datavorous/entrance-exam-dataset huggingface dataset:
Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof $)(x)=(\operatorname{gogof})(x)$, where $(f \circ g)(x)=f(g(x))$ is

Options:
A
$\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$

B
$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$

C
$\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

D
$2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$

The correct answer is:
$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$

Solution:
$f(x)=x^2, g(x)=\sin x$
$(g \circ f)(x)=\sin x^2$
go(gof) $(x)=\sin \left(\sin x^2\right)$
$(f \circ g \circ g \circ f)(x)=\left(\sin \left(\sin x^2\right)\right)^2$
Again, $(g \circ f)(x)=\sin x^2$
(gogof) $(x)=\sin \left(\sin x^2\right)$
Given, $($ fogogof $)(x)=($ gogof $)(x)$
$\Rightarrow \quad\left(\sin \left(\sin x^2\right)\right)^2=\sin \left(\sin x^2\right)$
$\Rightarrow \sin \left(\sin x^2\right)\left\{\sin \left(\sin x^2\right)-1\right\}=0$
$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$
$\Rightarrow \sin x^2=0$ or $\sin x^2=\frac{\pi}{2}$
$\therefore \quad x^2=n \pi$
(i.e. not possible as $-1 \leq \sin \theta \leq 1$ )

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m$ and $n$ are integers, $m \neq 0, n>0$ and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1^{+}} g(x)=p$, then

Options:
A
$n=1, m=1$

B
$n=1, m=-1$

C
$n=2, m=2$

D
$n>2, m=n$

The correct answer is:
$n=2, m=2$

Solution:
Given, $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 < x < 2, m \neq 0, n$ are integers and $|x-1|=\left\{\begin{array}{l}x-1 ; x \geq 1 \\ 1-x ; x < 1\end{array}\right.$
The left hand derivative of $|x-1|$ at $x=1$ is $p=-1$.
Also,
$$
\begin{aligned}
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(1+h-1)^n}{\log \cos ^m(1+h-1)}=-1 \Rightarrow \lim _{h \rightarrow 0} \frac{h^n}{\log \cos ^m h}=-1 \\
& \Rightarrow \quad \lim _{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1
\end{aligned}
$$
[Using L' Hospital rule]
$$
\begin{array}{ll}
\Rightarrow & \quad \lim _{h \rightarrow 0} \frac{n \cdot h^{n-1}}{m \frac{1}{\cos h}(-\sin h)}=-1 \\
\Rightarrow \quad & \quad \lim _{h \rightarrow 0}\left(-\frac{n}{m}\right) \cdot \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=-1 \\
\Rightarrow \quad & \quad\left(\frac{n}{m}\right) \lim _{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1 \\
\Rightarrow & n=2 \text { and } \frac{n}{m}=1 \\
\therefore & \quad m=n=2
\end{array}
$$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)$ is equal to

Options:
A
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

B
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

C
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$

D
$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$

The correct answer is:
$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$

Solution:
Since, $\quad f(x)=e^{g(x)}$
$$
\begin{aligned}
\Rightarrow \quad e^{g(x+1)} & =f(x+1) \\
& =x f(x) \\
& =x e^{g(x)}
\end{aligned}
$$
and
$$
g(x+1)=\log x+g(x)
$$
$$
\Rightarrow \quad g(x+1)-g(x)=\log x
$$
Replacing $x$ by $x-\frac{1}{2}$, we get
$$
\begin{aligned}
g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) & =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\
\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) & =-\frac{4}{(2 x-1)^2}
\end{aligned}
$$
Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get
$$
g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .
$$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2008 (Paper 2)
Let $H_1, H_2, \ldots, H_n$ be mutually exclusive events with $P\left(H_i\right)>0, i=1,2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$.
Statement I $P\left(H_i / E\right)>P\left(E / H_i\right) P\left(H_i\right)$ for $i=1,2, \ldots, n$.
Statement II $\sum_{i=1}^n P\left(H_i\right)=1$

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement II

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is false, Statement II is true

Solution:
Statement : I If $P\left(H_i \cap E\right)=0$ for some $i$, then
$$
P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0
$$
If $P\left(H_i \cap E\right) \neq 0$ for $\forall i=1,2, \ldots . n$, then
$$
\begin{aligned}
P\left(\frac{H_i}{E}\right) & =\frac{P\left(H_i \cap E\right)}{P\left(H_i\right)} \times \frac{P\left(H_i\right)}{P(E)} \\
& \left.=\frac{P\left(\frac{E}{H_i}\right) \times P\left(H_i\right)}{P(E)}>P\left(\frac{E}{H_i}\right) \cdot P\left(H_i\right) \quad \text { [as } 0 < P(E) < 1\right]
\end{aligned}
$$
Hence, Statement I may not always be true.
Statement II
Clearly, $H_1 \cup H_2 \cup \ldots . \cup H_n=S$ (sample space)
$$
\Rightarrow \quad P\left(H_1\right)+P\left(H_2\right)+\ldots+P\left(H_n\right)=1 .
$$
Hence, Statement II is true.

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2007 (Paper 1)
Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$.
Then, for an arbitrary constant $C$, the value of $J-I$ equals

Options:
A
$\frac{1}{2} \log \left|\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right|+C$

B
$\frac{1}{2} \log \left|\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right|+C$

C
$\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$

D
$\frac{1}{2} \log \left|\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right|+C$

The correct answer is:
$\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$

Solution:
Since, $J=\int \frac{e^{3 x}}{1+e^{2 x}+e^{4 x}} d x$
$$
\begin{aligned}
\therefore J-I & =\int \frac{\left(e^{3 x}-e^x\right)}{1+e^{2 x}+e^{4 x}} d x=\int \frac{\left(u^2-1\right)}{1+u^2+u^4} d u \\
& =\int \frac{\left(1-\frac{1}{u^2}\right)}{1+\frac{1}{u^2}+u^2} d u=\int \frac{\left(1-\frac{1}{u^2}\right)}{\left(u+\frac{1}{u}\right)^2-1} d u \\
& =\int \frac{d t}{t^2-1} \\
& =\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C \\
& =\frac{1}{2} \log \left|\frac{u^2-u+1}{u^2+u+1}\right|+C=\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Advanced, JEE Advanced 2008 (Paper 2)
Let $k$ be a positive real number and let $\begin{aligned} A & =\left[\begin{array}{ccc}2 k-1 & 2 \sqrt{k} & 2 \sqrt{k} \\ 2 \sqrt{k} & 1 & -2 k \\ -2 \sqrt{k} & 2 k & -1\end{array}\right] \text { and } \\ B & =\left[\begin{array}{ccc}0 & 2 k-1 & \sqrt{k} \\ 1-2 k & 0 & 2 \sqrt{k} \\ -\sqrt{k} & -2 \sqrt{k} & 0\end{array}\right]\end{aligned}$
If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then $[k]$ is equal to
[Note : adj $M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$ ].

Options:

The correct answer is:
4

Solution:
$|A|=(2 k+1)^3,|B|=0$
But $\operatorname{det}(\operatorname{adj} A)=\operatorname{det}(\operatorname{adj} B)=10^6$
$$
\begin{aligned}
& \Rightarrow(2 k+1)^6=10^6 \\
& \Rightarrow k=\frac{9}{2} \Rightarrow[k]=4
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2010 (Paper 2)
Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

Options:
A
$y-x+3=0$

B
$y+3 x-33=0$

C
$y+x-15=0$

D
$y-2 x+12=0$

The correct answers are:
$y-x+3=0$
,
$y+3 x-33=0$
,
$y-2 x+12=0$

Solution:
Normal to $y^2=4 x$, is $y-m x-2 m-m^3$ which passes through $(9,6)$.
Now, $\quad 6=9 m-2 m-m^3$ $\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3$
$\therefore$ Equation of normals are
$y-x+3=0$
$y+3 x-33=0$ and $y-2 x+12=0$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $L=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$, $a>0$. If $L$ is finite, then

Options:
A
$a=2$

B
$a=1$

C
$L=\frac{1}{64}$

D
$L=\frac{1}{32}$

The correct answers are:
$a=2$
,
$L=\frac{1}{64}$

Solution:

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2009 (Paper 1)
Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N=N M$. If $P^T$ denotes the transpose of $P$, then $M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T$ is equal to

Options:
A
$M^2$

B
$-N^2$

C
$-M^2$

D
$M N$

The correct answers are:
$-M^2$

Solution:
Given, $M^T=-M, N^T=-N$
and $\quad M N=N M$
$$
\begin{aligned}
\therefore & M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T \\
& =M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T \cdot M^T \\
& =M^2 N\left(N N^{-1}\right)(-M)^{-1}\left(N^T\right)^{-1}(-M) \\
& =M^2 N\left(-M^{-1}\right)(-N)^{-1}(-M) \\
& =-M^2 N M^{-1} N^{-1} M \\
& =-M \cdot(M N) M^{-1} N^{-1} M \\
& =-M(N M) M^{-1} N^{-1} M \\
& =-M N\left(N M^{-1}\right) N^{-1} M \\
& =-M\left(N N^{-1}\right) M=-M^2
\end{aligned}
$$
Note Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2011 (Paper 1)
$$
\text { Let } M \text { be a } 3 \times 3 \text { matrix satisfying }
$$
$$
M\left[\begin{array}{l}
0 \\
1 \\
0
\end{array}\right]=\left[\begin{array}{c}
-1 \\
2 \\
3
\end{array}\right], M\left[\begin{array}{c}
1 \\
-1 \\
0
\end{array}\right]=\left[\begin{array}{c}
1 \\
1 \\
-1
\end{array}\right] \text { and } M\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]=\left[\begin{array}{c}
0 \\
0 \\
12
\end{array}\right]
$$
Then, the sum of the diagonal entries of $M$ is

Options:

The correct answer is:
9

Solution:
Let $M=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]$
$\therefore M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right], M\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{r}1 \\ 1 \\ -1\end{array}\right]$,
$M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{r}0 \\ 0 \\ 12\end{array}\right]$,
$$
\begin{aligned}
\Rightarrow & {\left[\begin{array}{l}
a_2 \\
b_2 \\
c_2
\end{array}\right]=\left[\begin{array}{r}
-1 \\
2 \\
3
\end{array}\right],\left[\begin{array}{l}
a_1-a_2 \\
b_1-b_2 \\
c_1-c_2
\end{array}\right]=\left[\begin{array}{r}
1 \\
1 \\
-1
\end{array}\right], } \\
& {\left[\begin{array}{l}
a_1+a_2+a_3 \\
b_1+b_2+b_3 \\
c_1+c_2+c_3
\end{array}\right]=\left[\begin{array}{r}
0 \\
0 \\
12
\end{array}\right] } \\
\Rightarrow & a_2=-1, b_2=2, c_2=3, a_1-a_2=1, \\
\Rightarrow & b_1-b_2=1, c_1-c_2=-1 \\
\Rightarrow \quad & a_1+a_2+a_3=0, b_1+b_2+b_3=0, \\
\therefore & c_1+c_2+c_3=12 \\
\therefore & a_1=0, b_2=2 \text { and } c_3=7
\end{aligned}
$$
Hence, sum of diagonal elements
$$
=0+2+7=9
$$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $O(0,0), P(3,4)$ and $Q(6,0)$ be the vertices of the $\triangle O P Q$. The point $R$ inside the $\triangle O P Q$ is such that the $\triangle O P R, \triangle P Q R, \triangle O Q R$ are of equal area. The coordinates of $R$ are

Options:
A
$\left(\frac{4}{3}, 3\right)$

B
$\left(3, \frac{2}{3}\right)$

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

D
$\left(\frac{4}{3}, \frac{2}{3}\right)$

The correct answer is:
$\left(3, \frac{4}{3}\right)$

Solution:
$$
\text { 7. Since, triangle is on isosceles, hence centroid is the desired point. }
$$
$\therefore$ Coordinates of $R\left(3, \frac{4}{3}\right)$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$.
Then, the value of $\mu$ for which the vector $\mathbf{P Q}$ is parallel to the plane $x-4 y+3 z=1$ is

Options:
A
$1 / 4$

B
$-1 / 4$

C
$1 / 8$

D
$-1 / 8$

The correct answer is:
$1 / 4$

Solution:
$\mathbf{O Q}=(1-3 \mu) \mathbf{i}+(\mu-1) \mathbf{j}+(5 \mu+2) \mathbf{k}$ and $\mathbf{O P}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$, where $O$ is origin.
$$
\begin{aligned}
& \text { Now, } \mathbf{P Q}=(1-3 \mu-3) \hat{\mathbf{i}}+(\mu-1-2) \hat{\mathbf{j}} \\
& +(5 \mu+2-6) \hat{\mathbf{k}} \\
& =(2-3 \mu) \hat{\mathbf{i}}+(\mu-3) \hat{\mathbf{j}}+(5 \mu-4) \hat{\mathbf{k}} \\
&
\end{aligned}
$$
$\because \mathbf{P Q}$ is parallel to the plane
$$
\begin{array}{rr}
& x-4 y+3 z=1 . \\
\therefore & -2-3 \mu-4 \mu+12+15 \mu-12=0 \\
\Rightarrow & 8 \mu=2 \Rightarrow \mu=\frac{1}{4}
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2009 (Paper 1)
Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $X$-axis at $(9,0)$, then the eccentricity of the hyperbola is

Options:
A
$\sqrt{\frac{5}{2}}$

B
$\sqrt{\frac{3}{2}}$

C
$\sqrt{2}$

D
$\sqrt{3}$

The correct answer is:
$\sqrt{\frac{3}{2}}$

Solution:
Equation of normal to hyperbola at $\left(x_1, y_1\right)$ is
$$
\begin{gathered}
\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\
\therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2
\end{gathered}
$$
It passes throught $(9,0)$.
$$
\begin{aligned}
& \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\
& \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\
& \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}
\end{aligned}
$$

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $P=\left[a_{i j}\right]$ be a $3 \times 3$ matrix and let $Q=\left[b_{i j}\right]$, where $b_{i j}=2^{i+j} a_{i j}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is 2 , then the determinant of the matrix $Q$ is

Options:
A $2^{10}$
B $2^{11}$
C $2^{12}$
D $2^{13}$

The correct answer is:
$2^{13}$

Solution:
$|Q|=\left|\begin{array}{lll}2^{2} a_{11} & 2^{3} a_{12} & 2^{4} a_{13} \\ 2^{3} a_{21} & 2^{4} a_{22} & 2^{5} a_{23} \\ 2^{4} a_{31} & 2^{5} a_{32} & 2^{6} a_{33}\end{array}\right|$

$=2^{2} \cdot 2^{3} \cdot 2^{4}\left|\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ 2 a_{21} & 2 a_{22} & 2 a_{23} \\ 2^{2} a_{31} & 2^{2} a_{32} & 2^{2} a_{33}\end{array}\right|$

$=2^{9} \cdot 2 \cdot 2^{2}\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$

$=2^{12} \times|P|=2^{12} \times 2=2^{13}$

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq-q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$, then $a$ quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is

Options:
A
$\left(p^3+q\right) x^2-\left(p^3+2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

B
$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

C
$\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x$
$$
+\left(p^3-q\right)=0
$$

D
$\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x$
$$
+\left(p^3-q\right)=0
$$

The correct answer is:
$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$
$$
+\left(p^3+q\right)=0
$$

Solution:
Sum of roots $=\frac{\alpha^2+\beta^2}{\alpha \beta}$ and product $=1$
Given, $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$
$$
\begin{aligned}
& \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\
& \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}
\end{aligned}
$$
and $\quad(\alpha+\beta)^2=p^2$
$$
\Rightarrow \quad \alpha^2+\beta^2+2 \alpha \beta=p^2
$$
From Eqs. (i) and (ii), we get
$$
\alpha^2+\beta^2=\frac{p^3-2 q}{3 p}
$$
and $\alpha \beta=\frac{p^3+q}{3 p}$
$\therefore$ Required equation is
$$
\begin{gathered}
x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\
\Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0
\end{gathered}
$$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}, 4 \hat{\mathbf{i}}, 3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ an $\mathrm{d}-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ respectively. The quadrilateral $P Q R S$ must be a

Options:
A
parallelogram, which is neither a rhombus nor a rectangle

B
square

C
rectangle, but not a square

D
rhombus, but not a square

The correct answer is:
parallelogram, which is neither a rhombus nor a rectangle

Solution:
$m_{P Q}=\frac{1}{6}, m_{S R}=\frac{1}{6}, \quad m_{R Q}=-3$, $m_{S P}=-3$
$\Rightarrow$ Parallelogram
But neither $P R=S Q$ nor $P R \perp S Q$.
$\therefore$ Parallelogram, which is neither a rhombus nor a rectangle.

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ be two sets. Then,

Options:
A
$P \subset Q$ and $Q-P \neq \Phi$

B
$Q \not \subset P$

C
$P \not \subset Q$

D
$P=Q$

The correct answer is:
$P=Q$

Solution:
$$
\begin{aligned}
& P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \\
& \Rightarrow \quad \cos \theta(\sqrt{2}+1)=\sin \theta \\
& \Rightarrow \quad \tan \theta=\sqrt{2}+1 \\
& Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} \\
& \Rightarrow \quad \sin \theta(\sqrt{2}-1)=\cos \theta \\
& \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\
& =(\sqrt{2}+1) \\
& \therefore \quad P=Q \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right), y_1 < 0, y_2 < 0$, be the end points of the latusrectum of the ellipse $x^2+4 y^2=4$. The equations of parabolas with latusrectum $P Q$ are

Options:
A
$x^2+2 \sqrt{3} y=3+\sqrt{3}$

B
$x^2-2 \sqrt{3} y=3+\sqrt{3}$

C
$x^2+2 \sqrt{3} y=3-\sqrt{3}$

D
$x^2-2 \sqrt{3} y=3-\sqrt{3}$

The correct answers are:
$x^2-2 \sqrt{3} y=3+\sqrt{3}$
,
$x^2+2 \sqrt{3} y=3-\sqrt{3}$

Solution:
The equation $x^2+4 y^2=4$ represents an ellipse with 2 and 1 as semi-major and semi-minor axes and eccentricity $\frac{\sqrt{3}}{2}$. Thus, the ends of latusrect are $\left(\sqrt{3}, \frac{1}{2}\right)$ and $\left(\sqrt{3},-\frac{1}{2}\right),\left(-\sqrt{3},-\frac{1}{2}\right)$ and $\left(\sqrt{3},-\frac{1}{2}\right)$.
According to the question, we consider only $P\left(-\sqrt{3},-\frac{1}{2}\right)$ and $Q\left(\sqrt{3},-\frac{1}{2}\right)$.
Now, $\quad P Q=2 \sqrt{3}$
Thus, the coordinates of the vertex of the parabolas are $A\left(0, \frac{-1+\sqrt{3}}{2}\right)$ and $A^{\prime}\left(0, \frac{-1-\sqrt{3}}{2}\right)$ and corresponding equations are and
$$
(x-0)^2=-4 \cdot \frac{\sqrt{3}}{2}\left(y+\frac{1-\sqrt{3}}{2}\right)
$$
i.e., $(x-0)^2=4 \cdot \frac{\sqrt{3}}{2}\left(y-\frac{-1-\sqrt{3}}{2}\right)$ and
$$
\begin{aligned}
& x^2+2 \sqrt{3} y=3-\sqrt{3} \\
& x^2-2 \sqrt{3} y=3+\sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let $p(x)$ be a polynomial of degree 4 having extremum at $x=1,2$ and $\lim _{x \rightarrow 0}\left[1+\frac{p(x)}{x^2}\right]=2$. Then, the value of $p(2)$ is

Options:

The correct answer is:
0

Solution:
Let $p(x)=a x^4+b x^3+c x^2+d x+e$
$$
\Rightarrow \quad p^{\prime}(x)=4 a x^3+3 b x^2+2 c x+d
$$
$$
\therefore \quad p^{\prime}(1)=4 a+3 b+2 c+d=0
$$
and $p^{\prime}(2)=32 a+12 b+4 c+d=0$
Since, $\lim _{x \rightarrow 0}\left(1+\frac{p(x)}{x^2}\right)=2 \quad$ [given]
$$
\lim _{x \rightarrow 0} \frac{a x^4+b x^3+(c+1) x^2+d x+e}{x^2}=2
$$
$$
\begin{array}{rlrl}
\Rightarrow & & x+1=2, d & =0, e=0 \\
\Rightarrow & c & =1
\end{array}
$$
From Eqs. (i) and (ii)
$$
\begin{array}{rlrl}
& & 4 a+3 b=-2 & \text { and } 32 a+12 b=-4 \\
\Rightarrow & & a=\frac{1}{4} \text { and } b=-1 \\
\therefore & & p(x)=\frac{x^4}{4}-x^3+x^2 \\
\Rightarrow & & p(2)=\frac{16}{4}-8+4
\end{array}
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2009 (Paper 2)
Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p^{\prime}(0)$ is

Options:

The correct answer is:
9

Solution:
Since, $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$ and $p(x)$ is a real polynomial of least degree.

Hence, let $p^{\prime}(x)=k(x-1)(x-3)=k\left(x^{2}-4 x+3\right)$

$\Rightarrow p(x)=k\left(\frac{x^{3}}{3}-2 x^{2}+3 x\right)+c$

Now, $p(1)=6$ and $p(3)=2$

$\begin{array}{l}

\Rightarrow \frac{4}{3} k+C=6 \text { and } 0+C=2 \Rightarrow k=3 \\

\therefore \quad p^{\prime}(x)=3(x-1)(x-3) \Rightarrow p^{\prime}(0)=9

\end{array}$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $P Q R$ be a triangle of area $\Delta$ with $a=2, b=\frac{7}{2}$ and $c=\frac{5}{2} ;$ where $a, b$, and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}$ equals.

Options:
A $\frac{3}{4 \Delta}$
B $\frac{45}{4 \Delta}$
C $\left(\frac{3}{4 \Delta}\right)^{2}$
D $\left(\frac{45}{4 \Lambda}\right)^{2}$

The correct answer is:
$\left(\frac{3}{4 \Delta}\right)^{2}$

Solution:
$\frac{2 \sin P-\sin 2 P}{2 \sin P+\sin 2 P}=\frac{2 \sin P-2 \sin P \cos P}{2 \sin P+2 \sin P \cos P}=\frac{1-\cos P}{1-\sin P}$

$=\frac{2 \sin ^{2} \frac{P}{2}}{2 \cos ^{2} \frac{P}{2}}=\tan ^{2} \frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}$ where $s=\frac{a+b+c}{2}$

$=\frac{(s-b)^{2}(s-c)^{2}}{s(s-a)(s-b)(s-c)}=\frac{(a+c-b)^{2}(a+b-c)^{2}}{16 . \Delta^{2}}$

$=\frac{\left(2+\frac{5}{2}-\frac{7}{2}\right)^{2}\left(2+\frac{7}{2}-\frac{5}{2}\right)^{2}}{16 \Delta^{2}}=\frac{1 \times 9}{16 \Delta^{2}}=\left(\frac{3}{4 \Delta}\right)^{2}$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $S=\{1,2,3,4\}$. The total number of unordered pairs of disjoint subsets of $S$ is equal to

Options:
A
25

B
34

C
42

D
41

The correct answer is:
41

Solution:
Let $A \cap B=\phi, A, B \subset S$
$$
\begin{aligned}
& 3^4=\frac{81+1}{2}=41 \\
\Rightarrow \quad & \frac{3^4+1}{2}=41
\end{aligned}
$$

Topic: Mathematics, Sets and Relations, JEE Advanced, JEE Advanced 2010 (Paper 2)