datavorous/entrance-exam-dataset huggingface dataset:
The point P(2,1) is shifted through a distance of 32 units measured perpendicular to the line x-y=1 in the direction of decreasing ordinates, to reach at Q. The image of Q with respect to te line y+x=1 is

Options:
A 3,-4
B -3,2
C 0,-1
D 5,-2

The correct answer is:
3,-4

Solution:
PQ=32
⇒xQ=2+32⋅12=5
and yQ=1-32⋅12=-2
⇒Q≡5,-2
Distance of Q from the line x+y=1 is
5-2-12=2⇒QR=22
⇒xR=5-22⋅12=3 and yR=-2-22⋅12=-4
Thus, R≡3,-4

Topic: Mathematics, Straight Lines, JEE Main
The point P(-26,3) lies on the hyperbola x2a2-y2 b2=1 having eccentricity 52. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points Q and R respectively, then QR is equal to:

Options:
A 43
B 6
C 36
D 63

The correct answer is:
63

Solution:
As point P(-26,3) lies on hyperbola.
24a2-3b2=1 ...1

Given, eccentricity e=52

⇒e=a2+b2a2=52

⇒4a2+4b2=5a2
⇒a2=4b2 ...2

On solving equations 1 and 2, we get
⇒6b2-3b2=1
⇒b2=3 & a2=12
Hyperbola x212-y23=1
2x12-2y·y'3=0

y'|(-26,3)=-12

Equation of tangent is (y-3)=-12(x+26)
at x=0, y=-3, Q(0,-3)
Equation of normal is (y-3)=2(x+26)
at x=0, y=53
R(0,53)

Apply distance formula,
QR=63

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The point Pa,b undergoes the following three transformations successively:
a reflection about the line y=x.
b translation through 2 units along the positive direction of x- axis.
c rotation through angle π4 about the origin in the anti-clockwise direction.

If the co-ordinates of the final position of the point P are -12,72, then the value of 2a+b is equal to:

Options:
A 13
B 9
C 5
D 7

The correct answer is:
9

Solution:
Image of Aa,b along y=x is P'b,a.

Translating it 2 units along positive direction of x-axis it becomes P"b+2,a.

Now, applying Rotation theorem

-12+72i=b+2+aicosπ4+isinπ4

Multiplying the complex numbers we get,

-12+72i=b+22-a2+ib+22+a2

⇒b-a+2=-1 ...1

and b+2+a=7 ...2

⇒a=4;b=1

⇒2a+b=9

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (27 Jul Shift 2)
The point represented by 2+i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 22 units in the south-west wards direction. Then its new position in the Argand plane is at the point represented by :

Options:
A 1+i
B 2+2i
C -2-2i
D -1-i

The correct answer is:
1+i

Solution:
Let P2+i & S(z)
By Rotation theorem, rotating about 3+3i, we get
z-3+3i3+i-3+3i=222e-iπ4
⇒ z-3-3i-2i=1-i
⇒ z-3-3i=-2i-2⇒ z=1+i

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2016 (09 Apr Online)
The point which has the same power with respect to each of the circles x2+y2-8x+40=0, x2+y2-5x+16=0 and x2+y2-8x+16y+160=0 is

Options:
A −8,−152
B 8,−152
C 8,152
D −8,152

The correct answer is:
8,−152

Solution:
Given circle equations

x2+y2-8x+40=0---(I)

x2+y2-5x+16=0----(II)

x2+y2-8x+16y+160=0----III

Power of point with respect to circle =S1

For point P8, -152

With respect to equation I

=64+2254-64+40=3854

With respect to equation II

= 64+2254-40+16=3854

With respect to third circle

=3854

The remaining points in other options are not having same power.

So required point is 8, -152.

Topic: Mathematics, Circle, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
The point, which is at the shortest distance from the line x+y=7 and lying on an ellipse x2+2y2=6, has coordinates ( a,b ) then the value of a b is

Options:

The correct answer is:
2

Solution:
Equation of the ellipe is x 2 6 + y 2 3 =1
The tangent at the point of shortest distance from the line x+y=7 should be parallel to the given line
Any point on the given ellipse is (6cosâ¡Î¸, 3sinâ¡Î¸)
Equation of the tangent is
xcosâ¡Î¸6+ysinâ¡Î¸3=1. It is parallel to x+y=7
âµ Equation of the tangent at P( acosÎ¸,bsinÎ¸ ) on ellipse is xcosÎ¸ a + ysinÎ¸ b =1
âcosâ¡Î¸6=sinâ¡Î¸3
âcosâ¡Î¸2=sinâ¡Î¸1 âsinÎ¸= 1 3 ,cosÎ¸= 2 3
The required point is (2, 1)
Hence, ( a,b )=( 2,1 )
therefore, a b =2 .

Topic: Mathematics, Ellipse, JEE Main
The points 0,83,1,3 and 82,30

Options:
A form an obtuse angled triangle
B form an acute angled triangle
C lie on a straight line
D form a right angled triangle

The correct answer is:
lie on a straight line

Solution:
The points, A0,83,B1,3,C82,30 are collinear, as slope of AB=slope of BC, i.e.,

3-831-0= 30-382-1=13

Hence, option (Lie on a straight line) is correct.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2015 (10 Apr Online)
The points -2,-1,1,0,4,3 and 1,2 are

Options:
A collinear
B concyclic
C the vertices of a parallelogram
D the vertices of a rectangle

The correct answer is:
the vertices of a parallelogram

Solution:
Midpoint of -2,-1 & 4,3 is 1,1
Midpoint of 1,0 & 1,2 is 1,1
Hence, -2,-1,1,0,4,3 and 1,2 are the vertices of a parallelogram.
Now, the distance between -2,-1 and 4,3 is 52 and the distance between 1,0 & 1,2 is 2
Because 52â 2, hence, points do not form a rectangle

Topic: Mathematics, Straight Lines, JEE Main
The points \( (3,-2,-1),(-1,1,2),(2,3,-4) \) and \( (4,5, \lambda) \) are coplanar when \( \lambda= \)

Options:
A \( 0 \)
B \( \frac{-146}{17} \)
C \( 1 \)
D \( \frac{-17}{9} \)

The correct answer is:
\( \frac{-146}{17} \)

Solution:
The equation of a plane passing through three points A(x1, y1, z1), B(x2, y2, z2) and C(x3, y3, z3) can be obtained using:

x-x1y-y1z-z1x2-x1y2-y1z2-z1x3-x1y3-y1z3-z1=0

The points given in the question are (3,-2,-1), (-1, 1, 2) and (2, 3,-4). So, the equation passing through these can be found as

x-3y+2z+1-1-31+22+12-33+2-4+1=0x-3y+2z+1-433-15-3=0x-3-9-15-y+212+3+z+1-20+3=0-24x+72-15y-30-17z-17=024x+15y+17z=25

The given points are coplanar if the fourth point (4, 5, λ) satisfies the above equation, so the value of λ can be found as

(24×4)+(15×5)+17λ=2596+75+17λ=2517λ=-146λ=-14617

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The points A3,6 and B lie on the parabola y2=4ax, such that the chord AB subtends 90Â° at the origin, then the length of the chord AB is equal to

Options:
A 1513 units
B 1217 units
C 917 units
D 910 units

The correct answer is:
1513 units

Solution:
Since the chord subtends a right angle at the origin,

∴t1t2=-4

Now, the point A lies on the parabola,

∴36=4a3⇒a=3
i.e. A3,6=3t12,6t1
⇒t1=1⇒t2=-4
So, B≡48,-24
Hence, the length of the chord AB=1513 units

Topic: Mathematics, Parabola, JEE Main
The points at which crossing over has taken place between homologous chromosomes are called

Options:
A Protein axis
B Synaptonemal complexes
C Chiasmata
D Centromeres

The correct answer is:
Chiasmata

Solution:
The separation of the homologous chromosomes is not completed in deplotene stage. They remain attached at one or more points where crossing over has occurred. These points of attachment are called as chiasmata.

Topic: Biology, Cell Cycle and Cell Division, NEET
The points at which the function $f(x)=\frac{x+1}{x^2+x-12}$ is discontinuous, are

Options:
A $-3,4$
B $3,-4$
C $-1,-3,4$
D $-1,3,4$

The correct answer is:
$3,-4$

Solution:
$f(x)=\frac{x+1}{(x-3)(x+4)}$. Hence the points are $3,-4$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The points in the set $\left\{z \in C: \arg \left(\frac{z-2}{z-6 i}\right)=\frac{\pi}{2}\right\}$ (where $C$ denotes the set of all complex numbers) lie on the curve which is a

Options:
A circle
B pair of lines
C parabola
D hyperbola

The correct answer is:
circle

Solution:
Given that, $\arg \left(\frac{z-2}{z-6 i}\right)=\frac{\pi}{2}$
$$
\therefore \quad \arg (z-2)-\arg (z-6 i)=\frac{\pi}{2}
$$
Let $z=x+i y$
$$
\begin{aligned}
& \Rightarrow \arg [(x-2)+i y]-\arg [x+i(y-6)]=\frac{\pi}{2} \\
& \Rightarrow \quad \tan ^{-1} \frac{y}{x-2}-\tan ^{-1} \frac{y-6}{x}=\frac{\pi}{2} \\
& \Rightarrow \quad\left(\frac{\frac{y}{x-2}-\frac{y-6}{x}}{1+\frac{y}{x-2} \cdot \frac{y-6}{x}}\right)=\tan \frac{\pi}{2} \\
& \Rightarrow \quad 1+\frac{y}{x-2} \cdot \frac{y-6}{x}=0 \\
& \Rightarrow \quad x(x-2)+y(y-6)=0
\end{aligned}
$$
This is an equation of circle in diametric form.

Topic: Mathematics, Complex Number, JEE Main
The points of discontinuity of the function $f(x)=\frac{1}{x-1}$ if $0 \leq x \leq 2$ $=\frac{x+5}{x+3} \quad$ if $\quad 2 < x \leq 4$ in its domain are

Options:
A x=2 only
B x=1, x=2
C x=4 only
D x=0, x=2

The correct answer is:
x=1, x=2

Solution:
(B)
$\begin{aligned}
f(x) &=\frac{1}{x-1}, \text { if } 0 \leq x \leq 2 \Rightarrow f(x) \text { is not defined at } x=1 \\
&=\frac{x+5}{x+3}, \text { if } 2 < x \leq 4
\end{aligned}$
$\lim _{x \rightarrow 2^{-}} f(x)=\frac{1}{2-1}=1 \text { and } \lim _{x \rightarrow 2^{+}} f(x)=\frac{7}{5}$
Thus $f(x)$ is not continuous at $x=2$

Topic: Mathematics, Continuity and Differentiability, MHT CET, MHT CET 2020 (16 Oct Shift 1)
The points of intersection of the line $4 x-3 y-10=0$ and the circle $x^2+y^2-2 x+4 y-20=0$ are

Options:
A $(-2,-6),(4,2)$
B $(2,6),(-4,-2)$
C $(-2,6),(-4,2)$
D None of these

The correct answer is:
$(-2,-6),(4,2)$

Solution:
$\text { Substituting } x=\frac{3 y+10}{4}$ in equation of circle, we get a quadratic in $y$. Solving, we get two values of $y$ as 2 and -6 from which we get value of $x$.

Topic: Mathematics, Circle, JEE Main
The points on the curve at which the gradient is zero are

Options:
A $(0,2),(2,16)$
B $(0,-2),(2,-16)$
C $(2,-16),(-2,16)$
D $(2,16),(-2,-16)$

The correct answer is:
$(2,16),(-2,-16)$

Solution:
$\frac{d y}{d x}=12-3 x^2=0 \Rightarrow x= \pm 2$
Hence points are $(2,16)$ and $(-2,-16)$.

Topic: Mathematics, Application of Derivatives, JEE Main
The points on the curve fx=x1-x2, where the tangent to it has slope equal to unity, are x1,y1,x2,y2 and x3,y3. Then, x1+x2+x3 is equal to

Options:
A 0
B 32
C 34
D 3

The correct answer is:
0

Solution:
f ′ x = 1− x 2 1−x −2x 1− x 2 2
⇒1+x2=1+x4-2x2
⇒x4-3x2=0
⇒x=0,±3
⇒x1+x2+x3=0

Topic: Mathematics, Application of Derivatives, JEE Main
The points on the curve $y^2=x+\sin x$ at which the normal is parallel to the Y-axis lie on

Options:
A a line parallel to $Y$-axis
B a circle with centre at origin
C a parabola
D a pair of lines bisecting the angle between the coordinate axes

The correct answer is:
a parabola

Solution:
The given curve is $y^2=x+\sin x$ ...(i)
$\Rightarrow \frac{d y}{d x}=\frac{1+\cos x}{2 y}$
Now, slope of the normal is
$m=-\frac{1}{\left(\frac{d y}{d x}\right)}=-\frac{2 y}{1+\cos x}$
$\therefore \quad$ Normal is parallel to $y$-axis
$\therefore \quad m=\frac{1}{0}$
$\begin{aligned} & \Rightarrow-\frac{2 y}{1+\cos x}=\frac{1}{0} \Rightarrow 1+\cos x=0 \\ & \Rightarrow \cos x=-1 \Rightarrow \sin x=\sqrt{1-(-1)^2}=\sqrt{1-1}=0\end{aligned}$
$\Rightarrow \sin x=0$ ...(ii)
From eqns. (i) and (ii) :
$y^2=x+0 \Rightarrow y^2=x$ which is a parabola.

Topic: Mathematics, Application of Derivatives, JEE Main
The points on the curve y=x2 which are closest to the point P0,1 are

Options:
A ±2,2
B ±12,14
C ±12,12
D ±14,116

The correct answer is:
±12,12

Solution:
Let, the general point on the curve be Qα,α2
∴ PQ2=α2+α2-12=D(say)
⇒D=α4-α2+1
⇒dDdα=4α3-2α
For PQ to be minimum, α2=12⇒α=±12
∴ Q is ±12,12

Topic: Mathematics, Application of Derivatives, JEE Main
The points on the parabola $y^2=36 x$ whose ordinate is three times the abscissa are

Options:
A $(0,0),(4,12)$
B $(1,3),(4,12)$
C $(4,12)$
D None of these

The correct answer is:
$(0,0),(4,12)$

Solution:
$(4,12)$ (d) None of these $y_1=3 x_1$. According to given condition $9 x_1^2=36 x_1$
$\Rightarrow x_1=4,0 \Rightarrow y_1=12,0$
Hence the points are $(0,0)$ and $(4,12)$.

Topic: Mathematics, Parabola, JEE Main
The Poisson's ratio of a material is $0.1$. If the longitudinal strain of a rod of this material is $10^{-3}$, then the percentage change in the volume of the rod will be

Options:
A $0.008 \%$
B $0.08 \%$
C $0.8 \%$
D $8 \%$

The correct answer is:
$0.08 \%$

Solution:
If $\alpha$ and $\beta$ be the longitudinal strain and lateral strain respectively, then
$\alpha=\frac{l_{2}-l_{1}}{l_{1}}=10^{-3}=0.001$
$\Rightarrow \quad \frac{l_{2}}{l_{1}}=1.001$
Poisson's ratio, $\sigma=\frac{\text { Lateral strain }}{\text { Longitudinal strain }}$
$=\frac{\beta}{\alpha}$
$\begin{array}{ll}\Rightarrow & \beta=\sigma \alpha=0 \\ \text { But, } & \beta=\frac{r_{1}-r_{2}}{r_{1}}\end{array}$
$\therefore \quad \frac{r_{1}-r_{2}}{n_{1}}=10^{-4}=0.0001$
$\Rightarrow \quad \frac{r_{2}}{r_{1}}=0.9999$
Hence, percentage increase in vol
$=\frac{V_{2}-V_{1}}{V_{1}} \times 100$
$=\frac{\pi r_{2}^{2} l_{2}-\pi r_{1}^{2} l_{1}}{\pi r_{1}^{2} l_{1}} \times 100$
$=\left[\left(\frac{r_{2}}{r_{1}}\right)^{2}\left(\frac{l_{2}}{l_{1}}\right)^{2}-1\right] \times 100$
$=\left[(0.9999)^{2} \times(1.0001)^{2}-1\right] \times 100$
$=0.0008 \times 100=0.08 \%$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The polar co-ordinates of the point, whose Cartesian co-ordinates are $(-2 \sqrt{3}, 2)$, are

Options:
A $\left(4,\left(\frac{11 \pi}{12}\right)\right)$
B $\left(4,\left(\frac{5 \pi}{6}\right)\right)$
C $\left(4,\left(\frac{3 \pi}{4}\right)\right)$
D $\left(4,\left(\frac{2 \pi}{3}\right)\right)$

The correct answer is:
$\left(4,\left(\frac{5 \pi}{6}\right)\right)$

Solution:
$\begin{aligned} & x=-2 \sqrt{3} \text { and } y=2 \\ & r=\sqrt{x^2+y^2}=\sqrt{12+4}=4 \\ & \theta=\pi-\tan ^{-1}\left|\frac{2}{-2 \sqrt{3}}\right|=\pi-\tan ^{-1} \frac{1}{\sqrt{3}}=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The polarity of (i) (CH 3) 2 P (CF 3) 3 and (ii) (CH 3) 3 P (CF 3) 2 respectively are?

Options:
A (i) Non Polar (ii) Polar
B (i) Polar (ii) Non Polar
C (i) Non Polar (ii) Non Polar
D (i) Polar (ii) Polar

The correct answer is:
(i) Polar (ii) Non Polar

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The polars drawn from $(-1,2)$ to the circles $S_1 \equiv x^2+y^2+6 y+7=0$ and $S_2 \equiv x^2+y^2+6 x+1=0$, are

Options:
A Parallel
B Equal
C Perpendicular
D Intersect at a point

The correct answer is:
Intersect at a point

Solution:
The equation of polar to circle (i) is $x-5 y+13=0$ and equation of polar to circle (ii) is $x+y-1=0$ Clearly, polars intersect at a point.

Topic: Mathematics, Circle, JEE Main
The pole of the line $L x+m y+n=0$ w.r.t. the parabola $y^{2}=4 a x$ will be

Options:
A $\left(\frac{-n}{l}, \frac{-2 a m}{l}\right)$
B $\left(\frac{-n}{l}, \frac{2 a m}{l}\right)$
C $\left(\frac{n}{l}, \frac{-2 a m}{l}\right)$
D $\left(\frac{n}{l}, \frac{2 a m}{l}\right)$

The correct answer is:
$\left(\frac{n}{l}, \frac{-2 a m}{l}\right)$

Solution:
Tangent of the parabola $y^{2}=4 a x$ at point $P\left(x_{1}, y_{1}\right)$ is
$$
y y_{1}=2 a\left(x+x_{1}\right)
$$
$$
\Rightarrow \quad 2 a x_{1}-y y_{1}+2 \alpha x=0 \ldots(\mathrm{i})
$$
which is also equation of the polar of the parabola $y^{2}=4 a x$. Same as the line
$$
b x+m y+n=0...(ii)
$$
On comparing both lines, we get
$$
\frac{2 a}{l}=\frac{-y}{m}=\frac{2 a x}{n}
$$
Taking first two parts
$$
\left(y=-\frac{2 a m}{l}\right)
$$
Taking first and last parts,
$$
\left(x=\frac{n}{l}\right)
$$
$\therefore$ Required pole of the line is $\left\{\frac{n}{l}, \frac{-2 a m}{l}\right\} .$

Topic: Mathematics, Parabola, JEE Main
The pole of the straight line $9 x+y-28=0$ with respect to circle $2 x^2+2 y^2-3 x+5 y-7=0$, is

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

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

Solution:
$\begin{aligned} & x^2+y^2-\frac{3}{2} x+\frac{5}{2} y-\frac{7}{2}=0 \\ & \Rightarrow\left(x-\frac{3}{4}\right)^2+\left(y+\frac{5}{4}\right)^2-\frac{9}{16}-\frac{25}{16}-\frac{7}{2}=0 \\ & \Rightarrow\left(x-\frac{3}{4}\right)^2+\left(y+\frac{5}{4}\right)^2-\frac{45}{8}=0\end{aligned}$
Put $X=x-\frac{3}{4}$ and $Y=y+\frac{5}{4}$, we get the equation of circle $X^2+Y^2-\frac{45}{8}=0$ and the line $9 X+Y-\frac{45}{2}=0$
Hence pole$\equiv\left(\frac{9 \times \frac{45}{8}}{\frac{45}{2}}, \frac{1 \times \frac{45}{8}}{\frac{45}{2}}\right) \equiv\left(\frac{9}{4}, \frac{1}{4}\right)$
$x=\frac{9}{4}+\frac{3}{4} \text { and } y=\frac{1}{4}-\frac{5}{4}=-1$
Hence the pole is $(3,-1)$.

Topic: Mathematics, Circle, JEE Main
The pole of the straight line $x+2 y=1$ with respect to the circle $x^2+y^2=5$ is

Options:
A $(5,5)$
B $(5,10)$
C $(10,5)$
D $(10,10)$

The correct answer is:
$(5,10)$

Solution:
Let the pole be $\left(x_1, y_1\right)$, then the equation of polar is $x x_1+y y_1=a^2 \Rightarrow$ $x x_1+y y_1=5$ ...(i)
Since (i) and $x+2 y=1$ represents the same straight line .
$\therefore x_1=5$, $y_1=10 \Rightarrow$ Pole is $(5,10)$.

Topic: Mathematics, Circle, JEE Main
The pole of the straight line $x+4 y=4$ with respect to the ellipse $x^2+4 y^2=4$ is

Options:
A $(1,1)$
B $(1,4)$
C $(4,1)$
D $(4,4)$

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

Solution:
Equation of line is $x+4 y-4=0$
$$
l=1, m=4, n=-4
$$
and equation of ellipse is
$$
\begin{aligned}
\Rightarrow & x^2+4 y^2=4 \\
\Rightarrow & \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& a^2=4, b^2=1
\end{aligned}
$$
Pole of the straight line with respect to ellipse
$$
=\left(\frac{-a^2 l}{n}, \frac{-b^2 m}{n}\right)=\left(\frac{-4.1}{-4}, \frac{-1.4}{-4}\right)=(1,1)
$$

Topic: Mathematics, Ellipse, JEE Main
The polio virus enters the human body through

Options:
A Mosquito bite
B Tick bite
C Saliva and secretions from the nose
D Contaminated food and water

The correct answer is:
Contaminated food and water

Solution:
Polioviruses are more stable than most other viruses and can remain infectious for a relatively long period in water and food. The primary mode of transmission is ingestion of water contaminated with the virus. Its primary area of multiplication is the throat and small intestine. Next, the virus invades the tonsils and lymph nodes of the neck and ileum. From the lymph nodes the virus enters to blood and by penetrating the capillary walls enters the central nervous system. The virus displays high affinity for particularly motor nerve cells and multiplies within the cytoplasm of motor nerve cells. Hence, the cells die resulting in paralysis.

Topic: Biology, Human Health and Diseases, NEET
The pollen grains lose their viability in 30 minutes in

Options:
A Triticum aestivum
B Allium cepa
C Atropa belladonna
D Solanum nigrum

The correct answer is:
Triticum aestivum

Solution:
In some cereals such as rice and wheat, pollen grains lose viability within 30 minutes of their release, and in some members of Rosaceae, Leguminosae, and Solanaceae, they maintain viability for months.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The pollen tube usually enters the embryo sac

Options:
A through one of the synergids
B by directly penetrating the egg
C between one synergid and central cell
D by knocking off the antipodal cells.

The correct answer is:
through one of the synergids

Solution:
After entering the wall of the embryo sac the pollen tube during its entry into the embryo sac passes through the nucellar cells and the synergids, e.g. Fagopyrum. The pollen tubes passes between the egg and one synergid. In Cardiospermum, it passes between embryo sac wall and synergid. In Oxalis, it passes directly through a synergid. In Viola, it enters a synergid and makes its way through the base of the synergid. In normal cases only one synergid destroys by the impact of the pollen tube and the other remains intact until sometimes afterwards.

Topic: Biology, Sexual Reproduction in Flowering Plants, AIIMS, AIIMS 2004
The polymer containing strong intermolecular forces e.g. hydrogen bonding, is

Options:
A
teflon

B
nylon 6,6

C
polystyrene

D
natural rubber

The correct answer is:
nylon 6,6

Solution:
nylon 6,6 is a polymer of adipic acid and hexamethylene diamine

Topic: Chemistry, Polymers, JEE Main, JEE Main 2010
The polymer formed on heating Novolac with formaldehyde is:

Options:
A Polyester
B Bakelite
C Nylon- 6,6
D Melamine

The correct answer is:
Bakelite

Solution:
Novolac on heating with formaldehyde undergoes cross-linking to form an infusible solid mass called Bakelite. It is a thermosetting polymer that cannot be reused or remoulded. Bakelite is used for making combs, phonograph records, electrical switches, and handles of various utensils

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The polymer used for optical lenses is :

Options:
A
polypropylene

B
polyvinyl chloride

C
polythene

D
polymethyl methacrylate

The correct answer is:
polymethyl methacrylate

Solution:
Polymethyl methacrylate is hard, fairly rigid. It is used for optical lenses.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2013 (22 Apr Online)
The polymer, which can be stretched and retains its original status on releasing the force is

Options:
A Nylon- 6, 6
B Buna-N
C Terylene
D Bakelite

The correct answer is:
Buna-N

Solution:
Synthetic rubber is any vulcanisable rubber like polymer, which is capable of getting stretched to twice its length. However, it returns to its original shape and size as soon as the external stretching force is released. Thus, synthetic rubbers are either homopolymers of 1, 3 - butadiene derivatives or copolymers of 1, 3 - butadiene or its derivatives with another unsaturated monomer.

Buna-N is an example of synthetic rubber.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The polynomial equation of degree 4 having real coefficients with three of its roots as \(2 \pm \sqrt{3}\) and \(1+2 i\), is

Options:
A \(x^4-6 x^3-14 x^2+22 x+5=0\)
B \(x^4-6 x^3-19 x+22 x-5=0\)
C \(x^4-6 x^3+19 x-22 x+5=0\)
D \(x^4-6 x^3+14 x^2-22 x+5=0\)

The correct answer is:
\(x^4-6 x^3+14 x^2-22 x+5=0\)

Solution:
It is given that, the polynomial equation of degree 4 having real coefficients with three of its roots as \(2 \pm \sqrt{3}\) and \(l+2 i\), so the remaining root is \(1-2 i\). Now, the quadratic equation whose roots as \(2 \pm \sqrt{3}\) is
\(x^2-4 x+1=0 \text {, and }\)
the quadratic equation whose roots as \(1 \pm 2 i\), is
\(x^2-2 x+5=0\)
So, the required polynomial equation is
\(\begin{aligned}
\left(x^2-4 x+1\right)\left(x^2-2 x+5\right) & =0 \\
\Rightarrow \quad x^4-6 x^3+14 x^2-22 x+5 & =0
\end{aligned}\)
Hence, option (4) is correct.

Topic: Mathematics, Quadratic Equation, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
The polynomial equation $x^{3}-3 a x^{2}+\left(27 a^{2}+9\right) x+2016=0$ has -

Options:
A exactly one real root for any real a
B three real roots for any real a
C three real roots for any $a \geq 0$, and exactly one real root for any a $ < 0$
D three real roots for any a $\leq 0$, and exactly one real root for any a $>0$

The correct answer is:
exactly one real root for any real a

Solution:
$$
\begin{aligned}
\mathrm{f}^{\prime}(\mathrm{x}) &=3 \mathrm{x}^{2}-6 \mathrm{ax}+27 \mathrm{a}^{2}+9 \\
&=3\left[\mathrm{x}^{2}-2 \mathrm{ax}+9 \mathrm{a}^{2}+3\right]=3\left((\mathrm{x}-\mathrm{a})^{2}+8 \mathrm{a}^{2}+3\right)
\end{aligned}
$$
$\therefore f^{\prime}(x)$ is $+$ ve for $x \in R$ so $f(x)$ is monotonic $\uparrow$ for $\mathrm{x} \in \mathrm{R}$.

Topic: Mathematics, Quadratic Equation, JEE Main
The polysaccharide that is not made up of only one type of monosaccharide is

Options:
A glycogen
B cellulose
C inulin
D all of the above

The correct answer is:
inulin

Solution:
Polysaccharide having only one type of monosaccharide are called homopolymers. Polymers of glucose are called as Glucans. Glycogen, cellulose, and starch are polymers of glucose. Inulin is the naturally occurring polysaccharide, composed mainly of fructose units, and it can have a terminal glucose unit. So, inulin, which contains terminal glucose units is a heteropolymer, and inulin which has no terminal glucose units is a homopolymer.

Topic: Chemistry, Biomolecules, NEET
The population P=Pt at time t of a certain species follows the differential equation dPdt=0.5P-450. If P0=850, then the time at which population becomes zero is:

Options:
A loge9
B 2loge18
C loge18
D 12loge18

The correct answer is:
2loge18

Solution:
dPdt=0.5P-450

⇒ ∫0tdpP-900=∫0tdt2

⇒ InPt-9000t=t20t

⇒ InPt-900-InP0-900=t2

⇒ InPt-900-In50=t2

for Pt=0

⇒ In90050=t2⇒t=2In18

Topic: Mathematics, Differential Equations, JEE Main
The population P=Pt at time t of a certain species follows the differential equation dPdt=0.5P-450. If P0=850, then the time at which population becomes zero is:

Options:
A loge9
B 2loge18
C loge18
D 12loge18

The correct answer is:
2loge18

Solution:
dPdt=0.5P-450

⇒ ∫0tdpP-900=∫0tdt2

⇒ InPt-9000t=t20t

⇒ InPt-900-InP0-900=t2

⇒ InPt-900-In50=t2

for Pt=0

⇒ In90050=t2⇒t=2In18

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (24 Feb Shift 1)