datavorous/entrance-exam-dataset huggingface dataset:
Let $A B C$ and $A B C^{\prime}$ be two non-congruent triangles with sides $A B=4$, $A C=A C^{\prime}=2 \sqrt{2}$ and angle $B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is

Options:

The correct answer is:
4

Solution:
In $\triangle A B C$, by sine rule,
$$
\begin{aligned}
& \frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{\circ}}=\frac{4}{\sin C} \\
\Rightarrow \quad & C=45^{\circ}, C^{\prime}=135^{\circ} \\
\Rightarrow & A=180^{\circ}-\left(45^{\circ}+30^{\circ}\right)=105^{\circ}
\end{aligned}
$$
When $\quad C^{\prime}=135^{\circ}$, then
$$
\begin{aligned}
& \quad A=180^{\circ}-\left(135^{\circ}+30^{\circ}\right)=15^{\circ} \\
& \text { Area of } \triangle A B C=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(105^{\circ}\right) \\
& =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \\
& \text { Area of } \triangle A B C^{\prime}=\frac{1}{2} A B \times A C \sin A \\
& =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(15^{\circ}\right)=2(\sqrt{3}-1)
\end{aligned}
$$
Difference of areas of triangles
$$
=|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4
$$
$$
A D=2, D C=2
$$
Difference of areas of $\triangle A B C$ and $\triangle A B C^{\prime}$
$$
\begin{aligned}
& =\text { Area of } \triangle A C C^{\prime} \\
& =\frac{1}{2} A D \times C C^{\prime}=\frac{1}{2} \times 2 \times 4=4
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2009 (Paper 2)
Let $A B C$ be a triangle such that $\angle A C B=\frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a=x^2+x+1, b=x^2-1$ and $c=2 x+1$ is (are)

Options:
A
$-(2+\sqrt{3})$

B
$1+\sqrt{3}$

C
$2+\sqrt{3}$

D
$4 \sqrt{3}$

The correct answers are:
$1+\sqrt{3}$

Solution:
Using, $\cos C=\frac{a^2+b^2-c^2}{2 a b}$
$$
\begin{gathered}
\Rightarrow \frac{\sqrt{3}}{2}=\frac{\left(x^2+x+1\right)^2+\left(x^2-1\right)^2-(2 x+1)^2}{2\left(x^2+x+1\right)\left(x^2-1\right)} \\
\Rightarrow(x+2)(x+1)(x-1) x+\left(x^2-1\right)^2 \\
=\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right) \\
\Rightarrow x^2+2 x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right) \\
\Rightarrow(2-\sqrt{3}) x^2+(2-\sqrt{3}) x-(\sqrt{3}+1)=0
\end{gathered}
$$
$\Rightarrow x=-(2+\sqrt{3})$ and $x=1+\sqrt{3}$
But, $x=-(2+\sqrt{3}) \Rightarrow c$ is negative. $\therefore \quad x=1+\sqrt{3}$ is the only solution.
Hence, (b) is the correct option.

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $A B C D$ be a quadrilateral with area 18 , with side $A B$ parallel to the side $C D$ and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is

Options:
A
3

B
2

C
$\frac{3}{2}$

D
1

The correct answer is:
2

Solution:
$$
\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6
$$
Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle
$$
\begin{aligned}
(x-r)^2+(y-r)^2 & =r^2 \\
2 \alpha & =3 r \text { and } \alpha=6 \\
r & =2
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{2}(x+2 x) \times 2 r & =18 \\
x r & =6 \\
\tan \theta & =\frac{x-r}{r} \\
\tan \left(90^{\circ}-\theta\right) & =\frac{2 x-r}{r} \\
\frac{x-r}{r} & =\frac{r}{2 x-r} \times(2 x-3 r)=0
\end{aligned}
$$
$$
x=\frac{3 r}{2}
$$
From Eqs. (i) and (ii), we get
$$
r=2
$$

Topic: Mathematics, Circle, JEE Main
Let $A B C D$ be a quadrilateral with area 18 , with side $A B$ parallel to the side $C D$ and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is

Options:
A
3

B
2

C
$\frac{3}{2}$

D
1

The correct answer is:
2

Solution:
$$
\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6
$$
Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle
$$
\begin{aligned}
(x-r)^2+(y-r)^2 & =r^2 \\
2 \alpha & =3 r \text { and } \alpha=6 \\
r & =2
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{2}(x+2 x) \times 2 r & =18 \\
x r & =6 \\
\tan \theta & =\frac{x-r}{r} \\
\tan \left(90^{\circ}-\theta\right) & =\frac{2 x-r}{r} \\
\frac{x-r}{r} & =\frac{r}{2 x-r} \times(2 x-3 r)=0
\end{aligned}
$$
$$
x=\frac{3 r}{2}
$$
From Eqs. (i) and (ii), we get
$$
r=2
$$

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $\mathbf{A}$ be vector parallel to line of intersection of planes $P_1$ and $P_2$ through origin. $P_1$ is parallel to the vectors $2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $P_2$ is parallel to $\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$, then the angle between vector $\mathbf{A}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ is

Options:
A
$\frac{\pi}{2}$

B
$\frac{\pi}{4}$

C
$\frac{\pi}{6}$

D
$\frac{3 \pi}{4}$

The correct answers are:
$\frac{\pi}{4}$
,
$\frac{3 \pi}{4}$

Solution:
Let vector $\mathbf{A O}$ be parallel to line of intersection of planes $P_1$ and $P_2$ through, i.e.
$$
\begin{aligned}
& {[(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})] \times[(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})]=54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) .} \\
& \therefore \text { Angle between } 54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and }(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\
& \Rightarrow \cos \theta=\pm\left(\frac{54+108}{3.54 \cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}} \\
& \therefore \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2006
Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. A vector coplanar to $\mathbf{a}$ and $\mathbf{b}$ has a projection along c of magnitude $\frac{1}{\sqrt{3}}$, then the vector is

Options:
A
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

B
$4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}$

C
$2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$

D
None of these

The correct answer is:
$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

Solution:
Let vector $\mathbf{r}$ be coplanar to $\mathbf{a}$ and $\mathbf{b}$.
$$
\begin{array}{ll}
\therefore & \mathbf{r}=\mathbf{a}+t \mathbf{b} \\
\Rightarrow & \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t)
\end{array}
$$
The projection of $\mathbf{r}$ on $\mathbf{c}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ $\Rightarrow|2-t|=\pm 1 \Rightarrow t=1$ or 3
When, $t=1$ we have $\mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
When, $t=3$ we have $\mathbf{r}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2006
Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be three vectors. A vector $\mathbf{v}$ in the plane of $\mathbf{a}$ and $\mathbf{b}$ whose projection of $\mathbf{c}$ is $\frac{1}{\sqrt{3}}$, is given by

Options:
A
$\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

B
$-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$

C
$3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

D
$\hat{\mathbf{i}}+3 \hat{\mathbf{k}}-3 \hat{\mathbf{k}}$

The correct answer is:
$3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

Solution:
Let $\mathbf{v}=\mathbf{a}+\lambda \mathbf{b}$
$$
\mathbf{v}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+(1+\lambda) \hat{\mathbf{k}}
$$
Projection of $\mathbf{v}$ on $\mathbf{c}=\frac{1}{\sqrt{3}}$
$$
\begin{array}{lc}
\Rightarrow & \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}} \\
\Rightarrow & \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\
\Rightarrow & 1+\lambda-1+\lambda-1-\lambda=1 \\
\Rightarrow & \lambda-1=1 \Rightarrow \lambda=2 \\
\therefore & \mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ be three given vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}$ and $\mathbf{r} \cdot \mathbf{a}=0$, then the value of $\mathbf{r} \cdot \mathbf{b}$ is

Options:

The correct answer is:
9

Solution:
$$
\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=0 \Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b} \\
& \text { or } \quad \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
&
\end{aligned}
$$
Given, $\mathbf{r} \cdot \mathbf{a}=0$, taking dot product with a for Eq. (i).
Now, $\mathbf{r} \cdot \mathbf{a}=\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{a} \cdot \mathbf{b}$
$\therefore \quad \lambda=\frac{-\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \quad[\because \vec{r} \cdot \vec{a}=0] \ldots$ (i)
From Eqs. (i) and (ii), we get
$$
\mathbf{r}=\mathbf{c}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{b}
$$
Taking dot with $\mathbf{b}$, we get
$$
\mathbf{r} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}}(\mathbf{b} \cdot \mathbf{b})
$$
where, $\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}} \\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{array}\right]$
$$
=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $B_n=1-A_n$. Find a least odd natural number $n_0$, so that $B_n>A_n, \forall n \geq n_0$.

Options:

The correct answer is:
7

Solution:
$B_n=1-A_n>A_n \Rightarrow A_n < \frac{1}{2}$ Now, $A_n=\frac{\frac{3}{4}\left(1-\left(-\frac{3}{4}\right)^n\right)}{1+\frac{3}{4}} < \frac{1}{2} \Rightarrow\left(-\frac{3}{4}\right)>-\frac{1}{6}$
Obviously, it is true for all even values of $n$. But, for
$$
\begin{aligned}
n & =1, & \frac{3}{4} & < \frac{1}{6} \\
n & =3, & \frac{27}{64} & < \frac{1}{6} \\
n & =5, & \frac{243}{1024} & >\frac{1}{6}
\end{aligned}
$$
which is true for $n=7$.
Obviously, $n_0=7$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2006
Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 .

Question: Which of the following is correct?

Options:
A $a_{17}=a_{16}+a_{15}$
B $c_{17} \neq c_{16}+c_{15}$
C $b_{17} \neq b_{16}+c_{16}$
D $a_{17}=c_{17}+b_{16}$

The correct answer is:
$a_{17}=a_{16}+a_{15}$

Solution:
By recurring formula, $a_{17}=a_{16}+a_{15}$ is correct

Also $c_{17} \neq c_{16}+c_{15}$

$\Rightarrow a_{15} \neq a_{14}+a_{13}\left(\because c_{n}=a_{n-2}\right)$

$\therefore$ Incorrect

Similarly, other parts are also incorrect.

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 .

Question: The value of $b_{6}$ is

Options:
A 7
B 8
C 9
D 11

The correct answer is:
8

Solution:
$\because a_{n}=$ number of all $n$ digit +ve integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 .

and $b_{n}=$ number of such $n$ digit integers ending with 1 $c_{n}=$ number of such $n$ digit integers ending with 0

Clearly, $a_{n}=b_{n}+c_{n}\left(\because a_{n}\right.$ can end with 0 or 1)

Also $b_{n}=a_{n-1}$ and $c_{n}=a_{n-2}[\because$ if last digit is 0 , second last has to be 1]

$\therefore$ We get $a_{n}=a_{n-1}+a_{n-2}, n \geq 3$

Also $a_{1}=1, a_{2}=2$,

Now by this recurring formula, we get

$\begin{aligned} a_{3} &=a_{2}+a_{1}=3 \\ a_{4} &=a_{3}+a_{2}=3+2=5 \\ a_{5} &=a_{4}+a_{3}=5+3=8 \\ \text { Also } b_{6} &=a_{5}=8 \end{aligned}$

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy $y(2)=\frac{2}{\sqrt{3}}$
Statement $1 y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$
Statement $2 y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$

Options:
A
Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.

B
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is false, Statement 2 is true

The correct answer is:
Statement 1 is true, Statement 2 is false.

Solution:
$\because \frac{d y}{d x}=\frac{y \sqrt{y^2-1}}{x \sqrt{x^2-1}} \Rightarrow \int \frac{d y}{y \sqrt{y^2-1}}=\int \frac{d x}{x \sqrt{x^2-1}}$ $\Rightarrow \quad \sec ^{-1} y=\sec ^{-1} x+C$ At $\quad x=2, y=\frac{2}{\sqrt{3}}$ $$ \frac{\pi}{6}=\frac{\pi}{3}+C \Rightarrow C=-\frac{\pi}{6} $$
Now, $y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$
$$
\begin{aligned}
& =\cos \left[\cos ^{-1} \frac{1}{x}-\cos ^{-1} \frac{\sqrt{3}}{2}\right] \\
& =\cos \left[\cos ^{-1}\left(\frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^2}} \sqrt{1-\frac{3}{4}}\right)\right] \\
\Rightarrow \quad \frac{1}{y} & =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}} \\

Topic: Mathematics, Differential Equations, JEE Advanced, JEE Advanced 2008 (Paper 2)
Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is

Options:
A
1

B
2

C
3

D
4

The correct answer is:
3

Solution:
$\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^8-\beta^8\right)}{2\left(\alpha^9-\beta^9\right)}$ $=\frac{\alpha^8\left(\alpha^2-2\right)-\beta^8\left(\beta^2-2\right)}{2\left(\alpha^9-\beta^9\right)}$ $\left[\because \alpha\right.$ is root of $x^2-6 x-2=0$ $\left.\alpha^2-2=6 \alpha\right]$
Also, $\beta$ is root of $x^2-6 x-2=0$
$$
\begin{aligned}
& \Rightarrow \quad \beta^2-2=6 \beta \\
& =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. A fair die is thrown three times. If $r_1, r_2$ and $r_3$ are the numbers obtained on the die, then the probability that $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is

Options:
A
$\frac{1}{18}$

B
$\frac{1}{9}$

C
$\frac{2}{9}$

D
$\frac{1}{36}$

The correct answer is:
$\frac{2}{9}$

Solution:
Sample space A dice is thrown thrice, $n(s)=6 \times 6 \times 6$.
Favorable events $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ ie, $\left(r_1, r_2, r_3\right)$ are ordered 3-triples which can take values, $\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}$ ie, 8 ordered pairs and each can be arranged in 3 ! ways $=6$
$$
\begin{array}{ll}
\therefore & n(E)=8 \times 6 \\
\Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}
\end{array}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $\omega$ be the complex number $\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}$. Then the number of distinct complex number $z$ satisfying $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=0$ is equal to

Options:

The correct answer is:
1

Solution:
Let $A=\left[\begin{array}{ccc}1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega\end{array}\right]$
Now, $\quad A^2=\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$
and $\operatorname{Tr}(A)=0,|A|=0$
$\therefore \quad A^3=0$
$\Rightarrow\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=|A+z I|=0$
$\Rightarrow \quad z^3=0$
$\Rightarrow z=0$, the number of $z$ satisfying the given equation is 1 .

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $\alpha, \beta$ be the roots of the equation $x^2-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^2-q x+r=0$. Then, the value of $r$ is

Options:
A
$\frac{2}{9}(p-q)(2 q-p)$

B
$\frac{2}{9}(q-p)(2 p-q)$

C
$$
\frac{2}{9}(q-2 p)(2 q-p)
$$

D
$\frac{2}{9}(2 p-q)(2 q-p)$

The correct answer is:
$\frac{2}{9}(2 p-q)(2 q-p)$

Solution:
The equation $x^2-p x+r=0$ has roots $(\alpha, \beta)$ and the equation $x^2-q x+r=0$ has roots
$$
\begin{array}{ll}
\left(\frac{\alpha}{2}, 2 \beta\right) . & \\
\Rightarrow & r=\alpha \beta \text { and } \alpha+\beta=p \text { and } \frac{\alpha}{2}+2 \beta=q \\
\Rightarrow & \beta=\frac{2 q-p}{3} \text { and } \alpha=\frac{2(2 p-q)}{3} \\
\Rightarrow & \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q)
\end{array}
$$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2007 (Paper 1)
Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11 / 25$ and the probability of none of them occurring is $2 / 25$. If $P(T)$ denotes the probability of occurrence of the event $T$, then

Options:
A
$P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$

B
$P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$

C
$P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$

D
$P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$

The correct answers are:
$P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$
,
$P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$

Solution:
$$
P(E \cup F)-P(E \cap F)=\frac{11}{25}
$$
(i.e. only $E$ or only $F$ )
Neither of them occurs $=\frac{2}{25}$ $\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25}$
From Eq. (i), we get
$$
P(E)+P(F)-2 P(E \cap F)=\frac{11}{25}
$$
From Eq. (ii), we get
$$
\begin{gathered}
(1-P(E))(1-P(F))=\frac{2}{25} \\
\Rightarrow 1-P(E)-P(F)+P(E) \cdot P(F)=\frac{2}{25}
\end{gathered}
$$
(iv)
From Eqs. (iii) and (iv), we get
$$
\begin{aligned}
& P(E)+P(F)=\frac{7}{5} \text { and } P(E) \cdot P(F)=\frac{12}{25} \\
& \therefore \quad P(E) \cdot\left\{\frac{7}{5}-P(E)\right\}=\frac{12}{25}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad(P(E))^2-\frac{7}{5} P(E)+\frac{12}{25}=0 \\
& \Rightarrow \quad\left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0 \\
& \therefore \quad P(E)=\frac{3}{4} \text { or } \frac{4}{5} \Rightarrow P(F)=\frac{4}{5} \text { or } \frac{3}{5}
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $E^c$ denotes the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)>0$ and $P(E \cap F \cap G)=0$. Then, $P\left(E^c \cap F^c \mid G\right)$ equals

Options:
A
$P\left(E^c\right)+P\left(F^c\right)$

B
$P\left(E^c\right)-P\left(F^c\right)$

C
$P\left(E^c\right)-P(F)$

D
$P(E)-P\left(F^c\right)$

The correct answer is:
$P\left(E^c\right)-P(F)$

Solution:
$$
\text { } \begin{aligned}
P\left(\frac{E^c \cap F^c}{G}\right) & =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\
& =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\
& =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\
& =1-P(E)-P(F) \\
& =P\left(E^c\right)-P(F) .
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $\omega=e^{i \pi / 3}$ and $a, b, c, x, y, z$ be non-zero complex numbers such that $a+b+c=x, a+b \omega+c \omega^2=y, a+b \omega^2+c \omega=z$.
Then, the value of $\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2}$ is

Options:

The correct answer is:
3

Solution:
Here, $\omega=e^{i 2 \pi / 3}$, then only integer solution exists.
Then, $\frac{\left|x^2\right|+\left|y^2\right|+\left|z^2\right|}{\left|a^2\right|+\left|b^2\right|+\left|c^2\right|}=3$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x}$, where $b$ is a constant such that $0 < b < 1$. Then,

Options:
A
$f$ if not invertible on $(0,1)$

B
$f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$

C
$f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$

D
$f^{-1}$ is differentiable on $(0,1)$

The correct answers are:
$f$ if not invertible on $(0,1)$

Solution:
Here, $f(x)=\frac{b-x}{1-b x}$ where, $0 < b < 1,0 < x < 1$ For function to be invertible it should be one-one onto.
$\therefore$ Check range :
Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$
$$
\begin{aligned}
& \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\
& \Rightarrow \quad x=\frac{b-y}{1-b y}
\end{aligned}
$$
where, $0 < x < 1$
$$
\begin{gathered}
\therefore \quad 0 < \frac{b-y}{1-b y} < 1 \\
\frac{b-y}{1-b y}>0 \text { and } \frac{b-y}{1-b y} < 1 \\
\quad+\quad-\quad+ \\
\Rightarrow \quad b \\
y < b \text { or } y>\frac{1}{b} \\
\frac{(b-1)(y+1)}{1-b y} < -1 < y < \frac{1}{b}
\end{gathered}
$$
From Eqs. (i) and (ii), we get $y \in\left(-1, \frac{1}{b}\right) \subset$ Codomain
Thus, $f(x)$ is not invertible.

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $f:(-1,1) \rightarrow I R$ be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}$ for $\theta \in\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value (s) of $f\left(\frac{1}{3}\right)$ is (are)

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

The correct answers are:
$1-\sqrt{\frac{3}{2}}$, $1+\sqrt{\frac{3}{2}}$

Solution:
Given : $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}=\frac{2 \cos ^{2} \theta}{2 \cos ^{2} \theta-1}$

$=\frac{1+\cos 2 \theta}{\cos 2 \theta}=1+\frac{1}{\cos 2 \theta}$

Let $\cos 4 \theta=\frac{1}{3} \Rightarrow 2 \cos ^{2} 2 \theta-1=\frac{1}{3} \Rightarrow \cos 2 \theta=\pm \sqrt{\frac{2}{3}}$

$\therefore f(\cos 4 \theta)=1+\frac{1}{\cos 2 \theta}=1 \pm \sqrt{\frac{3}{2}}$ or $f\left(\frac{1}{3}\right)=1 \pm \sqrt{\frac{3}{2}}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x \in[-1,2]$. Let $R_1=\int_{-1}^2 x f(x) d x$ and $R_2$ be the area of the region bounded by $y=f(x), x=-1, x=2$ and the $X$-axis. Then,

Options:
A
$R_1=2 R_2$

B
$R_1=3 R_2$

C
$2 R_1=R_2$

D
$3 R_1=R_2$

The correct answer is:
$2 R_1=R_2$

Solution:
Here, $R_1=\int_{-1}^2 x f(x) d x$
Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$$
\begin{aligned}
& R_1=\int_{-1}^2(1-x) f(1-x) d x, \\
\therefore \quad & \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right]
\end{aligned}
$$
Given, $R_2$ is area bounded by
$$
\begin{aligned}
& f(x), x-1 \text { and } x=2 \\
\therefore & R_2=\int_{-1}^2 f(x) d x
\end{aligned}
$$
Adding Eqs. (i) and (ii), we get
$$
2 R_1=\int_{-1}^2 f(x) d x
$$
$\therefore$ From Eqs. (iii) and (iv), we get
$$
2 R_1=R_2
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2011 (Paper 2)