datavorous/entrance-exam-dataset huggingface dataset:
The number of solutions of the equation cot2⁡sin⁡x+3=1 in 0,3π is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
cot2⁡sin⁡x+3=1=cot2⁡π4
⇒sin⁡x+3=nπ±π4
Also, 2≤sin⁡x+3≤4
⇒sin⁡x+3=π-π4,π+π4
⇒sin⁡x=3π4-3 or 5π4-3

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation cotâ¡xcosâ¡x-1=cotâ¡x-cosâ¡x, âxâ0,2Ï is equal to

Options:

The correct answer is:
2

Solution:
cotx⋅cosx−cotx+ cosx−1 =0
⇒cot⁡x+1cos⁡x-1=0
⇒cot⁡x=-1 or cos⁡x=1 (Not possible as sin⁡x≠0)
Hence, 2 solutions are possible in 0,2π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation \( \cot x \cos x-1=\cot x-\cos x, \forall x \in[0,2 \pi] \) is equal to

Options:

The correct answer is:
2

Solution:
cotx⋅cosx−cotx+cosx−1=0

⇒cotx+1cosx-1=0

⇒cotx=-1 or cosx=1 (Not possible as sinx≠0)

Hence, 2 solutions are possible in 0,2π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation cot⁡x=cot⁡x+cosecx in 0,10π is/are

Options:

The correct answer is:
5

Solution:
If x lies in 1st or 3rd quadrant then cotâ¡x=cotâ¡x, thus equation becomes
cotâ¡x=cotâ¡x+1sinâ¡xâ1sinâ¡x=0, not possible
If x lies in 2nd or 4th quadrant, then cotâ¡x=-cotâ¡x, thus equation becomes
-cotâ¡x=cotâ¡x+1sinâ¡x
â-2cotâ¡x=1sinâ¡xâ-2cosâ¡xsinâ¡x=sinâ¡x
âsinâ¡x1+2cosâ¡x=0âcosâ¡x=-12âµsinâ¡xâ 0
âx=2Ï3 is the only solution
(In 2nd quad.)
In 0,2Ï there is only one solution
In 0,10Ï, we get, five solutions

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x},(0 \leq x \leq 2 \pi)$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
With the help of group of cot $x$, we know that When $x \in\left(0, \frac{\pi}{2}\right] \cup\left(\pi, \frac{3 \pi}{2}\right]$, then $\cot x \geq 0$
$\therefore \quad|\cot x|=\cot x+\frac{1}{\sin x}$ $\Rightarrow \cot x=\cot x+\frac{1}{\sin x} \Rightarrow \frac{1}{\sin x}=0$
$\Rightarrow$ No solution exist. $\quad \text{...(i)}$
When $x \in\left(\frac{\pi}{2}, \pi\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$, then $\cot x < 0$
$\begin{aligned} & &|\cot x|=\cot x+\frac{1}{\sin x} \\ \therefore \quad &-\cot x &=\cot x+\frac{1}{\sin x} \\ \Rightarrow \quad &-2 \cot x &=\frac{1}{\sin x} \\ \Rightarrow \quad & & \cos x=\frac{-1}{2} \Rightarrow x=\frac{2 \pi}{3} \quad \text{...(ii)} \end{aligned}$
$\therefore$ From Eqs. (i) and (ii), only one solution exists.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation log2cosâ¡Î¸2 + log4cosâ¡Î¸16cosâ¡Î¸=2 in the interval 0,2Ï is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
For domain, cos⁡θ>0⇒θ∈0,π2∪3π2,2π
Also, if cos⁡θ>0⇒4cos⁡θ≠1
⇒log2cos⁡θ2+log216cos⁡θlog24cos⁡θ=2
⇒log2cosθ2+log216+log2cosθlog24-log2cosθ=2
Let, log2cosθ=t

⇒t2+4+t2-t=2
⇒2t2-t3+4+t=4-2t⇒t3-2t2-3t=0
⇒t=0,3,-1 .
t=0⇒log2cosθ=0⇒cosθ=1⇒θ=0
t=-1⇒log2cosθ=-1⇒cosθ=12⇒θ=π3,5π3.
t=3⇒log2cosθ=3⇒cosθ=8 (not possible)
⇒ Number of solutions are 3

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation log2sin⁡x(1+cos⁡x)=2 in the interval 0,5π is

Options:

The correct answer is:
3

Solution:
log2sin⁡x1+cos⁡x=2
1+cos⁡x=2sin⁡x2=2sin2⁡x
1+cos⁡x=2-2cos2⁡x
2cos2⁡x+cos⁡x-1=0
⇒cos⁡x+12cos⁡x-1=0
⇒cos⁡x=12 or cosx=-1 (rejected)
⇒x=π3,5π3 in 0,2π
But at x=5π3, 2sin⁡x is negative
⇒x=π3 in 0,2π
⇒x=π3, π3+2π, π3+4π
⇒3 solutions in 0,5π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation log4x-1=log2x-3 is ______.

Options:

The correct answer is:
1

Solution:
log4x-1=log2x-3

⇒12log2x-1=log2x-3

⇒log2x-11/2=log2x-3

⇒x-11/2=x-3

⇒x-1=x2+9-6x

⇒x2-7x+10=0

⇒x-2x-5=0

⇒x=2,5

But x≠2 because it is not satisfying the domain of given equation i.e log2x-3→ its domain x>3 finally x is 5

∴ No. of solutions =1.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The number of solutions of the equation log(x+1)2x2+7x+5+log(2x+5)(x+1)2-4=0, x>0, is

Options:

The correct answer is:
1

Solution:
We have,

log(x+1)2x2+7x+5+log(2x+5)(x+1)2-4=0, x>0

⇒logx+12x+5x+1+2log2x+1x+1=4

⇒logx+12x+5+logx+1x+1+2log2x+1x+1=4

⇒logx+12x+5+1+2log2x+1x+1=4

⇒logx+12x+5+2log2x+1x+1=3

⇒logx+12x+5+2logx+12x+5=3

Put log(x+1)(2x+5)=t, then

t+2t=3

⇒t2-3t+2=0

⇒t=1, 2

Then,

log(x+1)(2x+5)=1 and log(x+1)(2x+5)=2

⇒2x+5=x+1 and 2x+5=x+12

x=-4 (rejected) as x>0

And,

2x+5=x+12

⇒x2=4

⇒x=2,-2 (rejected)

So, x=2

Number of solution is 1.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The number of solutions of the equation sin-1x=sin⁡x-1 is/are

Options:
A one
B two
C three
D zero

The correct answer is:
two

Solution:
∵sin⁡x-1=1sin⁡x=cosec⁡x
Now, from graphs of sin-1⁡x,cosec⁡x
Clearly, both graph intersects at two points
∴ two solutions

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of solutions of the equation sin-1x2+13+cos-1x2-23=x2 for x∈[-1,1], and [x] denotes the greatest integer less than or equal to x, is :

Options:
A 2
B 0
C 4
D Infinite

The correct answer is:
0

Solution:
Given equation is

sin-1x2+13+cos-1x2-23=x2

Now, sin-1x2+13 is defined if -1≤x2+13≤1

⇒-1≤x2+13<2

⇒-43≤x2<53

⇒0≤x2<53 ...1

Also, and cos-1x2-23 is defined if -1≤x2-23≤1

⇒-1≤x2-23<2

⇒-13≤x2<83

⇒0≤x2<83 ...2

So, from (1) and (2) we can conclude 0≤x2<53

Case -I: If 0≤x2<23

⇒sin-1(0)+cos-1(-1)=x2

⇒0+π=x2

⇒x2=π but π∉0,23

⇒ No value of 'x'

Case - II: If 23≤x2<53

⇒sin-1(1)+cos-1(0)=x2

⇒π2+π2=x2

⇒x2=π but π∉23,53

⇒ No value of 'x'

So, number of solutions of the equation is zero.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The number of solutions of the equation, $\sin ^{-1} x=2 \tan ^{-1} x$ (in principal values) is :

Options:
A
1

B
4

C
2

D
3

The correct answer is:
1

Solution:
Given equation is
$$
\sin ^{-1} x=2 \tan ^{-1} x
$$
Now, this equation has only one solution.
$$
\begin{aligned}
& \therefore \quad \text { LHS }=\sin ^{-1} 1=\frac{\pi}{2} \\
& \text { and } \text { RHS }=2 \tan ^{-1} 1=2 \times \frac{\pi}{4}=\frac{\pi}{2}
\end{aligned}
$$
Also, $x=1$ gives angle value as $\frac{\pi}{4}$ and $\frac{5 \pi}{4}$ $\frac{5 \pi}{4}$ is outside the principal value.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2013 (22 Apr Online)
The number of solutions of the equation $\sin 2 x-2 \cos x+4 \sin x=4$ in the interval $[0,5 \pi]$ is :

Options:
A
3

B
5

C
4

D
6

The correct answer is:
3

Solution:
$$
\text { } \begin{aligned}
& \sin 2 x-2 \cos x+4 \sin x=4 \\
& \Rightarrow 2 \sin x \cdot \cos x-2 \cos x+4 \sin x-4=0 \\
& \Rightarrow(\sin x-1)(\cos x-2)=0
\end{aligned}
$$
$$
\begin{aligned}
& \because \cos x-2 \neq 0, \therefore \sin x=1 \\
& \therefore \quad x=\frac{\pi}{2}, \frac{5 \pi}{2}, \frac{9 \pi}{2}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2013 (23 Apr Online)
The number of solutions of the equation sin3⁡xcos⁡x+sin2⁡xcos2⁡x+cos3⁡xsin⁡x+1=0 in the interval 0,2π is equal to

Options:
A 0
B 2
C 4
D 8

The correct answer is:
0

Solution:
The given equation can be written as sin⁡xcos⁡xsin2⁡x+cos2⁡x+sin2⁡xcos2⁡x+1=0
Substituting t=sin⁡xcos⁡x in this equation, we get t2+t+1=0.
This equation has no real roots.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation sin x=cos2 x in the interval 0, 10 is ______.

Options:

The correct answer is:
4

Solution:
Given, sinx=cos2x

⇒sinx=1-sin2x

⇒ sin2x+sinx-1=0

⇒ sinx=-1±52

⇒ sinx=5-12

Now plotting the graph we get,

⇒ Now by graph number of solutions in 0,10=4

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The number of solutions of the equation sinxâsin2xâsin3xâsin4xâsin5x=0 in 0,Ï is equal to

Options:

The correct answer is:
11

Solution:
Note sin⁡4x=0 contains all solutions of sin⁡x=0, sin⁡2x=0
sin⁡4x=0⇒x=0,π4,π2,3π4,π
sin⁡3x=0⇒x=0,π3,2π3,π
sin⁡5x=0⇒x=0,π5,2π5,3π5,4π5,π
Hence, the number of solutions =11

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation tan2x-sec10x+1=0 for xâ0,20 is equal to

Options:

The correct answer is:
6

Solution:
sec2x-sec10x=0
sec2x1-sec8x=0
1-sec8x=0 as sec2x≥1
⇒sec8x=1⇒cos8x=1⇒cosx=±1
⇒x=nπ, ∀n∈Ι
⇒x=π,2π,3π,4π,5π,6π∈0,20
So, the required number of roots =6

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation \( \tan ^{2} x-\sec ^{10} x+1=0 \) for \( x \in(0,20) \) is equal to

Options:

The correct answer is:
6

Solution:
sec2x-sec10x=0
sec2x1-sec8x=0
1-sec8x=0 as sec2x≥1
⇒sec8x=1⇒cos8x=1⇒cosx=±1
⇒x=nπ, ∀n∈Ι
⇒x=π,2π,3π,4π,5π,6π∈0,20
So, the required number of roots =6

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation tan⁡x+sec⁡x=2cos⁡x lying in the interval 0,5π is

Options:

The correct answer is:
6

Solution:
tan⁡x+sec⁡x=2cos⁡x
⇒sin⁡x+1=2cos2⁡x=2-2sin2⁡x
⇒2sin2⁡x+sin⁡x-1=0
⇒sin⁡x=12,-1
sin⁡x=-1⇒cos⁡x=0 and hence tan⁡x, sec⁡x is not defined
So, sin⁡x=12

Hence, in 0,5π there are 6 solutions.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation tan⁡xsin⁡x-1=tan⁡x-sin⁡x, ∀x∈0,2π is equal to.

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
tan⁡xsin⁡x-tan⁡x+sin⁡x-1=0
⇒tan⁡x+1sin⁡x-1=0
⇒tan⁡x=-1 or sin⁡x=1 (Not possible as x≠π2 )
Hence, for tanx=-1 there are 2 solutions in 0,2π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solution(s) of the equation $\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}$ is/are

Options:
A 2
B 0
C 3
D 1

The correct answer is:
0

Solution:
Given equation is
$$
\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}
$$
On squaring both sides, we get
$$
(x+1)+(x-1)-2 \sqrt{x^{2}-1}=4 x-1
$$
$\Rightarrow \quad 2 x-2 \sqrt{x^{2}-1}=4 x-1$
$\Rightarrow \quad-2 \sqrt{x^{2}-1}=2 x-1$
Again, squaring both sides, we get
$$
4\left(x^{2}-1\right)=4 x^{2}+1-4 x
$$
$\Rightarrow \quad-4=+1-4 x$
$\Rightarrow \quad 4 x=5$
$\Rightarrow$
$$
x=\frac{5}{4}
$$
But, when we put $x=\frac{5}{4}$ in Eq. (i), we get
$$
\sqrt{\frac{5}{4}+1}-\sqrt{\frac{5}{4}-1}=\sqrt{4 \times \frac{5}{4}-1}
$$
$\Rightarrow$
$\sqrt{\frac{9}{4}}-\sqrt{\frac{1}{4}}=\sqrt{5-1} \Rightarrow \frac{3}{2}-\frac{1}{2}=2$
$\Rightarrow 1=2,$ which is not true.
Hence, no value of $x$ satisfy the given equation.

Topic: Mathematics, Quadratic Equation, JEE Main
The number of solutions of the equation z3+3zÂ¯2z=0 (where, z is a complex number) are

Options:
A 2
B 3
C 6
D 5

The correct answer is:
5

Solution:
Let, z=reiθ
⇒r3ei3θ+3re-i2θ=0
⇒r2ei5θ=-3
⇒r2=3 and ei5θ=-1
⇒r=3 and θ=π5+2k5 where, k=0,1,2,3,4
⇒5 solutions

Topic: Mathematics, Complex Number, JEE Main
The number of solutions of the pair of equations $2 \sin ^2 \theta-\cos 2 \theta=0$ and $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is

Options:
A
zero

B
one

C
two

D
four

The correct answer is:
two

Solution:
$2 \sin ^2 \theta-\cos 2 \theta=0 \Rightarrow \sin ^2 \theta=\frac{1}{4}$
Also, $2 \cos ^2 \theta=3 \sin \theta \Rightarrow \sin \theta=\frac{1}{2}$
$\Rightarrow$ Two solutions in $[0,2 \pi]$.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of x3+4x-1=0 in the interval x∈-2,1 is

Options:

The correct answer is:
1

Solution:
Let, fx=x3+4x-1
â´f'x=3x2+4>0, âxâR
Also, f-2=-17<0 & f1=4>0
As fx is continuous and increasing, it has exactly one root in the interval -2,1.

Topic: Mathematics, Application of Derivatives, JEE Main
The number of solutions, the equation sin4x+cos4x=sinxcosx has, in [π,5π] is/are

Options:

The correct answer is:
4

Solution:
sin4x+cos4x=sinxcosx
â(sin2x+cos2x)2-2sin2xcos2x=sinxcosx
â2sin2xcos2x+sinxcosx-1=0
â2sinxcosx-1sinxcosx+1=0
âsin2x=1 â¦(i)
xâÏ,5Ïâ2xâ2Ï,10Ï
from (i), 2x=5Ï2,9Ï2,13Ï2,17Ï2
âx=5Ï4,9Ï4,13Ï4,17Ï4
â The number of solutions =4

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions to sinπsin2θ+sinπcos2θ=2cosπ2cosθ satisfying 0≤θ≤2π is

Options:
A 1
B 2
C 4
D 7

The correct answer is:
7

Solution:
The given equation

sinπsin2θ+sinπcos2θ=2cosπ2cosθ

⇒2sinπsin2θ+cos2θ2

cosπsin2θ-cos2θ2=2cosπ2cosθ

⇒cosπ2cos2θ=cosπ2cosθ

⇒π2cos2θ=2nπ±π2cosθ, n∈ Integers

⇒cos2θ=4n±cosθ,n∈I.

Case I

If cos2θ=4n+cosθ⇒cos2θ-cosθ=4n

The above equation will hold only if n=0,

so cos2θ=cosθ

⇒2cos2θ-cosθ-1=0⇒cosθ=1,-12

⇒θ=2kπ,2kπ±2π3,k∈I

∵ θ∈0,2π

So, θ=0,2π,2π3,4π3

Case II

If cos2θ=4n-cosθ

⇒cos2θ+cosθ=4n

The above equation will hold only if n=0,

so cos2θ+cosθ=0

⇒2cos2θ+cosθ-1=0

⇒cosθ=-1,12

⇒θ=2m+1π,2mπ±π3,m∈I.

∵ θ∈0,2π

So, θ=π,π3,5π3

∴ There are total 7 solutions.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions to the equation 2esin⁡x-3-e2sin⁡x=esin⁡x-e2sin⁡x-1 in 0,2π is

Options:
A 0
B 2
C 4
D 3

The correct answer is:
2

Solution:
Putting esinâ¡x=t, we get,
2t-3-t2=t-t2-1
t2-2t+3=t2-t+1
t2-2t+3>0, âtâR and t2-t+1>0, âtâR as their discriminant<0
ât2-2t+3=t2-t+1ât=2
âesinâ¡x=2âsinâ¡x=lnâ¡2â0,1
Hence, from the graph, we get, in 0,2Ï there are two solutions

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions to the equation $\cos ^{4} x+\frac{1}{\cos ^{2} x}=\sin ^{4} x+\frac{1}{\sin ^{2} x}$ in the interval $[0,2 \pi]$ is :

Options:
A 6
B 4
C 2
D 0

The correct answer is:
4

Solution:
$\cos ^{4} x-\sin ^{4} x=\frac{1}{\sin ^{2} x}-\frac{1}{\cos ^{2} x}$
$\left(\cos ^{2} x-\sin ^{2} x\right)=\frac{\left(\cos ^{2} x-\sin ^{2} x\right)}{\sin ^{2} x \cos ^{2} x}$
$\cos 2 x=\frac{4 \cos 2 x}{\sin ^{2} 2 x}$
$\cos 2 x\left(1-4 \operatorname{cosec}^{2} 2 x\right)=0$
$\cos 2 x=0$
$2 x=2 n \pi \pm \frac{\pi}{2}$
$x=n \pi \pm \frac{\pi}{4}$
At $n=0, x=\frac{\pi}{4}$
$n=1 ; \quad x=\frac{5 \pi}{4}, \frac{3 \pi}{4}$
$n=2, x=\frac{7 \pi}{4}$

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions to x+y+z=10, where 1â¤x,y,zâ¤6 and x,y,zâN, is equal to

Options:
A 35
B 36
C 27
D 66

The correct answer is:
27

Solution:
x+y+z=10
Number of solutions is equal to the coefficient of x10 in the expansion of x+x2+x3+.â¦â¦+x63=x31-x61-x3
or the coefficient of x7 in the expansion of 1-x631-x-3=1-3x61-x-3
=C7+3-17-3C1+3-11
=C97-3C31=36-9=27

Topic: Mathematics, Permutation Combination, JEE Main
The number of solutions $(x, y, z)$ to the system of equations $x+2 y+4 z=9,4 y z+2 x z+x y=13, x y z=3$
Such that at least two of $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are integers is $-$

Options:
A 3
B 5
C 6
D 4

The correct answer is:
5

Solution:
$\begin{array}{l}
x+2 y+4 z=9 \\
2 x y+8 y z+4 x z=26 \\
(x)(2 y)(4 z)=24
\end{array}$
Say roots of $\mathrm{P}^{3}-9 \mathrm{P}^{2}+26 \mathrm{P}-24=0$ are $\mathrm{x}, 2 \mathrm{y}$ and $4 \mathrm{z}$
Here $(P-2)\left(P^{2}-7 P+12\right)=0$
$(P-2)(P-3)(P-4)=0$
now $4 \mathrm{z}=4 ; 2 \mathrm{y}=2 ; \mathrm{x}=3$ then $(3,1,1)$
$x=2 ; 2 y=4 ; 4 z=3$ then $\left(2,2, \frac{3}{4}\right)$
$\mathrm{x}=3 ; 2 \mathrm{y}=4 ; 4 \mathrm{z}=2$ then $\left(3,2, \frac{1}{2}\right)$
$\mathrm{x}=2 ; 2 \mathrm{y}=3 ; 4 \mathrm{z}=4$ then $\left(2, \frac{3}{2}, 1\right)$
$\mathrm{x}=4 ; 2 \mathrm{y}=2 ; 4 \mathrm{z}=3$ then $\left(4,1, \frac{3}{4}\right)$
Hence there are five solutions.

Topic: Mathematics, Determinants, JEE Main
The number of sp3 hybridised carbons in an acyclic neutral compound with molecular formula C4H5N is

Options:

The correct answer is:
1

Solution:
Degree of unsaturationDU=2C+2+N-H-X2

DU=8+2+1-52=3

DU = Degree of unsaturation. It gives the information of number of pi bonds and rings in a compound. Given that acyclic compound, hence, the structure of the molecule is as follows,

or CH2=C=CH=CH=NHZero sp3 carbon

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The number of sp hybridized atoms in pseudohalogen cyanogen is/are

Options:

The correct answer is:
4

Solution:
Pseudohalogen cyanogen is (CN)2
:Nsp≡Csp–Csp≡Nsp:
All atoms are sp hybridized in it.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of species from the following which are paramagnetic and with bond order equal to one is
H2,He2+,O2+,N22-,O22-,F2,Ne2+,B2

Options:

The correct answer is:
1

Solution:
Paramagnetic molecules are those molecules in which unpaired electrons are present, while diamagnetic molecules are those in which unpaired electrons are absent.

The diatomic species with an odd number of electrons and those diatomic species that contain 10 or 16 electrons are paramagnetic. Among the given species O2+(15) and B2(10) are paramgnetic species.
The molecular orbital electronic configuration of B2 is.

B2:σ1s2, σ*1s2, σ2s2, σ*2s2, π2px1, π2py1

Bond order=Bonding electrons-antibonding electrons2

=6-42=22=1

Hence. the answer is 1

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The number of species from the following which have square pyramidal structure is _______

PF5,BrF4-,IF5,BrF5,XeOF4,ICl4-

Options:

The correct answer is:
3

Solution:
Among the given, IF5;BrF5 and XeOF4 have squre pyramidal structure. VSEPR theory helps us to predict the shape of molecules from the number of electron pairs and lone pairs of electrons that surrounds the central atoms.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (06 Apr Shift 1)
The number of species having a square planar shape from the following is

XeF4, SF4, SiF4, BF4-, BrF4-, CuNH342+, FeCl42-, PtCl42-

Options:

The correct answer is:
4

Solution:
XeF4 is sp3d2 hybridised with two lone pairs of electrons, hence, it is square planar.

SF4 is sp3d hybridised with one lone pairs of electrons, hence, it is See saw.

SiF4 is sp3 hybridised with zero lone pairs of electrons, hence, it is Tetrahedral

BF4- is sp3 hybridised with zero lone pairs of electrons, hence, it is Tetrahedral

[Cu(NH3)4] 2+ is dsp2 hybridised and is Square planar

[FeCl4] 2– is sp3 hybridised and is Tetrahedral

[PtCl4] 2– is dsp2 hybridised and is Square planar

BrF4- is sp3d2 with Square planar geometry.

So, 4 square planer shape compounds are present.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The number of species having non-pyramidal shape among the following is:

A SO3

B NO3-

C PCl3

D CO32-

Options:

The correct answer is:
3

Solution:
Hence non-pyramidal species are SO3,NO3-and CO32-.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The number of species trom the following carrying a single lone pair on central atom Xenon is
XeF5+,XeO3,XeO2 F2,XeF5-,XeO3 F2,XeOF4,XeF4

Options:

The correct answer is:
4

Solution:
The central Xe atom in XeO3 has three bonding domains and one lone pair of electrons. Hence, the electron geometry is tetrahedral and molecular geometry is pyramidal.

XeO2 F2 is trigonal bipyramidal and the shape is a see-saw.

XeOF4 is in square pyramidal shape.

So, XeF5+,XeO3,XeO2 F2 and XeOF4 have single lone pair on central atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The number of spheres contained (i) in one body centred cubic unit cell and (ii) in one face cepntred cubic unit cell, is

Options:
A In (i) 2 and in (ii) 4
B In (i) 3 and in (ii) 2
C In (i) 4 and in (ii) 2
D In (i) 2 and in (ii) 3

The correct answer is:
In (i) 2 and in (ii) 4

Solution:
The number of spheres in one body centred cubic and in one face centred cubic unit cell is 2 and 4 respectively.

Topic: Chemistry, Solid State, JEE Main
The number of square matrices of order 5 with entries from the set {0,1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is

Options:
A 225
B 120
C 150
D 125

The correct answer is:
120

Solution:
Given,

A 5×5 matrix whose each entry is either 0 or 1, is such that sum of entries of each column as well as each row is 1,

Taking a possible case we get,

1000001000001000001000001

Now arranging the above case we get,

1000001000001000001000001→5 ways→4 ways→3 ways→2 ways→1 way because if we fix 1 at first row and first column then in second row first column 1 cannot come and vice-versa.

So total number of such matrices will be 5×4×3×2×1=120

Topic: Mathematics, Matrices, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of statement(s) correct from the following for Copper is/are

(A) CuII complexes are always paramagnetic

(B) CuI complexes are generally colourless

(C) CuI is easily oxidized

(D) In Fehling solution, the active reagent has CuI

Options:

The correct answer is:
3

Solution:
Cu(I) have fulfilled d-subshell and does not have any unpaired electrons so, its complexes will be colourless mostly. and it will be diamagnetic. But Cu(II) has one unpaired electron, so, it will be paramagnetic.

A,B,C are correct and D is incorrect because Fehling solution has CuII

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (27 Jun Shift 1)
The number of statement/s which are the characteristics of physisorption is ___________.

A. It is highly specific in nature

B. Enthalpy of adsorption is high

C. It decreases with increase in temperature

D. It results into unimolecular layer

E. No activation energy is needed

Options:

The correct answer is:
2

Solution:
Physisorption is reversible in nature. It has low adsorption enthalpy nearly 20 to 40 kJ/mol. It results in a multimolecular layer. It decreases with increase in temperature. Activation energy is less or zero in physisorption.

C and E are correct. Hence, the correct answer is 2.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of stereo isomers possible for a compound of the molecular formula is CH3-CH=CH-CHNH2-CH3 is

Options:

The correct answer is:
4

Solution:
As about the double bond, two geometrical isomers are possible and the compound is having one chiral carbon so there are two optical isomers. It means it has total four stereo isomers.
CH3-CH=CH-*CHNH2-CH3
Total G.I = 2n=21=2
Total O.I = 2n=21=2
Here n = number of double bonds or number of chiral carbon atoms.

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of stereo isomers possible for the molecule,

Options:

The correct answer is:
16

Solution:
It has three chiral carbons
So number of enantiomers =23=8 .
Since each pair gives cis and trans isomers (around double bond).
So total number of stereoisomers =8×2=16

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of stereoisomers formed in a reaction of ±PhC=OCOHCNPh with HCN is

Options:

The correct answer is:
3

Solution:
Aldehydes and ketones undergo reaction with HCN to produce cyanohydrins. The reaction progresses very slowly by using pure hydrogen cyanide. Hence, base as a catalyst helps to speed up the reaction. This is an example of nucleophilic addition reaction.

here we are given racemic mixture of optically active compound. The attacking nucleophile can attack on carbonyl carbon from either side forming a new chiral carbon. So, in product there will be two chiral carbon but it is symmetrical. So, number of stereoisomers will be 3 only.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The number of stereoisomers obtained by bromination of trans-2 butene is

Options:
A 4
B 1
C 2
D 3

The correct answer is:
1

Solution:
Stereoisomerism â when isomers have the same structural formula but differ in relative arrangements of atoms or groups in space

Topic: Chemistry, Hydrocarbons, JEE Main
The number of stereoisomers obtained by bromination of trans-2-butene is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
As addition of Br2 is anti and trans-2-butene is used so meso form of 2,3-di-bromo-butane is formed which has 2 chiral carbon atoms.

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of stereoisomers possible for 1,2 -dimethyl cyclopropane is:

Options:
A Two
B One
C Four
D Three

The correct answer is:
Three

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (26 Aug Shift 2)