datavorous/entrance-exam-dataset huggingface dataset:
If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of the hyperbola passing through the focii of the ellipse and $e_1 e_2=1$, then equation of the hyperbola is

Options:
A
$\frac{x^2}{9}-\frac{y^2}{16}=1$

B
$\frac{x^2}{16}-\frac{y^2}{9}=-1$

C
$\frac{x^2}{9}-\frac{y^2}{25}=1$

D
None of these

The correct answers are:
$\frac{x^2}{16}-\frac{y^2}{9}=-1$

Solution:
The eccentricity of $\frac{x^2}{16}+\frac{y^2}{25}=1$ is $e_1=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ $\therefore \quad e_2=\frac{5}{3}$
$\Rightarrow$ Foci of ellipse $(0, \pm 3)$
$\therefore$ Equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=-1$
Hence (b) is the correct answer.

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2006
If $f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$ for all $x \in(0, \infty)$, then

Options:
A $f$ has a local maximum at $x=2$
B $f$ is decreasing on $(2,3)$
C there exists some $c \in(0, \infty)$, such that $f^{\prime \prime}(c)=0$
D $f$ has a local minimum at $x=3$

The correct answers are:
$f$ has a local maximum at $x=2$, $f$ is decreasing on $(2,3)$, there exists some $c \in(0, \infty)$, such that $f^{\prime \prime}(c)=0$, $f$ has a local minimum at $x=3$

Solution:
$f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$

$\Rightarrow f^{\prime}(x)=e^{x^{2}} \cdot(x-2)(x-3)$

Put $f^{\prime}(x)=0 \Rightarrow x=2,3$

$f^{\prime \prime}(x)=e^{x^{2}} \cdot 2 x\left(x^{2}-5 x+6\right)+e^{x^{2}}(2 x-5)$

$f^{\prime \prime}(2)=-$ ve and $f^{\prime \prime}(3)=+$ ve

$\therefore \quad x=2$ is a point of local maxima.

and $x=3$ is a point of local minima.

Also for $x \in(2,3), f^{\prime}(x) < 0$

$\Rightarrow \quad f$ is decreasing on $(2,3)$.

Also we observe $f^{\prime \prime}(0) < 0$ and $f^{\prime \prime}(1)>0$

$\therefore \quad$ There exists some $C \in(0,1)$ such that $f^{\prime \prime}(C)=0$

Hence all the options are correct.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2012 (Paper 2)
If $f^{\prime \prime}(x)=-f(x)$, where $f(x)$ is a continuous double differentiable function and $g(x)=f^{\prime}(x)$. If $F(x)=\left(f\left(\frac{x}{2}\right)\right)^2+\left(g\left(\frac{x}{2}\right)\right)^2$ and $F(5)=5$, then $F(10)$ is

Options:
A
0

B
5

C
10

D
25

The correct answer is:
5

Solution:
As, $f(x)=-f(x) \Rightarrow \frac{d}{d x}\left(f^{\prime}(x)\right)=-f(x)$ $\Rightarrow \quad g^{\prime}(x)=-f(x)$ and $f^{\prime}(x)=g(x)$ where, $\therefore \quad F^{\prime}(x)=2\left(f\left(\frac{x}{2}\right)\right) \cdot f^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}+2\left(g\left(\frac{x}{2}\right)\right) \cdot g^{\prime}\left(\frac{x}{2}\right) \cdot \frac{1}{2}=0$ [using Eq. (1)] $\therefore F(x)$ is constant $\Rightarrow F(10)=F(5)=5$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2006
If $f(x)$ is cubic polynomial which has local maximum at $x=-1$. If $f(2)=18, f(1)=-1$ and $f^{\prime}(x)$ has local minimum at $x=0$, then

Options:
A
the distance between $(-1,2)$ and $(a, f(a))$ where $x=a$ is the point of local minima is $2 \sqrt{5}$

B
$f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$

C
$f(x)$ has local minima at $x=1$

D
the value of $f(0)=5$

The correct answers are:
$f(x)$ is increasing for $x \in[1,2 \sqrt{5}]$
,
$f(x)$ has local minima at $x=1$

Solution:
Since $f(x)$ has local maxima at $x=-1$ and $f^{\prime}(x)$ has local minima at $x=0$.
$$
\begin{aligned}
f \wedge(x) & =\lambda x \\
f^{\prime}(x) & =\lambda \frac{x^2}{2}+c \\
\frac{\lambda}{2}+c & =0 \Rightarrow \lambda=-2 c
\end{aligned}
$$
$$
\left[f^{\prime}(-1)=0\right]
$$
Again, On integrating both sides, we get
$$
\begin{aligned}
& f(x)=\lambda \frac{x^3}{6}+c x+d \\
& f(2)=\lambda\left(\frac{8}{6}\right)+2 c+d=18 \\
& f(1)=\frac{\lambda}{6}+c+d=-1
\end{aligned}
$$
and
$\therefore$ Using Eqs. (i), (ii) and (iii), we get
$$
\begin{aligned}
f(x) & =\frac{1}{4}\left(19 x^3-57 x+34\right) \\
\therefore \quad f^{\prime}(x) & =\frac{1}{4}\left(57 x^2-57\right) \\
& =\frac{57}{4}(x-1)(x+1), \text { using number line rule }
\end{aligned}
$$
$\therefore f(x)$ is increasing for $[1,2 \sqrt{5}]$ and $f(x)$ is increasing for $[1,2 \sqrt{5}]$ and $f(x)$ has local maximum at $x=-1$ and $f(x)$ has local minimum at $C x=1$.
Also,
$$
f(0)=\frac{34}{4} \text {. }
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2006
If $f(x)$ is twice differentiable function such that $f(a)=0, f(b)=2, f(c)=1, f(d)=2$, $f(e)=0$, where $a < b < c < d < e$, then the minimum number of zeros of $g(x)=\left\{f^{\prime}(x)\right\}^2+f^{\prime \prime}(x) \cdot f(x)$ in the interval $[a, e]$ is

Options:

The correct answer is:
6

Solution:
Let, $g(x)=\frac{d}{d x}\left[f(x) \cdot f^{\prime}(x)\right]$ to get the zero of $g(x)$ we take function $h(x)=f(x) \cdot f^{\prime}(x)$ between any two roots of $h(x)$ there lies atleast one root of $h^{\prime}(x)=0$
$$
\begin{aligned}
& \Rightarrow \quad g(x)=0 \\
& \Rightarrow \quad h(x)=0 \\
& \Rightarrow \quad f(x)=0 \text { or } f^{\prime}(x)=0 \\
& \text { If } f(x)=0 \text { has } 4 \text { minimum solutions, } \\
& f^{\prime}(x)=0 \text { has } 3 \text { minimum solutions, } \\
& h(x)=0 \text { has } 7 \text { minimum solutions, then } \\
& h^{\prime}(x)=g(x)=0 \text { has } 6 \text { minimum solutions. } \\
&
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2006
If $f(x)=\min \left\{1, x^2, x^3\right\}$, then

Options:
A
$f(x)$ is continuous everywhere

B
$f(x)$ is continuous and differentiable everywhere

C
$f(x)$ is not differentiable at two points

D
$f(x)$ is not differentiable at one point

The correct answers are:
$f(x)$ is continuous everywhere
,
$f(x)$ is not differentiable at one point

Solution:
Here, $f(x)=\min .\left\{1, x^2, x^3\right\}$ which could be graphically shown as
$$
\therefore \quad f(x)=\left\{\begin{array}{c}
1, x \geq 1 \\
x^3, x < 1
\end{array}\right.
$$
$$
\Rightarrow f(x) \text { is continuous for } x \in R \text { and not differentiable at } x=1 \text { due to sharp edge. }
$$

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2006
If $f(x)=\left\{\begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array}\right.$, then

Options:
A
$f(x)$ is continuous at $x=-\frac{\pi}{2}$

B
$f(x)$ is not differentiable at $x=0$

C
$f(x)$ is differentiable at $x=1$

D
$f(x)$ is differentiable at $x=-\frac{3}{2}$

The correct answers are:
$f(x)$ is continuous at $x=-\frac{\pi}{2}$
,
$f(x)$ is not differentiable at $x=0$
,
$f(x)$ is differentiable at $x=1$
,
$f(x)$ is differentiable at $x=-\frac{3}{2}$

Solution:
$f(x)=\left\{\begin{array}{cc}-x-\frac{\pi}{2}, & x \leq-\frac{\pi}{2} \\ -\cos x, & -\frac{\pi}{2} < x \leq 0 \\ x-1, & 0 < x \leq 1 \\ \log x, & x>1\end{array}\right.$
Continuity at $x=-\frac{\pi}{2}$,
$$
\begin{aligned}
& f\left(-\frac{\pi}{2}\right)=-\left(-\frac{\pi}{2}\right)-\frac{\pi}{2}=0 \\
& \mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}-\cos \left(-\frac{\pi}{2}+h\right)=0
\end{aligned}
$$
$\therefore$ Continuous at $x=0$
Continuity at $x=0 \Rightarrow f(0)=-1$ $\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0}(0+h)-1=-1$
$\therefore$ Continuous at $x=0$
Continuity at $x=1 ; f(1)=0$ $\mathrm{RHL} \Rightarrow \lim _{h \rightarrow 0} \log (1+h)=0$
$\therefore$ Continuous at $x=1$
Here, $f^{\prime}(x)=\left\{\begin{array}{cc}-1, & x \leq-\frac{\pi}{2} \\ \sin x, & -\frac{\pi}{2} < x \leq 0 \\ 1, & 0 < x \leq 1 \\ \frac{1}{x}, & x>1\end{array}\right.$
Differentiable at $x=0$, $\mathrm{LHD}=0, \mathrm{RHD}=1$
$\therefore \quad$ Not differentiable at $x=0$
Differentiable at $x=1$, $\mathrm{LHD}=1, \mathrm{RHD}=1$
$\therefore$ Differentiable at $x=1$
Also, for $x=-\frac{3}{2} \Rightarrow f(x)=-x-\frac{3}{2}$
$\therefore$ Differentiable at $x=-\frac{3}{2}$

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2011 (Paper 2)
If $I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x ; n=0,1,2$, $\ldots$, then

Options:
A
$I_n=I_{n+2}$

B
$\sum_{m=1}^{10} I_{2 m+1}=10 \pi$

C
$\sum_{m=1}^{10} I_{2 m}=0$

D
$I_n=I_{n+1}$

The correct answers are:
$I_n=I_{n+2}$
,
$\sum_{m=1}^{10} I_{2 m+1}=10 \pi$
,
$\sum_{m=1}^{10} I_{2 m}=0$

Solution:
(a)
$$
\begin{aligned}
& I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x \\
& I_n=\int_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^x\right) \sin x} d x \\
& {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\
& 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\sin x} d x
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 I_n=2 \int_0^\pi \frac{\sin n x}{\sin x} d x \\
& \Rightarrow \quad I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x \\
& \text { Now, } \quad I_{n+2}-I_n \\
& =\int_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x} d x \\
& =\int_0^\pi \frac{2 \cos (n+1) x \sin x}{\sin x} d x \\
& =2\left[\frac{\sin (n+1) x}{(n+1)}\right]_0^\pi=0 \\
& \Rightarrow \quad I_{n+2}=I_n \\
& \text { (b) As } I_3=I_5=\ldots=I_{21} \\
& \therefore \sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int_0^\pi \frac{\sin 3 x}{\sin x} d x \\
& =10 \int_0^\pi\left(3-4 \sin ^2 x\right) d x \\
& =10[3 x-2 x+\sin 2 x]_0^\pi=10 \pi \\
& \text { (c) As } \quad I_2=I_4=\ldots=I_{20} \\
& \therefore \quad \sum_{m=1}^{10} I_{2 m}=10 \int_0^\pi \frac{\sin 2 x}{\sin x} d x \\
& =20[\sin x]_0^\pi=0 \\
&
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2009 (Paper 2)
If $\lim _{x \rightarrow 0}\left[1+x \log \left(1+b^2\right)\right]^{1 / x}=2 b \sin ^2 \theta b>0$ and $\theta \in(-\pi, \pi]$, then the value of $\theta$ is

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

B
$\pm \frac{\pi}{3}$

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

D
$\pm \frac{\pi}{2}$

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

Solution:
Here, $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}$
$$
\begin{aligned}
& \text { Given, } \quad\left[1^{\infty} \text { from }\right] \\
& \Rightarrow \quad e^{\lim _{x \rightarrow 0}\left\{x \log \left(1+b^2\right)\right\} \cdot \frac{1}{x}} \\
& \Rightarrow \quad e^{\log \left(1+b^2\right)}=(1+b)^2 \\
& \lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta \\
& \Rightarrow \quad\left(1+b^2\right)=2 b \sin ^2 \theta \\
& \therefore \quad \sin ^2 \theta=\frac{1+b^2}{2 b} \quad \ldots \text { (ii) }
\end{aligned}
$$
By $A M \geq G M$,
$$
\frac{b+\frac{1}{b}}{2} \geq\left(b \cdot \frac{1}{b}\right)^{1 / 2} \Rightarrow \frac{b^2+1}{2 b} \geq 1
$$
From Eqs. (ii) and (iii), we get $\sin ^2 \theta=1$
$$
\Rightarrow \quad \theta=\pm \frac{\pi}{2} \text { as } \theta \in(-\pi, \pi]
$$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2011 (Paper 2)
If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$, then

Options:
A $a=1, b=4$
B $a=1, b=-4$
C $a=2, b=-3$
D $a=2, b=3$

The correct answer is:
$a=1, b=-4$

Solution:
Given: $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$

$\begin{array}{l}

\Rightarrow \lim _{x \rightarrow \infty} \frac{x^{2}+x+1-a x^{2}-a x-b x-b}{x+1}=4 \\

\Rightarrow \lim _{x \rightarrow \infty} \frac{(1-a) x^{2}+(1-a-b) x+(1-b)}{x+1}=4

\end{array}$

For this limit to be finite $1-a=0 \Rightarrow a=1$ then given limit reduces to

$\lim _{x \rightarrow \infty} \frac{-b x+(1-b)}{x+1}=4 \Rightarrow \lim _{x \rightarrow \infty} \frac{-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}=4$

$\Rightarrow-b=4$ or $b=-4, \therefore a=1, b=-4$

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2012 (Paper 1)
If $P$ is a $3 \times 3$ matrix such that $P^{T}=2 P+I$, where $P^{T}$ is the transpose of $P$ and $I$ is the $3 \times 3$ identity matrix, then there exists a column matrix $\quad X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right] \neq\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$ such that

Options:
A $P X=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
B $P X=X$
C $P X=2 X$
D $P X=-X$

The correct answer is:
$P X=-X$

Solution:
$P^{T}=2 P+I$

$\begin{array}{l}

\Rightarrow P=2 P^{T}+I \Rightarrow P=2(2 P+I)+I \\

\Rightarrow P=4 P+3 I \Rightarrow P+I=0 \\

\Rightarrow P X+X=0 \Rightarrow P X=-X

\end{array}$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2012 (Paper 2)
If $r, s, t$ are prime numbers and $p, q$ are the positive integers such that LCM of $p, q$ is $r^2 s^4 t^2$, then the number of ordered pairs $(p, q)$ is

Options:
A
252

B
254

C
225

D
224

The correct answer is:
225

Solution:
Since, $r, s, t$ are prime numbers.
$\therefore$ Selection of $p$ and $q$ are as under
$\therefore$ Total number of ways to select $s=9$.
Similarly, the number of ways to select $t=5$.
$\therefore$ Total number of ways $=5 \times 9 \times 5=225$.

Topic: Mathematics, Permutation Combination, JEE Main
If $r, s, t$ are prime numbers and $p, q$ are the positive integers such that LCM of $p, q$ is $r^2 s^4 t^2$, then the number of ordered pairs $(p, q)$ is

Options:
A
252

B
254

C
225

D
224

The correct answer is:
225

Solution:
Since, $r, s, t$ are prime numbers.
$\therefore$ Selection of $p$ and $q$ are as under
$\therefore$ Total number of ways to select $s=9$.
Similarly, the number of ways to select $t=5$.
$\therefore$ Total number of ways $=5 \times 9 \times 5=225$.

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2006
If the adjoint of a $3 \times 3$ matrix $P$ is $\left[\begin{array}{lll}1 & 4 & 4 \\ 2 & 1 & 7 \\ 1 & 1 & 3\end{array}\right]$, then the possible value(s) of the determinant of $P$ is (are)

Options:
A $-2$
B $-1$
C 1
D 2

The correct answers are:
$-2$, 2

Solution:
We know for a third order matrix $P$,

$|\operatorname{Adj} P|=|P|^{2}$

Where

$\begin{aligned}

&|\operatorname{Adj} P|=1(3-7)-4(6-7)+4(2-1)=4 \\

\therefore \quad &|P|^{2}=4 \mathrm{P}|P|=2 \text { or }-2

\end{aligned}$

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2012 (Paper 2)
If the angles $A, B$ and $C$ of a triangle are in an arithmetic progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively, then the value of the expression $\frac{a}{c} \sin 2 C+\frac{c}{a} \sin 2 A$ is

Options:
A
$\frac{1}{2}$

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

C
1

D
$\sqrt{3}$

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

Solution:
Since, $A, B, C$ are in $\mathrm{AP}$
$$
\begin{aligned}
& \Rightarrow \quad 2 B=A+C \text { ie, } \angle B=60^{\circ} \\
& \therefore \frac{a}{c}(2 \sin C \cos C)+\frac{c}{a}(2 \sin A \cos A) \\
& =2 k(a \cos C+c \cos A) \\
& \quad\left[\begin{array}{l}
\text { using, } \\
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\frac{1}{k}
\end{array}\right] \\
& =2 k(b) \\
& =2 \sin B
\end{aligned}
$$
$$
=\sqrt{3}
$$
$$
\text { [using, } b=a \cos C+c \cos A]
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2010 (Paper 1)
If the bond length of $\mathrm{CO}$ bond in carbon monoxide is $1.128 \mathrm{~A}^{\circ}$, then what is the value of $\mathrm{CO}$ bond length in $\mathrm{Fe}(\mathrm{CO})_5$ ?

Options:
A
$1.15 Å$

B
$1.128 Å$

C
$1.72 Å$

D
$1.118 Å$

The correct answers are:
$1.15 Å$

Solution:
The metal (Fe) makes the back bonding to $\mathrm{CO}$. The metal to ligand bonding creates a synergic effect which strengthens the bond $\mathrm{CO}$ and the metal ( $\mathrm{Fe}$ ). Which results in the contraction of $\mathrm{CO}$ bond length (Synergic means working together towards a common end. This is a bond formed by a ligand to metal ion which has a $\sigma$-bond from ligand to metal ion and the back donation of electron density from the metal to the $\pi^*$ orbitals on the ligand. Thus enhancing the acceptor strength of the p-orbital. Hence up to a point the effect of $\sigma$-bond formation strengthens the $\pi$-bonding or vice-versa. Due to this synergic effect bond order decreases.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2006
If the distance between the plane $A x-2 y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$, then $|d|$ is

Options:

The correct answer is:
6

Solution:
Equation of plane containing the given lines is $\left|\begin{array}{ccc}x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=0$
$$
\begin{array}{rr}
\Rightarrow & (x-1)(-1)-(y-2)(-2) \\
& +(z-3)(-1)=0 \\
\Rightarrow & -x+1+2 y-4-z+3=0 \\
\Rightarrow & -x+2 y-z=0 \quad \ldots(i)
\end{array}
$$
Given plane is
$$
x-2 y+z=d
$$
Eqs. (i) and (ii) are parallel.
Clearly, $\quad A=1$
Now, distance between plane
$$
\begin{aligned}
& =\left|\frac{d}{\sqrt{1+4+1}}\right|=\sqrt{6} \\
& \Rightarrow \quad|d|=6 \\
&
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2010 (Paper 1)
If the distance of the point $P(1,-2,1)$ from the plane $x+2 y-2 z=\alpha$, where $\alpha>0$, is 5 , then the foot of the perpendicular from $P$ to the plane is

Options:
A
$\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$

B
$\left(\frac{4}{3},-\frac{4}{3}, \frac{1}{3}\right)$

C
$\left(\frac{1}{3}, \frac{2}{3}, \frac{10}{3}\right)$

D
$\left(\frac{2}{3},-\frac{1}{3}, \frac{5}{2}\right)$

The correct answer is:
$\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)$

Solution:
$$
\text { Distance of point } P \text { from plane }=5
$$
$$
\begin{gathered}
\therefore=\left|\frac{1-4-2-\alpha}{3}\right| \\
\alpha=10
\end{gathered}
$$
Foot of perpendicular
$$
\begin{aligned}
\quad \frac{x-1}{1} & =\frac{y+2}{2}=\frac{z-1}{-2}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad x & =\frac{8}{3}, y=\frac{4}{3}, z=-\frac{7}{3}
\end{aligned}
$$
Thus, the foot of the perpendicular is
$$
A\left(\frac{8}{3}, \frac{4}{3},-\frac{7}{3}\right)
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2010 (Paper 2)
If the function $f(x)=x^3+e^{\frac{x}{2}}$ and $g(x)=f^{-1}(x)$, then the value of $g^{\prime}(1)$ is

Options:

The correct answer is:
2

Solution:
Given, $g\{f(x)\}=x$
$$
\begin{array}{lc}
\Rightarrow & g^{\prime}\{f(x)\} f^{\prime}(x)=1 \\
\text { If } & f(x)=1 \Rightarrow x=0, f(0)=1
\end{array}
$$
Substitute $x=0$ in Eq. (i), we get
$$
\begin{aligned}
& g^{\prime}(1)=\frac{1}{f^{\prime}(0)} \\
& \Rightarrow \quad g^{\prime}(1)=2 \\
& {\left[\because f^{\prime}(x)=3 x^2+\frac{1}{2} e^{x / 2} \Rightarrow f^{\prime}(0)=\frac{1}{2}\right]} \\
&
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Advanced, JEE Advanced 2009 (Paper 2)
If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that

Options:
A
linear momentum of the system does not change in time

B
kinetic energy of the system does not change in time

C
angular momentum of the system does not change in time

D
potential energy of the system does not change in time

The correct answers are:
linear momentum of the system does not change in time

Solution:
On a system of particles if,
$$
\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext} .}=0
$$
then $\quad \overrightarrow{\mathbf{P}}_{\text {system }}=$ constant
No other conclusions can be drawn.

Topic: Physics, Center of Mass Momentum and Collision, JEE Advanced, JEE Advanced 2009 (Paper 1)
If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, then the plane (s) containing these two lines is (are)

Options:
A $y+2 z=-1$
B $y+z=-1$
C $y-z=-1$
D $y-2 z=-1$

The correct answers are:
$y+z=-1$, $y-z=-1$

Solution:
Given that lines are coplanar.

$\therefore\left|\begin{array}{ccc}

x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\

a_{1} & b_{1} & c_{1} \\

a_{2} & b_{2} & c_{2}

\end{array}\right|=\left|\begin{array}{ccc}

2 & 0 & 0 \\

2 & k & 2 \\

5 & 2 & k

\end{array}\right|=0 \Rightarrow k=\pm 2$

For $k=2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}\right|=0 \Rightarrow y-z+1=0$

For $k=-2$, equation of the plane is given by $\left|\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}\right|=0 \Rightarrow y+z+1=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2012 (Paper 2)
If the sum of first $n$ terms of an $\mathrm{AP}$ is $\mathrm{cn}^2$, then the sum of squares of these $n$ terms is

Options:
A
$\frac{n\left(4 n^2-1\right) c^2}{6}$

B
$\frac{n\left(4 n^2+1\right) c^2}{3}$

C
$\frac{n\left(4 n^2-1\right) c^2}{3}$

D
$\frac{n\left(4 n^2+1\right) c^2}{6}$

The correct answer is:
$\frac{n\left(4 n^2-1\right) c^2}{3}$

Solution:
Let $S_n=c^2$
$$
\begin{aligned}
& S_{n-1}=c(n-1)^2=c n^2+c-2 c n \\
& \therefore T_n=2 c n-c \quad\left(\because T_n=S_n-S_{n-1}\right) \\
& T_n^2=(2 c n-c)^2=4 c^2 n^2+c^2-4 c^2 n \\
& \therefore \text { Sum }=\Sigma T_n^2=\frac{4 c^2 \cdot n(n+1)(2 n+1)}{6} \\
& =\frac{2 c^2 n(n+1)(2 n+1)+3 n c^2-6 c^2 n(n+1)}{3} \\
& =\frac{n c^2\left[4 n^2+6 n+2+3-6 n-6\right]}{3} \\
& =\frac{n c^2\left(4 n^2-1\right)}{3}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2009 (Paper 2)
If the wavelength of the $n^{\text {th }}$ line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element $(Z=11)$. Find the value of $n$.

Options:

The correct answer is:
24

Solution:
$n^{\text {th }}$ line of Lyman series means transition from $(n+1)$ th state to first state.
$$
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left[1-\frac{1}{(n+1)^2}\right] \\
& \text { de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)} \\
& \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r} \\
&
\end{aligned}
$$
or
Equating (i) and (ii), we get
$$
\left(\frac{n+1}{2 \pi r}\right)=R Z^2\left[\frac{n(n+2)}{(n+1)^2}\right]
$$
Now, as
$$
\begin{aligned}
& r \propto \frac{n^2}{Z} \\
& r=\frac{(n+1)^2}{11} r_0
\end{aligned}
$$
$$
\therefore \quad r=\frac{(n+1)^2}{11} r_0
$$
Substituting in equations (iii), we get
$$
\frac{11}{2 \pi r_0}=\frac{R(11)^2(n)(n+2)}{(n+1)}
$$
or
$$
(n+1)=\left(1.09 \times 10^7\right)(11)(2 \pi) \times\left(0.529 \times 10^{-10}\right)\left(n^2+2 n\right)
$$
Solving this equation we get, $n=24$

Topic: Physics, Atomic Physics, JEE Advanced, JEE Advanced 2006
If $w=\alpha+i \beta$, where $\beta \neq 0$ and $z \neq 1$ satisfies the condition that $\left(\frac{w-\bar{w} z}{1-z}\right)$ is purely real, then the set of values of $z$ is

Options:
A
$|z|=1$ and $z \neq 2$

B
$|z|=1$ and $z \neq 1$

C
$z=\bar{Z}$

D
None of these

The correct answer is:
$|z|=1$ and $z \neq 1$

Solution:
Let $z_1=\frac{w-\bar{w}_z}{1-z}$, be purely real.
$$
\begin{array}{rlrl}
\Rightarrow & z_1 & =\bar{z}_1 \\
& \therefore & \frac{w-\bar{w} z}{1-z} & =\frac{\bar{w}-w \bar{z}}{1-\bar{z}} \\
\Rightarrow & & w-w \bar{z}-\bar{w} z+\bar{w} z \cdot \bar{z} & =\bar{w}-z \bar{w}-w \bar{z}+w z \cdot \bar{z} \\
\Rightarrow & & (w-\bar{w})+(\bar{w}-w)|z|^2 & =0 \\
\Rightarrow & & (w-\bar{w})\left(1-|z|^2\right) & =0 \\
\Rightarrow & & |z|^2 & =1 \\
\Rightarrow & & |z| & =1 \text { and } z \neq 1 .
\end{array}
$$
$$
\text { [as, } w-\bar{w} \neq 0 \text {, since } \beta \neq 0 \text { ] }
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2006
If $x^2-10 a x-11 b=0$ have roots $c$ and $d$ and $x^2-10 c x-11 d=0$ have roots $a$ and $b$, then, find $a+b+c+d$.

Options:

The correct answer is:
1210

Solution:
Here, $a+b=10 c$
$$
c+d=10 a
$$
On subtracting Eq. (ii) from Eq. (i), we get, $\quad(a-c)+(b-d)=10(c-a)$ $\Rightarrow \quad(b-d)=11(c-a)$
Since, $c$ is the root of $x^2-10 a x+11 b=0$
$$
\Rightarrow \quad c^2-10 a c-11 b=0
$$
Similarly, $a$ is the root of
$$
\begin{aligned}
\Rightarrow & x^2-10 c x+11 d=0 \\
& a^2-10 c a-11 d=0
\end{aligned}
$$
On subtracting Eq. (v) from Eq. (iv), we get
$$
\begin{aligned}
& \left(c^2-a^2\right)=11(b-d) \\
& \therefore \quad(c+a)(c-a)=11 \times 11(c-a) \quad\{\text { from (i) \& (ii) } \\
& \Rightarrow \quad c+a=121 . \\
& \therefore \quad a+b+c+d=10 c+10 a \\
& =10(c+a)=1210 \\
&
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Advanced, JEE Advanced 2006