datavorous/entrance-exam-dataset huggingface dataset:
The frequency of vibration r of a mass m suspended from a spring of spring constant k is given by a relation of this type $\mathrm{f}=\mathrm{Cm}^{\mathrm{x}} \mathrm{K}^{\mathrm{y}}$; where c is a dimensionless quantity. The value of x and y are

Options:
A $x=\frac{1}{2}, y=\frac{1}{2}$
B $x=-\frac{1}{2}, y=-\frac{1}{2}$
C $x=\frac{1}{2}, y=-\frac{1}{2}$
D $x=-\frac{1}{2}, y=\frac{1}{2}$

The correct answer is:
$x=-\frac{1}{2}, y=\frac{1}{2}$

Solution:
The correct option is $\mathbf{D}$
$x=-\frac{1}{2}, y=\frac{1}{2}$
The frequency of oscillation spring mass system $=\frac{1}{2 \pi}$ $\sqrt{\frac{k}{m}}$
Where, $\mathrm{k}=$ spring constant
and $\mathrm{m}=$ mass with spring
And given frequency, $\mathrm{f}=\mathrm{cm}^x \mathrm{k}^y$
Comparing above two frequency we get
$x=-\frac{1}{2}, y=\frac{1}{2}$

Topic: Physics, Units and Dimensions, JEE Main
The frequency of $X$-rays; $\gamma$-rays and ultraviolet rays are respectively $a, b$ and $c$ then

Options:
A
$a < b ; b>c$
B
$a>b ; b>c$

C
$a < b < c$
D
$a=b=c$

The correct answer is:
$a < b ; b>c$

Solution:
Frequency range of $\gamma$-ray, $b=10^{18}-10^{23} \mathrm{~Hz}$
Frequency range of X-ray, $a=10^{16}-10^{20} \mathrm{~Hz}$
Frequency range of ultraviolet ray, $c=10^{15}-10^{17} \mathrm{~Hz}$
$\therefore a < b ; b>c$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2012 (26 May Online)
The frequency radiation emitted when the electron falls from $\mathrm{n}=4$ to $n=1$ in a hydrogen atom will be (Given ionization energy of $\mathrm{H}=2.18 \times 10^{-18} \mathrm{~J} \mathrm{atom}^{-1}$ and $h$ $\left.=6.625 \times 10^{-34} \mathrm{Js}\right)$ :

Options:
A $1.54 \times 10^{15} \mathrm{~s}^{-1}$
B $1.03 \times 10^{15} \mathrm{~s}^{-1}$
C $3.08 \times 10^{15} \mathrm{~s}^{-1}$
D $2.00 \times 10^{15} \mathrm{~s}^{-1}$

The correct answer is:
$3.08 \times 10^{15} \mathrm{~s}^{-1}$

Solution:
$\mathrm{E}=h \mathrm{v}$ or $\mathrm{v}=\frac{\mathrm{E}}{h}$
For $\mathrm{H}$ atom,
$$
\begin{aligned}
E & =\frac{-21.76 \times 10^{-19}}{n^2} \mathrm{~J} \mathrm{~atm}^{-1} \\
\Delta E & =-21.76 \times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)
\end{aligned}
$$
$$
\begin{aligned}
& =20.40 \times 10^{-19} \mathrm{~J} \mathrm{~atm}^{-1} \\
v & =\frac{20.40 \times 10^{-19}}{6.625 \times 10^{-34}} \\
& =3.079 \times 10^{15} \mathrm{~s}^{-1} .
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, JEE Main
The frequency $v$ of the radiation emitted by an atom when an electron jumps from one orbit to another is given by $=\mathrm{k} \delta \mathrm{E}$, where $\mathrm{k}$ is a constant and $\delta \mathrm{E}$ is the change in energy level due to the transition. Then dimension of $\mathrm{k}$ is

Options:
A $\mathrm{ML}^{2} \mathrm{~T}^{-2}$
B the same dimension of angular momentum
C $\mathrm{ML}^{2} \mathrm{~T}^{-1}$
D $\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}$

The correct answer is:
$\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}$

Solution:
Hint:
$v=k \delta E$
$[k]=\left[\frac{v}{\delta E}\right]=\frac{\left[T^{-1}\right]}{\left[M L^{2} T^{-2}\right]}$
$[k]=\left[M^{-} L^{-2} T^{1}\right]$

Topic: Physics, Units and Dimensions, JEE Main
The friction coefficient between the board and the floor shown in diagram is μ. Find the maximum force that the man can exert on the rope, so that the board does not slip on the floor.

Options:
A μ m + M g 2 + μ
B Î¼ m + M g 1 + Î¼
C μ m + M g 2 - μ
D μ m + M g 1 - μ

The correct answer is:
μ m + M g 1 + μ

Solution:
The forces acting on the system are shown in the diagram
For vertical equilibrium of the point P
T = F ....(i)

And for the vertical equilibrium of the system
R + T = (m + M)g, i.e., R = (m + M)g - T ....(ii)
Now the system will not move horizontally till
T < fL
i.e., T < Î¼[m + Mg - T] fL = Î¼R
which on simplification gives
T<Î¼m+Mg1+Î¼
So in the light of equation (i), we get
Fmax=Î¼m+Mg1+Î¼

Topic: Physics, Laws of Motion, JEE Main
The friction coefficient between the board and the floor shown in diagram is μ. Find the maximum force that the man can exert on the rope, so that the board does not slip on the floor.

Options:
A μm+Mg2+μ
B μm+Mg1+μ
C μm+Mg2-μ
D μm+Mg1-μ

The correct answer is:
μm+Mg1+μ

Solution:
The forces acting on the system are shown in the diagram
For vertical equilibrium of the point P
T = F ....(i)

And for the vertical equilibrium of the system
R + T = (m+M)g, i.e., R = (m+M)g - T ....(ii)
Now the system will not move horizontally till
T < fL
i.e., T < μ[m+Mg - T] fL = μR
which on simplification gives
T<μm+Mg1+μ
So in the light of equation (i), we get
Fmax=μm+Mg1+μ

Topic: Physics, Laws of Motion, JEE Main
The fringe widths are found to be ω1 and ω2 respectively if a Young's double slit experiment is performed in medium of refractive index n1 and n2 respectively. The correct statement is

Options:
A ω1>ω2 if n1>n2
B ω1>ω2 if n1<n2
C ω1=ω2 if n1<n2
D ω1ω2=n1n2

The correct answer is:
ω1>ω2 if n1<n2

Solution:
Fringe width of Y.D.S.E. in a medium is given by, ω=Dλdn
For a particular Y.D.S.E. set up and light, D, d & λ remains constant, therefore
ω∝1n
So, we can conclude, if n2>n1, ω1>ω2.

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
The front wall of a drawer in a cabinet is a provider with two symmetrical handles. The distance between the handles is l and the length (i.e. depth) of the drawer is a. The maximum value of the coefficient of friction between drawer and cabinet, for which drawer can be pulled out by applying a force on one handle perpendicular to the face of the drawer is: (Neglect the weight of the drawer)

Options:
A al
B 2al
C a2l
D None of these

The correct answer is:
al

Solution:
If a force is applied only on one handle, as shown in figure, the drawer will have a tendency to rotate clockwise. That is, the right and left corners (shown in figure), will press against the side of the cabinet and produce some friction.
Balancing force,
F=2μN …….(i)
Balancing torque (about right side corner):
Fx+Na=μN(2x+l) ……..(ii)
Solving gives,
μ=al

Topic: Physics, Laws of Motion, JEE Main
The fruit fly, Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal theory of inheritance by Morgan and his colleagues because

Options:
A it reproduces parthenogenetically
B a single mating produces two young flies
C smaller female is easily recognisable from larger male
D it completes life cycle in about two weeks

The correct answer is:
it completes life cycle in about two weeks

Solution:
The fruit fly, Drosophila melanogaster was found to be very suitable for experimental verification of chromosomal theory of inheritance by Morgan and his colleagues because
- it completes its life cycle in a very short period (ie, about two weeks).
- a single mating produces many offsprings.
- larger female is easily recognisable from smaller male.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2010 (Mains)
The fruit fly has 8 chromosomes (2n) in each cell. During the interphase of mitosis if the number of chromosomes at G1 phase is 8. What would be the number of chromosomes after the S phase?

Options:
A 8
B 16
C 4
D 32

The correct answer is:
8

Solution:
The number of chromosomes remains the same after S-phase, but the amount of DNA becomes double.

S phase is the phase of the cell cycle in which DNA is replicated, occurring between the G1 phase and G2 phase. Since accurate duplication of the genome is critical to successful cell division, the processes that occur during S-phase are tightly regulated and widely conserved.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2021
The fruit is chambered, developed from inferior ovary and has seeds with succulent testa in

Options:
A Cucumber
B Pomegranate
C Orange
D Guava

The correct answer is:
Pomegranate

Solution:
This forms edible part of pomegranate.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2008 (Mains)
The function defined by $f(x)=\left\{\begin{array}{lr}|x-3| ; & x \geq 1 \\
\frac{1}{4} x^2-\frac{3}{2} x+\frac{13}{4} ; x \lt 1\end{array}\right.$ is

Options:
A Continuous at $x=1$
B Continuous at $x=3$
C Differentiable at $x=1$
D All the above

The correct answer is:
All the above

Solution:
Since $|x-3|=x-3$, if $x \geq 3=-x+3$, if $x \lt 3$
$\therefore$ The given function can be defined as
$f(x)=\left\{\begin{array}{rl} \frac{1}{4} x^2-\frac{3}{2} x+\frac{13}{4}, & x \lt 1 \\ 3-x, & 1 \leq x \lt 3 \\ x-3, & x \geq 3\end{array}\right.$
Now proceed to check the continuity and differentiability at $x=1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function f:0,7→0,70 where fx=x3-12x2+45x, is

Options:
A one-one & onto
B many-one & onto
C one-onto & into
D many-one & into

The correct answer is:
many-one & onto

Solution:
f'x=3x2-24x+45
=3x2-8x+15=3x-3x-5
∴f'x changes its sign for x∈0,7
∴fx is many-one
Now f0=0, f5=50,
f3=54, f7=70
∴Range∈0,70=co-domain
∴fx→onto

Topic: Mathematics, Functions, JEE Main
The function f:-∞,1→0,e5 defined as fx=ex3-3x+2 is

Options:
A Many one and onto
B Many one and into
C One-one and onto
D One-one and into

The correct answer is:
Many one and into

Solution:
Given, fx=ex3-3x+2
Let, gx=x3-3x+2
⇒ g ′ x =3 x 2 −3=3 x 2 −1
⇒g'x≥0, for x∈−∞,−1 and

g'x≤0 for x∈−1,1
∴gx is increasing as well as decreasing in different intervals
∴fx is many-one
Now, Range of fx is 0,e4, but co-domain is 0,e5.
∴fx is into function

Topic: Mathematics, Functions, JEE Main
The function f defined by fx=x3-3x2+5x+7 is:

Options:
A Decreasing in R
B Increasing in R
C Increasing in 0,Â â and decreasing in -â,Â 0
D Decreasing in 0,Â â and increasing in -â,Â 0

The correct answer is:
Increasing in R

Solution:
We have fx=x3-3x2+5x+7
f′x=3x2-6x+5>0

D=b2-4ac=36-4·3·5=-14 -ve

Then, f'x>0.
Hence, fx is increasing in R.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2017 (09 Apr Online)
The function f:N-{1}→N; defined by fn= the highest prime factor of n, is :

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

The correct answer is:
neither one-one nor onto

Solution:
Given: f:N-{1}→N

f(n)= The highest prime factor of n.

Since, the only prime factor of 2 is 2.

⇒f(2)=2

Also, factors of 4 are 1,2 and 4. Out of these, the highest prime factor is 2.

⇒f(4)=2

Similarly the highest prime factor of 16, 32, 64, 128,... is 2

So, the given function is many one.

Now, no composite number will be image of any number.

Example: 4 is not image of any element

So, the given function fx is into.

Hence many one and into

Or, neither one-one nor onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The function f :N→I defined by fx=x-5x5 , where N is the set of natural numbers and x denotes the greatest integer less than or equal to x, is:

Options:
A one-one but not onto
B one-one and onto
C neither one-one nor onto
D onto but not one-one

The correct answer is:
neither one-one nor onto

Solution:
Given, f( x )=x−5[ x 5 ]
Hence, f( x )=5( x 5 −[ x 5 ] )
=5{ x 5 }
Since, x is a natural number.
Hence, x5=0,15,25,35,45
Hence, f( x )=0 , when x=5n, n∈N
=1 when x=5n+1, n∈N
=2 when x=5n+2, n∈N
=3 when x=5n+3, n∈N
=4 when x=5n+4, n∈N
Hence, fx is neither one-one nor onto.

Topic: Mathematics, Functions, JEE Main, JEE Main 2017 (09 Apr Online)
The function $f:R \sim\{0\} \rightarrow R$ given by
$f(x)=\frac{1}{x}-\frac{2}{e^{2 x}-1}$
can be made continuous at $x=0$ by defining $f(0)$ as

Options:
A
$2$

B
$-1$

C
$0$

D
$1$

The correct answer is:
$1$

Solution:
$\lim _{x \rightarrow 0} \frac{1}{x}-\frac{2}{e^{2 x}-1}$
$\lim _{x \rightarrow 0} \frac{e^{2 x}-1-2 x}{x\left(e^{2 x}-1\right)}$
$\lim _{x \rightarrow 0} \frac{2 e^{2 x}-2}{\left(e^{2 x}-1\right)+2 x e^{2 x}}$
$\lim _{x \rightarrow 0} \frac{4 e^{2 x}}{4 e^{2 x}+4 x e^{2 x}}=1$.

Topic: Mathematics, Limits, JEE Main, JEE Main 2007
The function f :R→-12,12 defined as fx=x1+x2, is:

Options:
A Invertible
B Injective but not surjective
C Surjective but not injective
D Neither injective nor surjective

The correct answer is:
Surjective but not injective

Solution:
Let, y=x1+x2
⇒y+yx2-x=0
⇒x2y-x+y=0
⇒x=+1± 1-4y22y
⇒1-4y2 ≥0
⇒4y2 ≤1
⇒ So, y∈-12,12
Here, Co-Domain=Range

Hence, the function is surjective.
Now, for injective function

⇒x11+x12=x21+x22
⇒x1+x2x22-x2-x2x12=0
⇒x1-x2+x1x2x2-x1=0
⇒x2-x1 x1x2-1=0
⇒x2=x1 or x1=1x2
Hence, the function is not injective.

Topic: Mathematics, Functions, JEE Main, JEE Main 2017 (02 Apr)
The function f:R→R defined as fx=x2-x+1x2+x+1 is

Options:
A injective as well as surjective
B injective but not surjective
C surjective but not injective
D neither injective nor surjective

The correct answer is:
neither injective nor surjective

Solution:
fx=x2−x+1x2+x+1

⇒f'x=2x2−1x2+x+12

f'x is positive in -∞,-1∪1,∞ and negative in -1,1
Hence, fx is many-one.
For Range of fx, let y=x2-x+1x2+x+1
⇒y-1x2+y+1x+y-1=0
⇒ since x∈R,D≥0

⇒y+12-4y-12≥0

⇒-3y2+10y-3≥0

⇒y∈13,3
Hence fx is into

Topic: Mathematics, Functions, JEE Main
The function $f: R \rightarrow R$ defined by $f(x)=(x-1)(x-2)(x-3)$ is

Options:
A One-one but not onto
B Onto but not one-one
C Both one-one and onto
D Neither one-one nor onto

The correct answer is:
Onto but not one-one

Solution:
We have $f(x)=(x-1)(x-2)(x-3)$
and $f(1)=f(2)=f(3)=0 \Rightarrow f(x)$ is not one-one.
For each $y \in R$, there exists $x \in R$ such that $f(x)=y$. Therefore $f$ is onto. Hence $f: R \rightarrow R$ is onto but not one-one.

Topic: Mathematics, Functions, JEE Main
The function $f: R \rightarrow R, f(x)=x^2, \forall x \in R$ is

Options:
A Injection but not suriection
B Surjection but not injection
C Injection as well as surjection
D Neither injection nor suriection

The correct answer is:
Neither injection nor suriection

Solution:
Since the mapping is many-one into.

Topic: Mathematics, Functions, JEE Main
The function f:R→R is defined as fx=3-x. From the following statements,

I. f is one-one

II. f is onto

III. f is a decreasing function

the true statements are

Options:
A only I, II
B only II, III
C only I, III
D I, II, III

The correct answer is:
only I, III

Solution:
Since, f:RâR such that fx=3-x
Let y1 and y2 be two elements of f(x) such that y1=y2
â 3-x1=3-x2 â x1=x2
Since, if two images are equal, then their elements are equal, therefore it is one-one function.
Since, f(x) is positive for every value of x, therefore f(x) is into.
On differentiating w.r.t. x, we get dydx=-3-xlogâ¡3<0 for every value of x.
â´ It is decreasing function.
â´ Statement I and III are true.

Topic: Mathematics, Functions, JEE Main
The function $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements of it
I. $f$ is one-one
II. $f$ is onto
III. $f$ is a decreasing function
Out of these, true statement are

Options:
A Only I, II
B Only II,III
C Only I,III
D I,II,III

The correct answer is:
Only I, II

Solution:
Since, $f: R \rightarrow R$ such that $f(x)=3^{-x}$.
Let $y_1$ and $y_2$ be two elements of $f(x)$ such that
$$
\begin{aligned}
& y_1=y_2 \\
& \Rightarrow \quad 3^{-x_1}=3^{-x_2} \Rightarrow x_1=x_2 \\
&
\end{aligned}
$$
Since, if two images are equal, then their elements are equal, therefore it is one-one function
Let
$$
y=3^{-x}
$$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=-3^{-x} \log 3 < 0$ for every value of $x$
$\therefore$ It is decreasing function
$\therefore$ Statement I and II are true.

Topic: Mathematics, Functions, JEE Main
The function $f: R \rightarrow R$ is defined by $f(x)=\cos ^2 x+\sin ^4 x$ for $x \in R$, then $f(R)$ is equal to

Options:
A $\left(\frac{3}{4}, 1\right]$
B $\left[\frac{3}{4}, 1\right)$
C $\left[\frac{3}{4}, 1\right]$
D $\left(\frac{3}{4}, 1\right)$

The correct answer is:
$\left[\frac{3}{4}, 1\right]$

Solution:
We have,
$$
\begin{aligned}
& f(x)=\cos ^2 x+\sin ^4 x=1-\sin ^2 x+\sin ^4 x \\
& =1-\sin ^2 x\left(1-\sin ^2 x\right)=1-\sin ^2 x \cos ^2 x \\
& =1-\frac{1}{4} \sin ^2 2 x
\end{aligned}
$$
$$
\text { Here, } 0 \leq \sin ^2 2 x \leq 1 \Rightarrow 0 \leq \frac{\sin ^2 2 x}{4} \leq \frac{1}{4}
$$
$$
\begin{aligned}
& \Rightarrow \quad 0 \geq-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4} \\
& 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq-\frac{1}{4}+1 \\
& \Rightarrow \quad 1 \geq 1-\frac{\sin ^2 2 x}{4} \geq \frac{3}{4} \\
&
\end{aligned}
$$
Range of $f=\left[\frac{3}{4}, 1\right]$

Topic: Mathematics, Functions, JEE Main
The function \( f: R \rightarrow R \) is defined as \( f(x)=3^{-x} \). From the following statements,
I. \( f \) is one-one
II. \( f \) is onto
III. \( f \) is a decreasing function
the true statements are

Options:
A only I, II
B only II, III
C only I, III
D I, II, III

The correct answer is:
only I, III

Solution:
Since, f:R→R such that fx=3-x
Let y1 and y2 be two elements of f(x) such that y1=y2
⇒ 3-x1=3-x2 ⇒ x1=x2

Since, if two images are equal, then their elements are equal, therefore it is one-one function.
Since, f(x) is positive for every value of x, therefore f(x) is into.

On differentiating w.r.t. x, we get dydx=-3-xlog⁡3<0 for every value of x.
∴ It is decreasing function.
∴ Statement I and III are true.

Topic: Mathematics, Functions, JEE Main
The function $\text { f: R->R, }$ $f(x)=\frac{x^2+2 x-15}{x^2-4 x+9}, x \in \mathbb{R}$ is

Options:
A one-one but not onto.
B both one-one and onto.
C onto but not one-one.
D neither one-one nor onto.

The correct answer is:
neither one-one nor onto.

Solution:
$f(x)=\frac{(x+5)(x-3)}{x^2-4 x+9}$
$\begin{aligned} & \text { Let } g(x)=x^2-4 x+9 \\ & D < 0 \\ & g(x)>0 \text { for } x \in R\end{aligned}$
$\therefore\left[\begin{array}{l}
\mathrm{f}(-5)=0 \\
\mathrm{f}(3)=0
\end{array}\right.$
So, $\mathrm{f}(\mathrm{x})$ is many-one.
again,
$\begin{aligned}
& y x^2-4 x y+9 y=x^2+2 x-15 \\
& x^2(y-1)-2 x(2 y+1)+(9 y+15)=0 \\
& \text { for } \forall x \in R \Rightarrow D \geq 0 \\
& D=4(2 y+1)^2-4(y-1)(9 y+15) \geq 0 \\
& 5 y^2+2 y+16 \leq 0 \\
& (5 y-8)(y+2) \leq 0
\end{aligned}$

Topic: Mathematics, Functions, JEE Main, JEE Main 2024 (06 Apr Shift 1)
The function fx=1-sinx+cosx1+sinx+cosx is not defined at x=π. The value of f(π), so that f(x) is continuous at x=π, is

Options:
A − 1 2
B 1 2
C −1
D 1

The correct answer is:
−1

Solution:
limx→π (1+cosx)-sinx(1+cosx)+sinx
=limx→π2cos2x2−2sinx2cosx22cos2x2+2sinx2cosx2
=limx→π 1−tan x21+tanx2
=limx→πtan⁡π4-x2=-1
Since, f(x) is continuous at x=π
∴ fπ=limx→π fx=-1

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=2sin⁡x-ex, ∀x∈0,π has

Options:
A no local maximum
B 2 local minima
C 1 local maximum
D 1 local minimum

The correct answer is:
1 local maximum

Solution:
f â² x =2cosxâ e x
â´fx has a local maximum

Topic: Mathematics, Application of Derivatives, JEE Main
The function $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}},$ where $[x]$ denotes the greatest integer $\leq x$, is

Options:
A continuous for all values of $x$
B discontinuous at $x=\frac{\pi}{2}$
C not differentiable for some values of $x$
D discontinuous at $x=-2$

The correct answer is:
continuous for all values of $x$

Solution:
Given, $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}}$
Since. $\left[x-\frac{\pi}{2}\right]$ is an integer for all $x$, therefore $\pi\left[x-\frac{\pi}{2}\right]$ is an integral multiple of $\pi$ for all $x$. Hence, $\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}=0$ for all $x$
Also, $2+[x]^{2} \neq 0$ for all $x$
Hence, $f(x)=0$ for all $x$ Hence, $f(x)$ is continuous and derivable for all $x$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=2x3-3a+bx2+6abx has a local maximum at x=a, if

Options:
A a>b
B a<b
C a>0
D a<0

The correct answer is:
a<b

Solution:
f ′ x =6 x 2 − a+b x+ab
=6x-ax-b
i.e. fx has local maximum at x=a if a<b

Topic: Mathematics, Application of Derivatives, JEE Main
The function fx=2x+3x23, x∈R, has

Options:
A exactly one point of local minima and no point of local maxima
B exactly one point of local maxima and no point of local minima
C exactly one point of local maxima and exactly one point of local minima
D exactly two points of local maxima and exactly one point of local minima

The correct answer is:
exactly one point of local maxima and exactly one point of local minima

Solution:
Given: f(x)=2x+3(x)23

⇒f'x=2+2x-13

⇒f'x=21+1x13

⇒f'x=2x13+1x13

The function changes from positive to negative at x=1 and from negative to positive at x=0

So, the point of local maxima M is at x=-1 and the point of local minima m at x=0

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2024 (29 Jan Shift 2)
The function $f(x)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at

Options:
A
x = 2

B
x = −2

C
x = 0

D
x = 1

The correct answer is:
x = 2

Solution:
$\frac{x}{2}+\frac{2}{x}$ is of the form $x+\frac{1}{x} \geq 2$ \& equality holds for $x=1$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2006
The function, fx=3x-7x23, x∈R, is increasing for all x lying in

Options:
A -∞,0∪1415,∞
B -∞,0∪37,∞
C -∞,1415
D -∞,-1415∪0,∞

The correct answer is:
-∞,0∪1415,∞

Solution:
fx=3x-7x23

For fx is increasing f'x>0.

Differentiate fx

f'x=3x23+3x-723x13

f'x=9x+23x-73x13

f'x=15x-143x13

Now f'x>0 ⇒15x-143x13>0

x∈-∞, 0∪1415, ∞

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The function fx=π4+tan-1x,x≤112x-1,x>1 is :

Options:
A continuous on R-1 and differentiable on R--1,1.
B both continuous and differentiable on R-1
C continuous on R--1and differentiable on R--1,1
D both continuous and differentiable on R--1

The correct answer is:
continuous on R-1 and differentiable on R--1,1.

Solution:
fx=π4+tan-1x,x≤112x-1,x>1

Making graph

Clearly, function is continuous on R–1 and differentiable on R––1,1.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The function fx=4x3-3x26-2sinx+2x-1cosx:

Options:
A increases in 12,∞
B decreases in -∞,12
C decreases in 12,∞
D increases in -∞,12

The correct answer is:
increases in 12,∞

Solution:
fx=4x3-3x26-2sinx+2x-1cosx

f'x=2x2-x-2cosx+2cosx-sinx2x-1.

=2x-1x-sinx

for x>0,x-sinx>0

x<0,x-sinx<0

for x∈-∞,0∪12,∞,f'x≥0

for x∈0,12,f'x≤0

⇒ fx increases in 12,∞

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The function $f(x)=\frac{x+1}{9 x+x^{3}}$ is

Options:
A discontinuous at exactly two points.
B continuous for all real values of $x$.
C discontinuous at exactly three points.
D discontinuous at exactly one point.

The correct answer is:
discontinuous at exactly one point.

Solution:
(C)
$f(x)=\frac{x+1}{9 x+x^{3}}=\frac{x+1}{x\left(9+x^{2}\right)}$
The function is discontinuous at exactly one point i.e. $x=0$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function $f(x)=a \sin |x|+b e^{| x \mid} \quad$ is differentiable at $x=0$ when

Options:
A $3 a+b=0$
B $3 a-b=0$
C $a+b=0$
D $a-b=0$

The correct answer is:
$a+b=0$

Solution:
Given, $f(x)=a \sin |x|+b e^{|x|}$
We know that $\sin |x|$ and $e^{|x|}$ is not
differentiable at $x=0$.
Therefore, for $f(x)$ to differentiable at $x=0$, we
must have $a=b=0$.
$\therefore$
$$
a+b=0
$$

Topic: Mathematics, Continuity and Differentiability, JEE Main
The function fx=esin⁡x+cos⁡x ∀x∈0,2π attains local extrema at x=α and x=β, then α+β is equal to

Options:
A π
B 2π
C 3π2
D π2

The correct answer is:
3π2

Solution:
f ′ x = e sinx+cosx ⋅ cosx−sinx
â´fx attains local extrema at x=Ï4,5Ï4
i.e. Î±+Î²=6Ï4=3Ï2

Topic: Mathematics, Application of Derivatives, JEE Main
The function fx=ex3-6x2+10 attains local extremum at x=a and x=b a<b, then the value of a+b is equal to

Options:

The correct answer is:
4

Solution:
f â² x = e x 3 â6 x 2 +10 3 x 2 â12x
Thus, fx attains extremum at
x=0 & x=4
i.e. a=0 & b=4
âa+b=4

Topic: Mathematics, Application of Derivatives, JEE Main