, then $\cos 2 \theta=\frac{a}{b}$
$$
\begin{aligned}
& \tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right] \\
& =\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)=2\left(\frac{1+\tan ^2 \theta}{1-\tan ^2 \theta}\right)=\frac{2}{\cos 2 \theta}=\frac{2}{a / b}=\frac{2 b}{a}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2009