datavorous/entrance-exam-dataset huggingface dataset:
Extraction of metal from the ore cassiterite involves

Options:
A
carbon reduction of an oxide ore

B
self-reduction of a sulphide ore

C
removal of copper impurity

D
removal of iron impurity

The correct answers are:
carbon reduction of an oxide ore
,
removal of iron impurity

Solution:
The important ore of tin is cassiterite $\left(\mathrm{SnO}_2\right)$. Tin is extracted from cassiterite ore by carbon reduction method in a blast furnace. $\mathrm{SnO}_2+2 \mathrm{C} \rightarrow \mathrm{Sn}+2 \mathrm{CO}$
The product often contain traces of iron which is removed by blowing air through the melt to oxidise to $\mathrm{FeO}$ which then floats to the surface.
$2 \mathrm{Fe}+\mathrm{O}_2 \rightarrow 2 \mathrm{FeO}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Advanced, JEE Advanced 2011 (Paper 1)
Figure shows three resistor configurations $R_1, R_2$ and $R_3$ connected to $3 \mathrm{~V}$ battery. If the power dissipated by the configuration $R_1, R_2$ and $R_3$ is $P_1, P_2$ and $P_3$, respectively, then

Options:
A
$P_1>P_2>P_3$

B
$P_1>P_3>P_2$

C
$P_2>P_1>P_3$

D
$P_3>P_2>P_1$

The correct answer is:
$P_2>P_1>P_3$

Solution:
Applying $P=\frac{V^2}{R}$
$R_1=1 \Omega, R_2=0.5 \Omega$ and $R_3=2 \Omega, V_1=V_2=V_3=3$ volt
$\therefore \quad P_1=\frac{(3)^2}{1}=9 \mathrm{~W}$
$\quad P_2=\frac{(3)^2}{0.5}=18 \mathrm{~W}$ and $P_3=\frac{(3)^2}{2}=4.5 \mathrm{~W}$
$\therefore \quad P_2>P_1>P_3$
$\therefore \quad$ correct option is (c).

Topic: Physics, Current Electricity, JEE Main
Figure shows three resistor configurations $R_1, R_2$ and $R_3$ connected to $3 \mathrm{~V}$ battery. If the power dissipated by the configuration $R_1, R_2$ and $R_3$ is $P_1, P_2$ and $P_3$, respectively, then

Options:
A
$P_1>P_2>P_3$

B
$P_1>P_3>P_2$

C
$P_2>P_1>P_3$

D
$P_3>P_2>P_1$

The correct answer is:
$P_2>P_1>P_3$

Solution:
Applying $P=\frac{V^2}{R}$
$R_1=1 \Omega, R_2=0.5 \Omega$ and $R_3=2 \Omega, V_1=V_2=V_3=3$ volt
$\therefore \quad P_1=\frac{(3)^2}{1}=9 \mathrm{~W}$
$\quad P_2=\frac{(3)^2}{0.5}=18 \mathrm{~W}$ and $P_3=\frac{(3)^2}{2}=4.5 \mathrm{~W}$
$\therefore \quad P_2>P_1>P_3$
$\therefore \quad$ correct option is (c).

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2008 (Paper 1)
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.

Options:

None

Solution:
Chimps are more evolved than humans (The Times of India):
Chimpanzees are more evolved than humans, a study suggests. There is no doubt that humans are the more advanced species. But a comparison of 14,000 human and chimpanzee genes shows the forces of natural selection have and the greatest impact on our ape cousins.
The researchers' discovery challenges the common assumption that our large brains and high intelligence were the gifts of natural selection. Humans and chimps followed different evolutionary paths from a common ape ancestor about 5 million years ago. Both underwent changes as the fittest survived to pass their genes on to future generations. But the US study shows that humans possess a 'substantially smaller' number of positively-selected genes than chimps.

Topic: Biology, Evolution
$$
\text { Find the time constant for the given } R C \text { circuits in correct order. }
$$
$R_1=1 \Omega, R_2=2 \Omega, C_1=4 \mu \mathrm{F}, C_2=2 \mu \mathrm{F}$

Options:
A
$18,4,8 / 9$

B
$18,8 / 9,4$

C
$4,18,8 / 9$

D
$4,8 / 9,18$

The correct answer is:
$18,8 / 9,4$

Solution:
$$
\begin{aligned}
\tau & =C R \\
\tau_1 & =\left(C_1+C_2\right)\left(R_1+R_2\right)=18 \mu \mathrm{s} \\
\tau_2 & =\left(\frac{C_1 C_2}{C_1+C_2}\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=\frac{8}{6} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{s} \\
\tau_2 & =\left(C_1+C_2\right)\left(\frac{R_1 R_2}{R_1+R_2}\right)=(6)\left(\frac{2}{3}\right)=4 \mu \mathrm{s}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Advanced, JEE Advanced 2006
For $0 < \theta < \frac{\pi}{2}$, the solution(s) of $\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right]$ $\operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}$ is/are

Options:
A
$\frac{\pi}{4}$

B
$\frac{\pi}{6}$

C
$\frac{\pi}{12}$

D
$\frac{5 \pi}{12}$

The correct answers are:
$\frac{\pi}{12}$
,
$\frac{5 \pi}{12}$

Solution:
For $0 < \theta < \frac{\pi}{2}$,
$$
\begin{array}{r}
\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right] \operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right) \\
=4 \sqrt{2} \\
\Rightarrow \sum_{m=1}^6 \frac{\sin \left[\theta+\frac{m \pi}{4}-\left(\theta+\frac{(m-1) \pi}{4}\right)\right]}{\sin \frac{\pi}{4}\left\{\sin \left(\theta+\frac{(m-1) \pi}{4}\right) \sin \left(\theta+\frac{m \pi}{4}\right)\right\}} \\
=4 \sqrt{2}
\end{array}
$$
$$
\begin{array}{r}
\Rightarrow \sum_{m=1}^6 \frac{\cot \left(\theta+\frac{(m-1) \pi}{4}\right)-\cot \left(\theta+\frac{m \pi}{4}\right)}{1 / 2} \\
=4 \sqrt{2}
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad \sum_{m=1}^6\left[\cot \left(\theta+\frac{(m-1) \pi}{4}\right)\right. \\
& \left.-\cot \left(\theta+\frac{m \pi}{4}\right)\right]=4 \\
& \Rightarrow \cot (\theta)-\cot \left(\theta+\frac{\pi}{4}\right)+\cot \left(\theta+\frac{\pi}{4}\right) \\
& -\cot \left(\theta+\frac{2 \pi}{4}\right)+\ldots+\cot \left(\theta+\frac{5 \pi}{4}\right) \\
& -\cot \left(\theta+\frac{6 \pi}{4}\right)=4 \\
& \Rightarrow \quad \cot \theta-\cot \left(\frac{3 \pi}{2}+\theta\right)=4 \\
& \Rightarrow \quad \cot \theta+\tan \theta=4 \\
& \Rightarrow \quad \tan ^2 \theta-4 \tan \theta+1=0 \\
& \Rightarrow \quad(\tan \theta-2)^2-3=0 \\
& \Rightarrow(\tan \theta-2+\sqrt{3})(\tan \theta-2-\sqrt{3})=0 \\
& \Rightarrow \quad \tan \theta=2-\sqrt{3} \text { or } \\
& \tan \theta=2+\sqrt{3} \\
& \Rightarrow \theta=\frac{\pi}{12} ; \theta=\frac{5 \pi}{12} \Rightarrow \theta \in\left(0, \frac{\pi}{2}\right) \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2009 (Paper 2)
For a dilute solution containing $2.5 \mathrm{~g}$ of a non-volatile non-electrolyte solute in $100 \mathrm{~g}$ of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( $\mathrm{mm}$ of $\mathrm{Hg}$ ) of the solution is (take $K_{b}=0.76 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Options:
A 724
B 740
C 736
D 718

The correct answer is:
724

Solution:
From Raoult's law,

$\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }+\text { No. of moles of solute }}$

When the concentration of solute is much lower than the concentration of solvent,

$\begin{array}{l}

\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }} \\

\frac{760-p}{760}=\frac{2.5 / m}{100 / 18}...(i)

\end{array}$

From elevation in boiling point, $\Delta T_{b}=K_{b} \times m$

$\begin{array}{l}

2=0.76 \times m \\

m=\frac{2}{0.76}...(ii)

\end{array}$

From(i) and (ii), $p=724 \mathrm{~mm}$

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2012 (Paper 2)
For a first order reaction, $A \rightarrow P$, the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation, $\log k=-(2000) / T+6.0$
The pre-exponential factor $A$ and the activation energy $\left(E_a\right)$, respectively, are

Options:
A
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $9.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$6.0 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$1.0 \times 10^6 \mathrm{~s}^{-1}$ and $38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Comparing the slope and intercept of the given equation with the following Arrhenius equation :
$$
\log k=-\frac{E_a}{2303 R T}+\log A
$$
Hence, $\log A=6$ i.e. $A=10^6 \mathrm{~s}^{-1}$.
Comparing slope gives $E_a=38.3 \mathrm{~kJ} / \mathrm{mol}$.

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2009 (Paper 2)
For an ideal gas, consider only P-V work in going from an initial state $\mathrm{X}$ to the final state $\mathrm{Z}$. The final state $\mathrm{Z}$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct?

[Take $\Delta \mathrm{S}$ as change in entropy and w as work done].

Options:
A $\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}$
B $\mathrm{w}_{x \rightarrow z}=\mathrm{w}_{x \rightarrow y}+\mathrm{w}_{y \rightarrow z}$
C $\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}$
D $\Delta \mathrm{S}_{x \rightarrow y \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}$

The correct answers are:
$\Delta \mathrm{S}_{x \rightarrow z}=\Delta \mathrm{S}_{x \rightarrow y}+\Delta \mathrm{S}_{y \rightarrow z}$, $\mathrm{w}_{x \rightarrow y \rightarrow z}=\mathrm{w}_{x \rightarrow y}$

Solution:
$\Delta S_{X \rightarrow Z}=\Delta S_{X \rightarrow Y}+\Delta S_{Y \rightarrow Z}$ [Entropy is a state function, hence additive]

$w_{X \rightarrow Y \rightarrow Z}=w_{X \rightarrow Y}$ [Work done in $Y \rightarrow Z$ is zero because it is an isochoric process].

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2012 (Paper 1)
For any real number $x$, let $[x]$ denotes the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.$.
Then the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x$ is

Options:

The correct answer is:
4

Solution:
Given, $f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.$ $f(x)$ and $\cos \pi x$ both are periodic with period 2 and both are even.
$$
\begin{aligned}
\therefore \quad & \int_{-10}^{10} f(x) \cos \pi x d x \\
& =2 \int_0^{10} f(x) \cos \pi x d x
\end{aligned}
$$
$$
\begin{aligned}
& =10 \int_0^2 f(x) \cos \pi x d x \\
& \text { Now, } \quad \int_0^1 f(x) \cos \pi x d x \\
& =\int_0^1(1-x) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \text { and } \int_1^2 f(x) \cos \pi x d x \\
& =\int_1^2(x-1) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \therefore \quad \int_{-10}^{10} f(x) \cos \pi x d x \\
& =-20 \int_0^1 u \cos \pi u d u=\frac{40}{\pi^2} \\
& \Rightarrow \quad \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x=4
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
For any real number $x$, let $[x]$ denotes the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10,10]$ by $f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.$.
Then the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x$ is

Options:

The correct answer is:
4

Solution:
Given, $f(x)=\left\{\begin{array}{cc}x-[x] & \text { if }[x] \text { is odd } \\ 1+[x]-x & \text { if }[x] \text { is even }\end{array}\right.$ $f(x)$ and $\cos \pi x$ both are periodic with period 2 and both are even.
$$
\begin{aligned}
\therefore \quad & \int_{-10}^{10} f(x) \cos \pi x d x \\
& =2 \int_0^{10} f(x) \cos \pi x d x
\end{aligned}
$$
$$
\begin{aligned}
& =10 \int_0^2 f(x) \cos \pi x d x \\
& \text { Now, } \quad \int_0^1 f(x) \cos \pi x d x \\
& =\int_0^1(1-x) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \text { and } \int_1^2 f(x) \cos \pi x d x \\
& =\int_1^2(x-1) \cos \pi x d x=-\int_0^1 u \cos \pi u d u \\
& \therefore \quad \int_{-10}^{10} f(x) \cos \pi x d x \\
& =-20 \int_0^1 u \cos \pi u d u=\frac{40}{\pi^2} \\
& \Rightarrow \quad \frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x d x=4
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2010 (Paper 1)
If $\frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5}$, then

Options:
A
$\tan ^2 x=\frac{2}{3}$

B
$\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$

C
$\tan ^2 x=\frac{1}{3}$

D
$\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{2}{125}$

The correct answers are:
$\tan ^2 x=\frac{2}{3}$
,
$\frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125}$

Solution:
$$
\begin{aligned}
& \frac{\sin ^4 x}{2}+\frac{\cos ^4 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad \frac{\sin ^4 x}{2}+\frac{\left(1-\sin ^2 x\right)^2}{3}=\frac{1}{5} \\
& \Rightarrow \frac{\sin ^4 x}{2}+\frac{1+\sin ^4 x-2 \sin ^2 x}{3}=\frac{1}{5} \\
& \Rightarrow \quad 5 \sin ^4 x-4 \sin ^2 x+2=\frac{6}{5} \\
& \Rightarrow \quad 25 \sin ^4 x-20 \sin ^2 x+4=0 \\
& \Rightarrow \quad\left(5 \sin ^2 x-2\right)^2=0 \\
& \Rightarrow \sin ^2 x=\frac{2}{5}, \cos ^2 x=\frac{3}{5}, \tan ^2 x=\frac{2}{3} \\
& \therefore \quad \frac{\sin ^8 x}{8}+\frac{\cos ^8 x}{27}=\frac{1}{125} \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2009 (Paper 1)
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors in space given by $\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}}$ and $\overrightarrow{\mathbf{b}}=\frac{2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}}$, then the value of $(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]$ is

Options:

The correct answer is:
5

Solution:
From the given information, it is clear that
$$
\begin{gathered}
\overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}} \\
\Rightarrow \quad|\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=1, \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 \\
\text { Now, }(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]
\end{gathered}
$$
$$
\begin{aligned}
& =(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot\left[a^2 \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}+2 b^2 \cdot \overrightarrow{\mathbf{a}}\right. \\
& =[2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}] \cdot[\overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{a}}]=4 \overrightarrow{\mathbf{a}}^2+\overrightarrow{\mathbf{b}}^2 \\
& =4 \cdot 1+1=5 \quad \text { [as } a \cdot b=0]
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2010 (Paper 1)
If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$, then a possible value of $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

Options:
A 0
B 3
C 4
D 8

The correct answer is:
4

Solution:
Given that $\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}$ $\Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=\overrightarrow{0}$

But $\vec{a}+\vec{b} \neq 0$ and $2 \hat{i}+3 \hat{j}+4 \hat{k} \neq 0$

$\therefore(\vec{a}+\vec{b}) \|(2 \hat{i}+3 \hat{j}+4 \hat{k})$.

Let $\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$

Also given that $|\vec{a}+\vec{b}|=\sqrt{29} \Rightarrow \lambda=\pm 1$

$\therefore \vec{a}+\vec{b}=\pm(2 \hat{i}+3 \hat{j}+4 \hat{k})$

So, $(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2012 (Paper 2)
If $a, b$ and $c$ are the sides of a $\triangle A B C$ such that $x^2-2(a+b+c) x+3 \lambda(a b+b c+c a)=0$ has real roots, then

Options:
A
$\lambda < \frac{4}{3}$
B
$\lambda>\frac{5}{3}$

C
$\lambda \in\left(\frac{4}{3}, \frac{5}{3}\right)$

D
$\lambda \in\left(\frac{1}{3}, \frac{5}{3}\right)$

The correct answer is:
$\lambda < \frac{4}{3}$

Solution:
Since, roots are real, $D \geq 0$
$$
\begin{array}{ll}
\Rightarrow & 4(a+b+c)^2-12 \lambda(a b+b c+c a) \geq 0 \\
\Rightarrow & (a+b+c)^2 \geq 3 \lambda(a b+b c+c a) \\
\Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a)(3 \lambda-2) \\
\Rightarrow & 3 \lambda-2 \leq \frac{a^2+b^2+c^2}{a b+b c+c a} \\
\text { where, } a^2+b^2 \geq 2 a b, b^2+c^2 \geq 2 b c \text { and } c^2+a^2 \geq 2 c a \\
\Rightarrow & a^2+b^2+c^2 \geq(a b+b c+c a) \\
\therefore & \frac{a^2+b^2+c^2}{a b+b c+c a} \geq 1
\end{array}
$$
Also, $\quad \cos A=\frac{b^2+c^2-a^2}{2 b c} < 1$
$$
\Rightarrow \quad b^2+c^2-a^2 < 2 b c
$$
Similarly, $c^2+a^2-b^2 < 2 c a$ and $a^2+b^2-c^2 < 2 a b$
$$
\Rightarrow a^2+b^2+c^2 < 2(a b+b c+c a) \text { or } \frac{a^2+b^2+c^2}{a b+b c+c a} < 2
$$
From Eqs. (i), (ii) and (iii)
$$
\begin{array}{lrl}
\therefore & 3 \lambda-2 & < 2 \\
\Rightarrow & \lambda < \frac{4}{3}
\end{array}
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2006
If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors satisfying $|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9$, then $|2 \vec{a}+5 \vec{b}+5 \vec{c}|$ is

Options:

The correct answer is:
3

Solution:
$\begin{array}{l}

\because|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \\

|\vec{a}-\vec{b}|^{2}+|\vec{b}-\vec{c}|^{2}+|\vec{c}-\vec{a}|^{2}=9 \\

\Rightarrow 2\left(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}\right)-2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=9 \\

\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}

\end{array}$

Also,

$\begin{array}{l}

|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\

=1+1+1+2 \times\left(-\frac{3}{2}\right)=0 \\

\Rightarrow \vec{a}+\vec{b}+\vec{c}=0 \Rightarrow(\vec{b}+\vec{c})=-\vec{a} \\

\therefore|2 \vec{a}+5(\vec{b}+\vec{c})|=|2 \vec{a}-5 \vec{a}|=|-3 \vec{a}|=3

\end{array}$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2012 (Paper 1)
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{d}$ are the unit vectors such that $(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$ and $\mathbf{a} \cdot \mathbf{c}=\frac{1}{2}$, then

Options:
A
$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar

B
$\mathbf{b}, \mathbf{d}$ are non-parallel

C
$\mathbf{a}, \mathbf{b}, \mathbf{d}$ are non-coplanar

D
a, d are parallel and $\mathbf{b}, \mathbf{c}$ are parallel

The correct answer is:
$\mathbf{a}, \mathbf{b}, \mathbf{d}$ are non-coplanar

Solution:
Let angle between $\mathbf{a}$ and $\mathbf{b}$ be $\theta_1, \mathbf{c}$ and $\mathbf{d}$ be $\theta_2$ and $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c} \times \mathbf{d}$ be $\theta$.
Since, $(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$
$\Rightarrow \quad \sin \theta_1 \cdot \sin \theta_2 \cdot \cos \theta=1$
$\Rightarrow \theta_1=90^{\circ}, \theta_2=90^{\circ}, \theta=0^{\circ}$
$\Rightarrow \mathbf{a} \perp \mathbf{b}, \mathbf{c} \perp \mathbf{d},(\mathbf{a} \times \mathbf{b}) \|(\mathbf{c} \times \mathbf{d})$
So, $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})$
and $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})$
$\Rightarrow \quad(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{c}$
and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{d}$
$\Rightarrow\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=0$ and $\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{d}\end{array}\right]=0$
$\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{a}, \mathbf{b}, \mathbf{d}$ are coplanar vectors, so options (a) and (b) are incorrect.
Let $\quad \mathbf{b} \| \mathbf{d} \Rightarrow \mathbf{b}=\pm \mathbf{d}$
As $\quad(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$
$(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{b})=\pm 1$
$\Rightarrow \quad[\mathbf{a} \times \mathbf{b} \mathbf{c} \mathbf{b}]=\pm 1$
$\Rightarrow \quad[\mathbf{c} \mathbf{b} \mathbf{a} \times \mathbf{b}]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot[\mathbf{b} \times(\mathbf{a} \times \mathbf{b})]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot[\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b}]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=\pm 1 \quad[\because \mathbf{a} \cdot \mathbf{b}=0]$
which is a contradiction, so option (c) is correct.
Let option (d) be correct.
$$
\begin{array}{ll}
\Rightarrow & \mathbf{d}=\pm \mathbf{a} \text { and } \mathbf{c}=\pm \mathbf{b} \\
\text { As } & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1 \\
\Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{a})=\pm 1
\end{array}
$$
which is a contradiction, so option (d) is incorrect.
Alternatively option (c) and (d) may be observed from the given figure.

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{d}$ are the unit vectors such that $(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$ and $\mathbf{a} \cdot \mathbf{c}=\frac{1}{2}$, then

Options:
A
$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar

B
$\mathbf{b}, \mathbf{d}$ are non-parallel

C
$\mathbf{a}, \mathbf{b}, \mathbf{d}$ are non-coplanar

D
a, d are parallel and $\mathbf{b}, \mathbf{c}$ are parallel

The correct answer is:
$\mathbf{a}, \mathbf{b}, \mathbf{d}$ are non-coplanar

Solution:
Let angle between $\mathbf{a}$ and $\mathbf{b}$ be $\theta_1, \mathbf{c}$ and $\mathbf{d}$ be $\theta_2$ and $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c} \times \mathbf{d}$ be $\theta$.
Since, $(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$
$\Rightarrow \quad \sin \theta_1 \cdot \sin \theta_2 \cdot \cos \theta=1$
$\Rightarrow \theta_1=90^{\circ}, \theta_2=90^{\circ}, \theta=0^{\circ}$
$\Rightarrow \mathbf{a} \perp \mathbf{b}, \mathbf{c} \perp \mathbf{d},(\mathbf{a} \times \mathbf{b}) \|(\mathbf{c} \times \mathbf{d})$
So, $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})$
and $\mathbf{a} \times \mathbf{b}=k(\mathbf{c} \times \mathbf{d})$
$\Rightarrow \quad(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{c}$
and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d}=k(\mathbf{c} \times \mathbf{d}) \cdot \mathbf{d}$
$\Rightarrow\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{c}\end{array}\right]=0$ and $\left[\begin{array}{lll}\mathbf{a} & \mathbf{b} & \mathbf{d}\end{array}\right]=0$
$\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}$ and $\mathbf{a}, \mathbf{b}, \mathbf{d}$ are coplanar vectors, so options (a) and (b) are incorrect.
Let $\quad \mathbf{b} \| \mathbf{d} \Rightarrow \mathbf{b}=\pm \mathbf{d}$
As $\quad(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1$
$(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{b})=\pm 1$
$\Rightarrow \quad[\mathbf{a} \times \mathbf{b} \mathbf{c} \mathbf{b}]=\pm 1$
$\Rightarrow \quad[\mathbf{c} \mathbf{b} \mathbf{a} \times \mathbf{b}]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot[\mathbf{b} \times(\mathbf{a} \times \mathbf{b})]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot[\mathbf{a}-(\mathbf{b} \cdot \mathbf{a}) \mathbf{b}]=\pm 1$
$\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=\pm 1 \quad[\because \mathbf{a} \cdot \mathbf{b}=0]$
which is a contradiction, so option (c) is correct.
Let option (d) be correct.
$$
\begin{array}{ll}
\Rightarrow & \mathbf{d}=\pm \mathbf{a} \text { and } \mathbf{c}=\pm \mathbf{b} \\
\text { As } & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{d})=1 \\
\Rightarrow & (\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{a})=\pm 1
\end{array}
$$
which is a contradiction, so option (d) is incorrect.
Alternatively option (c) and (d) may be observed from the given figure.

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2009 (Paper 1)