datavorous/entrance-exam-dataset huggingface dataset:
If $P Q$ is a double ordinate of hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 .$ Such that $O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola, then the eccentricity ' $e$ ' of the hyperbola satisfies

Options:
A $1 < e < \frac{2}{\sqrt{3}}$
B $e=\frac{2}{\sqrt{3}}$
C $e=\frac{\sqrt{3}}{2}$
D $e>\frac{2}{\sqrt{3}}$

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

Solution:
Let $\mathrm{P}(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta), \mathrm{Q}(\mathrm{a} \sec \theta,-\mathrm{b} \tan \theta)$ be end points of double ordinates and $(0,0)$ is the centre of the hyperbola. So, $\mathrm{PQ}=2 \mathrm{~b} \tan \theta$
$\mathrm{OQ}=\mathrm{OP}=\sqrt{\mathrm{a}^{2} \sec ^{2} \theta+\mathrm{b}^{2} \tan ^{2} \theta}$
Since, $\mathrm{OQ}=\mathrm{OP}=\mathrm{PQ}$
$\therefore 4 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta$
$\Rightarrow 3 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta$
$\Rightarrow 3 b^{2} \sin ^{2} \theta=a^{2}$
$\Rightarrow 3 a^{2}\left(e^{2}-1\right) \sin ^{2} \theta=a^{2}$
$\Rightarrow 3\left(e^{2}-1\right) \sin ^{2} \theta=1$
$\Rightarrow \frac{1}{3\left(e^{2}-1\right)}=\sin ^{2} \theta < 1,\left(\because \sin ^{2} \theta < 1\right)$
$\Rightarrow \frac{1}{e^{2}-1} < 3 \Rightarrow e^{2}-1>\frac{1}{3} \Rightarrow e^{2}>\frac{4}{3}$
$\therefore e>\frac{2}{\sqrt{3}}$

Topic: Mathematics, Hyperbola, JEE Main
If pressure and temperature of an ideal gas are doubled and volume is halved, the number of molecules of gas

Options:
A Become half
B Become two times
C Become four times
D Remain constant

The correct answer is:
Become two times

Solution:
Applying ideal gas equation, we have
pV=nRT
Where, p= pressure of the gas
V= volume of the gas
n= number of moles
R= gas constant
T= temperature of the gas
∴ n1=p1V1RT1
and n2=p2V2RT2
⇒ n2n1=p2V2p1V1×T1T2=p2p1V2V1T1T2
n2n1=2122
n2=2n1

Topic: Physics, Thermodynamics, NEET
If pressure at half the depth of a lake is equal to $2 / 3$ pressure at the bottom of the lake then what is the depth of the lake

Options:
A $10 \mathrm{~m}$
B $20 \mathrm{~m}$
C $60 \mathrm{~m}$
D $30 \mathrm{~m}$

The correct answer is:
$20 \mathrm{~m}$

Solution:
Pressure at half the depth $=P_0+\frac{h}{2} d g$
Pressure at the bottom $=P_0+h d g$
According to given condition
$P_0+\frac{h}{2} d g=\frac{2}{3}\left(P_0+h d g\right)$
$\Rightarrow 3 P_0+\frac{3 h}{2} d g=2 P_0+2 h d g$
$\Rightarrow h=\frac{2 P_o}{d g}=\frac{2 \times 10^5}{10^3 \times 10}=20 \mathrm{~m}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
If pressure of real gas $\mathrm{O}_{2}$ in a container is given by $\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$, then the mass of the gas in the container is

Options:
A $32 \mathrm{gm}$
B $16 \mathrm{gm}$
C $4 \mathrm{gm}$
D $64 \mathrm{gm}$

The correct answer is:
$16 \mathrm{gm}$

Solution:
$\mathrm{P}=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}-\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]=\frac{\mathrm{RT}}{2 \mathrm{~V}-\mathrm{b}}$
$2\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\mathrm{RT}$
$\left[\mathrm{P}+\frac{\mathrm{a}}{4 \mathrm{~b}^{2}}\right]\left[\mathrm{V}-\frac{\mathrm{b}}{2}\right]=\frac{\mathrm{RT}}{2}$
$\therefore \mathrm{n}=1 / 2 \mathrm{~mole}$
$\therefore$ Mass of $\mathrm{O}_{2}=\frac{1}{2} \times 32=16 \mathrm{gm}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formula of force is

Options:
A P V 2 T 2
B P −1 V 2 T −2
C PV T 2
D P −1 V T 2

The correct answer is:
P V 2 T 2

Solution:
Let F∝ P x V y T z by substituting the following dimensions : [P]=[M L −1 T −2 ] [V]=[L T −1 ], [T]=[T] and comparing the dimension of both sides x=1, y=2, z=2 , so F=P V 2 T 2

Topic: Physics, Units and Dimensions, JEE Main
If progesterone receptors are disrupted then

Options:
A secondary sexual characters would not be maintained.
B pituitary gonadotropins will not be secreted.
C the uterus will not be able to support pregnancy.
D the uterine endometrium will not begin to proliferate.

The correct answer is:
the uterus will not be able to support pregnancy.

Solution:
After zygote formation, embryonic development begins. This includes cleavage, morula formation, blastula formation, implantation, gastrulation and organogenesis. Implantation happens on 8th day after fertilisation and this is considered to be the beginning of pregnancy. This continues till parturition begins. This is maintained because the shedding of the endometrium is prevented as progesterone and estrogen levels are maintained and this is due to secretion of human chorionic gonadotrophin. Thus if progesterone receptors are disrupted then uterus will not be able to support pregnancy as it will not respond to progesterone.

Topic: Biology, Chemical Coordination and Integration, NEET
If $P S Q$ is the focal chord of the parabola $y^2=8 x$ such that $S P=6$. Then the length $S Q$ is

Options:
A $6$
B $4$
C $3$
D None of these

The correct answer is:
$3$

Solution:
Since the semi-latus rectum of a parabola is the harmonic mean between the segments of any focal chord of a parabola, therefore $S P, 4, S Q$ are in H.P.
$\Rightarrow 4=2 \cdot \frac{S P \cdot S Q}{S P+S Q}$ $\Rightarrow4=\frac{2(6)(S Q)}{6+S Q} \Rightarrow S Q=3$

Topic: Mathematics, Parabola, JEE Main
If $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$, and $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0 ; \mathrm{p}, \mathrm{q}$ and r being real numbers, then which of the following is not possible?

Options:
A $\alpha=\beta=\mathrm{p}$
B $\alpha \neq \beta=\mathrm{p}$
C $\alpha=\beta \neq \mathrm{p}$
D $\beta \neq \alpha=\mathrm{p}$

The correct answer is:
$\alpha=\beta=\mathrm{p}$

Solution:
Given equation is $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$
$\Rightarrow p x^{2}+q x+r=p x^{2}-p(\alpha+\beta) x+\alpha \beta p$
$\Rightarrow \alpha \beta \mathrm{p}=\mathrm{r}$ and $\mathrm{q}=-(\alpha+\beta) \mathrm{p}$...(1)
Also given that $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0$
Putting value of $q$ and $r$ from $(1)$ $\Rightarrow p^{3}-p^{2}(\alpha+\beta)+\alpha \beta p=0$
$\Rightarrow \mathrm{p}^{2}-\mathrm{p}(\alpha+\beta)+\alpha \beta=0$
$\Rightarrow(p-\alpha)(p-\beta)=0 \Rightarrow \alpha=\beta=p$

Topic: Mathematics, Quadratic Equation, JEE Main
If Q(0,-1,-3) is the image of the point P in the plane 3x-y+4z=2 and R is the point (3,-1,-2), then the area (in sq. units) of ΔPQR is

Options:
A 914
B 912
C 213
D 652

The correct answer is:
912

Solution:
Given, equation of plane is 3x-y+4z-2=0.
Distance, PM=1-12-29+1+16=132
PR=9+1=10
∴RM=10-132=72
∴ Area of ∆PQR =212×PM×RM
=212×132×72
=912

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If $\mathrm{Q}, \mathrm{E}$ and $\mathrm{W}$ denote respectively the heat added, change in internal energy and the work done in a closed cycle process, then

Options:
A $\mathrm{Q}=0$
B $W=0$
C $\mathrm{Q}=\mathrm{W}=0$
D $\mathrm{E}=0$

The correct answer is:
$\mathrm{E}=0$

Solution:
From the $1^{\text {st }}$ law of thermodynamics, $\mathrm{Q}=\mathrm{E}+\mathrm{W}$
For cyclic process, $E=0, \therefore Q=W \neq 0$

Topic: Physics, Thermodynamics, NEET, NEET 2008 (Mains)
If qf is the free charge on the capacitor plates and qb is the bound charge on the dielectric slab of dielectric constant k placed between the capacitor plates, then bound charge qb can be expressed as :

Options:
A qb=qf1-1k
B qb=qi1-1k
C qb=qf1+1k
D qb=qf1+1k

The correct answer is:
qb=qi1-1k

Solution:
When a dielectric is inserted in a capacitor,

qb=qi1-1k

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If $\mathrm{q}$ is false and $\mathrm{p} \wedge \mathrm{q} \leftrightarrow \mathrm{r}$ is true, then which one of the following statements is a tautology?

Options:
A $(\mathrm{p} \vee \mathrm{r}) \rightarrow(\mathrm{p} \wedge \mathrm{r})$
B $\quad(\mathrm{p} \wedge \mathrm{r}) \rightarrow(\mathrm{p} \vee \mathrm{r})$
C $\mathrm{p} \wedge \mathrm{r}$
D $\mathrm{p} \vee \mathrm{r}$

The correct answer is:
$\quad(\mathrm{p} \wedge \mathrm{r}) \rightarrow(\mathrm{p} \vee \mathrm{r})$

Solution:
$q$ is false and $[(p \wedge q) \leftrightarrow r]$ is true As $(p \wedge q)$ is false

$[$ False $\leftrightarrow r]$ is true

Hence $r$ is false

Option (1): says $p \vee r$,

Since $r$ is false

Hence $(p \vee r)$ can either be true or false

Option (2): says $(p \wedge r) \rightarrow(p \vee r)$

$(p \wedge r)$ is false

Since, $F \rightarrow T$ is true and

$F \rightarrow F$ is also true

Hence, it is a tautology

Option (3): $(p \vee r) \rightarrow(p \wedge r)$

i.e. $(p \vee r) \rightarrow F$

It can either be true or false

Option (4): $(p \wedge r)$

Since, $r$ is false

Hence, $(p \wedge r)$ is false.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If ∑r=025Cr 50C25-r 50-r=KC25 50, then K is equal to

Options:
A 225
B 225-1
C 224
D 252

The correct answer is:
225

Solution:
Given:

∑r=025Cr 50C25-r 50-r=KC25 50

LHS

=∑r=025 50Cr 50-rC25-r
=∑r=02550!r!50-r!50-r!25-r!25!
=∑r=025 50!  25!r!25-r!25!25!
= 50C25 ∑r=025 25Cr
= 50C25 ×225

=KC25 50
∴K=225

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Jan Shift 2)
If ∑r=025Cr 50.C25-r 50-r=KC25 50, then K is equal to

Options:
A 225
B 225-1
C 224
D 252

The correct answer is:
225

Solution:
LHS =∑r=025 50Cr 50-rC25-r
=∑r=02550!r!50-r!50-r!25-r!25!
=∑r=025 50!  25!r!25-r!25!25!
= 50C25 ∑r=025 25Cr
= 50C25 ×225
⇒K=225

Topic: Mathematics, Binomial Theorem, JEE Main
If ∑r=150tan-112r2=p, then the value of tanp is :

Options:
A 100
B 5150
C 5051
D 101102

The correct answer is:
5051

Solution:
∑r=150tan-112r2=∑r=150tan-124r2

=∑r=150tan-121+4r2-1

=∑r=150tan-121+2r-12r+1

=∑r=150tan-1(2r+1)-(2r-1)1+(2r+1)(2r-1)

=∑r=150[tan-12r+1-tan-12r-1]

So, ∑r=150tan-112r2=tan-13-tan-11+tan-15-tan-13+tan-17-tan-15...........tan-1101-tan-199=tan-1101-tan-11

=tan-1101-11+101

=tan-1100102=p

So, tanp=100102=5051

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (26 Aug Shift 2)
If $\alpha \in \mathbb{R}-\{-1\}$ and $\mathrm{f}(\mathrm{x})=|(|\mathrm{x}|+\alpha)(|\mathrm{x}|-1)|$, then the number of points at which $f(x)$ is not differentiable, is

Options:
A 3 , when $\alpha < 0$
B 5 , when $\alpha>0$
C 4 , when $\alpha>0$
D 5 , when $\alpha < 0$

The correct answer is:
5 , when $\alpha < 0$

Solution:
Since, $\alpha \in R-\{-1\}$
$\begin{aligned} & \text { and } \mathrm{f}(\mathrm{x})=|(|x|+\alpha)(|x|-1)| \\ & |x|+\alpha=0 \Rightarrow|x|=-\alpha \Rightarrow \alpha < 0 \\ & |x|-1=0 \Rightarrow x= \pm 1\end{aligned}$
Now, graph of $f(x)$ is
It is clear from graph that there are 5 points at which $\mathrm{f}(\mathrm{x})$ is not differentiable and $\alpha < 0$

Topic: Mathematics, Continuity and Differentiability, JEE Main
If r1 is the radius of first orbit of hydrogen atom, then the radii of second, third and fourth orbits in terms of r1 are

Options:
A r12,r13,r14
B 8r1,27r1,64r1
C 4r1,9r1,16r1
D 2r1,6r1,8r1

The correct answer is:
4r1,9r1,16r1

Solution:
rn=n2Z5.29×10-11m
Z=1 for H-atom
So r1=5.29×10-11m=4r1
r2=4×5.29×10-11m=4r1
r3=9×5.29×10-11m=9r1
r4=16×5.29×10-11m=16r1

Topic: Chemistry, Structure of Atom, JEE Main
If ∑r=1n2r+1=440 , then n =…..

Options:
A 20
B 22
C 21
D 19

The correct answer is:
20

Solution:
We have, âr=1n2r+l=440
⇒3+5+7â¦+2n+1=440
⇒n22Ã3+n-12=440
⇒n3+n-1=440
⇒nn+2=440
⇒n=20

Topic: Mathematics, Sequences and Series, JEE Main
If ∑r=1ntr=16nn+1n+2, ∀n≥1, then the value of limn→∞∑r=1n1tr is equal to

Options:
A 2
B 3
C 32
D 6

The correct answer is:
2

Solution:
We have, for n≥1, tn=∑r=1ntr-∑r=1n-1tr
=16nn+1n+2-16n-1nn+1
=16nn+1n+2-n+1
=12nn+1.
⇒1tr=2rr+1=21r-1r+1
⇒∑r=1n1tr=21-1n+1
⇒limn→∞∑r=1n1tr=2

Topic: Mathematics, Limits, JEE Main
If ∆r1 represents the difference in radii of stationary orbits for n=3 and n=4 in a hydrogen atom and ∆r2 represents the difference in radii of stationary orbits for n=8 and n=9, then the value of ∆r1∆r2 is

Options:
A 0.71
B 0.41
C 2.43
D 14.82

The correct answer is:
0.41

Solution:
Use the relationship between the radius of an orbit and quantum number n of that orbit.
r4-r3r9-r8=r1×42-r1×32r1×92-r1×82=16-981-64=0.41

Topic: Physics, Atomic Physics, JEE Main
If $R_1=\{(x, y) \mid y=2 x+7$, where $x \in R$ and $-5 \leq x \leq 5\}$ is a relation. Then, find the domain and range of $\mathrm{R}_1$.

Options:

None

Solution:
Given, $R_1=\{x, y) \mid y=2 x+7$, where $x \in R$
Domin of $R_1=\{-5 \leq x \leq 5, x \in R\}=[-5,5]$
Since $y=2 x+7$
When $x=-5$, then, $y=2(-5)+7=-3$
When $x=5$, then, $y=2(5)+7=17$
$\therefore$ Range of $R_1=\{-3 \leq y \leq 17, y \in R\}=[-3,17]$

Topic: Mathematics, Relations and Functions (Class 11)
If Îr=2r-1 mCr1m2-12mm+1sin2â¡m2 sin2â¡msin2â¡m+1 , then the value of âr=0mÎ is

Options:
A 1
B 3
C 2
D 0

The correct answer is:
0

Solution:
Îr=2r-1 mCr1m2-12mm+1sin2m2sin2â¡msin2â¡m+1
â´ âr=0mÎr= âr=0m2r-1âr=0m mCrâr=0m1m2-12mm+1sin2â¡m2sin2â¡msin2â¡m+1
=m2-12mm+1m2-12mm+1sin2â¡m2sin2â¡msin2â¡m+1
=0 ( âµ two rows are identical)

Topic: Mathematics, Determinants, JEE Main
If $R_3=\{(x,|x|) \mid x$ is a real number $\}$ is relation, then find domain and range of $R_3$.

Options:

None

Solution:
Given $R_3=\{(x,|x|) \mid x$ is real number $\}$
Clearly, domain $\left(R_3\right)=R$
Since, image of any real number under $R_3$ is positive real number or zero.
$\therefore \quad$ Range $\left(R_3\right)=R^{+} \cup\{0\}$ or $[0, \infty)$

Topic: Mathematics, Relations and Functions (Class 11)
If $\mathrm{R}=\{(a, \mathrm{~b}) / \mathrm{b}=a-1, a \in \mathrm{Z}, 5 < a < 9\}$, then the range of $\mathrm{R}$ is

Options:
A $\{7,8,9\}$
B $\{5,6,7\}$
C $\{6,7,8\}$
D $\{5,6,7,8,9\}$

The correct answer is:
$\{5,6,7\}$

Solution:
(A)
Given $\mathbf{R}=\{(\mathbf{a}, \mathbf{b}), \mathbf{b}=\mathbf{a}-1, \mathbf{a} \in \mathbf{z}, 5 < \mathbf{a} < 9\}$
$\therefore \quad a=6,7,8$
$a=6, b=6-1=5$
$a=7, b=7-1=6$
$a=8, b=8-1=7$
$R=\{(6,5),(7,6),(8,7)\}$
Range of $R=\{5,6,7\}$

Topic: Mathematics, Functions, JEE Main
If $R$ and $r$ are the radii of the circumcircle and incircle of a regular polygon of $n$ sides, each side being of length a, then $2(R+r)$ equals

Options:
A $\frac{a}{4} \cot \frac{\pi}{2 n}$
B a $\cot \frac{\pi}{n}$
C $\frac{a}{2} \cot \frac{\pi}{2 n}$
D $\operatorname{a~} \cot \frac{\pi}{2 n}$

The correct answer is:
$\operatorname{a~} \cot \frac{\pi}{2 n}$

Solution:
We know that $R=\frac{1}{2}$ a $\operatorname{cosec}\left(\frac{\pi}{n}\right)$ and $r=\frac{1}{2}$ a $\cot \left(\frac{\pi}{n}\right)$. so that
$2(R+r)=a\left(\frac{1}{\sin \left(\frac{\pi}{n}\right)}+\frac{\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}\right)=a \frac{1+\cos \left(\frac{\pi}{n}\right)}{\sin \left(\frac{\pi}{n}\right)}$
$=a \frac{2 \cos ^2\left(\frac{\pi}{2 n}\right)}{2 \sin \left(\frac{\pi}{2 n}\right) \cos \left(\frac{\pi}{2 n}\right)}=a \cot \frac{\pi}{2 n}$

Topic: Mathematics, Properties of Triangles, JEE Main
If $\vec{r}$ and $\vec{s}$ arenon-zero constant vectors and the scalar $b$ is chosen such that $|\vec{r}+b \vec{s}|$ is minimum, then the value of $|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}$ is equal to

Options:
A $2|\vec{r}|^{2}$
B $|\vec{r}|^{2} / 2$
C $3|\vec{r}|^{2}$
D $|\vec{r}|^{2}$

The correct answer is:
$|\vec{r}|^{2}$

Solution:
For minimum value $|\vec{r}+b \vec{s}|=0$. Let $\vec{r}$ and $\vec{s}$ are anti-parallel so $b \vec{s}=-\vec{r}$

$\therefore|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$

Topic: Mathematics, Vector Algebra, JEE Main
If α,β∈R are such that 1-2i (here i2=-1) is a root of z2+αz+β=0, then α-β is equal to:

Options:
A -7
B 7
C -3
D 3

The correct answer is:
-7

Solution:
∵α,β∈R⇒ other root is 1+2i

α=-( sum of roots )=-1-2i+1+2i=-2

β=product of roots =1-2i1+2i=5

∴α-β=-7

option 1

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (25 Feb Shift 2)
If \( R \) be a relation defined as \( a R b \) iff \( |a-b|>0 \), then the relation is

Options:
A Reflexive
B Symmetric
C Transitive
D Symmetric and transitive

The correct answer is:
Symmetric

Solution:
Since, R is defined as aRb iff a-b>0.
Reflexive : aRa iff a-a>0
Which is not true. So, R is not reflexive.
Symmetric : aRb iff a-b>0
Now, bRa iff b-a>0
⇒ a-b>0 ⇒ aRb
Thus, R is symmetric.
Transitive : aRb iff a-b>0
bRc iff b-c>0

aRb & bRa ⇒ aRa which is not correct, hence relation is not transitive.

Topic: Mathematics, Sets and Relations, JEE Main
If $R$ denotes the set of all real number, then the function $f: R \rightarrow R$ defined $f(x)=|x|$ is

Options:
A one-one only
B onto only
C both one-one and onto
D neither one-one nor onto

The correct answer is:
neither one-one nor onto

Solution:
We have,
$$
\begin{array}{lrlrl}
& & f(x) & =|x| \\
\text { Now, } & f(1) & =|1|=1 \\
\text { and } & f(-1) & =|-1|=1 \\
& \therefore & f(1) & =f(-1) \text { but } 1 \neq-1
\end{array}
$$
So, $f(x)$ is not one-one.
Again, $|x| \geq 0, \forall x \in R$
So, range of $f(x)=[0, \infty]$
$\therefore$ Range $\neq$ Co-domain
So, $f(x)$ is not onto.

Topic: Mathematics, Functions, COMEDK, COMEDK 2013
If $R$ denotes the set of all real numbers then the function $f: R \rightarrow R$ defined $f(x)=[x]$

Options:
A One-one only
B Onto only
C Both one-one and onto
D Neither one-one nor onto

The correct answer is:
Neither one-one nor onto

Solution:
Let $f\left(x_1\right)=f\left(x_2\right) \Rightarrow\left[x_1\right]=\left[x_2\right] \Rightarrow x_1=x_2$
{For example, if $x_1=1.4, x_2=1.5$, then [1.4]=[1.5] =1}
$\therefore f$ is not one-one.
Also, $f$ is not onto as its range I (set of integers) is a proper subset of its co-domain $R$.

Topic: Mathematics, Functions, JEE Main
If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth r below and a height r above the earth surface is:
(Given : r<RE)

Options:
A 1+rRE-r2RE2-r3RE3
B 1+rRE+r2RE2+r3RE3
C 1-rRE-r2RE2-r3RE3
D 1+rRE-r2RE2+r3RE3

The correct answer is:
1+rRE-r2RE2-r3RE3

Solution:
ghgd=GMRE+r2GMRE3RE-r=RE3RE+r2RE-r
gdgh=RE+r2RE-rRE3=RE3-REr2+rRE2-r3RE3=1-r2RE2+rRE-rRE3

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If Rez-12z+i=1, where z=x+iy, then the point x,y lies on a

Options:
A circle whose centre is at -12,-32
B straight line whose slope is -23
C straight line whose slope is 32
D circle whose diameter is 52

The correct answer is:
circle whose diameter is 52

Solution:
Let z=x+iy
z-12z+i=x-1+iy2x+iy+i=x-1+iy2x+2y+1i×2x-2y+1i2x-2y+1i
Rez+12z+i=2xx-1+y2x+12x2+2y+12=1
⇒2x2+2y2-2x+y=4x2+4y2+4y+1
⇒2x2+2y2+2x+3y+1=0
⇒x2+y2+x+32y+12=0

Circle with center -12, -34
r=14+916-12=4+9-816=54 ⇒ d=52

where, d=diameter of the given circle.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If r→⋅i^=2r→⋅j^=4r→⋅k^ and |r→|=84 , then the value of r→⋅2i^-3j^+k^ may be

Options:
A 0
B 2
C 4
D 6

The correct answer is:
6

Solution:
r→⋅i^=2r→⋅j^=4r→⋅k^
Let r→=xi^+yj^+zk→
x=2y=4z
|r→|=84⇒4z2+2z2+z2=84
z=2;y=4;x=8
r→=8i^+4j^+2k^
r→⋅2i^-3j^+k^=16 –12+2=6

Topic: Mathematics, Vector Algebra, JEE Main
If r→ is a unit vector satisfying r→×a→=b→, |a→|=2 and |b→|=3, then one such r→=

Options:
A 142a→+b→×a→
B 14a→-2b→×a→
C 13a→-b→×
D 14a→-b→×a→

The correct answer is:
14a→-b→×a→

Solution:
Given, r→×a→=b→

⇒r→×a→×a→=b→×a→

⇒-a→·a→r→+a→·r→a→=b→×a→

⇒r→=14a→·r→a→-b→×a→ ...i

Let, the angle between r→ and a→ is θ

⇒r→×a→=b→

⇒1×2sinθ=3

⇒θ=π3

Now, r→·a→=1×2cosπ3=1

Put, r→·a→=1 in the Equation i

⇒r→=14a→-b→×a→

Topic: Mathematics, Vector Algebra, JEE Main
If λ∈R is such that the sum of the cubes of the roots of the equation x2+2-λx+10-λ=0 is minimum, then the magnitude of the difference of the roots of this equation is :

Options:
A 42
B 20
C 25
D 27

The correct answer is:
25

Solution:
x2+2-λx+10-λ=0

⇒α+β=λ-2& αβ=10-λ ..........i

Let roots are α and β.

⇒ α3+β3=α+β3-3αβ(α+β)

=λ-23-310-λ(λ-2)

= λ3-6λ2+12λ-8-3(10λ-λ2-20+2λ)

= λ3-3λ2-24λ+52

dzdλ=3λ2-6λ-24=3λ2-2λ-8 (where, z=α3+β3)

For maximum and critical points, derivative must be zero.

⇒ λ2-2λ-8=0

⇒λ-4λ+2=0

⇒λ=-2, 4

Now, d2zdλ2=6λ-6

For λ=-2, d2zdλ2<0⇒α3+β3 is maximum and

For (λ=4), d2zdλ2>0⇒α3+β3 is minimum.

⇒Equation will be x2-2x+6=0.

Using quadratic formula, we get

x=2±-22-4×1×62×1=2±-202=2±25i2=1±5i

Thus, difference of roots is α-β=25

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2018 (15 Apr)
If $\lambda \in \mathrm{R}$ is such that the sum of the cubes of the roots of the equation, $x^2+(2-\lambda) x+(10-\lambda)=0$ is minimum, then the magnitude of the difference of the roots of this equation is

Options:
A
20

B
$2 \sqrt{5}$

C
$2 \sqrt{7}$

D
$4 \sqrt{2}$

The correct answer is:
$2 \sqrt{5}$

Solution:
Let, the roots of the equation, $x^2+(2-\lambda) x+(10-\lambda)=0$ are $\alpha$ and $\beta$.
Also roots of the given equation are
$$
\frac{\lambda-2 \pm \sqrt{4-4 \lambda+\lambda^2-40+4 \lambda}}{2}=\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}
$$
The magnitude of the difference of the roots is $\left|\sqrt{\lambda^2-36}\right|$
So, $\alpha^3+\beta^3=\frac{(\lambda-2)^3}{4}+\frac{3(\lambda-2)\left(\lambda^2-36\right)}{4}$
$$
\begin{aligned}
=\frac{(\lambda-2)\left(4 \lambda^2-4 \lambda-104\right)}{4} \\
=&(\lambda-2)\left(\lambda^2-\lambda-26\right)=f(\lambda)
\end{aligned}
$$
As $f(\lambda)$ attains its minimum value at $\lambda=4$.
Therefore, the magintude of the difference of the

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If R is the least value of a such that the function fx = x2 + ax + 1 is increasing on [1,2] and S is the greatest value of a such that the function f(x) = x2 + ax + 1 is decreasing on [1, 2], then the value of |R-S| is

Options:

The correct answer is:
2

Solution:
If f'(x)=2x+a≥0∀x∈[1,2] ⇒a≥-2x∀x∈[1,2]

amin=R=-2 If

amax=S=-4

|R-S|=2

Alternate :

If f(x) is increasing for all x∈[1,2] then

-a2≤1⇒a≥-2⇒R=-2

If f(x) is decreasing for x∈[1,2], then

-a2≥2⇒a≤-4⇒S=-4

|R-S|=2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (31 Aug Shift 1)
If r is the orbital radius and v is the orbital velocity of an electron in a hydrogen atom, then its magnetic dipole moment is

Options:
A

evr 2π

B evr
C evr2
D evr

The correct answer is:
evr2

Solution:
The magnetic dipole moment of a revolving electron: The electron of change (–e) performs uniforms circular motion around a stationary heavy nucleus of charge +Ze. This constitutes a current I.

∴ I= e T ...(i)

Where, T is the time period of revolution. Let r be the orbital radius of electron and v the orbital speed. Then

T= 2πr υ ...(ii)

Substituting the value of T in (i), we get

I= e 2πr υ = eυ 2πr

The magnetic moment (m) associated with this circular current is

m=Iπ r 2 [ ∵ m → =I A → ]

∴ m= eυ 2πr ⋅π r 2 = eυr 2

Topic: Physics, Magnetic Properties of Matter, JEE Main
If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g=π2 m s-2, then the length of the second's pendulum at a height h=2R from the surface of earth will be:

Options:
A 29 m
B 19 m
C 49 m
D 89 m

The correct answer is:
19 m

Solution:
Acceleration due to gravity at the given height, g'=GM3R2=19GMR2=19g, where g is acceleration due to gravity at the surface of the Earth.

Now, time period is given by, T=2πlg'

Also the time period of second pendulum is 2 sec.

Therefore, T=2 sec

Then, 2=2πlg×9

⇒l=19 m

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If R is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by 20% is

Options:
A 52-1R
B 52+1R
C 52-1R
D (52+1)R

The correct answer is:
52-1R

Solution:
If g and gh denote the acceleration due to gravity at the Earth's surface and at the height h above the Earth's surface respectively, then
ghg=R2(R+h)2
But according to the equation
gh=g-20100g=80100g=45g
∴45=R2R+h2   or  45=RR+h
or 25=RR+h   or  2R+2h=5R
or 2h=5-2R  or  h=5-2R2=52-1R

Topic: Physics, Gravitation, JEE Main