datavorous/entrance-exam-dataset huggingface dataset:
If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \mathrm{~N}$, then the increase in length is $0.5 \mathrm{~mm}$. The ratio of the energy of the wire and the work done in displacing it through $1.5 \mathrm{~mm}$ by the weight is

Options:
A $\frac{1}{3}$
B $\frac{1}{4}$
C $\frac{1}{2}$
D 1

The correct answer is:
$\frac{1}{2}$

Solution:
Work done in stretching a wire
$W=\frac{1}{2} F l=\frac{1}{2} \times 10 \times 0.5 \times 10^{-3}=2.5 \times 10^{-3} \mathrm{~J}$
Work done to displace it through $1.5 \mathrm{~mm}$
$W=F \times l=5 \times 10^{-3} \mathrm{~J}$
The ratio of above two work $=1: 2$

Topic: Physics, Mechanical Properties of Solids, JEE Main
If one mole of the polyatomic gas is having two vibrational modes and \( \beta \) is the ratio of molar specific heats for polyatomic
gas \( \left(\beta=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{v}}}\right) \) then the value of \( \beta \) is :

Options:
A \( 1.02 \)
B \( 1.2 \)
C \( 1.25 \)
D 1. \( 35 \)

The correct answer is:
\( 1.2 \)

Solution:
f=4+3+3=10, assuming non-linear,

β=CpCv=1+2f=1210=1.2

Topic: Physics, Kinetic Theory of Gases, JEE Main
If one of the diameter of the circle, given by the equation $x^{2}+y^{2}+4 x+6 y-12=0,$ is a chord of a circle $S,$ whose centre is $(2,-3$ ), the radius of $S$ is

Options:
A $\sqrt{41}$ unit
B $3 \sqrt{5}$ unit
C $5 \sqrt{2}$ unit
D $2 \sqrt{5}$ unit

The correct answer is:
$\sqrt{41}$ unit

Solution:
Given, equation of circle is $x^{2}+y^{2}+4 x+6 y-12=0$
Centre $C(-2,-3)$ and radius $=\sqrt{(-2)^{2}+(-3)^{2}+12}=5$
$$
x^{2}+y^{2}+4 x+6 y-12=0
$$
$$
\begin{aligned}
C_{1} C_{2} &=\sqrt{\left(2+{2})^{2}+(-3+3)^{2}\right.} \\
&=\sqrt{(4)^{2}+(0)^{2}}=4
\end{aligned}
$$
$\therefore$ Radius of circle, $S$
$$
=\sqrt{(4)^{2}+({5})^{2}}=\sqrt{16+25}=\sqrt{41} \text { unit }
$$

Topic: Mathematics, Circle, JEE Main
If one of the diameters of the circle, given by the equation, x2+y2-4x+6y-12=0, is a chord of a circle S, whose centre is at -3, 2, then the radius of S is

Options:
A 5
B 10
C 52
D 53

The correct answer is:
53

Solution:
The centre of the given circle x2+y2-4x+6y-12=0 is 2, -3 and the radius is 5.

âââââââ

The distance between the centres 52 and radius is 5. The triangle OPQ is a right angled triangle
OQ= 522+52= 532=53
Hence answer is 53

Topic: Mathematics, Circle, JEE Main, JEE Main 2016 (03 Apr)
If one of the diameters of the circle x2+y2-22x -62y+14=0 is a chord of the circle x-222 +y-222=r2, then the value of r2 is equal to

Options:

The correct answer is:
10

Solution:
Given that, one of the diameter of the circle x-22+y-322=6 is a chord of another circle x-222+y-222=r2.

Let, C and C1 be the centres of the circles x-22+y-322=6 and x-222+y-222=r2 respectively.

So, C2, 32 & C122, 22

The position of the two circles as shown in figure.

Now, CC1=22-22+ 22-322=2+2=2

It is clear from the above diagram, ∆ACC1 is a right angled triangle.⇒r2=AC2+CC12=62+22 ⇒r2=10

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If one of the diameters of the circle x2+y2-2x-6y+6=0 is a chord of another circle C'', whose center is at 2,1, then its radius is_____.

Options:

The correct answer is:
3

Solution:
x2+y2-2x-6y+6=0
centre 1,3

radius =2
Distance between 1,3 and 2,1 is 5

∴ 52+22=r2

⇒ r=3

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If one of the diameters of the circle x2+y2-10x+4y+13=0 is a chord of another circle C, whose center is the point of intersection of the lines 2x+3y=12 and 3x-2y=5, then the radius of the circle C is

Options:
A 20
B 4
C 6
D 32

The correct answer is:
6

Solution:
Given,

S1≡x2+y2-10x+4y+13=0 with radius R=4 and its one diameter is chord of other circle S2 whose centre is given by intersection of 2x+3y=12 & 3x-2y=5 which is 3,2

Now, plotting the diagram we get,

Now, from above circle the distance between the centres is given by l=5-32+-2-22=20

Hence, radius of S2 will be, r2=202+42=20+16=36

⇒r=6

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (31 Jan Shift 1)
If one of the diameters of the circle \( x^{2}+y^{2}-2 x-6 y+6=0 \) is a chord of another circle \( { }^{\prime} C^{\prime} \), whose center is at \( (2,1) \), then its radius is

Options:

The correct answer is:
3

Solution:
x2+y2+2x-6y+6=0
center 1,3

radius =2
distance between 1,3 and 2,1 is 5

∴ 52+22=r2

⇒ r=3

Topic: Mathematics, Circle, JEE Main
If one of the lines given by $6 x^2-x y+4 c y^2=0$ is $3 x+4 y=0$, then $c$ equals

Options:
A
1

B
$-1$

C
3

D
$-3$

The correct answer is:
$-3$

Solution:
$m_1+m_2=\frac{1}{4 c}, m_1 m_2=\frac{6}{4 c}$ and $m_1=-\frac{3}{4}$ Hence c $=-3$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2004
If one of the lines of $m y^2+\left(1-m^2\right) x y-m x^2=0$ is a bisector of the angle between the lines $x y=0$, then $m$ is

Options:
A
$-1 / 2$

B
$-2$

C
$1$

D
$2$

The correct answer is:
$1$

Solution:
Equation of bisectors of lines $x y=0$ are $y=\pm x$
put $y=\pm x$ in $m y^2+\left(1-m^2\right) x y-m x^2=0$, we get $\left(1-m^2\right) x^2=0$
$\Rightarrow m=\pm 1$.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2007
If one of the roots of the equation 76x2-252x2-252x2-2537=0 is x=3, then the sum of all other five roots is

Options:
A 0
B -3
C -6
D -8

The correct answer is:
-3

Solution:
Let the roots of expansion of the given determinant is a,b,c,d,e and 3 (given)
∵ the given determinant represents a 6degree polynomial and contains even power of x

∴coefficient of x5=0
⇒3+ the sum of other roots =0
⇒ the sum of other roots =-3

Topic: Mathematics, Determinants, JEE Main
If one real root of the quadratic equation $81 x^{2}+k x+256=0$ is cube of the other root, then a value of $\mathrm{k}$ is :

Options:
A -81
B 100
C 144
D -300

The correct answer is:
-300

Solution:
Let $\alpha$ and $\beta$ be the roots of the equation, $81 x^{2}+k x+256=0$

Given $(\alpha)^{\frac{1}{3}}=\beta \Rightarrow \alpha=\beta^{3}$

$\because \quad$ Product of the roots $=\frac{256}{81}$

$\therefore \quad(\alpha)(\beta)=\frac{256}{81}$

$\Rightarrow \quad \beta^{4}=\left(\frac{4}{3}\right)^{4} \Rightarrow \beta=\frac{4}{3} \Rightarrow \alpha=\frac{64}{27}$

$\because$ Sum of the roots $=-\frac{k}{81}$

$\therefore \quad \alpha+\beta=-\frac{k}{81} \Rightarrow \frac{4}{3}+\frac{64}{27}=-\frac{k}{81}$

$\Rightarrow \quad k=-300$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If one root is greater than 2 and the other root is less than 2 for the equation x2-k+1x+k2+k-8=0, then the value of k lies between

Options:
A -2,2
B -2,4
C -2,0
D -2,3

The correct answer is:
-2,3

Solution:
Opening upward parabola, 2 lies between the roots.
⇒f2<0
4-k+12+k2+k-8<0
k2+k-8-2k-2+4<0
k2-k-6<0
k-3k+2<0⇒k∈-2,3

Topic: Mathematics, Quadratic Equation, JEE Main
If one root of the equation \((l-\mathrm{m}) \mathrm{x}^2+l \mathrm{x}+1=0\) is double of the other and if \(l\) is real then the greatest value of \(\mathrm{m}\) is \((l \neq \mathrm{m})\) :

Options:
A \(\frac{1}{3}\)
B \(\frac{8}{9}\)
C \(\frac{9}{8}\)
D 3

The correct answer is:
\(\frac{9}{8}\)

Solution:
Let the roots of the equation be \(\alpha\) and \(2 \alpha\), then
\(\alpha+2 \alpha=\frac{l}{\mathrm{~m}-l} \text { and } \alpha \cdot 2 \alpha=\frac{1}{l-\mathrm{m}}\)
\(\Rightarrow \alpha=\frac{l}{3(\mathrm{~m}-l)} \text { and } \alpha^2=\frac{1}{2(l-\mathrm{m})} \text {. }\)
Eliminating \(\alpha\), we get
\(\frac{l^2}{9(\mathrm{~m}-l)^2}=\frac{1}{2(l-\mathrm{m})} \Rightarrow 2 l^2-9 l+9 \mathrm{~m}=0\)
Since \(l\) is real, so \((-9)^2-4 \cdot 2 \cdot 9 \mathrm{~m} \geq 0\)
\(\Rightarrow \mathrm{m} \leq \frac{9}{8}\)

Topic: Mathematics, Quadratic Equation, JEE Main
If one root of the equation $x^2+p x+12=0$ is 4 , while the equation $x^2+p x+q=0$ has equal roots, then the value of ' $q$ ' is

Options:
A
$\frac{49}{4}$

B
4

C
3

D
12

The correct answer is:
$\frac{49}{4}$

Solution:
Since 4 is one of the root of $x^2+p x+12=0 \Rightarrow 16+4 p+12=0 \Rightarrow p=-7$ and equation $x^2+p x+q=0$ has equal roots
$$
\Rightarrow D=49-4 q=0 \Rightarrow q=\frac{49}{4}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2004
If one root of the equation x 2 +px+q=0 is the square of the other, then

Options:
A p 3 + q 2 −q( 3p+1 )=0
B p 3 + q 2 +q( 1+3p )=0
C p 3 + q 2 +q( 3p−1 )=0
D p 3 + q 2 +q( 1â3p )=0

The correct answer is:
p 3 + q 2 +q( 1−3p )=0

Solution:
x2+px+q=0
Let, α , α 2 are the roots of the given equation.
⇒α+α2=-p
and α · α 2 = q
⇒ α 3 = q
∵ α + α 2 = - p
⇒ α + α 2 3 = - p 3
⇒α6+α3+3α.α2α+α2=-p3
⇒ q 2 + q + 3 α 3 α + α 2 = - p 3
⇒ q 2 + q + 3 q - p = - p 3
⇒p3+q2+q-3pq=0
⇒ p 3 + q 2 + q 1 - 3 p = 0

Topic: Mathematics, Quadratic Equation, JEE Main
If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is:

Options:
A CATTAGCT
B TACGAAGT
C GTACTTAC
D ATGCGACT

The correct answer is:
TACGAAGT

Solution:
A DNA molecule consists of two long polynucleotide chains composed of four types of nucleotide subunits. Each of these chains is known as a DNA chain, or a DNA strand. Hydrogen bonds between the base portions of the nucleotides hold the two chains together.

Adenine base pairs with thymine with 2 hydrogen bonds and cytosine base pairs with guanine with 3 hydrogen bonds.

As we know that in DNA molecule , adenine (A) always pairs with thymine (T) and cytosine (C) always pairs with guanine (G).

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If one were to apply the Bohr model to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the nth level will be:

Options:
A nhqBπm
B nhqB4πm
C nhqB2πm
D nhqB8πm

The correct answer is:
nhqB4πm

Solution:
For a charge q moving in a uniform magnetic field B
fm=mv2R
qVB= mv2R
⇒ mv2=qVBR
⇒ 12 mv2=qVBR2
⇒ Energy=qVBR2 (1)
By Bohr's quantization condition,
Angular momentum L=nh2π
⇒ mvR=nh2π
⇒ vR=nh2π m (2)
Put (2) in (1)
⇒ Energy=qB2 nh2 π m
= qB nh4 π m

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2015 (10 Apr Online)
If only $2 \%$ of the main current is to be passed through a galvanometer of resistance $G$, then the resistance of shunt should be

Options:
A $G / 50$
B $G / 49$
C $50 G$
D $49 G$

The correct answer is:
$G / 49$

Solution:
For shunt resistance connected to galvanometer, current is given by
$I_{g}=\left(\frac{S}{S+G}\right) I$
$\frac{2}{100} I=\left(\frac{S}{S+G}\right) I$
$\begin{aligned} \Rightarrow & S+G &=50 S \\ \text { or } & S &=\frac{G}{49} \end{aligned}$

Topic: Physics, Current Electricity, COMEDK, COMEDK 2016
If only $2 \%$ of the main current is to be passed through a galvanometer of resistance $G$, then the resistance of shunt will be

Options:
A $\frac{G}{50}$
B $\frac{G}{49}$
C $50 \mathrm{G}$
D $49 G$

The correct answer is:
$\frac{G}{49}$

Solution:
$i_g=2 \%$ of $i=\frac{i}{50} \Rightarrow S=\frac{G}{(n-1)}=\frac{G}{(50-1)}=\frac{G}{49}$

Topic: Physics, Current Electricity, JEE Main
If only \(\frac{I^{\text {th }}}{51}\) of the main current is to be passed through a galvanometer then the shunt required is \(R_1\) and if only \(\frac{1^{\text {th }}}{11}\) of the main voltage is to be developed across the galvanometer, then the resistance required \(R_2\). Then \(\frac{R_2}{R_1}\)

Options:
A \(\frac{1}{500}\)
B \(\frac{50}{9}\)
C \(\frac{500}{3}\)
D 500

The correct answer is:
500

Solution:
According to the question,
Case I
If only \(\frac{1}{51}\) th of the main current \(i\) is to be passed through galvanometer \(G\) then the shunt required is main current, \(i=51\)
\(\begin{aligned}
& i_g=1 \\
& \therefore \quad R_1=\frac{G}{i-i_g} \Rightarrow R_1=\frac{G}{51-1}=\frac{G}{50} \quad \ldots .(i) \\
\end{aligned}\)
Case II
If only \(\frac{1}{11}\) th of the main voltage is developed across the \(G\) then the resistance required, \(R_2\).
\(\begin{aligned}
& R_2=G\left(V_G-1\right) \\
& R_2=G(11-1)=10 G \quad \ldots (ii)
\end{aligned}\)
Now, from Eqn. (i) and (ii), we get
\(\begin{aligned}
& \frac{R_2}{R_1}=\frac{10 G}{\frac{G}{50}}=500 \\
& \therefore \frac{R_2}{R_1}=500
\end{aligned}\)

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
If only the 4th term in the expansion of 2+3x810 has the greatest numerical value, then the integral values of x are

Options:
A -3,-2,2,3
B -2,-1,1,2
C -3,3
D -3,-2,-1,0,1,2,3

The correct answer is:
-3,3

Solution:
T3<T4>T5
âT4T3>1 and T5T4<1
Tr+1=Cr10210-r3x8r
T 4 T 3 = C 10 3 2 7 3x 8 3 C 10 2 2 8 3x 8 2 = x 2
T 5 T 4 = C 10 4 2 6 3x 8 4 C 10 3 2 7 3x 8 3 = 21 64 x
x2>1âx>2 or x<-2
2164x<1â-6421<x<6421
Taking intersection, we get,

xâ-6421,-2âª2,6421

Hence, integral values of x=-3,+3

Topic: Mathematics, Binomial Theorem, JEE Main
If osmotic pressure of a solution is $2 \mathrm{~atm}$ at $273 \mathrm{~K}$, then at $546 \mathrm{~K}$, the osmotic pressure is

Options:
A $0.5 \mathrm{~atm}$
B $1 \mathrm{~atm}$
C $2 \mathrm{~atm}$
D $4 \mathrm{~atm}$

The correct answer is:
$4 \mathrm{~atm}$

Solution:
$\pi \propto T$; if $T$ is doubled $\pi$ is also doubled.

Topic: Chemistry, Solutions, JEE Main
If \( \overrightarrow{\mathrm{m}}, \overrightarrow{\mathrm{n}} \) are non-parallel unit vectors and \( \overrightarrow{\mathrm{r}} \) is a vector which is perpendicular to \( \overrightarrow{\mathrm{m}} \) and \( \overrightarrow{\mathrm{n}} \) such that \( |\overrightarrow{\mathrm{r}}|=5 \) and \( |\overrightarrow{\mathrm{m}}+\overrightarrow{\mathrm{n}}|^{2}=2+4|\overrightarrow{\mathrm{m}} \times \overrightarrow{\mathrm{n}}| \), then the value of \( |[\overrightarrow{\mathrm{m}} \overrightarrow{\mathrm{n}} \overrightarrow{\mathrm{r}}]|^{2} \) is equal to

Options:
A \( 7 \)
B \( \frac{21}{5} \)
C \( 5 \)
D \( \frac{10}{3} \)

The correct answer is:
\( 5 \)

Solution:
r→ is parallel to m→ × n→
Let θ be angle between m→ and n→
∴ m→+n→2=2+4 m→×n→
1+1+2cos⁡θ=2+4×sin⁡θ
⇒tan⁡θ=12
m→ n→ r→2=r→.m→×n→2
= r→2m→2n→2sin2θ
=25×15=5

Topic: Mathematics, Vector Algebra, JEE Main
If $P=(0,1,0), Q=(0,0,1)$, then the projection of $P Q$ on the plane $x+y+z=3$ is

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

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

Solution:
Direction ratios of $P Q$
$$
=(0-0,0-1,1-0)=(0,-1,1)
$$
Direction cosine $=\left(|0|,\left|-\frac{1}{\sqrt{2}}\right|,\left|\frac{1}{\sqrt{2}}\right|\right)$
The given plane is $x+y+z=3$
Direction ratios of the plane are $(1,1,1)$ Length of the projection
$$
\begin{aligned}
& =0.1+1 \cdot \frac{1}{\sqrt{2}}+1 \cdot \frac{1}{\sqrt{2}} \\
& =\frac{2}{\sqrt{2}}=\sqrt{2}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If P=λ011 and Q=4031 such that P2=Q, then P3 is equal to

Options:
A 8071
B 10180
C 7081
D 6041

The correct answer is:
8071

Solution:
P2=Î»011Î»011=Î»20Î»+11=4031
So, Î»=2
Hence, P3=20114031=8071

Topic: Mathematics, Matrices, JEE Main
If P0 and PS are the vapour pressures of the solvent and solution respectively, n1 and n2 are the mole fractions of the solvent and solute respectively, then

Options:
A PS=P0n1
B PS=P0n2
C P0=PSn2
D PS=P0n1n2

The correct answer is:
PS=P0n1

Solution:
According Raoult’s law, P0-PSP0=nn+N; or
P0-PSP0=n2 (as given) or 1-PSP0=n2
Or PSP0=1-n2=n1 or PS=P0×n1

Topic: Chemistry, Solutions, JEE Main
If P=10121, then P50 is:

Options:
A 10251
B 15001
C 12501
D 10501

The correct answer is:
10251

Solution:
P=10121

P2=1012110121=1011

P3=101110121=10321

P4=1012110321=1021

⋮

∴ P50=1012+49121=10251

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right] Q=P P^{T}$, then the value of the
determinant of $Q$ is

Options:
A 2
B -2
C 1
D 0

The correct answer is:
2

Solution:
Given, $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right]$
$$
\begin{array}{l}
\therefore \quad Q=P P^{T}=\left[\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
2 & 3 \\
1 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 \times 1+2 \times 2+1 \times 1 & 1 \times 1+2 \times 3+1 \times 1 \\
1 \times 1+3 \times 2+1 \times 1 & 1 \times 1+3 \times 3+1 \times 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1+4+1 & 1+6+1 \\
1+6+1 & 1+9+1
\end{array}\right]=\left[\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right] \\
\therefore \text { The determintant of } Q=\left|\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right| \\
=66-64=2
\end{array}
$$

Topic: Mathematics, Matrices, JEE Main
If P11+2·P22+3·P33+…+15·P1515=Prq-s, 0≤s≤1, then Cr-sq+s is equal to

Options:

The correct answer is:
136

Solution:
We have,

P11+2·P22+3·P33+…+15·P1515=Prq-s

where, 0≤s≤1

Now,

1P1+2·2P2+3⋅3P3+…+15·15P15

=1!+2·2!+3·3!+……+15·15!

=(2!-1!)+(3!-2!)+(4!-3!)+……+(16!-15!)

∵n·n!=n+1-1n!=n+1!-n!

=16!−1!=Prq−s

⇒q=r=16 & s=1

So,

Cr-sq+s=C1517=17×162=136

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (26 Aug Shift 1)
If $P=(1,1), Q=(3,2)$ and $R$ is a point on $x$-axis then the value of $P R+R Q$ will be minimum at

Options:
A $\left(\frac{5}{3}, 0\right)$
B $\left(\frac{1}{3}, 0\right)$
C $(3,0)$
D $(1,0)$

The correct answer is:
$\left(\frac{5}{3}, 0\right)$

Solution:
Let co-ordinate of $R(x, 0)$
Given $P(1,1)$ and $Q(3,2)$
$\begin{aligned}P R+R Q & =\sqrt{(x-1)^2+(0-1)^2}+\sqrt{(x-3)^2+(0-2)^2} \\& =\sqrt{x^2-2 x+2}+\sqrt{x^2-6 x+13}\end{aligned}$
For minimum value of $P R+R Q, \quad \frac{d}{d x}(P R+R Q)=0$
$\begin{aligned}& \Rightarrow \frac{d}{d x}\left(\sqrt{x^2-2 x+2}\right)+\frac{d}{d x}\left(\sqrt{x^2-6 x+13}\right)=0 \\
& \Rightarrow \frac{(x-1)}{\sqrt{x^2-2 x+2}}=-\frac{(x-3)}{\sqrt{x^2-6 x+13}}\end{aligned}$
Squaring both sides, $\frac{(x-1)^2}{\left(x^2-2 x+2\right)}=\frac{(x-3)^2}{x^2-6 x+13}$
$\Rightarrow 3 x^2-2 x-5=0 \Rightarrow(3 x-5)(x+1)=0, \quad x=\frac{5}{3},-1$
Also $1 \lt x \lt 3 . \therefore \quad R=(5 / 3,0)$.

Topic: Mathematics, Application of Derivatives, JEE Main
If P1=1-w2+w24-w38+..........â and P2=1-w22 {where w is non-real root of the equation x3=1 }, then P1P2 is equal to

Options:

The correct answer is:
1

Solution:
P1=11+w2 ∵ sum of infinite G.P a+aα+aα2...=a1-αHere, α∈-1,1
=22+w+21+1+w=21-w2 ∵1+w+w2=0
Now, P1P2=21-w2×1-w22=1

Topic: Mathematics, Complex Number, JEE Main
If P=1α3133244 is the adjoint of a 3×3 matrix A and A=4, then α is equal to

Options:
A 5
B 0
C 4
D 11

The correct answer is:
11

Solution:
Given P=adjA
⇒P=adjA=A2 ∵ adjA=An-1 where n=3

Now, P=1α3133244=1×12−12−α×4−6+3×4−6=2α-6

⇒2α-6=16

⇒ α = 11

Topic: Mathematics, Matrices, JEE Main, JEE Main 2013 (07 Apr)
If $P_1$ and $P_2$ are two points on the ellipse $\frac{x^2}{4}+y^2=1$ at which the tangents are parallel to the chord joining the points $(0,1)$ and $(2,0)$, then the distance between $P_1$ and $P_2$ is

Options:
A
$2 \sqrt{2}$

B
$\sqrt{5}$

C
$2 \sqrt{3}$

D
$\sqrt{10}$

The correct answer is:
$\sqrt{10}$

Solution:
Any tangent on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by
$$
y=m x \pm \sqrt{a^2 m^2+b^2}
$$
Here $a=2, b=1$
$$
\begin{aligned}
& m=\frac{1-0}{0-2}=-\frac{1}{2} \\
& c=\sqrt{4\left(-\frac{1}{2}\right)^2+1^2}=\sqrt{2}
\end{aligned}
$$
So, $y=-\frac{1}{2} x \pm \sqrt{2}$
For ellipse : $\frac{x^2}{4}+\frac{y^2}{1}=1$
We put $y=-\frac{1}{2} x+\sqrt{2}$
$$
\begin{aligned}
& \therefore \frac{x^2}{4}+\left(-\frac{x}{2}+\sqrt{2}\right)^2=1 \\
& \frac{x^2}{4}+\left(\frac{x^2}{4}-2\left(\frac{x}{2}\right) \sqrt{2}+2\right)=1 \\
& \Rightarrow x^2+2 \sqrt{2} x+2=0 \\
& \text { or } x^2-2 \sqrt{2} x+2=0 \\
& \Rightarrow x=\sqrt{2} \text { or }-\sqrt{2}
\end{aligned}
$$
If $x=\sqrt{2}, y=\frac{1}{\sqrt{2}}$ and $x=-\sqrt{2}, y=-\frac{1}{\sqrt{2}}$
$\therefore$ Points are $\left(\sqrt{2}, \frac{1}{\sqrt{2}}\right),\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$
$$
\left.\therefore P_1 P_2=\sqrt{\left\{\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)\right\}^2+\left\{\sqrt{2}-(-\sqrt{2}\}^2\right.}\right)
$$
$$
=\sqrt{\left(\frac{2}{\sqrt{2}}\right)^2+\left(2 \sqrt{2}^2\right.}=\sqrt{2+8}=\sqrt{10}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2012 (12 May Online)
If $P_{1}$ and $P_{2}$ be the length of perpendiculars from the origin upon the straight lines $x \sec \theta+y \operatorname{cosec} \theta$ $=a$ and $x \cos \theta-y \sin \theta=a \cos 2 \theta$ respectively,

then the value of $4 \mathrm{P}_{1}^{2}+\mathrm{P}_{2}^{2}$

Options:
A $a^{2}$
B $2 a^{2}$
C $a^{2} / 2$
D $3 a^{2}$

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

Solution:
We have $P_{1}=$ length of perpendicular from (0,0) on $x \sec \theta+y \operatorname{cosec} \theta=a$

$\begin{aligned} \text { i.e. } P_{1} &=\left|\frac{a}{\sqrt{\sec ^{2} \theta+\cos e c^{2} \theta}}\right|=|a \sin \theta \cos \theta| \\ &=\left|\frac{a}{2} \sin 2 \theta\right| \text { or } 2 P_{1}=|a \sin 2 \theta| \end{aligned}$

$\mathrm{P}_{2}=$ Length of the perpendicular from (0,0) on $x \cos \theta-y \sin \theta=a \cos 2 \theta$

$P_{2}=\left|\frac{a \cos 2 \theta}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\right|=|a \cos 2 \theta|$

Now, $4 P_{1}^{2}+P_{2}^{2}=a^{2} \sin ^{2} 2 \theta+a^{2} \cos ^{2} 2 \theta=a^{2}$

Topic: Mathematics, Straight Lines, JEE Main
If $P=\left(\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right),$ then $P^{5}$ is equal to

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

The correct answer is:
$P$

Solution:
Given, $P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)$
$\begin{aligned} P^{2} &=P \cdot P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right) \\ &=\left(\begin{array}{ccc}4+2-4 & -4-6+8 & -8 & -8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9\end{array}\right) \\ &=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)=P \\ \therefore P^{4}=P^{2}=P \\ \Rightarrow P^{5}=P^{2}=P \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
If $\mathrm{P}=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta$ then

Options:
A $\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$
B $\mathrm{P} \geq \frac{1}{2}$
C $\quad 2 \leq \mathrm{P} \leq 3$
D $-\frac{\sqrt{13}}{6} \leq \mathrm{P} \leq \frac{\sqrt{13}}{6}$

The correct answer is:
$\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$

Solution:
Hints: $P=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta=\frac{1}{2} \sin ^2 \theta+\frac{1}{3}\left(1-\sin ^2 \theta\right)=\frac{1}{3}+\frac{1}{6} \sin ^2 \theta$
$$
\begin{aligned}
& 0 \leq \sin ^2 \theta \leq 1 \Rightarrow \frac{1}{3} \leq \frac{1}{3}+\frac{1}{6} \sin ^2 \theta \leq \frac{1}{3}+\frac{1}{6} \\
& \Rightarrow \frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Main
If P=3212-1232, A=1101 and Q=PAPT, then PT Q2015 P is :

Options:
A 0201500
B 2015012015
C 1201501
D 2015102015

The correct answer is:
1201501

Solution:
P=3212-1232 PT=32-121232
PPT=PTP=I
Q2015=PAPT PAPT ......(2015 terms)
=PA2015 PT ...i
∴ PT Q2015 P=PTP A2015PTP=A2015 from..i
Given A=1101
⇒ A2=1101 1101=1201
⇒ A3=1201 1101=1301
∴ A2015=1201501

Topic: Mathematics, Matrices, JEE Main
If P=3212-1232, A=1101 and Q=PAPT, then PT Q2015 P is :

Options:
A 0201500
B 2015012015
C 1201501
D 2015102015

The correct answer is:
1201501

Solution:
P=3212-1232 PT=32-121232
PPT=PTP=I
Q2015=PAPT PAPT ......(2015 terms)
=PA2015 PT ...i
∴ PT Q2015 P=PTP A2015PTP=A2015 from..i
Given A=1101
⇒ A2=1101 1101=1201
⇒ A3=1201 1101=1301
∴ A2015=1201501

Topic: Mathematics, Matrices, JEE Main, JEE Main 2016 (09 Apr Online)
If $\mathrm{P}(3,2,6), \mathrm{Q}(1,4,5)$ and $\mathrm{R}(3,5,3)$ are the vertices of $\Delta \mathrm{PQR}$, then $\mathrm{m} \angle \mathrm{PQR}$ is

Options:
A $90^{\circ}$
B $50^{\circ}$
C $70^{\circ}$
D $30^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
We have $P \equiv(3,2,6) ; Q \equiv(1,4,5)$ and $R \equiv(3,5,3)$
d.r. of $\mathrm{PQ}$ are $-2,2,-1$ and d.r. of $\mathrm{QR}$ are $2,1,-2$
We find that : $(-2)(2)+2(1)+(-1)(-2)=-4+2+2=0$
$\therefore \mathrm{PQ} \perp \mathrm{QR} \Rightarrow \mathrm{m} \angle \mathrm{PQR}=90^{\circ}$

Topic: Mathematics, Vector Algebra, JEE Main