datavorous/entrance-exam-dataset huggingface dataset:
The number of points, at which the function fx=2x+1-3x+2+x2+x-2,x∈R is not differentiable, is

Options:

The correct answer is:
2

Solution:
fx=2x+1-3x+2+x2+x-2

=2x+1-3x+2+x+2x-1

=2x+1+x+2x-1-3

Critical points are x=-12,-2,-1

but x=-2 is making a zero twice in product so, points of non differentiability are x=-12 and x=-1

∴ Number of points of non-differentiability =2

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The number of points at which the function $\mathrm{f}(\mathrm{x})=\frac{1}{\log |\mathrm{x}|}$ is discontinuous is:

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
The function $\log |x|$ is not defined at $x=0$, so, $\mathrm{x}=0$ is a point of discontinuity. Also for $\mathrm{f}(\mathrm{x})$ to be defined, $\log |\mathrm{x}| \neq 0 \Rightarrow \mathrm{x} \neq\pm 1$

Hence, 0,1,-1 are three points of discontinuity.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of points at which the function fx=x-0.5+x-1+tan⁡x is not differentiable in the interval (0, 2) is/are

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
The curve will have sharp corner at x = 0.5 and at x = 1 and a point of discontinuity at x= Ï 2.
Hence, total number of points where the function is not differentiable is 3.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices 0, 0, 0, 41 and 41, 0 is

Options:
A 780
B 901
C 861
D 820

The correct answer is:
780

Solution:
Px1, y1 lies inside the triangle ⇒ x1, y1∈N
x1+y1<41
∴ 2≤x1+y1≤40
Number of points inside = Number of solutions of the equation
x1+y1=n
2≤n≤40
x1≥1, y1≥1
x1-1+y1-1=n-2
(Number of non - negative integral solutions of x1+x2+.....+xn=r is n+r-1Cr)
Number of solutions =Cn-22+n-2-1

=n-1Cn-2
=n-1
We have 2≤n≤40
∴ Number of solutions =1+2+......+39

= 3 9 × 4 0 2
= 7 8 0

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2015 (04 Apr)
The number of points in the interval $(0,2)$ at which $f(x)=|x-0.5|+|x-1|+\tan x$ is not differentiable is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Given, $f(x)=|x-0.5|+|x-1|+\tan x$
Clearly $f(x)$ is not differentiable at $x=0.5,1$ and $\frac{\pi}{2}$ for $x \in(0,2)$.
$\therefore$ Number of Non differentiable points of $f(x)$ are 3 .
$\therefore$ Hence, option (c) is correct.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of points of discontinuity of \( f(x)=\left[x^{3}+1\right] \) in \( (1,2) \) is/are (where [.] denotes Greatest Integer Function)

Options:
A \( 1 \)
B \( 6 \)
C \( 5 \)
D \( 4 \)

The correct answer is:
\( 6 \)

Solution:
fx=x3+1

By property of GIF, we know that x is discontinuous at every integral point.

So we have to just check that for how many values of x∈1,2, x3+1 is taking integral values.

1<x<2⇒ 1<x3<8⇒ 2<x3+1<9

As x3+1∈2,9, then integer lying in this range are

x3 +1=3,4,5,6,7,8⟵6 points are there at which fx is becoming discontinuous.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of points of intersection z-4+3i|=2 and z+z-4=6,z∈C is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Clearly, z-4+3 i=2 represents a circle with centre 4, 3 and radius 2 units and z+z-4=6 represents an ellipse with foci 0, 0 and 4, 0 and vertices are 5, 0 and -1, 0 .

Observe that, z=4+i lies on circle and 4+i+4+i-4=1+17<6.

Hence, the point z=4+i lies inside the ellipse.

So, the graphs of the curves are as shown in figure.

It is clear from the above diagram, there exist two complex numbers which satisfy both the given curves.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The number of point(s) on the curve y3=12y-3x2 where a tangent is vertical is/are

Options:

The correct answer is:
2

Solution:
y3=12y-3x2
3y2dydx=12dydx-6x
dydx=2x4-y2
Tangent is vertical so
4-y2=0⇒y=±2
At y=2⇒8=24-3x2
x=±43
At y=-2⇒-8=-24-3x2
x2 is negative so x∉R
So, the required points are 43,2 & -43,2
Hence, the number of points is 2

Topic: Mathematics, Application of Derivatives, JEE Main
The number of points where fx=x2-3x-4 is non-differentiable is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Graph of fx=x2-3x-4 is
From the graph, we see that fx is non-differentiable at x=0,-4,4

Topic: Mathematics, Continuity and Differentiability, JEE Main
The number of points, where the curve f(x)=e8x-e6x-3e4x-e2x+1,x∈ℝ cuts x-axis, is equal to............

Options:

The correct answer is:
2

Solution:
Given,

e8x-e6x-3e4x-e2x+1=0

⇒e4x-e2x-3-1e2x+1e4x=0

⇒e4x+1e4x+2-e2x+1e2x=5

⇒e2x+1e2x2-e2x+1e2x=3

Now let e2x+1e2x=t

So, the equation becomes

t2-t-5=0

⇒t=1+212, ignoring negative sign as exponential function are positive,

Now e2x+1e2x=1+212

⇒e4x-1+212e2x+1=0 which is quadratic equation in e2x with upward parabola,

Now by A.M≥G.M we get, e2x+1e2x≥2

So, y=1+212>2, hence it will cut at two distinct point,

Hence, there will be two solution.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The number of points where the function fx=2x2-3x-7 if x⩽-14x2-1 if -1<x<1x+1+x-2 if x⩾1, where t denotes the greatest integer ⩽t, is discontinuous is ______

Options:

The correct answer is:
7

Solution:
Here both the modulus function is continuous in its domain.

At x=-1,

limx→-1-2x2-3x-7=2 & limx→-1+4x2-1=2

Hence fx is continuous at x=-1

At x=1.

limx→1-4x2-1=2 & limx→1+x+1+x-2=3

Hence, fx is discontinuous at x=1

For x∈-1,1 the function fx is discontinuous when 4x2-1=0/ 1/ 2

i.e when x∈±12,±12,±32

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The number of polynomials of the form x3+ax2+bx+c âa,b,câ1,2,3,.....,10 which are divisible by x2+1 is equal to

Options:
A 5
B 10
C 20
D 100

The correct answer is:
10

Solution:
fx=x3+ax2+bx+c is divisible by x2+1 or x+ix-i
So, fi=0 and f-i=0
fi=0⇒i3+ai2+bi+c=0
⇒-i-a+bi+c=0
⇒c-a+ib-1=0⇒c=a, b=1
f-i=0⇒-i3+a-i2+b-i+c=0
⇒i-a-ib+c=0
⇒c-a-ib-1=0⇒c=a, b=1
b=1 and c=a
⇒10 polynomials are possible

Topic: Mathematics, Quadratic Equation, JEE Main
The number of positive divisors of 4896 is

Options:
A 32
B 34
C 36
D 38

The correct answer is:
36

Solution:
We have, $4896=2^{5} \times 3^{2} \times 17^{1}$
So, $T(a)=(1+5)(1+2)(1+1)=36$

Topic: Mathematics, Binomial Theorem, JEE Main
The number of positive integers k such that the constant term in the binomial expansion of 2x3+3xk12,x≠0 is 28·l, where l is an odd integer, is ______.

Options:

The correct answer is:
2

Solution:
For 2x3+3xk12

Tr+1=Cr122x312-r·3xkr=Cr12·212-r·3r·x36-3r-kr

For the term to be independent of x,

36-3r-kr=0 ⇒r=36k+3

Here k can be 0,1,3,6,9 for r to be a whole number.

But for Cr12·212-r·3r to be in the form of 28·l

Only k=3, 6 satisfies the above relation.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The number of positive integral solutions of the equation x3+1x2yx2zxy2y3+1y2zxz2yz2z3+1=11 is/are

Options:
A 0
B 3
C 6
D 12

The correct answer is:
3

Solution:
Multiplying R1 by x, R2 by y and R3 by z and dividing the determinant by xyz, we get,
1xyzx4+xx3yx3zxy3y4+yy3zxz3yz3z4+z=11
âxyzxyzx3+1x3x3y3y3+1y3z3z3z3+1=11
Use R1âR1+R2+R3
x3+y3+z3+1111y3 y3+1y3z3z3z3+1=11
âx3+y3+z3=10 (as the value of the determinant is 1)
Hence, all the possible solutions are (2, 1, 1), (1, 2, 1), (1, 1, 2)

Topic: Mathematics, Determinants, JEE Main
The number of positive integral solutions of the inequality x+y+z≤20 is

Options:
A 1008
B 1028
C 1108
D 1140

The correct answer is:
1140

Solution:
The number of solutions (positive integer) of x+y+z=r is r-1C2
∴ The desired number of solutions is
∑r=320 r-1C2
=2C2+3C2+4C2+........+19C2
=3C3+3C2+4C2+........+19C2
=4C3+4C2+........+18C2+19C2
=20C3 (Adding in the same way)
=20×19×1833×2=1140

Topic: Mathematics, Permutation Combination, JEE Main
The number of possible aromatic benzenoid isomers for $\mathrm{C}_6 \mathrm{H}_4 \mathrm{Cl}_2$ are

Options:
A $2$
B $3$
C $4$
D $5$

The correct answer is:
$3$

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of possible enatiomeric pair(s) produced from the bromination of I and II, respectively, are

Options:
A 0,1
B 1,0
C 0,2
D 1,1

The correct answer is:
0,1

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The number of possible isomeric products formed when 3-chloro-1-butene reacts with HCl through carbocation formation is.........

Options:

The correct answer is:
4

Solution:
3-chloro-1-butene undergo protonation and gives secondary carbocation according Markonikov's addition in the first step. In the second step the lone pair on the chlorine attacks the positively charged carbon to form cyclic intermediate. Now the nucleophile in the reaction i.e., chloride ion attacks cyclic intermediate and forms two structural products as shown below.

2,2-dichlorobutane exit in single isomer only, but 2,3-dichlorobutane exhibits three isomers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The number of possible mono bromo products are (excluding stereo isomers)

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hybridized metal atom, respectively, is:
Note: A and B are unidentate neutral and unidentate monoanionic ligands, respectively.

Options:
A 0 and 2
B 2 and 2
C 0 and 0
D 0 and 1

The correct answer is:
0 and 0

Solution:
If the complex MA2B2 is  sp3 hybridised then the shape of this complex is tetrahedral this structure is optically inactive due to the presence of plane of symmetry
Optical isomers = 0 (b) If the complex MA 2 B 2  is   dsp 2 hybridised then the shape of this complex is square planar.
Both isomers are optically inactive due to the presence of plane of symmetry. Optical isomers =0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (07 Jan Shift 2)
The number of possible resonance structures for $\mathrm{CO}_3^{2-}$ is

Options:
A 2
B 3
C 6
D 9

The correct answer is:
3

Solution:
There are three resonance structure of $\mathrm{CO}_3^{2-}$ ion

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of possible sterio isomers in Ma3b3 is equal to

(Here a, b are different monodentate ligands).

Options:

The correct answer is:
2

Solution:
Ma3b3 can have only two geometrical isomers as follows

Topic: Chemistry, Coordination Compounds, JEE Main
The number of possible straight lines passing through the point $(2,3)$, while forming a triangle with coordinate axes enclosing an area 12 sq. units is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Let the equation of line forming a triangle having area $12 \mathrm{sq}$. units is
$$
\frac{x}{a}+\frac{y}{b}=1
$$
So,
$$
|a b|=24
$$
$\because$ Line (i) passes through point $(2,3)$, so
$$
\frac{2}{a}+\frac{3}{b}=1
$$
from Eqs. (ii) and (iii), we get
$$
|2 b+3 a|=24
$$
If $a$ and $b$ are positive, then
$$
\begin{array}{rlrl}
& & a b=24 \text { and } 3 a+2 b=24 \\
\Rightarrow & a\left(\frac{24-3 a}{2}\right)=24 \\
\Rightarrow & 3 a^2-24 a+48=0 \\
& \because \text { Discriminant }=24^2-4 \times 3 \times 48=0 \\
\therefore & a=4 \text { and } b=6
\end{array}
$$
If $a$ is positive and $b$ is negative then $a b=-24$ and $3 a+2 b=24$.
$$
\Rightarrow \quad 2\left(\frac{24-3 a}{2}\right)=-24 \Rightarrow 3 a^2-24 a-48=0
$$
$\because$ Discriminant $=24^2+4 \times 3 \times 48>0$
$$
\begin{array}{ll}
\therefore \quad a=\frac{24 \pm \sqrt{2 \times 24^2}}{6}=4+4 \sqrt{2} \quad \because a>0 \\
\text { and } \quad b=6-6 \sqrt{2}
\end{array}
$$
If $a$ is negative and $b$ is positive, then $a b=-24$ and $3 a+2 b=24$
$$
\Rightarrow \quad a=4-4 \sqrt{2} \text { and } b=6+6 \sqrt{2}
$$
And at the last it is not possible that $a$ and $b$ both are negative.
So, 3 triangles are possible.

Topic: Mathematics, Straight Lines, JEE Main
The number of primary, secondary and tertiary carbons in the following, structure are respectively

Options:
A 6, 3, 3
B 3, 6, 3
C 3, 6, 2
D 3, 2, 1

The correct answer is:
3, 6, 3

Solution:
1o carbon =3, 2o carbons =6, 3o carbons =3 ,

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of quadratic equations that are unchanged by squaring their roots is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
Î±+Î²=Î±2+Î²2 & Î±Î²=Î±2Î²2âÎ±Î²=0 or 1 .
If Î±Î²=0, then let, Î±=0âÎ²=0 or 1 .
If Î²= 1 Î± ,
Î±+ 1 Î± = Î± 2 + 1 Î± 2 = Î±+ 1 Î± 2 â2
â Î±+ 1 Î± 2 â Î±+ 1 Î± â2=0âÎ±+ 1 Î± =2 or â1âÎ±=1 or Ï,â Ï 2
Hence number of such equations are four 0,â0 ,â 0,â1 ,â 1,â1 â&â Ï,â Ï 2

Topic: Mathematics, Quadratic Equation, JEE Main
The number of quadratic polynomials ax2+2bx+c which satisfy the following conditions is k

(i) a,b,c are distinct

(ii) a,b,c∈1,2,3,4,…2001,2002

(iii) x+1 divides ax2+2bx+c.

Then k105 is equal to

Options:

The correct answer is:
20.02

Solution:
x=-1 is the root of the quadratic equation ax2+2bx+c=0
⇒a-2b+c=0⇒2b=a+c
⇒a and c both must be even or odd.
So, the total number of quadratic polynomials, k=C21001×2+C21001×2
=C10012×2×2=1001×10002×2×2
⇒k105=20.02

Topic: Mathematics, Permutation Combination, JEE Main
The number of radial and angular nodes in 4 d orbital are, respectively

Options:
A 1 & 2
B 3 & 2
C 1 & 0
D 2 & 1

The correct answer is:
1 & 2

Solution:
Number of Radial nodes =n-l-1

Number of angular nodes =l

Where,

l=Azimuthal quantum number n=principal quantum number

For 4d orbital, n=4 and l=2

In 4d orbital the number of Radial nodes=4-2-1=1

In 4d orbital the number of Angular nodes =l=2.

Thus, the number of radial and angular nodes in 4d orbital are 1 and 2 respectively.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The number of rational point(s) (a point (a, b) is rational, if a and b both are rational numbers) on the circumference of a circle having centre Ï, e is

Options:
A at most one
B at least two
C exactly two
D infinite

The correct answer is:
at most one

Solution:
Let two points ( x 1 , y 1 ) and ( x 2 , y 2 ) lie on the circumference of a circle with centre Ï , e where x 1 , y 1 , x 2 , y 2 â Q
Point x,y equidistant from x1, y1 & (x2, y2) will lie on the perpendicular bisector of this line segment.
y-y1+y22=-x2-x1y2-y1x-x1+x22
âpx+qy+r=0 where p, q, r âQ
Now it must pass through Ï,e â pÏ+qe+r=0
âp=q=r=0
There is no such line possible, hence atmost one point lies on the circle.

Topic: Mathematics, Circle, JEE Main
The number of rational terms in the binomial expansion of 414+516120is_______.

Options:

The correct answer is:
21

Solution:
We have,

414+516120

Tr+1=Cr120414120-r5r6

⇒Tr+1=Cr1202120-r25r6

For rational terms r=6λ, 0≤r≤120, hence

r=0, 6, 12, ..., 120

So, total number of rational terms are 21.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The number of reactions that give SO 2 gas:

(a)âCuâ+âconc . H 2 SO 4 â

(b) Znâ+âdil .H 2 SO 4 â

(c) S+ H 2 SO 4 â

(d) FeS 2 â+â O 2 â

(e) NaHSO 3 â+âdil .H 2 SO 4 â

(f)â H 2 SO 4 â+â PCl 5 â

Options:

The correct answer is:
4

Solution:
Cu+conc. H2SO4→CuSO4+SO2+H2O
Zn+dil.H2SO4→ZnSO4+H2S+H2O
S+H2SO4→3SO2+2H2O
FeS2+O2→Fe2+O3+SO2
NaHSO3+dil.H2SO4→NaHSO4+SO2+H2O
H2SO4+PCl5→POCl3+HSO3Cl+HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The number of real linear functions $f(x)$ satisfying $f(f(x))=x+f(x)$ is

Options:
A 0
B 4
C 5
D 2

The correct answer is:
2

Solution:
Let $f(x)=a x+b$
Given,
$$
\begin{aligned}
& f(f(x))=x+f(x) \\
& +b)+b=x+(a x+b) \\
& +a b+b=x+a x+b
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow & a(a x+b)+b & =x+(a x+b) \\
& \Rightarrow & a^2 x+a b+b & =x+a x+b \\
& \Rightarrow & x\left(a^2-a-1\right)+a b & =0 \\
& \therefore & a^2-a-1 & =0 \text { and } a b=0 \\
& \Rightarrow & a & \neq 0, b=0 \\
& \therefore & a & =\frac{1 \pm \sqrt{5}}{2}
\end{aligned}
$$
Now, $\quad f(x)=\left(\frac{1 \pm \sqrt{5}}{2}\right) x+0=\left(\frac{1 \pm \sqrt{5}}{2}\right) x$
$$
f(x)=\left(\frac{1+\sqrt{5}}{2}\right) x,\left(\frac{1-\sqrt{5}}{2}\right) x
$$
$\therefore$ It has two linear equation.

Topic: Mathematics, Functions, JEE Main
The number of real numbers $\lambda$ for which the equality $\frac{\sin (\lambda \alpha)}{\sin \alpha}-\frac{\cos (\lambda \alpha)}{\cos \alpha}=\lambda-1$, holds for all real $\alpha$ which are not integral multiples of $\pi / 2$ is

Options:
A 1
B 2
C 3
D Infinite

The correct answer is:
2

Solution:
$\frac{\sin (\lambda \alpha)}{\sin \alpha}-\frac{\cos (\lambda \alpha)}{\cos \alpha}=\lambda-1$
$\sin \lambda \alpha \cos \alpha-\cos \lambda \alpha \sin \alpha=(\lambda-1) \sin \alpha \cos \alpha$
$\sin (\lambda \alpha-\alpha)=\frac{(\lambda-1)}{2} \sin 2 \alpha$
so $=\lambda=1 ; 3$

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of real numbers λ for which the equality sin(λα)sinα-cos(λα)cosα=λ-1, holds for all real α which are not integral multiple of π/2 is

Options:
A 1
B 2
C 3
D Infinite

The correct answer is:
2

Solution:
We have, sinλαsinα-cosλαcosα=λ-1

sinλαcosα-cosλαsinα=λ-1sinαcosα

sinλ-1α=λ-1sinαcosα

sinλ-1α=λ-12sin2α

It is possible, when λ-1=0⇒λ=1

and λ-12=1⇒λ=3

∴ λ has only two solution.

Topic: Mathematics, Trigonometric Equations, JEE Main
The number of real roots of the equation 4xx2+x+3+5xx2-5x+3=-32 is equal to

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
x=0 is not a root.

Dividing both the numerators and denominators by x and putting y=x+3x, we get,
4y+1+5y-5=-32⇒y=-5,3
x+3x=-5⇒x2+5x+3=0→ two real roots
x+3x=3⇒x2-3x+3=0→ two imaginary roots
⇒ Total two real roots

Topic: Mathematics, Quadratic Equation, JEE Main
The number of real roots of the equation 5+2x-1=2x2x-2 is :

Options:
A 2
B 3
C 1
D 4

The correct answer is:
1

Solution:
Let 2x=t

5+t-1=t2-2t

⇒t-1gt=t2-2t-5ft

Since, 2x>0 ∀x∈R⇒t>0

From the graph, we have

It is clear from the graph that the two graphs have only one point of intersection.

So, number of real roots of the given equation is 1.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The number of real roots of the equation, e4x+e3x-4e2x+ex+1=0 is:

Options:
A 1
B 3
C 2
D 4

The correct answer is:
1

Solution:
Let ex=t∈0,∞
Given equation
t4+t3-4t2+t+1=0
t2+t-4+1t+1t2=0
t2+1t2+t+1t-4=0
Let t+1t=α
α2-2+α-4=0
α2+α-6=0
α2+α-6=0
α=-3,2⇒α=2⇒ex+e-x=2
x=0 is the only solution.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The number of real roots of the equation e6x-e4x-2e3x-12e2x+ex+1=0 is:

Options:
A 2
B 4
C 6
D 1

The correct answer is:
2

Solution:
e6x-e4x-2e3x-12e2x+ex+1=0

e3x2-2e3x+1-e4x+ex=12e2x

⇒e3x-12-exe3x-1=12e2x

⇒e3x-1e3x-1-ex=12e2x

⇒e3x-1-exe2x=12e3x-1

⇒ex-e-x-e-2x=12e3x-1

Assume, fx=ex-e-x-e-2x & gx=12e3x-1

fx=ex-e-x-e-2x

f'x=ex+e-x+2e-2x

Clearly, f'x>0∀x∈R ∵ex & e-x>0∀x∈R

So, here fx is increasing and f0=-1and clearly gx is decreasing function because it is reciprocal function.

From the graph, we can say that number of real roots =2

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The number of real roots of the equation $\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$ that are greater than or equal to one are

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
Given equation
$\sin \left[2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}\right]=0$
$\Rightarrow 2 \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=n \pi, n \in \mathbf{Z}$
$\Rightarrow \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=\frac{n \pi}{2}$
$\because \quad \cos ^{-1} x \in[0, \pi]$
$\therefore \quad \cos ^{-1}\left\{\cot \left(2 \tan ^{-1} x\right)\right\}=0, \frac{\pi}{2}, \pi$
$\Rightarrow \cot \left(2 \tan ^{-1} x\right)=1,0,-1 \Rightarrow 2 \tan ^{-1} x=m \pi+\frac{\pi}{4}$
$m \pi+\frac{\pi}{2}$ or $m \pi-\frac{\pi}{4}, m \in \mathbf{Z}$
$\Rightarrow \tan ^{-1} x=\frac{m \pi}{2} \pm \frac{\pi}{8}$ or $\frac{m \pi}{2}+\frac{\pi}{4}, m \in \mathbf{Z}$
$\because \quad \tan ^{-1} x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
$\therefore \tan ^{-1} x= \pm \frac{\pi}{8}, \frac{\pi}{2}-\frac{\pi}{8},-\frac{\pi}{2}+\frac{\pi}{8}, \pm \frac{\pi}{4}$
$\therefore \quad x= \pm \tan \frac{\pi}{8}, \pm \cot \frac{\pi}{8}, \pm \tan \frac{\pi}{4}$
$\Rightarrow \quad x= \pm(\sqrt{2}-1), \pm \frac{1}{\sqrt{2}-1}, \pm 1$
$= \pm(\sqrt{2}-1), \pm(\sqrt{2}+1), \pm 1$
The roots which are greater than or equal to one are $\sqrt{2}+1,1$.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of real roots of the equation $\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0,(x \neq 0)$ is

Options:
A $1$
B $0$
C $2$
D $3$

The correct answer is:
$0$

Solution:
Given equation,
$\frac{\left(x^2+1\right)^3}{x^3}+\frac{x^2+1}{3 x}=0(x \neq 0)$
$\Rightarrow \quad \frac{3\left(x^2+1\right)^3+x^2\left(x^2+1\right)}{3 x^3}=0$
$\Rightarrow \quad\left(x^2+1\right)\left[3\left(x^2+1\right)^2+x^2\right]=0$ ...(i)
$\begin{aligned} & \because \quad x^2 \geq 0 \\ & \Rightarrow \quad\left(x^2+1\right) \geq 1 \text { and } 3\left(x^2+1\right)^2+x^2 \geq 3\end{aligned}$
By using this result in Eq. (i), we can observe that
for no real value of $x$ will satisfy Eq. (i).
Hence, the number of real roots is zero.

Topic: Mathematics, Functions, JEE Main
The number of real roots of the equation xx-5x+2+6=0, is

Options:
A 5
B 4
C 6
D 3

The correct answer is:
3

Solution:
Given,

xx-5x+2+6=0

Now taking case 1 when x<-2 we get,

-x2+5x+10+6=0

⇒x2-5x-16=0

⇒x=5-892, ignoring x=5+892 as x<-2, so one solution is possible here,

Now taking case 2 when -2≤x<0 we get,

-x2-5x-10+6=0

⇒x2+5x+4=0

⇒x=-1, ignoring x=-4 as -2≤x<0

Now taking case 3, when x≥0 we get,

x2-5x-10+6=0

⇒x2-5x-4=0

⇒x=5+412, ignoring x=5-412 as x≥0

So, from all cases we get total 3 solutions.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The number of real roots of the equation \( x^{5}+3 x^{3}+4 x+30=0 \) is

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 5 \)

The correct answer is:
\( 1 \)

Solution:
Let fx=x5+3x3+4x+30
⇒f′x=5x4+9x2+4
⇒f′x>0 for all x∈R
Hence, the fx=0 has only one real root.

Topic: Mathematics, Quadratic Equation, JEE Main
The number of real solution of \( \tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2} \), is

Options:
A \( 0 \)
B \( 1 \)
C \( 2 \)
D \( \infty \)

The correct answer is:
\( 2 \)

Solution:
For xx+1 to be defined xx+1≥0 ...i

Also, for sin-1x, we know that x∈-1, 1

⇒x2+x+1≤1

⇒x2+x≤0 ...ii

From i and ii, we get

xx+1=0

⇒x=0,-1

When x=0,

L.H.S.=tan-10+sin-11=π2

And, when x=-1,

L.H.S.=tan-10+sin-11-1+1

=0+sin-1⁡1=π2

Thus, the number of solutions are 2.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of real solutions of cot-1⁡xx+3+sin-1⁡x2+3x+1=π2 is/are

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
0

Solution:
Let, fx=cot-1⁡xx+3+sin-1⁡x2+3x+1
For fx to be defined, we have
x2+3x≥0 and 0≤x2+3x+1≤1
Which is only possible if
x2+3x=0⇒x=0,-3
But these values do not satisfy the equation
∴ Number of solutions =0

Topic: Mathematics, Functions, JEE Main
The number of real solutions of cot-1⁡xx+4+cos-1⁡x2+4x+1=π2 is equal to

Options:
A 0
B 1
C 2
D Infinite

The correct answer is:
2

Solution:
Let, fx=cot-1â¡xx+4+cos-1â¡x2+4x+1
For fx to be defined, we have
x2+4xâ¥0 and 0â¤x2+4x+1â¤1
Which is only possible if
x2+4x=0âx=0,-4
Also these values satisfies the equation
â´ Number of solutions =2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is

Options:
A Zero
B One
C Two
D Infinite

The correct answer is:
Two

Solution:
$\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$
$\tan ^{-1} \sqrt{x(x+1)}$ is defined when
$x(x+1) \geq 0$
$\sin ^{-1} \sqrt{x^2+x+1}$ is defined when
$0 \leq x(x+1)+1 \leq 1 \text { or } 0 \leq x(x+1) \leq 0$
From (i) and (ii), $x(x+1)=0$
or $x=0$ and -1 .
Hence number of solution is 2 .

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main