datavorous/entrance-exam-dataset huggingface dataset:
The line $y=m x+c$ is a normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$, if $c=$

Options:
A $-\left(2 a m+b m^2\right)$
B $\frac{\left(a^2+b^2\right) m}{\sqrt{a^2+b^2 m^2}}$
C $-\frac{\left(a^2-b^2\right) m}{\sqrt{a^2+b^2 m^2}}$
D $\frac{\left(a^2-b^2\right) m}{\sqrt{a^2+b^2}}$

The correct answer is:
$-\frac{\left(a^2-b^2\right) m}{\sqrt{a^2+b^2 m^2}}$

Solution:
$a x \sec \theta-b y \operatorname{cosec} \theta=a^2-b^2$
$\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2$
$m x-y+c=0$
$\frac{a}{m \cos \theta}=\frac{-b}{(\sin \theta)(-1)}=\frac{b^2-a^2}{c}$
$\cos \theta=\frac{a c}{m\left(b^2-a^2\right)}$
$\sin \theta =\frac{b c}{b^2-a^2}$
$\sin ^2 \theta+\cos ^2 \theta =1 .$
$\frac{b^2 c^2}{\left(b^2-a^2\right)^2}+\frac{a^2 c^2}{m^2\left(b^2-a^2\right)^2}=1$
$\Rightarrow \frac{m^2-b^2 c^2+a^2 c^2}{m^2\left(b^2-a^2\right)^2}=1$
$c^2\left(b^2 m^2+a^2\right)=m^2\left(b^2-a^2\right)^2$
$c \sqrt{m^2 b^2+a^2}= \pm m\left(b^2-a^2\right)$
$c=\frac{m\left(b^2-a^2\right)}{\sqrt{m^2 b^2+a^2}}$

Topic: Mathematics, Ellipse, JEE Main
The line $y=m x+c$ touches the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if

Options:
A $c^2=a^2 m^2+b^2$
B $c^2=a^2 m^2-b^2$
C $c^2=b^2 m^2-a^2$
D $a^2=b^2 m^2+c^2$

The correct answer is:
$c^2=a^2 m^2-b^2$

Solution:
$y=m x+c$ touches the curve $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if $c^2=a^2 m^2-b^2$.

Topic: Mathematics, Hyperbola, JEE Main
The line $y=m x+c$ touches the parabola $x^2=4 a y$, if

Options:
A $c=-a m$
B $c=-a / m$
C $c=-a m^2$
D $c=a / m^2$

The correct answer is:
$c=-a m^2$

Solution:
$x^2=4 a(m x+c) \Rightarrow x^2-4 a m x-4 a c=0$
It touches, then $B^2-4 A C=0$ (discriminant \(\mathrm{D}=0\)).
$\Rightarrow 16 a^2 m^2+16 a c \Rightarrow a c=-a^2 m^2 \Rightarrow c=-a m^2$

Topic: Mathematics, Parabola, JEE Main
The line $y=m x+c$ will be a normal to the circle with radius $r$ and centre at $(a, b)$, if

Options:
A $a=m b+c$
B $b=m a+c$
C $r=m a-b+c$
D $r=m a-b$

The correct answer is:
$b=m a+c$

Solution:
Normal passes through centre, therefore $b=m a+c$.

Topic: Mathematics, Circle, JEE Main
The line y=x+1 meets the ellipse x24+y22=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then 3r2 is equal to

Options:
A 20
B 12
C 11
D 8

The correct answer is:
20

Solution:
Plotting the diagram of given data we have,

Let P be the point -1+p2,p2

Since it lies on the ellipse x24+y22=1, we get

-1+p224+p222=1 ⇒3p28-p22-34=0

Here p1-p22=2232-4-2=89+8=809

We know, p1-p22=PQ2

⇒ PQ2=809

Now r=PQ2⇒9r2=9PQ24=94×809=20

Topic: Mathematics, Ellipse, JEE Main
The line y=x+1 meets the ellipse x24+y22=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then 3r2 is equal to

Options:
A 20
B 12
C 11
D 8

The correct answer is:
20

Solution:
Plotting the diagram of given data we have,

Let P be the point -1+p2,p2

Since it lies on the ellipse x24+y22=1, we get

-1+p224+p222=1 ⇒3p28-p22-34=0

Here p1-p22=2232-4-2=89+8=809

We know, p1-p22=PQ2

⇒ PQ2=809

Now r=PQ2⇒9r2=9PQ24=94×809=20

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The linear combination of atomic orbitals to form molecular orbitals takes place only when the combining atomic orbitals

A. have the same energy B. have the minimum overlap C. have same symmetry about the molecular axis D. have different symmetry about the molecular axis

Choose the most appropriate from the options given below:

Options:
A A, B, C only
B A and C only
C B, C, D only
D B and D only

The correct answer is:
A and C only

Solution:
The given conditions should be satisfied by atomic orbitals to form molecular orbitals:

(a) The atomic orbital with one electron (completely half filled) can take part in combination or overlapping.

(b) During the time of overlap, the combining orbitals must have opposite spins of electrons.

(c) The combining atomic orbitals must have the same or nearly the same energy. This means that in a homonuclear molecule, the 1s-atomic orbital of an atom can combine with the 1s-atomic orbital of another atom, and not with the 2s-orbital.

(d) The combining atomic orbitals must have proper orientations to ensure that the overlap is maximum.

(e) The extent of overlapping should be large.

The combining atoms should have the same symmetry around the molecular axis for proper combination, otherwise, the electron density will be sparse.

Hence option B is the answer.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The linear mass density of a thin rod AB of length L varies from A to B as λx=λ01+xL, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod :

Options:
A 512ML2
B 718ML2
C 25ML2
D 37ML2

The correct answer is:
718ML2

Solution:
dm=λ.dv

dI=dm.x2

=λ01+xL.x2dx

I=λ0∫01x2+x3Ldx

=λ0x33+x44L0L

=λ0L33+L34

I=7λ0L312

m=∫0Lλdx=3λ0L2

Im=7L212×32=7L218

⇒ I=718mL2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The linear shape of \(\mathrm{CO}_2\) is due to

Options:
A \(s p^3\)-hybridisation of carbon
B \(s p\)-hybridisation of carbon
C \(p \pi-d \pi\) bonding between carbon and oxygen
D \(s p^2\)-hybridisation of carbon

The correct answer is:
\(s p\)-hybridisation of carbon

Solution:
The geometry of \(\mathrm{CO}_2\) is linear. As there are no non-bonding pairs, the spape is not even distorted or bent and remain a perfect \(180^{\circ}\) line. This is due to the overlapping of \(s\) and \(p\).
Orbitals of both atoms to form \(s p\) hybrid and \(p \pi-p \pi\) bonding between carbon and oxygen. Hence, option (b) is correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 1)
The linear system of equations
$\left.\begin{array}{l}8 x-3 y-5 z=0 \\ 5 x-8 y+3 z=0 \\ 3 x+5 y-8 z=0\end{array}\right\}$ has

Options:
A only zero solution
B only finite number of non-zero solutions
C no non-zero solution
D infinitely many non-zero solutions

The correct answer is:
infinitely many non-zero solutions

Solution:
We have,
$$
\begin{aligned}
& 8x-3 y-5 z=0 \\
& 5 x-8 y+3 z=0 \\
& 3 x+5 y-8 z=0 \\
&D=\left|\begin{array}{ccc}
8 & -3 & -5 \\
5 & -8 & 3 \\
3 & 5 & -8
\end{array}\right| \\
&=8(64-15)+3(-40-9)-5(25+24) \\
=& 8 \times 49+3 \times(-49)-5 \times 49 \\
=& 0
\end{aligned}
$$
$\therefore$ The system has infinitely many non-zero solutions.

Topic: Mathematics, Determinants, JEE Main
The linear velocity of a rotating body is given by $\vec{v}=\vec{\omega} \times \vec{r}$, where $\vec{\omega}$ is the angular velocity and $\vec{r}$ is the radius vector. The angular velocity of a body is $\vec{\omega}=\hat{i}-2 \hat{j}+2 \hat{k} \quad$ and the radius vector $\vec{r}=4 \hat{j}-3 \hat{k}$, then $|\vec{v}|$ is

Options:
A $\sqrt{29}$ units
B $\sqrt{31}$ units
C $\sqrt{37}$ units
D $\sqrt{41}$ units

The correct answer is:
$\sqrt{29}$ units

Solution:
The correct option is $\mathrm{1} \sqrt{29}$ units
We know that, $\vec{v}=\vec{\omega} \times \vec{r}=(\hat{i}-2 \hat{j}+2 \hat{k}) \times(4 \hat{j}-3 \hat{k})$
$\begin{aligned} \vec{v} & =\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 0 & 4 & -3\end{array}\right| \\ \vec{v} & =-2 \hat{i}+3 \hat{j}+4 \hat{k}\end{aligned}$
Thus, the magnitude of $\vec{v}$ is, $|\vec{v}|=\sqrt{(-2)^2+3^2+4^2}=\sqrt{29}$ units
Hence option 1 is the correct answer

Topic: Physics, Mathematics in Physics, JEE Main
The lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-4}{-\mathrm{k}}$ and $\frac{\mathrm{x}-1}{\mathrm{k}}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{1}$ are coplanar if

Options:
A
$\mathrm{k}=3$ or $-2$

B
$\mathrm{k}=0$ or $-1$

C
$\mathrm{k}=1$ or $-1$

D
$\mathrm{k}=0$ or $-3$

The correct answer is:
$\mathrm{k}=0$ or $-3$

Solution:
$\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{l}_1 & \mathrm{~m}_1 & \mathrm{n}_1 \\ \mathrm{l}_2 & \mathrm{~m}_2 & \mathrm{n}_2\end{array}\right|=0$
$\left|\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -\mathrm{k} \\ \mathrm{k} & 2 & 1\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}0 & 0 & -1 \\ 2 & 1+\mathrm{k} & -\mathrm{k} \\ \mathrm{k}+2 & 1 & 1\end{array}\right|=0$
$\mathrm{k}^2+3 \mathrm{k}^2=0 \Rightarrow \mathrm{k}(\mathrm{k}+3)=0$ or $\mathrm{k}=0$ or $-3$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
The lines $2 \mathrm{x}-3 \mathrm{y}=5$ and $3 \mathrm{x}-4 \mathrm{y}=7$ are diameters of a circle having area as 154 sq. units. Then the equation of the circle is

Options:
A
$x^2+y^2-2 x+2 y=62$

B
$x^2+y^2+2 x-2 y=62$

C
$x^2+y^2+2 x-2 y=47$

D
$x^2+y^2-2 x+2 y=47$

The correct answer is:
$x^2+y^2-2 x+2 y=47$

Solution:
$\pi \mathrm{r}^2=154 \Rightarrow \mathrm{r}=7$
For centre on solving equation
$2 x-3 y=5 \quad \& \quad 3 x-4 y=7 \text { or } x=1, y=1 \text { centre }=(1,-1)$
Equation of circle, $(\mathrm{x}-1)^2+(\mathrm{y}+1)^2=7^2$
$x^2+y^2-2 x+2 y=47$

Topic: Mathematics, Circle, JEE Main, JEE Main 2003
The lines $2 x+3 y=6,2 x+3 y=8$ cut the $X$-axis at $A$ and $B$, respectively. A line $L$ drawn through the point $(2,2)$ meets the $X$-axis as $C$ in such a way that abscissae of $A, B$ and $C$ are in arithmetic progression. Then, the equation of the line $L$ is

Options:
A $2 x+3 y=10$
B $8 x+2 y=10$
C $2 x-3 y=10$
D $8 x-2 y=10$

The correct answer is:
$2 x+3 y=10$

Solution:
Given lines are
$2 x+3 y=6$
and
$2 x+3 y=8$
These lines meet the $X$-axis at $A(3,0)$ and $B(4,0)$.
The line $L$ passes through $(2,2)$ cuts $X$-axis at $C$, such that $x$-coordinate of $A, B, C$ are in ap i.e. $3,4, x_1$
$\begin{array}{ll}\Rightarrow & 2.4=3+x_1 \\ \Rightarrow & x_1=8-3=5\end{array}$
To coordinates of $C$ are $(5,0)$.
A line $L$ passing through the points $(2,2)$ and $(5,0)$, is
$\begin{aligned} & y-0 & =\frac{0-2}{5-2}(x-5) \\ \Rightarrow & y & =\frac{-2}{3}(x-5) \\ \Rightarrow & 3 y & =-2 x+10 \\ \Rightarrow & 2 x+3 y & =10\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The lines $2 x+y-1=0, a x+3 y-3=0$ and $3 x+2 y-2=0$ are concurrent for

Options:
A All a
B $a=4$ only
C $-1 \leq a \leq 3$
D $a\gt0$ only

The correct answer is:
All a

Solution:
Given lines are concurrent, if $\left|\begin{array}{lll}2 & 1 & -1 \\ a & 3 & -3 \\ 3 & 2 & -2\end{array}\right|=0$
This is true for all values of a because $C_2$ and $C_3$ are identical.

Topic: Mathematics, Straight Lines, JEE Main
The lines $15 x-18 y+1=0, \quad 12 x+10 y-3=0$ and $6 x+66 y-11=0$ are

Options:
A Parallel
B Perpendicular
C Concurrent
D None of these

The correct answer is:
Concurrent

Solution:
$3(12 x+10 y-3)-2(15 x-18 y+1)=6 x+66 y-11=0$
Hence the lines are concurrent.

Topic: Mathematics, Straight Lines, JEE Main
The lines $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ are perpendicular to each other, if

Options:
A $a_1 b_2-b_1 a_2=0$
B $a_1 a_2+b_1 b_2=0$
C $a_1^2 b_2+b_1^2 a_2=0$
D $a_1 b_1+a_2 b_2=0$

The correct answer is:
$a_1 b_1+a_2 b_2=0$

Solution:
$\ \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0$
The equations of the lines are
$a_1 x+b_1 y+c_1=0$
Therefore, the slope $m_1=\left(-\frac{a_1}{b_1}\right)$
and $a_2 x+b_2 y+c_2=0$
Therefore, the slope $m_2=\left(-\frac{a_2}{b_2}\right)$
If the lines are perpendicular to each other, then
$\begin{aligned}
& \mathrm{m}_1 \times \mathrm{m}_2=-1 \\
& \Rightarrow\left(-\frac{\mathrm{a}_1}{\mathrm{~b}_1}\right) \times\left(-\frac{\mathrm{a}_2}{\mathrm{~b}_2}\right)=-1 \\
& \Rightarrow \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2=0
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
The lines joining the origin to the points of intersection of the line 4x+3y=24 with the circle x-32+y-42=25 are

Options:
A coincident
B perpendicular
C equally inclined to x-axis
D None of these

The correct answer is:
perpendicular

Solution:
Let the line intersects the circle at points A & B

Nw, the given line AB is a diameter as it passes through centre 3,4.

O0,0 is a point on the circle as 9+16=25.

∴∠AOB=π2 as angle in a semi-circle is π2
∴OA is perpendicular to OB.

Topic: Mathematics, Circle, JEE Main
The lines $L_1: y-x=0$ and $L_2: 2 x+y=0$ intersect the line $L_3: y+2=0$ at $P$ and $Q$ respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersect $L_3$ at $R$.
This question has Statement $-1$ and Statement $-2$. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement-1 : The ratio $P R: R Q$ equals $2 \sqrt{2}: \sqrt{5}$.
Statement-2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.

Options:
A
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$

B
Statement $-1$ is true, Statement- 2 is false.

C
Statement $-1$ is false, Statement- 2 is true.

D
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

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

Solution:
$$
P(-2,-2) ; Q=(1,-2)
$$
Equation of angular bisector $\overline{\mathrm{OR}}$ is $(\sqrt{5}+2 \sqrt{2}) \mathrm{x}=(\sqrt{5}-\sqrt{2}) \mathrm{y}$
$$
\therefore P R: R Q=2 \sqrt{2}: \sqrt{5}
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2011
The lines $p\left(p^2+1\right) x-y+q=0$ and $\left(p^2+1\right)^2 x+\left(p^2+1\right) y+2 q=0$ are perpendicular to a common line for

Options:
A
no value of $p$

B
exactly one value of $p$

C
exactly two values of $p$

D
more than two values of $p$

The correct answer is:
exactly one value of $p$

Solution:
Lines must be parallel, therefore slopes are equal $\Rightarrow p\left(p^2+1\right)=-\left(p^2+1\right) \Rightarrow p=-1$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2009
The lines r→=i^-j^+l2i^+k^ and r→=2i^-j^+mi^+j^-k^

Options:
A Do not intersect for any values of l and m
B Intersect for all values of l and m
C Intersect when l=2 and m=12
D Intersect when l=1 and m=2

The correct answer is:
Do not intersect for any values of l and m

Solution:
Lines r→=i^-j^+l2i^+k^ and r→=2i^-j^+mi^+j^-k^

Clearly both lines do not have same direction means lines are not parallel.

⇒Lines are skew lines or intersecting lines.

Let lines are skew lines and distance between lines is d if distance between them comes out zero then they will be intersecting lines.

For lines r→=a→+λp→, r→=b→+λq→

d=b→-a→·p→×q→p→×q→

Applying here

d=2i^-j^-i^-j^·2i^+k^×i^+j^-k^2i^+k^×i^+j^-k^

d=i^·-i^+3j^+2k^-i^+3j^+2k^=-114=114

⇒Lines are skew lines means do not intersect.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The lines represented by the equation x2+23xy+3y2-3x-33y-4=0, are

Options:
A perpendicular to each other
B parallel
C inclined at 45o to each other
D None of these

The correct answer is:
parallel

Solution:
The given equation can be written as ( x+ 3 y ) 2 −3( x+ 3 y )−4=0
⇒x+ 3 y+1=0, x+ 3 y=4
Hence, given equation represents two parallel lines

Topic: Mathematics, Straight Lines, JEE Main
The lines x-12=y+12=z-14 and x-31=y-62=z1 intersect each other at point

Options:
A -2,-4,5
B -2,-4,-5
C 2,4,-5
D 2,-4,-5

The correct answer is:
-2,-4,-5

Solution:
A General point on L1 is x-12=y+12=z-14=r
âPr=1+2r, -1+2r, 1+4r
Substituting the coordinates of this point in the equation of L2
1+2r-31=-1+2r2=1+4r1
â 2r-2=4r+1 âr= -32
So the intersection point is (-2, -4, -5)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The lines x-22=y-2=z-716 and x+34=y+23=z+21 intersect at the point P. If the distance of P from the line x+12=y-13=z-11 is l, then 14l2 is equal to _____.

Options:

The correct answer is:
108

Solution:
Given: x-22=y-2=z-716 and x+34=y+23=z+21

⇒x-21=y-1=z-78=λ let and x+34=y+23=z+21=k let

⇒x=λ+2=4k-3, y=-λ=3k-2 and z=8λ+7=k-2

⇒λ=-5+4k, λ=-3k+2, 8λ-k=-9

⇒-5+4k=-3k+2

⇒7k=7

⇒k=1, λ=-31+2=-1

⇒x=1, y=1, z=-1

⇒P≡(1,1,-1)

Projection of PQ≡2i^-2k^ on QR line with direction ratio 2i^+3j^+k^ is given by,

QR=4-24+9+1=214

Now, in triangle PQR we get,

∴l2=8-414=10814

⇒14l2=108

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The lines x=ay-1=z-2 and x=3y-2=bz-2,(ab≠0) are coplanar, if:

Options:
A b=1,a∈R-{0}
B

a=1, b∈R-{0}

C a=2, b=2
D a=2, b=3

The correct answer is:
b=1,a∈R-{0}

Solution:
Given lines can be written as

x+1a=y-01=z-1a

x+23=y-01=z-03b

We know that lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

Hence,

-10-1a1a313b=0

where, b≠0⇒a≠0, since ab≠0.

⇒-3 b-a-1a-3=0

⇒a-3b-a+3=0

⇒b=1

∴b=1,a∈R-0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The lines $x-y-2=0, x+y-4=0$ and $x+3 y=6$ meet in the common point :

Options:
A $(1,2)$
B $(2,2)$
C $(3,1)$
D $(1,1)$

The correct answer is:
$(3,1)$

Solution:
The equation of lines are
$x-y-2=0$ ...(i)
$x+y-4=0$ ...(ii)
and $\quad x+3 y=6$ ...(iii)
On solving Eqs. (i) and (ii), we get
$x=3, y=1$
On putting these value in Eq. (iii)
$3+3=6$
which is satisfied. Thus the required point is $(3,1)$.

Topic: Mathematics, Straight Lines, JEE Main
The lines $y=2 x$ and $x=-2 y$ are

Options:
A Parallel
B Perpendicular
C Equally inclined to axes
D Coincident

The correct answer is:
Perpendicular

Solution:
Since
$m_1 m_2=(2)\left(-\frac{1}{2}\right)=-1$

Topic: Mathematics, Straight Lines, JEE Main
The linking of antibiotic resistance gene with the plasmid vector became possible with

Options:
A DNA ligase
B endonucleases
C DNA polymerase
D exonucleases

The correct answer is:
DNA ligase

Solution:
DNA ligase is used to seal the nicks that remain in recombinant DNA molecule. Infact DNA ligase joins together the neighbouring nucleotides flanking a discontinuity in a DNA strand by forming a phosphodiester bond. A recombinant DNA molecule is a vector into which the desired DNA fragment has been inserted to enable its cloning in an appropriate host.
Endonucleases are the enzymes that produce internal cuts called cleavage in DNA molecules. A class of endonucleases cleaves DNA only within or near those sites, which has specific base sequences, such endonucleases are known as restriction endonucleases.
Exonucleases are enzymes that remove one or more nucleotide from the free ends.
DNA polymerase is the enzyme, which cause polymerization of nucleotides during DNA replication.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2008 (Screening)
The linking of antibiotic resistance gene with the plasmid vector became possible with

Options:
A Exonucleases
B DNA ligase
C Endouncleases
D DNA polymerase

The correct answer is:
DNA ligase

Solution:
Restriction endonucleases act as molecular scissors; but DNA ligases help in linking foreign DNA to plasmid.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2008 (Mains)
The load versus elongation graph for four wires of the same materials shown in the figure. The thinnest wire is represented by the line

Options:
A $O C$
B $O D$
C $\mathrm{OA}$
D $O B$

The correct answer is:
$\mathrm{OA}$

Solution:
It is clear from the graph that, for same value of load $(F)$, elongations $\mathrm{q}, l_{2}, l_{3}, l_{4}$ are such that
$$
l_{4}>l_{3}>l_{2}>l_{1}
$$
So, OA represents the thinnest wire.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The load versus elongation graphs for four wires of same length and made of the same material are shown in the figure. The thinnest wire is represented by the line

Options:
A
$O A$

B
$O C$

C
$O D$

D
$O B$

The correct answer is:
$O A$

Solution:
From the graph, it is clear that for the same value of load, elongation is maximum for wire $O A$. Hence $O A$ is the thinnest wire among the four wires.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The load versus elongation graphs for four wires of same length and made of the same material are shown in the figure. The thinnest wire is represented by the line

Options:
A
$O A$

B
$O C$

C
$O D$

D
$O B$

The correct answer is:
$O A$

Solution:
From the graph, it is clear that for the same value of load, elongation is maximum for wire $O A$. Hence $O A$ is the thinnest wire among the four wires.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2012 (07 May Online)
The local maximum value of the function, fx=2xx2, x>0, is

Options:
A 1
B 4ee4
C (e)2e
D (2e)1e

The correct answer is:
(e)2e

Solution:
f'(x)=0 for maximum value
Let y=2xx2

lny=x2ln2x

1yy'=2xln2x+x212x×-2x2

y'=(xy)2ln2x-1

y'=2xx2x2ln2x-1

2ln2x=1

2x=e12

x=2e-12

Then maximum value will be
f2e-12=22e-124e-1=e2e-1=e2e

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The lock of a safe consists of five discs each of which features the digits 0,â1,â2,....,9. The safe can be opened by dialing a special combination of the digits. If the work day lasts 13 hours and to dial one combination of digits takes 5 seconds, then number of days sufficient enough to open the safe, are

Options:
A 9
B 10
C 11
D 12

The correct answer is:
11

Solution:
Total time required=(total number of dials required to sure open the lock)×5s
=105×5s
=50000060×60×13days=10.7 days
Hence, 11 days are enough to open the safe.

Topic: Mathematics, Permutation Combination, JEE Main
The locus of a foot of perpendicular drawn to the tangent of parabola $y^2=4 a x$ from focus, is

Options:
A $x=0$
B $y=0$
C $y^2=2 a(x+a)$
D $x^2+y^2(x+a)=0$

The correct answer is:
$x=0$

Solution:
Tangent to parabola is, $y=m x+\frac{a}{m} \ldots(i)$
A line perpendicular to tangent and passing from focus $(a, 0)$ is, $y=-\frac{x}{m}+\frac{a}{m} \ldots(ii)$
Solving both lines (i) and (ii) $\Rightarrow x=0$

Topic: Mathematics, Parabola, JEE Main
The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y=\alpha x+\beta$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

Options:
A A parabola
B A hyperbola
C An ellipse
D A circle

The correct answer is:
A hyperbola

Solution:
If $y=m x+c$ is tangent to the hyperbola then $c^2=a^2 m^2-b^2$. Here $\beta^2=a^2 \alpha^2-b^2$. Hence locus of $P(\alpha, \beta)$ is $a^2 x^2-y^2=b^2$, which is a hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y=\alpha x+\beta$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

Options:
A
an ellipse

B
a circle

C
a parabola

D
a hyperbola

The correct answer is:
a hyperbola

Solution:
Tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$y=m x \pm \sqrt{a^2 m^2-b^2}$
Given that $y=\alpha x+\beta$ is the tangent of hyperbola
$\Rightarrow m=\alpha$ and $a^2 m^2-b^2=\beta^2$
$\therefore \mathrm{a}^2 \alpha^2-\mathrm{b}^2=\beta^2$
Locus is $a^2 x^2-y^2=b^2$ which is hyperbola.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2005
The locus of a point Pα,β moving under the condition that the line y=αx+β is a tangent to the hyperbola x21-y2b2=1 is a conic, with eccentricity equal to

Options:
A 1
B 2
C 12
D 2

The correct answer is:
2

Solution:
The tangent to the hyperbola x21-y2b2=1 is y=mxÂ±m2-b2
Given that, y=Î±x+Î² is a tangent to the hyperbola.
So, m=Î± and m2-b2=Î²2
â´Î±2-b2=Î²2
The locus is x2-y2=b2, which is a hyperbola having eccentricity =2

Topic: Mathematics, Hyperbola, JEE Main
The locus of a point $P$ which divides the line joining $(1,0)$ and $(2 \cos \theta, 2 \sin \theta)$ internally in the ratio $2: 3$ for all $\theta$, is a

Options:
A Straight line
B Circle
C Pair of straight lines
D Parabola

The correct answer is:
Circle

Solution:
Let the coordinates of the point $P$ which divides the line joining $(1,0)$ and $(2 \cos \theta, 2 \sin \theta)$ in the ratio $2: 3$
be ${ }^{(h, k)}$. Then, $h=\frac{4 \cos \theta+3}{5}$ and $k=\frac{4 \sin \theta}{5}$
$\begin{aligned}
& \Rightarrow \cos \theta=\frac{5 h-3}{4} \text { and } \sin \theta=\frac{5 k}{4} \\
& \Rightarrow\left(\frac{5 h-3}{4}\right)^2+\left(\frac{5 k}{4}\right)^2=1 \Rightarrow(5 h-3)^2+\left(5 k^2\right)=16
\end{aligned}$
Therefore locus of $(h, k)$ is $(5 x-3)^2+(5 y)^2=16$, which is a circle

Topic: Mathematics, Straight Lines, JEE Main