datavorous/entrance-exam-dataset huggingface dataset:
Consider the function f:12,1→R defined by fx=42x3-32x-1. Consider the statements
(I) The curve y=fx intersects the x-axis exactly at one point
(II) The curve y=fx intersects the x-axis at x=cosπ12

Then

Options:
A Only (II) is correct
B Both (I) and (II) are incorrect
C Only (I) is correct
D Both (I) and (II) are correct

The correct answer is:
Both (I) and (II) are correct

Solution:
Given: fx=42x3-32x-1

⇒f'x=122x2-32

⇒f'x=324x2-1

⇒f'x=322x-12x+1

So, the function is increasing in x∈12,1.

Thus, the function will intersect the x-axis only at one point.

Hence, statement I is correct.

Now, putting x=π12in the equation cos3x=4cos3π12-3cosπ12.

⇒cosπ4=4cos3π12-3cosπ12

⇒42cos3π12-32cosπ12-1=0

Now, comparing with fx=42x3-32x-1 we get.

cosπ12 is a solution of fx.

So, statement-II is also correct.

Hence, both the statements are correct.

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Consider the function $f: R \rightarrow\{0,1\}$ such that $[2010-I I]$ $f(x)=\left\{\begin{array}{l}1, \text { if } x \text { is rational } \\ 0, \text { if } x \text { is irrational }\end{array}\right.$
Which one of the following is correct?

Options:
A The function is one-one into
B The function is many-one into
C The function is one-one onto
D The function is many-one onto

The correct answer is:
The function is one-one onto

Solution:
Since, on taking a straight line parallel to $\mathrm{x}$ -axis, the group of given function intersect it at one point. $\therefore \quad \mathrm{f}(\mathrm{x})$ is one-one.
and as range of $\mathrm{f}(\mathrm{x})=\mathrm{Co}$ -domain $\therefore \mathrm{f}(\mathrm{x})$ is onto.
Hence, $\mathrm{f}(\mathrm{x})$ is one-one onto.

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 2)
Consider the function \( f: R \rightarrow R \) defined by \( f(x)=\left\{\begin{array}{c}\left(2-\sin \left(\frac{1}{x}\right)\right)|x|, x \neq 0 \\ 0\end{array}, x=0 .\right. \) Then \( f \) is :

Options:
A monotonic on \( (-\infty, 0) \cup(0, \infty) \)
B not monotonic on \( (-\infty, 0) \) and \( (0, \infty) \)
C monotonic on \( (0, \infty) \) only
D monotonic on \( (-\infty, 0) \) only

The correct answer is:
not monotonic on \( (-\infty, 0) \) and \( (0, \infty) \)

Solution:
fx=-x2-sin1xx<00x=0

f'x=-2-sin1x-x-cos1x·-1x2x<02-sin1x+x-cos1x-1x2x>0

f'x=-2+sin1x-1xcos1x , x<02-sin1x+1xcos1x , x>0

f'(x) is an oscillating function which is non-monotonic in (-∞,0)∪(0,∞).

Topic: Mathematics, Application of Derivatives, JEE Main
Consider the function fx=a7x-12-x2bx2-7x+12,x<32sin(x-3)x-[x],x>3b,x=3,where [x] denotes the greatest integer less than or equal to x. If S denotes the set of all ordered pairs (a, b) such that f(x) is continuous at x=3, then the number of elements in S is :

Options:
A 2
B Infinitely many
C 4
D 1

The correct answer is:
1

Solution:
Given: fx=a7x-12-x2bx2-7x+12,x<32sin(x-3)x-[x],x>3b,x=3

⇒f3-=a7x-12-x2b|x2-7x+12|

⇒f3-=-a(x-3)(x-4)b(x-3)(x-4)

⇒f3-=-ab ...i

Now, f3+=2limx→3+sin(x-3)x-3

⇒f3+=2 ...ii

Also, f(3) = b ...iii

It is given that fx is continuous at x=3.

Hence, f(3)=f(3+)=f(3-)

⇒b=2=-ab

⇒b=2, a=-4

Hence, there is only 1 elements in S, which is ordered pair (-4,2).

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Consider the function fx=cos-12x+sin-12x-1, then
(where [.] represents the greatest integer part function)

Options:
A Domain of fx is x∈–∞,0
B Range of fx is singleton
C fx is an even function
D fx is an odd function

The correct answer is:
Range of fx is singleton

Solution:
For fx to be defined 2x=0,1
â´2xâ0,2âxâââ,1
Range cos-10+sin-1-1,cos-11+sin-10â¡0
â´ fx=0 âxâââ,1
So, Domain is not symmetric about the origin
â´ fx is neither even nor odd

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Consider the function fx=cos-1â¡3x+sin-1â¡3x-1, then (where, â represents the greatest integer part of x)

Options:
A the domain of fx is xâ-â, 0
B the range of fx is singleton
C fx is an even function
D fx is an odd function

The correct answer is:
the range of fx is singleton

Solution:
For fx to be defined 3x=0, 1
∴3x∈0, 2⇒x∈-∞,log3⁡2
Range of fx=cos-1⁡0+sin-1⁡-1,cos-1⁡1+sin-1⁡0≡0
∴fx=0 ∀x∈-∞,log3⁡2
So, the domain is not symmetric about the origin
∴fx is neither even nor odd

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Consider the function fx=maxsin⁡x, cos⁡x, ∀x∈0,3π. If λ is the number of points at which fx is non-differentiable, then the value of λ35 is _______. If your answer is K, mark 10K as answer after rounding off to nearest integer.

Options:

The correct answer is:
432

Solution:
Using graph of sin⁡x&cos⁡x, we get,
fx is non differentiable at
x1=π4, x2=3π4, x3=5π4, x4=7π4, x5=9π4, x6=11π4
Hence, λ=6

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=maxsin⁡x, cos⁡x, ∀x∈0,3π. If λ is the number of points at which fx is non-differentiable, then the value of λ35 is

Options:

The correct answer is:
43.20

Solution:
Using graph of sin⁡x&cos⁡x, we get,
fx is non differentiable at
x1=π4, x2=3π4, x3=5π4, x4=7π4, x5=9π4, x6=11π4
Hence, λ=6

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=minx2-9, x2-1, then the number of points where fx is non-differentiable is/are

Options:
A 0
B 7
C 6
D 4

The correct answer is:
6

Solution:
Using the graph of y=x2-9,y=x2-1
Clearly, from the graph,
we can see that f(x) is non-differentiable at 6 points

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=min⁡x2-4,x2-1, then the number of points where fx is non-differentiable is/are

Options:
A 0
B 7
C 6
D 4

The correct answer is:
6

Solution:
Using the graph of y=x2-4,y=x2-1
Clearly, from the graph we can see fx is non-differentiable at 6 points.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=Pxsinx-2, x≠2, and fx=7, x=2where Px is a polynomial such that P"x is always a constant and P3=9. If fx is continuous at x=2, then P5 is equal to __________.

Options:

The correct answer is:
39

Solution:
We have, fx=Pxsinx-2,7,x≠2x=2

P"x= Constant

⇒Px is a 2 degree polynomial

Let Px=x-2ax+b

fx is continuous at x=2

f2+=f2-

Now, limx→2+Pxsinx-2=7

⇒limx→2+x-2ax+bsinx-2=7

⇒2a+b=7

⇒P3=3-23a+b=9

⇒3a+b=9

⇒a=2,b=3

Hence, P5=5-22.5+3

=3.13

=39.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider the function fx=sin⁡2xtan22x, x≠π4. The value of f( π 4 ) such that f is continuous at x= π 4 is

Options:
A e
B 1/e
C 2
D None of these

The correct answer is:
1/e

Solution:
fπ4 =limx→π4⁡sin⁡2xtan2⁡2x
=limx→π4⁡etan2⁡2xsin⁡2x-1= limx→π4⁡esin⁡2x-1cot2⁡2x
= elimx→π42 cos 2x − 02 cot 2x (−2cosec2 2x)=limx→π4  e−sin3 2x2=e−12

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=sinx-4.tan-11x-4x≠40x=4, then

Options:
A f(x) is continuous and differentiable at x=4
B fx is continuous but non differentiable at x=4
C fx is discontinuous but differentiable at x=4
D fx is discontinuous and non differentiable at x=4

The correct answer is:
fx is continuous but non differentiable at x=4

Solution:
Let us shift the graph of fx in the direction of the negative x-axis by 4 units
& let gx=fx+4
∵gx at x=0 will have the same behaviour as of fx at x=4
gx=sin⁡x. tan-11x0 x≠0 x=0
Continuity at x=0,
limx→0-gx=limx→0-sin⁡x.tan-11x=0×-π/2=0
limx→0+gx=limx→0+sin x.tan-11x=0×π/2=0
∴limx→0+gx=g0 ∴gx is continuous at x=0
⇒fx is continuous at x=4
Differentiability at x=0
g'0−=limh→0+g0−h−g0−h=limh→0+sin−htan−1−1h−0−h
⇒limh→0+-sin hhtan-11h=-1.π2=-π2
g'0+=limh→0+g0+h−g0h=limh→0+sin h.tan−11h−0h
⇒limh→0+sin hhtan-11h=π2
Clearly, g'0−≠g'0+
∴gx is non-differentiable at x=0
⇒fx is non-differentiable at x=4

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function fx=tan-13x-23+2x, âxâ¥0. If gx is the inverse function of fx, then the value of g â² Ï 4 is equal to

Options:

The correct answer is:
26

Solution:
fx=tan-1x-231+x.23=tan-1x-tan-123
⇒f'x=11+x2 ... 1
Now, g'fx=1f'x ∵ gfx=x}
Putting x=5, we get,
g'f5=1f'5⇒g'tan−135−23+25=111+52
⇒g'tan−11=26⇒g'π4=26

Topic: Mathematics, Differentiation, JEE Main
Consider the function :
$f(x)=[x]+|1-x|,-1 \leq x \leq 3$ where $[x]$ is the greatest integer function.
Statement 1: $f$ is not continuous at $x=0,1,2$ and 3
Statement 2:f(x)= $=\begin{array}{cc}-x, & -1 \leq x < 0 \\ 1-x, & 0 \leq x < 1 \\ 1+x, & 1 \leq x < 2 \\ 2+x, & 2 \leq x \leq 3\end{array}$

Options:
A
Statement 1 is true; Statement 2 is false,

B
Statement 1 is true; Statement 2 is true; Statement 2 is not correct explanation for Statement 1.

C
Statement 1 is true; Statement 2 is true; Statement It is a correct explanation for Statement 1.

D
Statement 1 is false; Statement 2 is true.

The correct answer is:
Statement 1 is true; Statement 2 is false,

Solution:
Let $f(x)=[x]+|1-x|,-1 \leq x \leq 3$ where $[x]=$ greatest integer function. $f$ is not continuous at $x=0,1,2,3$ But in statement-2 $f(x)$ is continuous at $x=3$.
Hence, statement-1 is true and 2 is false.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2013 (25 Apr Online)
Consider the function fx=x-2x2-3x+2, then the incorrect statement is

Options:
A fx is continuous at x=1
B fx is continuous at x=2
C fx is differentiable at x=1
D fx is differentiable at x=2

The correct answer is:
fx is differentiable at x=1

Solution:
∵ x-2 & x2-3x+2 both are continuous ∀x∈R
∴fx is continuous ∀x∈R
Now, x2-3x+2=x-1x-2
∴ fx=x-2x-1x-2:x∈-∞,1∪2,∞-x-2x-1x-2:x∈1,2
fx=x-1x-22:x∈-∞,1∪2,∞-x-1x-22:x∈1,2
f'x=x−12x−2+x−22: x∈−∞,1∪2,∞−x−12x−2−x−22: x∈1,2
Clearly, f ′ 2 − =f' 2 + =0⇒f x is differentiable at x=2
and f ′ 1 − =1, f ′ 1 + =−1⇒f x is non-differentiable at x=1

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the function $f(x)=|x-2|+|x-5|, x \in R$.
Statement $1$: $f^{\prime}(4)=0$
Statement $2$: $f$ is continuous in $[2,5]$, differentiable in $(2,5)$ and $f(2)=f(5)$.

Options:
A
Statement $1$ is false, statement $2$ is true

B
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

C
Statement $1$ is true, statement $2$ is true; statement $2$ is not a correct explanation for statement $1$

D
Statement $1$ is true, statement $2$ is false

The correct answer is:
Statement $1$ is true, statement $2$ is true; statement $2$ is a correct explanation for statement $1$

Solution:
$f(x) =7-2 x ; x < 2$
$=3$ ; $\quad2 \leq x \leq 5$
$=2 x-7 ; x>5$
$f(x)$ is constant function in $[2,5]$
$f$ is continuous in $[2,5]$ and differentiable in $(2,5)$ and $f(2)=f(5)$
by Rolle's theorem $f^{\prime}(4)=0$
$\therefore$ Statement $2$ and statement $1$ both are true and statement $2$ is correct explanation for statement $1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2012 (Offline)
Consider the function
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{3} & \sin \mathrm{x} & \cos \mathrm{x} \\ 6 & -1 & 0 \\ \mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}\end{array}\right|$ where $\mathrm{p}$ is a constant.
What is the value of $\mathrm{f}^{\prime}(0) ?$

Options:
A $\mathrm{p}^{3}$
B $3 \mathrm{p}^{3}$
C $6 \mathrm{p}^{3}$
D $-6 \mathrm{p}^{3}$

The correct answer is:
$-6 \mathrm{p}^{3}$

Solution:
$f(x)=\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$f^{\prime}(x)=\left|\begin{array}{ccc}3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|+\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 6 & -1 & 0 \\ 0 & 0 & 0\end{array}\right|$ $+\left|\begin{array}{ccc}x^{3} & \sin x & \cos x \\ 0 & 0 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$\begin{aligned} f^{\prime}(x) &=\left|\begin{array}{ccc}3 x^{2} & \cos x & -\sin x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right| \\ f^{\prime}(0) &=\left|\begin{array}{ccc}0 & \cos 0 & -\sin 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right| \\ &=-6 p^{3} \end{aligned}$

Topic: Mathematics, Determinants, JEE Main
Consider the function
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}\mathrm{x}^{3} & \sin \mathrm{x} & \cos \mathrm{x} \\ 6 & -1 & 0 \\ \mathrm{p} & \mathrm{p}^{2} & \mathrm{p}^{3}\end{array}\right|$ where $\mathrm{p}$ is a constant.
What is the value of p for which $\mathrm{f}^{\prime \prime}(0)=0$ ?

Options:
A $-\frac{1}{6}$ or 0
B $-1$ or 0
C $-\frac{1}{6}$ or 1
D $-1$ or 1

The correct answer is:
$-\frac{1}{6}$ or 0

Solution:
$f^{\prime \prime}(x)=\left|\begin{array}{ccc}6 x & -\sin x & -\cos x \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$f^{\prime \prime}(0)=\left|\begin{array}{ccc}0 & 0 & -\cos 0 \\ 6 & -1 & 0 \\ p & p^{2} & p^{3}\end{array}\right|$
$\quad=-1\left(6 p^{2}+p\right)$
$f^{\prime \prime}(0)=0$
$-\left(6 p^{2}+p\right)=0$
$p(6 p+1)=0$
$p=0$ or $p=-\frac{1}{6}$

Topic: Mathematics, Determinants, JEE Main
Consider the function fx=x3-xx2-6x+5, âxâR, then fx is

Options:
A discontinuous at x=1
B discontinuous at x=5
C non differentiable at x=1
D non differentiable at x=5

The correct answer is:
non differentiable at x=5

Solution:
fx=x3-xx-1x-5
=xx+1x-1x-1x-5
fx=x3-xx2-6x+5x∈-∞,1∪5,∞-x3-xx2-6x+5x∈1,5
∵x3-x, x2-6x+5 both are continuous∀x∈R

∴fx is also continuous ∀x∈R
Now,
f'x=x3−x2x−6+x2−6x+53x2−1x∈−∞,1∪5,∞−x3−x2x−6−x2−6x+53x2−1x∈1,5
Now, check differentiability
At x=1
f ′ 1 − = 1 3 −1 2−6 + 1−6+5 3−1 =0
f ′ 1 + =− 1−1 2−6 − 1−6+5 3−1 =0
∵f'1−=f1+ ∴ at x=1 differentiable
At x=5
f ′ 5 − = 125−5 10−6 +0
f ′ 5 + =− 125−5 10−6 −0
Clearly, f ′ 5 − ≠ f ′ 5 +
∴ non-differentiable at x=5

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the given chemical reaction :
Product " \(\mathrm{A}\) " is :

Options:
A picric acid
B acetic acid
C adipic acid
D oxalic acid

The correct answer is:
adipic acid

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Consider the given compounds
(1) CH3–CH2–NH2
(2) CH3–CH=NH
(3) CH3–C≡N
(4) C2H5–NH–C2H5
Arrange basicity of these compounds in decreasing order:

Options:
A 4 > 1 > 2 > 3
B 1 > 2 > 3 > 4
C 1 > 4 > 2 > 3
D 4 > 1 > 3 > 2

The correct answer is:
4 > 1 > 2 > 3

Solution:
Here the correct order of basic nature is as follows
C 2 H 5  – NH –  C 2 H 5  >  CH 3  –  CH 2  –  NH 2  >  CH 3  – CH = NH >  CH 3  – C≡N
(i) Secondary amine is more basic than primary amine.
(ii) CH3–CH=Nsp2H
iii   CH 3 −C¹≡ N sp

Topic: Chemistry, General Organic Chemistry, JEE Main
Consider the given diagram. An ideal gas is contained in a chamber (left) of volume $V$ and is at an absolute temperature $T$. It is allowed to rush freely into the right chamber of volume $V$ which is initially vacuum. The whole system is thermally isolated. What will be the final temperature of the equilibrium has been attained?

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

The correct answer is:
$T$

Solution:
Since, the whole system is isolated, this means, there is no transfer of heat between the system and the surrounding. Also, the right side container is initially vacuum, so the gas could easily rush there without any resistive force. As the right side container has no temperature. Thus, the temperature of the ideal gas would remain same even if it enters right side chamber. Therefore, final temperature attained at the equilibrium will be $T$

Topic: Physics, Thermodynamics, WBJEE, WBJEE 2019
Consider the given plot of enthalpy of the following reaction between A and B.
A+B→C+D. Identify the incorrect statement.

Options:
A Formation of A and B from C has highest enthalpy of activation.
B D is kinetically stable product.
C C is the thermodynamically stable product.
D Activation enthalpy to form C is 5 kJ mol-1 less than that to form D.

The correct answer is:
Activation enthalpy to form C is 5 kJ mol-1 less than that to form D.

Solution:
A+B→C+D
Activation enthalpy for C=20-0=20 kJ/mol
Activation enthalpy for D=15-5=10 kJ/mol

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Consider the given plots for a reaction obeying Arrhenius equation (0°C<T<300°C): ( K and Ea are rate constant and activetion energy, respectively )
(I)

(II)

Options:
A I is wrong but II is right
B Both I and II are wrong
C I is right but II is wrong
D Both I and II are correct

The correct answer is:
Both I and II are correct

Solution:
K=Ae-Ea/RT
K decreases exponentially with Ea
â´ The first graph is correct.
But temperature is given in oC
So, Kâ 0 at 0oC.
â´ For temperature between 0oCand 300oC, the graph will come like (II)

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Consider the given problem:
$\begin{aligned}
& 5 x+y \leq 100 \\
& x+y \leq 60 \\
& x \geq 0 \\
& y \geq 0
\end{aligned}$
If we solve the above linear equations by the graphical method of Linear Programming, then the following point will not form the the boundary of the feasible region.

Options:
A $(60,0)$
B $(20,0)$
C $(0,60)$
D $(10,50)$

The correct answer is:
$(60,0)$

Solution:
After equating each constraint to zero (0) we get the equations of lines on cartesian coordinate.
On comparing the inequality with $(0,0)$ and shading the common area we get the feasible region as follows,
Hence we can easily conclude that point $(60,0)$ and $(0,100)$ are outside the feasible region.

Topic: Mathematics, Linear Programming, JEE Main
Consider the given reaction, Identify X'' and Y'' :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Consider the given reaction, percentage yield of :

Options:
A C>A>B
B B>C>A
C A>C>B
D C>B>A

The correct answer is:
C>B>A

Solution:
% yield order ⇒C>B>A

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider the given reaction, the product 'X' is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Consider the given reaction. The total number of oxygen atoms present per molecule of the product (P) is _________.

Options:

The correct answer is:
1

Solution:
During ozonolysis, O3​ adds across C=C, and it produces ozonide.

The ozonide on hydrolysis produces carbonyl compounds. The reaction is used to locate double bond in alkene.

Hence total number of oxygen atom present per molecule of

is 1.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Consider the given statement about the molecule

1. Three carbon atoms are sp3 -hybridised

2. Three carbon atoms are sp2 -hybridised

3. Two carbon atoms are sp-hybridised

Of three statements

Options:
A 1, 2 and 3 are correct
B 1 and 2 are correct
C 2 and 3 are correct
D 1 and 3 are correct

The correct answer is:
1 and 3 are correct

Solution:
Here statements 1 and 3 are correct.
This molecule has four carbon atoms (i.e., -CH=CH- and -CH=CH2)sp2 -hybridised so statement 2 is wrong.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Consider the graph between compressibility factor Z and pressure P :

The correct increasing order of ease of liquefaction of the gases shown in the above graph is

Options:
A H 2 â<â N 2 â<â CH 4 â<â CO 2
B CO 2  <  CH 4  <  N 2  <  H 2
C H 2  <  CH 4  <  N 2  <  CO 2
D CH 4  <  H 2  <  N 2  <  CO 2

The correct answer is:
H 2  <  N 2  <  CH 4  <  CO 2

Solution:
Z<1 means, force of attraction dominating i.e., a is considerable, b can be negligible at low temperature and low pressure and liquification is more.

Topic: Chemistry, States of Matter, JEE Main
Consider the graph of Gibbs free energy G vs extent of reaction. The number of statement/s from the following which are true with respect to points (a), (b) and (c) is ______

A. Reaction is spontaneous at (a) and (b)

B. Reaction is at equilibrium at point (b) and non-spontaneous at point (c)

C. Reaction is spontaneous at (a) and non-spontaneous at (c)

D. Reaction is non-spontaneous at (a) and (b)

Options:

The correct answer is:
2

Solution:
For spontaneous process, ∆G=-ve, at equilibrium ∆G=0 and for non-spontaneous process ∆G=+ve.

The relation between the change in Gibbs reaction energy and Gibbs energy can be defined as the slope of the Gibbs energy plotted against the extent of reaction at constant pressure and temperature.

At point a: Slope=-ve

∆G=-ve

Hence, reaction is spontaneous at point a.

At point b: Slope=0

∆G=0

Hence. it is the equilibrium condition.

At point c: Slope=+ve

∆G=+ve

Hence, reaction is non-spontaneous at point c.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (06 Apr Shift 1)
Consider the greatest integer function, defined by f(x)=[x],0≤x<2. Then

Options:
A f is derivable at x=1
B f is not derivable at x=1
C f is derivable at x=2
D None of these

The correct answer is:
f is not derivable at x=1

Solution:
The given function is

fx=0,0≤x<11,1≤x<2

Now, LHD=limx→1-f(x)-f(1)x-1=limx→1-0-1x-1

Which does not exist

∴ f is not derivable at x=1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider the ground state of $\mathrm{Cr}$ atom $(Z=24)$. The number of electrons with the azimuthal quantum numbers $\mathrm{I}=1$ and 2 are respectively

Options:
A
12 and 4

B
16 and 5

C
16 and 4

D
12 and 5

The correct answer is:
12 and 5

Solution:
$24 \longrightarrow 1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^1 3 d^5$
$I=1 \rightarrow p \longrightarrow 12$
$I=2 \rightarrow d \longrightarrow 5$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2004
Consider the ground state of \( \mathrm{Cr} \) atom \( (\mathrm{Z}=24) \). The numbers of electrons with azimuthal quantum numbers, \( \mathrm{l}=1 \) and \( l=2 \), respectively are

Options:
A \( 12 \) and \( 4 \)
B \( 16 \) and \( 4 \)
C \( 16 \) and \( 5 \)
D \( 12 \) and \( 5 \)

The correct answer is:
\( 12 \) and \( 5 \)

Solution:
24Cr: 1s2 2s2 2p6 3s2 3p6 3d5 4s1
We know for p, l=1 and for d, l=2.
For l=1, the total number of electrons =12 2p6 3p6
For l=2, the total number of electrons =5 3d5

Topic: Chemistry, Structure of Atom, JEE Main
Consider the ground state of $(Z=24)$. The numbers of electrons with the azimuthal quantum numbers, $I=1$ and 2 are, respectively

Options:
A 16 and 4
B 12 and 5
C 12 and 4
D 16 and 5

The correct answer is:
12 and 5

Solution:
(We know that for $p$ the value of $l=1$ and for $d, l=2$ )
For $I=1$ total number of electron $=12$
For $I=2$ total number of electron $=5$.

Topic: Chemistry, Structure of Atom, JEE Main
Consider the hydrated ions of Ti2+, V2+, Ti3+ and Sc3+ . The correct order of their spin-only magnetic moments is:

Options:
A Sc3+<Ti3+<V2+<Ti2+
B Ti3+<Ti2+<Sc3+<V2+
C Sc3+<Ti3+<Ti2+<V2+
D V2+<Ti2+<Ti3+<Sc3+

The correct answer is:
Sc3+<Ti3+<Ti2+<V2+

Solution:
Magnetic moment μ=nn+2
More than value of n, more will be magnetic moment.
Ti+2 electronic configuration;
3d2 μ=22+2=2.82
Ti+3⇒3d1 μ=11+2=3=1.73
V+2⇒3d3 μ=33+2=3.87
Sc3+⇒Ar no unpaired electron. →μ=0
Hence Order of Magnetic Moment is V+2>Ti+2>Ti+3>Sc+3

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0, 1, 2, ……..n + 1. Where n is the principal quantum number. Then, the element with atomic number:

Options:
A 9 is the first alkali metal
B 13 has a half-filled valence subshell
C 8 is the first noble gas
D 6 has a 2p-valence subshell

The correct answer is:
13 has a half-filled valence subshell

Solution:
For n=1 value of l=0, 1, 2

For n=2 value of l=0, 1, 2, 3

So, according to n+l rule the filling order of subshells will be :

1s 1p 2s 1d 2p 3s 2d 3p 4s.....

(1) 1st alkali metal will have configuration 1s2 1p6 so atomic number will be 8.

(2) 1st alkali metal will have electronic configuration ⇒ 1s1 ⇒ Z=1

(3) Electronic configuration of CZ=6 ⇒ 1s2 1p4

(4) Z=13, Electronic configuration =1s2 1p6 2s2 1d3

So it will not have half-filled electronic configuration.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2020 (03 Sep Shift 2)