datavorous/entrance-exam-dataset huggingface dataset:
The number of real solutions of the equation 3x2+1x2-2x+1x+5=0, is

Options:
A 4
B 0
C 3
D 2

The correct answer is:
0

Solution:
Given equation is:

3x2+1x2-2x+1x+5=0

We know,

x+1x2=x2+ 2+1x2 (using algebraic identities)

⇒x2+1x2=x+1x2-2

So, we can write the given equation as

3x+1x2-2-2x+1x+5=0

Let x+1x=t

⇒3t2-2-2t+5=0⇒3t2-6-2t+5=0

⇒3t2-2t-1=0⇒3t2-3t+t-1=0

⇒3tt-1+(t-1)=0⇒3t+1t-1=0

Therefore, t=-13 or t=1

Case I

x+1x=1

⇒x2-x+1=0

Here, D=-3 i.e. no real solution

Case II x+1x=-13

⇒3x2+x+3=0

Here, D=-35 i.e. no real solution

Hence, we don't have real solutions.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of real solutions of the equation $\left(\frac{9}{10}\right)=-3+x-x^{2}$ is

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
0

Solution:
Let $f(x)=-3+x-x^{2}$
Then, $f(x) < 0$ for all $x$ because coefficient of $\mathrm{x}^{2} < 0$ and disc $ < 0 .$ Thus, LHS of the given equation is always positive whereas the RHS is always less than zero.
Hence, the given equation has no solution.
Alternate Solution :
Given, equation is
$$
\frac{9}{10}=-3+x-x^{2}
$$
Let $y=\frac{9}{10}$, therefore
$$
\begin{array}{l}
y=-3+x-x^{2} \\
y=-\left[x^{2}-x+\frac{1}{4}\right]-3+\frac{1}{4} \\
\Rightarrow y+\frac{11}{4}=-\left(x-\frac{1}{2}\right)^{2}
\end{array}
$$
It is clear from the graph that two curves do not intersect. Hence, no solution exists.

Topic: Mathematics, Quadratic Equation, JEE Main
The number of real solutions of the equation e4x+4e3x-58e2x+4ex+1=0 is _____.

Options:

The correct answer is:
2

Solution:
Given equation e4x+4e3x-58e2x+4ex+1=0

Let fx=e2xe2x+1e2x+4ex+1ex-58

Now let ex+1ex=t

i.e. t2+4t-58=0

t=-4±16+4×582

=-4±2622

t=-2+262 is possible and t=-2-262 is not possible as t≥2

i.e. ex+1ex=-2+262

e2x--2+262ex+1=0

For real solution, discriminant 248-862>0

Also -b2a=-2+2622=62-1>0

Hence, both roots of the equation are positive real roots.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The number of real solution(s) of the equation sin-1x2-5x+5+cos-14x-x2-3=Ï is/are

Options:
A one
B two
C zero
D infinite

The correct answer is:
one

Solution:
Since, x2-5x+5â¥0â0â¤sin-1x2-5x+5â¤Ï2
Since, 4x-x2-3â¥0â0â¤cos-14x-x2-3â¤Ï2
â´ For LHS to be Ï
sin-1x2-5x+5=Ï2 and cos-14x-x2-3=Ï2
So, x2-5x+5=1 & 4x-x2-3=0
âx2-5x+5=1 & x2-4x+3=0
âx=1,4 & x=1,3
Hence, the common value is x=1

Topic: Mathematics, Functions, JEE Main
The number of real solutions of the equation $x^2-3|x|+2=0$ is

Options:
A
3

B
2

C
4

D
1

The correct answer is:
4

Solution:
$x^2-3|x|+2=0 \quad$ or $(|x|-2)(|x|-1)=0$
$|x|=1,2$ or $x=\pm 1, \pm 2$ or $\therefore$ No. of solution $=4$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2003
The number of real solutions of the equation, x2-x-12=0 is:

Options:
A 2
B 3
C 1
D 4

The correct answer is:
2

Solution:
We have, x2-x-12=0

⇒x2-4x+3x-12=0

⇒x+3x-4=0

⇒x=4 as x≠-3

⇒x=±2

Hence, number of real solution is 2.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The number of real solutions of the equation x+14-8x-2+x+23-10x-2=3 are

Options:
A 2
B 4
C 8
D infinite

The correct answer is:
2

Solution:
The given equation is
x-2-42+x-2-52=3
âx-2-4+x-2-5=3
Put x-2=y

Then, the given equation becomes y-4+y-5=3
Case 1:y<4â-y+4-y+5=3
â-2y=-6ây=3
Case 2: 4â¤y<5ây-4-y+5=3

â1=3â no solution
Case 3: yâ¥5ây-4+y-5=3
â2y=12ây=6
y=3âx-2=3âx-2=9âx=11
y=6âx-2=6âx-2=36âx=38
âx=11 and 38 i.e. two solutions.

Topic: Mathematics, Quadratic Equation, JEE Main
The number of real solutions of the equation \(x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0\) is ______.

Options:

The correct answer is:
1

Solution:
Given,

xx2+3|x|+5|x-1|+6|x-2|=0

⇒x=0 or x2+3|x|+5|x-1|+6|x-2|=0

We know that sum of all positive terms can never be zero,

So, x2+3|x|+5|x-1|+6|x-2| can never be zero for any value of x,

So there is no value of x which satisfies the above equation
Hence, only solution is x=0, i.e. number of solution is 1 .

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The number of real solutions of x7+5x3+3x+1=0 is equal to _____.

Options:
A 0
B 1
C 3
D 5

The correct answer is:
1

Solution:
fx=x7+5x3+3x+1

For number of real solutions, we need to find the number of times the curve cuts x-axis.

Now f'x=7x6+15x2+3>0

∴ fx is strictly increasing function

i.e. when x→-∞,y→-∞ and when x→∞,y→∞

So, the curve will cut x-axis only once.

∴ number of real solution will be 1

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The number of real solutions x of the equation cos2xsin2x+11+x2=cos2x+sec2x is

Options:
A 0
B 1
C 2
D infinite

The correct answer is:
1

Solution:
We have,

cos2xsin2x+11+x2=cos2x+sec2x

LHS=cos2xsin2x+11+x2

cos2xsin2x+11+x2≤2

RHS=cos2x+sec2x≥2

LHS=RHS

∴ cos2x+sec2x=2 at x=0

Hence, only one solution.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of real values of $\alpha$ for which the system of equations
$
\begin{array}{l}
x+3 y+5 z=\alpha x \\
5 x+y+3 z=\alpha y
\end{array}
$
$3 x+5 y+z=\alpha z$
has infinite number of solutions is

Options:
A 1
B 2
C 4
D 6

The correct answer is:
1

Solution:
The system of equations are
$$
\begin{array}{l}
x+3 y+5 z=\alpha x \\
5 x+y+3 z=\alpha y \\
3 x+5 y+z=\alpha z \\
(1-\alpha) x+3 y+5 z=0 \\
5 x+(1-\alpha) y+3 z=0 \\
3 x+5 y+(1-\alpha) z=0
\end{array}
$$
For infinite number of solutions, we must have $\left|\begin{array}{ccc}1-\alpha & 3 & 5 \\ 5 & 1-\alpha & 3 \\ 3 & 5 & 1-\alpha\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{ccc}9-\alpha & 3 & 5 \\ 9-\alpha & 1-\alpha & 3 \\ 9-\alpha & 5 & 1-\alpha\end{array}\right|=0,$
$C_{1} \rightarrow C_{2}+C_{2}+C_{3}$
Taking $(9-\alpha)$ common from the first column, we get $\begin{array}{l}\Rightarrow \quad(9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 1 & 1-\alpha & 3 \\ 1 & 5 & 1-\alpha\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left|\begin{array}{ccc}1 & 3 & 5 \\ 0 & -\alpha-2 & -2 \\ 0 & 2 & -\alpha-4\end{array} \mid=0\right. \\ \Rightarrow & (9-\alpha)\left|\begin{array}{cc}\alpha+2 & 2 \\ -2 & \alpha+4\end{array}\right|=0 \\ \Rightarrow & (9-\alpha)\left(\alpha^{2}+6 \alpha+8+4\right)=0 \\ (\alpha-9)\left(\alpha^{2}+6 \alpha+12\right)=0\end{array}$
$\Rightarrow \alpha=9$ is the only real value of $\alpha$.

Topic: Mathematics, Determinants, JEE Main
The number of real values of λ for which the system of linear equations, 2x+4y-λz=0, 4x+λy+2z=0 and λx+2y+2z=0, has infinitely many solutions, is:

Options:
A 3
B 1
C 2
D 0

The correct answer is:
1

Solution:
For Infinite many solution Δ=0
i.e , 24-λ4λ2λ22=0

⇒ 22-2λ-48-2λ-λ8-λ2=0
⇒λ3+4λ-40=0

f(x)=x3+4x-40 is strictly increasing

i.e., equation has only one real root,
Also, ∆1=0, ∆2=0, ∆3=0
∴ Only one value of λ is possible

Topic: Mathematics, Determinants, JEE Main, JEE Main 2017 (08 Apr Online)
The number of real values of $t$ such that the system of homogeneous equations
$$
\begin{aligned}
t x+(t+1) y+(t-1) z & =0 \\
(t+1) x+t y+(t+2) z & =0 \\
(t-1) x+(t+2) y+t z & =0
\end{aligned}
$$
has non-trivial solutions is

Options:
A $3$
B $2$
C $1$
D None of these

The correct answer is:
None of these

Solution:
Given,
$$
\begin{aligned}
& t x+(t+1) y+(t-1) z=0 \\
& (t+1) x+t y+(t+2) z=0 \\
& (t-1) x+(t+2) y+t z=0
\end{aligned}
$$
Here,
Coefficient matrix, $A=\left[\begin{array}{ccc}t & t+1 & t-1 \\ t+1 & t & t+2 \\ t-1 & t+2 & t\end{array}\right]$ If $|A|=0$, then system of equations has non-trivial solution and it has infinite solutions.
$$
|A|=\left|\begin{array}{ccc}
t & t+1 & t-1 \\
t+1 & t & t+2 \\
t-1 & t+2 & t
\end{array}\right|=0
$$
Apply operation $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$, we get
$$
=\left|\begin{array}{ccc}
t & t+1 & t-1 \\
1 & -1 & 3 \\
-1 & 1 & 1
\end{array}\right|=0
$$
Apply $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1$, we get
$$
\left|\begin{array}{crr}
t & 1 & -1 \\
1 & -2 & 2 \\
-1 & 2 & 2
\end{array}\right|=0
$$
Expanding along $R_1$, we get
$$
\begin{array}{rlrl}
& & t(-4-4)-1(2+2)-1(2-2)=0 \\
& \Rightarrow & & -8 t-4=0 \\
\therefore & & t=\frac{-1}{2}
\end{array}
$$
So, no option is correct.

Topic: Mathematics, Determinants, JEE Main
The number of real values of x that satisfies the equation x4+4x3+12x2+7x-3=0 is

Options:

The correct answer is:
2

Solution:
Let, fx=x4+4x3+12x2+7x-3
∴ f ′ x =4 x 3 +12 x 2 +24x+7
f ″ x =12 x 2 +2x+2
i.e. f''x>0, ∀x∈R
Also, f0=-3
Hence, fx=0 has 2 real roots

Topic: Mathematics, Application of Derivatives, JEE Main
The number of relations, on the set 1, 2, 3 containing 1, 2 and 2, 3 which are reflexive and transitive but not symmetric, is _________.

Options:

The correct answer is:
4

Solution:
Given,

Set A=1,2,3

Now Cartesian product A×A=1,1,2,1,1,2,........,3,3

Now, given the relation is reflexive,

So, 1, 1, 2, 2, 3, 3 ∈ R

Also given 1, 2, 2, 3∈ R, 1, 3 must ∈ R

Now finding, Possible cases :

Case-1: All of 2, 1, 3, 2, 3, 1 ∉R→1 relation.

Case-2: Only one of 2, 1, 3, 2, 3, 1∈R→3 relations.

For example if relation is 1, 1, 2, 2, 3, 3,2,1 then it is reflexive as well as transitive as 2,2,2,1→2,1 is present in relation

Note that exactly two of 2, 1, 3, 2, 3, 1∈R is not possible because if two of these ∈R, third must ∈R to make relation transitive.

Total number of relations =4

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The number of resonance structures for Anthracene are

Options:

The correct answer is:
4

Solution:
4 resonance structures are possible for Anthracene.

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of roots of the equation 22x+1=2x-1 is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
LHS is always non negative.
So, RHS also has to be the same

So, x≥12.
Squaring both sides, we get,
42x+1=2x-12=4x2+1-4x
4x2-12x-3=0⇒x=12±144+482×4
x=3±9+32=3±232
⇒x=3+232 as x>12

Topic: Mathematics, Quadratic Equation, JEE Main
The number of roots of the equation, 81sin2x+81cos2x=30 in the interval 0,π is equal to :

Options:
A 3
B 4
C 8
D 2

The correct answer is:
4

Solution:
81sin2x+81cos2x=30

81sin2x+81181sin2x=30

81sin2x=t

t+81t=30

t2-30t+81=0

t-3t-27=0

81sin2x=31 or 81sin2x=33

34sin2x=31 or 34sin2x=33

sin2x=14 or sin2x=34

Hence, total 4 solution.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The number of roots of the equation $\cos ^{7} \theta-\sin ^{4} \theta=1$ that lie in the interval $[0,2 \pi]$ is-

Options:
A 2
B 3
C 4
D 8

The correct answer is:
2

Solution:
$\quad \cos ^{7} \theta=1+\sin ^{4} \theta \geq 1$ but $\cos ^{7} \theta \leq 1$,
so $\cos \theta=1 ; \sin \theta=0$
$$
=0,2 \pi
$$

Topic: Mathematics, Quadratic Equation, JEE Main
The number of roots of the equation sin-1x-cos-1x=sin-15x-3 is/are

Options:
A 3
B 1
C 2
D 0

The correct answer is:
1

Solution:
sin-1x-cos-1x=sin-15x-3
âÏ2-cos-1x-cos-1x=Ï2-cos-15x-3
â2cos-1x=cos-15x-3. Also xâ-1,1 â¦ 1
âcos-12x2-1=cos-15x-3 and 5x-3â-1,1, i.e., -1â¤5x-3â¤1
â2x2-1=5x-3, hence, xâ25,45
â2x2-5x+2=0âx=2 or 12
but, x=2 does not satisfy the equation 1

Hence, the given equation has only one root

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of roots of the equation tan⁡x+sec⁡x=2cos⁡x in 0,4π is

Options:
A 2
B 4
C 6
D 0

The correct answer is:
4

Solution:
tan⁡x+sec⁡x=2cos⁡x
Multiplying by cos⁡x ,
sin⁡x+1=2cos2⁡x=2-2sin2⁡x
⇒2sin2⁡x+sin⁡x-1=0⇒sin⁡x=-1,12
sin⁡x=12 at
x=π6,5π6 in 0,2π
sin⁡x=12→ four times in 0,4π
sin⁡x=-1⇒x=3π2 for which tanx and secx are not defined
⇒sin⁡x=-1 is not possible
⇒ Number of roots in 0,4π is 4.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of S=O and S-OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are:

Options:
A 2 and 2 and 2 and 4.
B 4 and 2 and 4 and 2.
C 4 and 2 and 2 and 4.
D 2 and 2 and 2 and 2.

The correct answer is:
4 and 2 and 4 and 2.

Solution:
As we know that, the peroxodisulfuric acid and pyrosulfuric acid are the examples of oxyacid and both contain sulfur oxygen and sulfur hydrogen bond. In the oxyacids, the oxygen is the main element.

Peroxodisulphuric acid
Pyrosulphuric acid

The number of S=O and S−OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are (4 and 2) and (4 and 2).

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2017 (08 Apr Online)
The number of sp2 hybrid orbitals in a molecule of benzene is:

Options:
A 24
B 6
C 18
D 12

The correct answer is:
18

Solution:
In benzene total six sp2 hybrid carbon atoms are present .Each carbon atom has 3sp2 hybrid orbitals. Therefore total sp2 hybrid orbitals are 18 in benzene.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (09 Jan Shift 2)
The number of $\mathrm{S}-\mathrm{S}$ bonds in $\mathrm{SO}_3, \mathrm{~S}_2 \mathrm{O}_3^{2-}$ $\mathrm{S}_2 \mathrm{O}_6^{2-}$ and $\mathrm{S}_2 \mathrm{O}_8^{2-}$ respectively are

Options:
A
$1,0,0,1$

B
1,0,1,0

C
$0,1,1,0$

D
$0,1,0,1$

The correct answer is:
$0,1,1,0$

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of $\mathrm{S}-\mathrm{S}$ bonds in $\mathrm{SO}_3, \mathrm{~S}_2 \mathrm{O}_3^{2-}$ $\mathrm{S}_2 \mathrm{O}_6^{2-}$ and $\mathrm{S}_2 \mathrm{O}_8^{2-}$ respectively are

Options:
A
$1,0,0,1$

B
1,0,1,0

C
$0,1,1,0$

D
$0,1,0,1$

The correct answer is:
$0,1,1,0$

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2012 (26 May Online)
The number of S – S bonds in sulphur trioxide trimer [ S3O9] is.

Options:

The correct answer is:
0

Solution:
The cyclic structure of sulphur trioxide trimer (S3O9) is

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of s-electrons present in an ion with 55 protons in its unipositive state is

Options:
A 8
B 9
C 12
D 10

The correct answer is:
10

Solution:
The atom with 55 protons is caesium(Cs). The electronic configuration of Cs is 1s22s22p63s23p64s23d104p65s24d105p66s1.

The mono positive ion means one electron is removed from outermost orbital. Hence, the sum of electrons in 1s, 2s, 3s, 4s, 5s orbitals are10.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only is:

Options:
A 77
B 82
C 42
D 35

The correct answer is:
77

Solution:
I First possiblity is 1,1,1,1,1,2,3

required number =7!5!=7×6=42

II Second possiblity is 1,1,1,1,2,2,2

required number =7!4!3!=7×6×56=35

Total =42+35=77

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is

Options:

The correct answer is:
240

Solution:
To find the number of 7 digit odd number formed by using the seven digits 1,2,2,2,3,3,5

We will take following cases,

Case 1- fixing 1 as unit place,

So total number of ways will be 6!3!2! as 2 is repeating three times and 3 is repeating two times

Case 2- fixing 3 unit place,

So total number of ways will be 6!3!

Case 3- fixing 5 as unit place,

So total number of ways will be 6!3!2!

Now adding all the cases we will get, total number of ways as 6!3!2!+6!3!+6!3!2!=60+120+60=240

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (30 Jan Shift 2)
The number of sigma and pi bonds in peroxodisulphuric acid are, respectively

Options:
A 9 and 4
B 11 and 4
C 4 and 8
D 4 and 9

The correct answer is:
11 and 4

Solution:
The structure of peroxodisulphuric acid $\left(\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_8\right)$ is
Hence, it contains $11 \sigma$ and $4 \pi$ bonds.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of sigma bonds in is _______.

Options:

The correct answer is:
10

Solution:
Numbers of σ bonds =10

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The number of significant figures in 0.00340 is _________

Options:

The correct answer is:
3

Solution:
Number of significant figures =3

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The number of significant figures in 50000.020×10-3 is

Options:

The correct answer is:
8

Solution:
50000.020×10-3 All digits are significant.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is

Options:

The correct answer is:
576

Solution:
Given word 'VOWELS'

Number of words Arranging all the letters = 6!

In the given word Consonants are W, L, S.

Treating all the consonants as one unit then number of units will be 4.

No of such arrangements =4!.3!

VOWELS all the consonants never come together =Total - all consonants come together =6!-4!×3!=576

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The number of solution of cos2⁡x+cos2⁡2x=2 in 0,20 is equal to

Options:

The correct answer is:
7

Solution:
For this to be true cos2â¡x=1 and cos2â¡2x=1
So, x=nÏ
Since xâ0,20, n=0,1,2,3,4,5,6

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solution of $\tan x+\sec x=2 \cos x$ in $[0,2 \pi)$ is

Options:
A
2

B
3

C
0

D
1

The correct answer is:
3

Solution:
The given equation is $\tan x+\sec x=2 \cos x \Rightarrow \sin x+1=2 \cos ^2 x$
$$
\begin{aligned}
& \Rightarrow \sin x+1=2\left(1-\sin ^2 x\right) \Rightarrow 2 \sin ^2 x+\sin x-1=0 \\
& \Rightarrow(2 \sin x-1)(\sin x+1)=0 \Rightarrow \sin x=\frac{1}{2},-1 \Rightarrow x=30^{\circ}, 150^{\circ}, 270^{\circ}
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2002
The number of solution of the equation ∑r=15cos⁡rx=0 lying in 0,π is

Options:

The correct answer is:
5

Solution:
cos⁡x+cos⁡2x+cos⁡3x+cos⁡4x+cos⁡5x=0
⇒sin⁡5x2sin⁡x2cos⁡3x=0
⇒sin⁡5x2=0 or cos⁡3x=0
but sin⁡x2≠0
If x∈0,π⇒5x2∈0,5π2
⇒ Two solutions 5x2=π,2π
If x∈0,π⇒3x∈0,3π
⇒3 solutions 3x=π2,3π2,5π2
So, total five solutions

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solution to the equation 2a+3b+6c=60, where a,b,c∈N, is equal to

Options:

The correct answer is:
36

Solution:
2a+3b+6c=60
Note that 3b,6c,60 are divisible by 3, hence 2a is divisible by 3
Similarly 2a,6c,60 are even numbers hence 3b is an even number
Therefore a=3k1,b=2k2
So the equation reduces to k1+k2+C=10, ∀k1,k2,C∈N
Hence, the number of solutions =C2=369

Topic: Mathematics, Permutation Combination, JEE Main
The number of solutions for the equation $x^{2}-5|x|+6=0$ is

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

The correct answer is:
$4$

Solution:
We have,
$$
\begin{aligned}
x^{2}-5|x|+6 &=0 \\
|x|^{2}-5|x|+6 &=0 \\
(|x|-3)(|x|-2) &=0 \\
|x| &=3,2 \\
\Rightarrow \quad \quad x &=\pm 3, \pm 2 \\
\therefore & \text { Number of solution is } 4 .
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
The number of solutions in the interval 0,π of the equation sin3⁡xcos⁡3x+sin⁡3xcos3⁡x=0 is equal to

Options:
A 7
B 6
C 5
D 4

The correct answer is:
5

Solution:
The given equation can be written as
3sin⁡x-sin⁡3x4cos⁡3x+sin⁡3x3cos⁡x+cos⁡3x4
sin⁡xcos3x⁡+cos⁡xsin⁡3x=0
So, sin⁡x+3x=0⇒sin⁡4x=0
⇒x=nπ4, n∈Z.
⇒x=0,π4,π2,3π4,π
⇒5 solutions

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of cosx=sinx, such that -4π≤x≤4π is

Options:
A 4
B 6
C 8
D 12

The correct answer is:
8

Solution:
The number of solutions will be the number of points of intersection of the graphs y=cosx & y=sinx

Plotting the two curves on the same axis, we get,

the number of intersection as 8

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$ is

Options:
A none
B one
C two
D infinite

The correct answer is:
one

Solution:
Since, $\frac{d y}{d x}=\frac{y+1}{x-1}$
$\Rightarrow \quad \frac{d y}{y+1}=\frac{d x}{x-1}$
After integrating on both sides, we have
$\log (y+1)=\log (x-1)-\log C$
$C(y+1)=(x-1)$
$C=\frac{x-1}{y+1}$
If $x=1$, then $y=2$, so $C=0$
Therefore, $x-1=0$
Hence, there is only one solution.

Topic: Mathematics, Differential Equations, JEE Main
The number of solutions of equation, \( \sin 5 x \cos 3 x=\sin 6 x \cos 2 x \), in the interval \( [0, \pi] \) are

Options:
A \( 3 \)
B \( 4 \)
C \( 5 \)
D \( 6 \)

The correct answer is:
\( 5 \)

Solution:
Given,

sin5xcos3x=sin6xcos2x

The given equation can be written as

2sin5xcos3x=2sin6xcos2x

⇒sin8x+sin2x=sin8x+sin4x; ∵2sinAcosB=sinA+B+sinA-B

⇒sin2x-sin4x=0

⇒-2sinxcos3x=0,

Hence, sinx=0 or cos3x=0.

x=nπn∈I or 3x=kπ+π2k∈I

Therefore, since x∈0,π, the given equation is satisfied, x=0, π, π6, π2 or 5π6. Hence number of solutions are 5.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of equations $x+y-z=0,3 x-y-z=0, x-3 y+z=0$ is

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
Infinite

Solution:
For the system of given homogeneous equations
$\Delta=\left|\begin{array}{rrr}1 & 1 & -1 \\ 3 & -1 & -1 \\ 1 & -3 & 1\end{array}\right|=1(-1-3)-1(3+1)-1(-9+1)$
$=-4-4+8=0 .$. There are infinite number of solutions.

Topic: Mathematics, Determinants, JEE Main
The number of solutions of $\sin 3 x=\cos 2 x$, in the interval $\left(\frac{\pi}{2}, \pi\right)$ is

Options:
A
3

B
4

C
2

D
1

The correct answer is:
1

Solution:
$\sin 3 x=\cos 2 x$
$$
\begin{aligned}
&\Rightarrow 3 \sin x-4 \sin ^3 x=1-2 \sin ^2 x \\
&\Rightarrow 4 \sin ^3 x-2 \sin ^2 x-3 \sin x+1=0 \\
&\Rightarrow \sin x=1, \frac{-2 \pm 2 \sqrt{5}}{8}
\end{aligned}
$$
In the interval $\left(\frac{\pi}{2}, \pi\right), \sin x=\frac{-2+2 \sqrt{5}}{8}$
So, there is only one solution.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The number of solutions of the equation 1+sin4x=cos23x, x∈-5π2,5π2 is:

Options:
A 5
B 7
C 3
D 4

The correct answer is:
5

Solution:
The given trigonometric equation can be written as
1-cos23x+sin4x=0

⇒ sin23x+sin4x=0

⇒sin3x=0 and sinx=0 , x∈-5π2,5π2

In given interval sinx=0 at x=-2π, -π, 0, π, 2π and these values are also satisfied by sin3x=0.
Hence total number of solutions is 5.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The number of solutions of the equation 3+cos⁡x2=4-2sin8⁡x in 0,9π is equal to

Options:
A 4
B 5
C 6
D 7

The correct answer is:
4

Solution:
-1≤cos⁡x≤1⇒ 2≤3+cos⁡x≤4
⇒4≤3+cos⁡x2≤16
Also, -1≤sin⁡x≤1
⇒0≤2sin8⁡x≤2
⇒2≤4-2sin8⁡x≤4 so, equality is possible when LHS=RHS=4
⇒cos⁡x=-1 and sin⁡x=0
⇒x=(2n+1)π
⇒x=π,3π,5π,7π

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of solutions of the equation 4sin2x−4cos3x+9−4cosx=0;x∈−2π,2π is:

Options:
A 1
B 3
C 2
D 0

The correct answer is:
0

Solution:
Given: 4sin2x-4cos3x+9-4cosx=0

⇒4-4cos2x-4cos3x+9-4cosx=0

⇒4cos3x+4cos2x+4cosx-13=0

⇒4cos3x+4cos2x+4cosx=13

We know that, -1≤cosx≤1

So, the maximum value of LHS is 4+4+4=12

But, RHS is 13.

So, no solutions are possible for the given equation.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2024 (01 Feb Shift 2)
The number of solutions of the equation 32tan2x+32sec2x=81, 0≤x≤π4 is :

Options:
A 0
B 2
C 1
D 3

The correct answer is:
1

Solution:
Given, 32tan2x+32sec2x=81, 0≤x≤π4

32tan2x+321+tan2x=81

32tan2x+32.32tan2x=81

33×32tan2x=81

32tan2x=2711

tan2x=ln322711
tanx=ln322711∈0,1
One solution in 0,π4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (31 Aug Shift 2)
The number of solutions of the equation cosx+π3cosπ3-x=14cos22x, x∈-3π,3π is:

Options:
A 8
B 5
C 6
D 7

The correct answer is:
7

Solution:
Given,

cosx+π3cosx-π3=14cos22x

⇒2cosx+π3cosx-π3=12cos22x

⇒cos2x+cos2π3=12cos22x

⇒cos2x+-12=12cos22x

⇒cos22x-2cos2x+1=0

⇒cos2x-12=0 or cos2x=1

⇒2x=-6π,-4π,-2π,0,2π,4π,6π

So, x∈-3π,-2π,-π,0,π,2π,3π

So total 7 solutions.

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2022 (24 Jun Shift 2)