datavorous/entrance-exam-dataset huggingface dataset:
During the kinetic study of the reaction, $2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}$, following results were obtained:

$$
\begin{array}{|c|c|c|c|}
\hline \text { Run } & {[\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l}
\text { Initial rate of } \\
\text { formation of } \\
\mathrm{D} / \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1}
\end{array} \\
\hline \text { I } & 0.1 & 0.1 & 6.0 \times 10^{-1} \\
\hline \text { II } & 0.3 & 0.2 & 7.2 \times 10^{-1} \\
\hline \text { III } & 0.3 & 0.4 & 28.8 \times 10^{-1} \\
\hline \text { IV } & 0.4 & 0.1 & 24.0 \times 10^{-1} \\
\hline
\end{array}
$$

Based on the above data which one of the following is correct?

Options:
A rate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]$
B rate $=k[\mathrm{~A}][\mathrm{B}]$
C rate $=k[\mathrm{~A}]^{2}[\mathrm{~B}]^{2}$
D rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

The correct answer is:
rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

Solution:
In case of (II) and (III) Keeping concentration of [A] constant, when the concentration of $[\mathrm{B}]$ is doubled, the rate quadruples. Hence it is second order with respect to $\mathrm{B}$. In case of $\mathrm{I} \&$ IV Keeping the concentration of [B] constant, when the concentration of $[\mathrm{A}]$ is increased four times, rate also increases four times. Hence, the order with respect to $\mathrm{A}$ is one. Hence Rate $=k[\mathrm{~A}][\mathrm{B}]^{2}$

Topic: Chemistry, Chemical Kinetics, JEE Main
During the meiotic division, at which stage the homologous chromosomes separate without the sister chromatids breaking apart?

Options:
A Metaphase I
B Metaphase II
C Anaphase I
D Anaphase II

The correct answer is:
Anaphase I

Solution:
Meiosis is a reductional division process where the cell undergoes two successive division processes to reduce chromosome number to half. It is divided into meiotic stage I and II. Anaphase I is part of the meiotic stage I, where the sister chromatids separate from its homologous counterpart. The chromatids centromere remains intact in this stage and do not separate. Their separation occurs in anaphase II in meiotic stage II.

Topic: Biology, Cell Cycle and Cell Division, NEET
During the melting of a slab of ice at $273 \mathrm{~K}$ at atmospheric pressure

Options:
A Positive work is done by ice-water system on the atmosphere
B Positive work is done on the ice-water system by the atmosphere
C The internal energy of the ice-water system increases
D The internal energy of the ice-water system decreases

The correct answer is:
The internal energy of the ice-water system increases

Solution:
Correct option is $\mathrm{3}$. the internal energy of the ice-water increases
By the sign convention the work done by the system is positive whereas, the work done on the system is negative.
From 1st law of thermodynamics that change in heat increase the internal energy.

Topic: Physics, Thermodynamics, JEE Main
During the melting of a slab of ice at 273 K at atmospheric pressure

Options:
A positive work is done by ice-water system on the atmosphere
B positive work is done on the ice-water system by the atmosphere
C the internal energy of the ice-water system decreases
D the internal energy of the ice-water system remains same

The correct answer is:
positive work is done on the ice-water system by the atmosphere

Solution:
Density of ice is less than that of water. Therefore, on melting, volume decreases. So, the work done on the ice-water system by the atmosphere is positive.

Topic: Physics, Thermal Properties of Matter, NEET
During the melting of a solid substance at constant pressure and temperature, what happens to the temperature?

Options:
A It decreases.
B It remains constant.
C It increases.
D It fluctuates.

The correct answer is:
It remains constant.

Solution:
During the melting of a solid substance at constant pressure and temperature, the temperature remains constant. This is because the heat is absorbed during the phase change without affecting the temperature until the phase change is complete.

Topic: Chemistry, Thermodynamics (C), NEET
During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1μg of 90Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically ____________.

Options:

The correct answer is:
23.03

Solution:
Radioactive disintegration is always a first order reaction for a first order reaction. Radioactive decay is the process by which an unstable atomic nucleus loses energy by radiation.

Kt=2.303  log A o A t

where A o = initial concentration
A t = concentration at time ‘t’
t 1 2 = 0.693 K K= 0.693 t 1 2 = 0.693 6.93 ⇒0.1 yent s −1 t 90% = 2.303 0.1 log 100 10 t 90% =23.03 years.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
During the nuclear explosion, one of the products is 90 Sr with half life of 6.93 years. If 1 âÎ¼âg of 90 Sr was absorbed in the bones of a newly born baby in place of Ca , how much time (in years) is required to reduce it by 90%, if it is not lost metabolically?
Report your answer by rounding it up to a nearest whole number.

Options:

The correct answer is:
23

Solution:
t90%t50%=lnâ¡10010lnâ¡2
t90%=ln10â¡ ln2â¡ Ãt50%=6.93Ãlnâ¡100.693=23.03 years

Topic: Chemistry, Chemical Kinetics, JEE Main
During the process of electro refining of copper, some metals present as impurity settle down as anode mud. These are:

Options:
A Sn and Ag
B Pb and Zn
C Ag and Au
D Fe and Ni

The correct answer is:
Ag and Au

Solution:
In anode mud, less electropositive elements are present. Ag and Au are less electropositive than Cu .

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’ These are

Options:
A
Sn and Ag

B
Pb and Zn

C
Ag and Au

D
Fe and Ni

The correct answer is:
Ag and Au

Solution:
No Solution Available

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2005
During the process of gene amplification using PCR, if a very high temperature is not maintained in the beginning, then which of the following steps of PCR will be affected first?

Options:
A Annealing
B Extension
C Denaturation
D Ligation

The correct answer is:
Denaturation

Solution:
Denaturation -94-95°C, annealing- 50-60°C, extension- 72°C. Since maximum temperature is required during denaturation, it will be affected the most.

The initial denaturation step is carried out at the beginning of PCR to separate the double-stranded template DNA into single strands so that the primers can bind to the target region and initiate extension.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2021
During the propagation of a longitudinal wave, in the region of compressions and rarefactions

Options:
A density varies
B density remains constant
C there is heat transfer
D Boyle's law is obeyed

The correct answer is:
density varies

Solution:
When a longitudinal wave propagates through solid liquid or gas then the particles of medium start oscillating about there mean position up and down in the same direction in which the wave travels. At any instant there are point in space where density is height called compression and the point where density is lowest called rare faction.

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
During the propagation of a nerve impulse, the action potential results from the movement of

Options:
A $\mathrm{Na}^{+}$ions from extracellular fluid to intracellular fluid
B $\mathrm{Ka}^{+}$ions from extracellular fluid to intracellular fluid
C $\mathrm{Na}^{+}$ions from intracellular fluid to extracellular fluid
D $\mathrm{Ka}^{+}$ions from intracellular fluid to extracellular fluid

The correct answer is:
$\mathrm{Na}^{+}$ions from extracellular fluid to intracellular fluid

Solution:
When the threshold stimulus is applied, the permeability of the membrane for sodium ions will change. Voltage gated sodium ion channels open and $\mathrm{Na}^{+}$ion flows form extracellular fluid to intracellular fluid, the outside becomes negative and inside positive. The membrane is depolarised \& when the two ends of this membrane is connected to oscilloscope a potential difference of $+30 \mathrm{mv}$, called as action potential is generated.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2008 (Mains)
During the propagation of electromagnetic wave in a particular medium :-

Options:
A Electric energy density is double of the magnetic energy density
B Electric energy density is half of the magnetic energy density
C Electric energy density is equal to the magnetic energy density
D Both electric and magnetic energy densities are zero

The correct answer is:
Electric energy density is equal to the magnetic energy density

Solution:
UVelec=12∈0E2 ....(i)
UVmag=12B2/μ0 ....(ii)
B = E C (for EMW) ....(iii)
and C= 1 μ 0 ε 0 .....(iv)
∴ devide eqn (i) by (ii)
(U/V)elec(U/V)mag=12ε0E212B2/μ0
= ( E B ) 2 μ 0 ε 0

by using eq (iii) & (iv)
= C 2 C 2 =1
⇒(U/V)elec.=(UV)mag

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2014 (06 Apr)
During the qualitative analysis of salt with cation y2+, addition of a reagent X to alkaline solution of the salt gives a bright red precipitate. The reagent X and the cation y2+ present respectively are :

Options:
A Dimethylglyoxime and Ni2+
B Dimethylglyoxime and Co2+
C Nessler's reagent and Hg2+
D Nessler's reagent and Ni2+

The correct answer is:
Dimethylglyoxime and Ni2+

Solution:
During the qualitative analysis of salt with cation Ni2+, addition of a reagent Dimethylglyoxime to alkaline solution of the salt gives a bright red precipitate according to the following reaction. So, The reagent X and the cation y2+ respectively are Dimethylglyoxime and Ni2+

Ni2++dmg→Ammonical SolutionNidmg2Red colur↓,

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 1)
During the qualitative analysis of SO32- using dilute H2SO4,SO2 gas is evolved which turns K2Cr2O7 solution (acidified with dilute H2SO4):

Options:
A Black
B Red
C Green
D Blue

The correct answer is:
Green

Solution:
On treating sulphite (SO32-) with warm dilute H2SO4, SO2 gas is evolved which is suffocating with the smell of burning sulphur.

Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2

The gas turns potassium dichromate paper acidified with dil. H2SO4, green.

K2Cr2O7 + H2SO4 + 3SO2 → K2SO4 + Cr2(SO4)3 + H2O

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (30 Jan Shift 1)
During the resting state of the neural membrane, if diffusion occurs due to concentration gradients, it would drive

Options:
A K+ into the cell
B K+ and Na+ out of the cell
C Na+ into the cell
D Na+ out of the cell

The correct answer is:
Na+ into the cell

Solution:
During the resting state, of neural membrane Na+ concentration is higher on the outer side, and K+ level is more within the cell. Voltage-gated channels maintain this concentration gradient. Hence, if diffusion is allowed, Na+ would enter the cell and K+ would leave.

Topic: Biology, Neural Control and Coordination, NEET
During transcription the template strand is the one with polarity

Options:
A Always 3' to 5' as the template strand
B Always 5' to 3' as the template strand
C Any of the strands can become template strand.
D Alternatively both strand work as template strand.

The correct answer is:
Always 3' to 5' as the template strand

Solution:
Transcription occurs in only one direction. Thus the strand that has the polarity 3'â5' acts as the template strand. The other strand which has the polarity 5'â3' is referred to as the coding strand.
Transcription whether prokaryotic or eukaryotic has three main events.
â¢ Initiation - Binding of RNA polymerase to double-stranded DNA; this step involves a transition to single strandedness in the region of binding; RNA polymerase binds at 5â end on a sequence of DNA called the promoter. Initiation is the most important step in gene expression.
â¢ Elongation - The covalent addition of nucleotides to the 3' end of the growing polynucleotide chain. This involves the development of a short stretch of DNA that is transiently single-stranded.
â¢ Termination - The recognition of the transcription termination sequence and the release of RNA polymerase. The promoter and terminator flank the structural gene in a transcription unit.

Topic: Biology, Molecular Basis of Inheritance, NEET
During urine formation, which of the following process helps in maintaining osmotic pressure in the uriniferous tubule?

Options:
A Active Na+ absorption, followed by absorption of Cl-
B Active Cl- absorption, followed by absorption of Na+
C Active secretion of Na+ into efferent arteriole followed by absorption of Cl- into efferent renal arteriole
D Active secretion of Cl- and absorption of N+a into efferent renal arteriole

The correct answer is:
Active Na+ absorption, followed by absorption of Cl-

Solution:
Osmotic pressure is a measure of the tendency to take water by osmosis. During urine formation, 2/3 of NaCl and water filtered in Bowman's capsule, is immediately absorbed by PCT.

Topic: Biology, Excretory Products and their Elimination, NEET
During urine formation, which of the following processes create high osmotic pressure in the uriniferous tubule?

Options:
A Active $\mathrm{Na}^{+}$absorption, followed by absorption of $\mathrm{Cl}^{-}$
B Active $\mathrm{Cl}^{-}$absorption, followed by absorption of $\mathrm{Na}^{+}$
C Active secretion of $\mathrm{Na}^{+}$into efferent arteriole followed by absorption of $\mathrm{Cl}^{-}$into efferent renal arteriole
D Active secretion of $\mathrm{Cl}^{-}$and absorption of $\mathrm{N}^{+}$a into efferent renal arteriole

The correct answer is:
Active $\mathrm{Na}^{+}$absorption, followed by absorption of $\mathrm{Cl}^{-}$

Solution:
- Active sodium absorption is followed by the absorption of chlorine. Sodium is the positively charged electrolyte in extracellular body fluid. The amount of sodium in the fluid influences its volume, which in turn determines blood volume and blood pressure.
- Most of the sodium is reabsorbed in the form of sodium bicarbonate and sodium chloride.
- About \(70 \%\) of the sodium reabsorption occurs in the proximal convoluted tubule. When sodium ion moves, water follows. The movement of water balances the osmotic pressure within or across the tubule walls, which maintains extracellular fluid volume.
- The fluid entering the late proximal tubule has been depleted of the essential substances. As bicarbonate being negatively charged was reabsorbed initially with sodium, chloride ion has been left behind in the tubule.
- Due to the extensive reabsorption of water in the early part of the tubule, chloride ions are highly concentrated and get reabsorbed. So, the correct option is A.

Topic: Biology, Excretory Products and their Elimination, JIPMER, JIPMER 2014
During water-gas shift reaction

Options:
A Carbon monoxide is oxidized to carbon dioxide.
B Water is evaporated in presence of catalyst.
C Carbon is oxidized to carbon monoxide.
D Carbon dioxide is reduced to carbon monoxide.

The correct answer is:
Carbon monoxide is oxidized to carbon dioxide.

Solution:
The reaction between CO of syn gas and steam in presence of catalyst is called water gas shift reaction.

CO+H2O→CO2+H2O

This reaction increases the yield of hydrogen obtained from coal gasification reaction as shown below.

C+H2O⟶CO+H2⏟syn gas

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2023 (15 Apr Shift 1)
During which of the following processes, does entropy decrease ?

(A) Freezing of water to ice at 0°C
(B) Freezing of water to ice at -10°C
(C) N2( g)+3H2( g)→2NH3( g)
(D) Adsorption of CO(g) and lead surface
(E) Dissolution of NaCl in water

Options:
A (A), (B),(C) and (D) only
B (B) and (C) only
C (A) and (E) only
D (A), (C) and (E) only

The correct answer is:
(A), (B),(C) and (D) only

Solution:
(A) Water ⟶0°C ice; ΔS=-ve

(B) Water ⟶-10°C ice; ΔS=-ve

(C) N2( g)+3H2( g)→2NH3( g);ΔS=-ve

(D) Adsorption; ΔS=-ve

(E) NaCl(s)→Na+(aq)+Cl-(aq);ΔS=+ve

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (17 Mar Shift 2)
During which phase of cell cycle amount of DNA in a cell remains at 4C level if the initial amount is denoted as 2C?

Options:
A G2 and G1
B G1 and S
C Only G2
D G2 and M

The correct answer is:
Only G2

Solution:
Cell cycle consists of interphase and mitotic (M) phase. Interphase is divided into G1, S and G2 stages. The mitotic phase is the actual division phase and is divided into karyokinesis and cytokinesis. DNA is seen in all phases. In the M phase both 4C and 2C of DNA is seen. In S phase the synthesis of DNA takes place and hence the amount of DNA doubles. Thus after S phase in G2 phase the DNA doubles i.e. it becomes 4C.

Topic: Biology, Cell Cycle and Cell Division, NEET
During which phase of the cell cycle the amount of DNA in a cell remains at 4C level if the initial amount is denoted as 2C?

Options:
A G0 and G1
B G1 and S
C Only G2
D G2 and M

The correct answer is:
G2 and M

Solution:
The sequential events that lead to formation of the new daughters by cell division and preceded by duplication of its genome is termed as cell cycle. Whole cell cycle is broadly divided into two parts -interphase and divisional phase. The interphase is divided into -
G1 phase - The cell grows and develops by been metabolically active but does not duplicate the genetic content.
S phase - In the synthesis phase, DNA duplication occurs. The genetic content is doubled in this phase. There is no increase in the chromosome number as such.
G2 phase - This phase follows S phase having double the genomic content. During this phase, all proteins required for the divisional process is synthesized and the growth of cell continues.
G0 phase - Some cells lose the ability to divide further into new cells. Such cells after G1 phase enter the G0 or quiescent stage of cell cycle.
After interphase, the cell enters the divisional phase which is symbolized as M phase for mitosis. The cell in divisional phase undergoes nuclear division, where the duplicated DNA content is separated into two new nuclei and then the whole cell divides to form new daughter cells.

Topic: Biology, Cell Cycle and Cell Division, NEET
During which phase(s) of cell cycle, amount of DNA in a cell remains at 4 C level if the initial amount is denoted as 2C?

Options:
A G0 and G1
B G1 and S
C Only G2
D G2 and M

The correct answer is:
G2 and M

Solution:
Explanation for the correct option:
1. At certain points during M-phase, both \(4 C\) and \(2 C\) DNA is present.
2. The S phase occurs when the amount of DNA in the cell doubles.
3. As a result, the amount of DNA present in the \(G 2\) phase is 4 C.
4. The cell's contents have been doubled, preparing it for further mitotic division.
5. Interphase is made up of the G1 phase (cell growth). S phase (DNA synthesis), and G2 phase (cell growth).
6. At the completion of the interphase, the mitotic phase begins, which consists of mitosis and cytokinesis and results in the production of two daughter cells.
Explanation for the incorrect option:
Option A:
1. The RNA, proteins and other macromolecules essential for the multiplication of cell organelles, spindle formation, and cell development are created during the G2 phase (Gap 2) as the cell prepares to enter the mitotic phase.
2. The G2 phase is characterized by fast cell growth and protein synthesis as the cell prepares for mitosis.
3. Surprisingly, the G2 phase is not required for all cell types, since some (especially immature Xenopus embryos and certain malignancies) progress directly from DNA replication to mitosis.
Option B:
1. During the G1 phase, the cell physically expands and increases the volume of both protein and organelles.
2. During the S phase, the cell duplicates its nucleosomes and copies its DNA to form two sister chromatids.
3. The G1/S transition is a step in the cell cycle that occurs at the junction of the G1 phase, during which the cell develops, and the S phase, during which DNA is reproduced.
Option C:
1. GO phase: resting state, also known as the gap phase.
2. Many cells spend the majority of their time in this phase, either resting or executing assigned tasks.
3. Chemotherapy is generally ineffective. G1 phase: gap 1 phase, also known as interphase.
Final answer: G2 and M phase(s) of the cell cycle, the amount of DNA in a cell remains at 4C level if the initial amount is denoted as 2C.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2014
During which phase(s) of the cell cycle, the amount of DNA in a cell remains at 4C level if the initial amount is denoted as 2C?

Options:
A G0 and G1
B G1  and S
C only G2
D G2 and M

The correct answer is:
only G2

Solution:
Cell cycle consists of Interphase and Mitotic phase. Interphase is divided into G1 , S and G2 stages. Mitotic Phase is the actual division phase and is divided into Karyokinesis and Cytokinesis. DNA is seen in all phases. In M phase both 4C and 2C of DNA is seen. In S phase the synthesis of DNA takes place and hence the amount of DNA doubles. Thus after S phase in G2 phase the DNA doubles i.e. it becomes 4C.

Topic: Biology, Cell Cycle and Cell Division, NEET
Dust, oolong and brick are varieties of

Options:
A coffee
B pepper
C tea
D lavang

The correct answer is:
tea

Solution:
Four main types of commercial tea are generally recognized in the trade. These are black tea (China, India, Sri Lanka), green tea (China, Japan and Taiwan), oolong tea (Taiwan) and brick tea (China). Dust or fannings are left over small part of tea leaves, which are also widely used.

Topic: Biology, Biodiversity and its Conservation, AIIMS, AIIMS 2012
E1, E2 and E3 are the emf of the following three galvanic cells respectively

(i) ZnsZn2+0.1 MCu2+1 MCus

(ii) ZnsZn2+1 MCu2+1 MCus

(iii) ZnsZn2+1 MCu2+0.1 MCus

Which of the following is true?

Options:
A E2>E1>E3
B E1>E2>E3
C E3>E1>E2
D E3>E2>E1

The correct answer is:
E1>E2>E3

Solution:
Key-idea

Use Nernst equation, i.e.

Eemf=E°cell-0.05912logZn2+Cu2+

For reaction i

E1=E°cell-0.05912log0.11

E1=E°cell+0.05912

For reaction ii

E2=E°cell-0.05912log11=0

E2=E°cell

For reaction iii

E3=E°cell-0.05912log10.1

E3=E°cell-0.05912

Hence, E1>E2>E3.

Topic: Chemistry, Electrochemistry, JEE Main
$\int e^{-2 x} \sin 3 x d x=$

Options:
A $\frac{1}{13} e^{-2 x}[\sin 3 x+\cos 3 x]+c$
B $-\frac{1}{13} e^{-2 x}[\sin 3 x+\cos 3 x]+c$
C $\frac{1}{13} e^{-2 x}[2 \sin 3 x+3 \cos 3 x]+c$
D $-\frac{1}{13} e^{-2 x}[2 \sin 3 x+3 \cos 3 x]+c$

The correct answer is:
$-\frac{1}{13} e^{-2 x}[2 \sin 3 x+3 \cos 3 x]+c$

Solution:
$\begin{aligned} & \text { Let } I=\int e^{-2 x} \sin 3 x d x \\ & =-\frac{e^{-2 x} \cos 3 x}{3}-\int \frac{2 e^{-2 x} \cos 3 x}{3} d x \\ & =-\frac{e^{-2 x} \cos 3 x}{3}-\frac{2}{3}\left[\frac{e^{-2 x} \sin 3 x}{3}+\int \frac{2 e^{-2 x} \sin 3 x}{3} d x\right]\end{aligned}$
$\begin{aligned} & \Rightarrow I=-\frac{e^{-2 x} \cos 3 x}{3}-\frac{2 e^{-2 x} \sin 3 x}{9}-\frac{4}{9} I \\ & \Rightarrow \frac{13}{9} I=-e^{-2 x}\left[\frac{3 \cos 3 x+2 \sin 3 x}{9}\right] \\ & \text { Hence } I=-\frac{1}{13} e^{-2 x}[3 \cos 3 x+2 \sin 3 x]\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
$\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=$

Options:
A e. $3^{-3 x}+c$
B $e^3 \log x+c$
C $\frac{x^3}{3}+c$
D None of these

The correct answer is:
$\frac{x^3}{3}+c$

Solution:
$\begin{aligned} & \int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x=\int \frac{x^5-x^4}{x^3-x^2} d x \\ &=\int \frac{x^4(x-1)}{x^2(x-1)} d x=\int x^2 d x=\frac{x^3}{3}+c\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main