datavorous/entrance-exam-dataset huggingface dataset:
The apparent coefficients of volume expansion of a liquid filled in vessel A and B of identical volumes are found to be Î³ 1 and Î³ 2, respectively. If Î± 1 be the coefficient of linear expansion of A then the coefficient of linear expansion of B will be

Options:
A γ 1 - γ 2 3 - α 1
B Î³ 2 - Î³ 1 3 + Î± 1
C γ 2 - γ 1 3 - α 1
D γ 1 - γ 2 3 + α 1

The correct answer is:
γ 2 - γ 1 3 + α 1

Solution:
Let V be the volume of the liquid and Δ T the rise in temperature. Since apparent expansion = true expansion - expansion of vessel
∴ Vγ1ΔT=VγΔT-V3α1ΔT
or, γ 1 = γ + 3 α 1 ...(i)
(for vessel A)
and γ 2 = γ + 3 α 2 ...(ii)
(for vessel B)
where γ is the coefficient of real expansion of the liquid.
Subtracting Eq. (ii) from Eq. (i),
γ 1 - γ 2 = 3 α 1 - α 2
or, α 1 - α 2 = γ 1 - γ 2 / 3
or, α2= γ2-γ1/3+α1

Topic: Physics, Thermal Properties of Matter, JEE Main
The apparent weight of the body, when it is travelling upwards with an acceleration of $2 \mathrm{~m} / \mathrm{s}^2$ and mass is $10 \mathrm{~kg}$, will be

Options:
A $198 \mathrm{~N}$
B $164 \mathrm{~N}$
C $140 \mathrm{~N}$
D $118 \mathrm{~N}$

The correct answer is:
$118 \mathrm{~N}$

Solution:
$R=m(g+a)=10 \times(9.8+2)=118 \mathrm{~N} \mid$

Topic: Physics, Laws of Motion, JEE Main
The appearance of colour in solid alkali metal halides is generally due to:

Options:
A
Schottky defect

B
Frenkel defect

C
Interstitial position

D
F-centres

The correct answer is:
F-centres

Solution:
The appearance of colour in alkali metal halides is generally due to F-centres. F centers in $\mathrm{NaCl}$ produce a band of optical absorption towards the blue end of visible
spectrum, thus the colour appears yellow under transmitted light. On the other hand, $\mathrm{KCl}$ with $\mathrm{F}$-centers appears magenta, and $\mathrm{KBr}$ appears blue.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2014 (11 Apr Online)
The appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because:

Options:
A nodes are shorter than internodes
B nodes have meristematic cells
C nodes are located near the soil
D nodes have non-photosynthetic cells

The correct answer is:
nodes have meristematic cells

Solution:
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because nodes have meristematic cells. Examples of vegetative propagules: (i) Leaf buds of Bryophyllum, (ii) Eyes of potato, (iii) Bulbil of Agave, (iv) Offset of water hyacinth, (v) Rhizome of Ginger.

Topic: Biology, Reproduction in Organisms, NEET
The appropriate reagent for the transformation

Options:
A Zn(Hg), HCl
B NH2NH2,OH-
C H2/Ni
D NaBH4

The correct answer is:
NH2NH2,OH-

Solution:
Both Zn(Hg), HCl (Clemmenson’s reduction) and NH2NH2 , OH- (Wolf-Kishner reduction) can reduce -C||O-CH3 to -CH2-CH3 group but as HCl may also bring about the dehydration of alcohol to give an alkene so the most appropriate reagent will be NH2NH2,OH- .

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The approximate height from the surface of earth at which the weight of the body becomes 13 of its weight on the surface of earth is :
[Radius of earth R=6400 km and 3=1.732]

Options:
A 3840 km
B 4685 km
C 2133 km
D 4267 km

The correct answer is:
4685 km

Solution:
The variation of the acceleration due to gravity with height h is given by,

g'=g1+hR2.

Since the weight at given height is 13 of the weight at the surface

⇒g'=g3⇒g1+hR2=g3⇒1+hR=3⇒h=R3-1=6400×1.732-1=4685 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The approximate pH of a 0.1 M aqueous H2S solution is ( K1 and K2 for H2S are 1Ã10-7 and 1.3Ã10-13 respectively at 25 Â° C)

Options:
A 4
B 2
C 8
D 6

The correct answer is:
4

Solution:
Second dissociation of H2S is negligible from data
H2S⇌H++HS-
[H+]=[HS-]=x
K1=[H+][HS-][H2S]
1×10-7=x20.1
So x=10-4
pH=-log⁡[H+]
=-log⁡10-4
=4log⁡10=4 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
The approximate value of $\sqrt[3]{28}$ is

Options:
A $3.0037$
B $3.037$
C $3.0086$
D $3.37$

The correct answer is:
$3.037$

Solution:
Let $y=\sqrt[3]{28}$
Taking log on both sides, we get
$=3.037$ (approximately)

Topic: Mathematics, Application of Derivatives, JEE Main
The approximate value of $\log _{10} 998$ is (given that $\log _{10} \mathrm{e}=0.4343$ )

Options:
A 3.0008686
B 1.9991314
C 2.0008686
D 2.9991314

The correct answer is:
2.9991314

Solution:
Let
$\begin{aligned}
\mathrm{f}(x)=\log _{10} x=\frac{\log _{\mathrm{e}} x}{\log _{\mathrm{e}} 10} & =\left(\log _{10} \mathrm{e}\right)\left(\log _{\mathrm{e}} x\right) \\
& =0.4343\left(\log _{\mathrm{e}} x\right)
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned}
\mathrm{f}^{\prime}(x) & =\frac{0.4343}{x} \\
\text { Let } x & =998 \\
& =1000-2=\mathrm{a}+\mathrm{h} \\
\therefore \quad \mathrm{a}= & 1000, \mathrm{~h}=-2 \\
\mathrm{f}(\mathrm{a}) & =\mathrm{f}(1000) \\
& =\log _{10}(1000) \\
& =3 \log _{10} 10 \\
\therefore \quad \mathrm{f}(\mathrm{a}) & =3
\end{aligned}$
$\begin{aligned}
& \text { Also, } \mathrm{f}^{\prime}(\mathrm{a})=\mathrm{f}^{\prime}(1000)=\frac{0.4343}{1000}=0.0004343 \\
& \mathrm{f}(\mathrm{a}+\mathrm{h}) \approx \mathrm{f}(\mathrm{a})+\mathrm{hf}^{\prime}(\mathrm{a}) \\
& \therefore \quad \log _{10}(998) \approx 3-2(0.0004343) \\
&
\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
The aqueous 0.01 M of [Cr(NH3)6]2[Co(NH3)(NO2)5]3 is expected to have ÎTf equal to
Given: K f of H 2 O is 1.86 K kg  mol −1 .

Options:
A 0.0186
B 0.0372
C 0.0558
D 0.093.

The correct answer is:
0.093.

Solution:
ΔTf=i×m×Kf
=5×0.01×1.86
=9.30×0.01=0.093

Topic: Chemistry, Solutions, JEE Main
The aqueous solution containing which one of the following ions will be colourless?
(Atomic number : $\mathrm{Sc}=21, \mathrm{Fe}=26, \mathrm{Ti}=$ $22, \mathrm{Mn}=25$)

Options:
A $\mathrm{Sc}^{3+}$
B $\mathrm{Fe}^{2+}$
C $\mathrm{Ti}^{3+}$
D $\mathrm{Mn}^{2+}$

The correct answer is:
$\mathrm{Sc}^{3+}$

Solution:
\(\begin{array}{|c|c|c|}
\hline \text { Ion } & \begin{array}{c}
\text { No. of } \\
\text { Electrons }
\end{array} & \begin{array}{c}
\text { Electronic } \\
\text { Configuration }
\end{array} & \text{ Electronic Configuration in box } & \begin{array}{c}
\text { No. of } \\
\text { Unpaired }
\end{array} & \begin{array}{c}
\text { Magnetic } \\
\text { Nature }
\end{array} & \begin{array}{c}
\text { Magnetic Mo- } \\
\text { ment (BM) } \\
\mu=\sqrt{n(\boldsymbol{n}+2)}
\end{array} \\
\hline \mathrm{Sc}^{3+} & 18 & {[\mathrm{Ar}] 3 d^0 4 \mathrm{~s}^0} & [\mathrm{Ar}] ~ \begin{array}{|l|l|l|l|l|}
\hline ~~ & ~~ & ~~ & ~~ & ~~ \\
\hline
\end{array} ~~ \begin{array}{|l|}
\hline ~~ \\
\hline
\end{array} & 0 & \text { Diamagnetic } & 0 \\ \\
\hline \mathrm{Ti}^{3+} & 22 & {[\mathrm{Ar}] 3 d^4 4 \mathrm{~s}^0} & [\mathrm{Ar}] ~ \begin{array}{|l|l|l|l|l|}
\hline \uparrow & ~~ & ~~ & ~~ & ~~ \\
\hline
\end{array} ~~ \begin{array}{|l|}
\hline ~~ \\
\hline
\end{array} & 1 & \text { Paramagnetic } & 1.73 \\ \\
\hline \mathrm{Mn}^{2+} & 23 & {[\mathrm{Ar}] 3 d^5 4 \mathrm{~s}^0} & [\mathrm{Ar}] ~ \begin{array}{|l|l|l|l|l|}
\hline \uparrow & \uparrow & \uparrow & \uparrow & \uparrow \\
\hline
\end{array} ~~ \begin{array}{|l|}
\hline ~~ \\
\hline
\end{array} & 5 & \text { Paramagnetic } & 5.92 \\ \\
\hline \mathrm{Fe}^{2+} & 24 & {[\mathrm{Ar}] 3 d^6 4 \mathrm{~s}^0} & [\mathrm{Ar}] ~ \begin{array}{|l|l|l|l|l|}
\hline \uparrow \downarrow & \uparrow & \uparrow & \uparrow & \uparrow \\
\hline
\end{array} ~~ \begin{array}{|l|}
\hline ~~ \\
\hline
\end{array} & 4 & \text { Paramagnetic } & 4.9 \\ \\
\hline
\end{array}\)
Related Theory
Because of the relatively high forces of attraction among the water molecules due to intermolecular hydrogen bonding, water has $a$ higher surface tension than most other liquids.
Related Theory
Magnetic properties arise from the spin and orbital angular momentum of the electrons contained in a compound. Compounds are diamagnetic when they contain no unpaired electrons.

Topic: Chemistry, d and f Block Elements, JEE Main
The Arctic tern is a bird which travels from the Arctic to the Antarctic region and then back again. Hence it sees two summers in a year. On an average, Arctic terns travel a distance of 70,000 km annually. This is an example of:

Options:
A Conformation
B Regulation
C Migration
D Suspension

The correct answer is:
Migration

Solution:
There are four major ways through which an organism can cope with stressful conditions:
By regulating the constant body temperature, by changing their body temperature with ambient temperature found in animals that cannot maintain constant body temperature known as conforming, some animals migrate to the comfortable area and return when the stressful period gets over, by forming thick walled spores that help to survive in stressful condition (e.g. in bacteria, fungi and lower plants) and this response is called 'suspend'. The response observed here is migration. During winter many animals, especially birds, take on long distance migration to friendlier environment.

Topic: Biology, Organisms and Populations, NEET
The area between the curve y=2x4-x2, the x-axis and the ordinates of the two minima of the curve is

Options:
A 11 60 sq.âunits
B 7 120 âsq.âunits
C 1 30 sq.âunits
D 7 90 âsq.âunits

The correct answer is:
7 120  sq. units

Solution:
The equation of the curve is y=2x4-x2=2x2-1 x 2
The curve is symmetrical about the y -axis.
Also, it is a polynomial of degree four having roots 0, 0,âÂ± 1 2 .
x=0 is repeated root. Hence, graph touches x -axis at (0,0) and intersects the x -axis at A â 1 2 , â0 â&âB 1 2 , â0
â dy dx =8 x 3 â2x
=2x(4 x 2 â1)=0
x=0,âx=Â± 1 2
d 2 y d x 2 >0 at x= 1 2 and at x= â1 2
So x = Â± 1 2 are the points of local minima
Thus, the graph of the curve is shown in diagram.
Here, yâ¤0 , as x varies from x=-12 to x=12
â´ The required area
=2 Area OCDO
=2â«012y dx
=2â«0122x4-x2 dx
=7120 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area between the curve $y=\sin ^2 x, \quad x$ - axis and the ordinates $x=0$ and $x=\frac{\pi}{2}$ is

Options:
A $\frac{\pi}{2}$
B $\frac{\pi}{4}$
C $\frac{\pi}{8}$
D $\pi$

The correct answer is:
$\frac{\pi}{4}$

Solution:
$\begin{aligned} \text{Required area } A & =\int_0^{\pi / 2} \sin ^2 x \cdot d x=\int_0^{\pi / 2}\left(\frac{1-\cos 2 x}{2}\right) d x a \\ & =\frac{1}{2}[x]_0^{\pi / 2}-\frac{1}{4}[\sin 2 x]_0^{\pi / 2}=\frac{\pi}{4}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area between the curves x=4y-y2 and x=0 is λ square units, then the value of 3λ is equal to

Options:
A 28
B 30
C 32
D 36

The correct answer is:
32

Solution:
The required area is (say A)
A=∫04x⋅dy=∫044y-y2dy
=2y2-y3304
=32-643=323
⇒3λ=32

Topic: Mathematics, Area Under Curves, JEE Main
The area between the parabola $y^2=4 a x$ and $x^2=8 a y$ is

Options:
A $\frac{8}{3} a^2$
B $\frac{4}{3} a^2$
C $\frac{32}{3} a^2$
D $\frac{16}{3} a^2$

The correct answer is:
$\frac{32}{3} a^2$

Solution:
Required area = $A=\int_0^{\left(a 2^{8 / 3}\right)} \sqrt{4 a x} d x-\int_0^{a 2^{8 / 3}} \frac{x^2}{8 a} d x=\frac{32 a^2}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area between the parabola $y=x^2$ and the line $y=x$ is

Options:
A $\frac{1}{6}$ sq. unit
B $\frac{1}{3}$ sq. unit
C $\frac{1}{2}$ sq. unit
D None of these

The correct answer is:
$\frac{1}{6}$ sq. unit

Solution:
Given curves are $y=x^2$ and $y=x$ On solving, we get $x=0, x=1$
Therefore, required area
$A=\int_0^1\left(x^2-x\right) d x$
$=\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_0^1=\frac{1}{3}-\frac{1}{2}=\frac{1}{6}$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded between the parabolas $x^2=\frac{y}{4}$ and $x^2=9 y$, and the straight line $y=2$ is

Options:
A
$20 \sqrt{2}$

B
$\frac{10 \sqrt{2}}{3}$

C
$\frac{20 \sqrt{2}}{3}$

D
$10 \sqrt{2}$

The correct answer is:
$\frac{20 \sqrt{2}}{3}$

Solution:
Required area
$\begin{aligned}
& A=2\left[\int_0^2\left(3 \sqrt{y}-\frac{\sqrt{y}}{2}\right) d y\right]=2 \int_0^2 \frac{5 \sqrt{y}}{2} d y \\
& =5\left[\frac{y^{3 / 2}}{3 / 2}\right]_0^2=\frac{10}{3}\left[2^{3 / 2}-0\right]=\frac{20 \sqrt{2}}{3}
\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2012 (Offline)
The area bounded by fx=sin2x and the x-axis from x=a to x=b, where f"a=f"b=0 (âa, bâ0,Ï), is

Options:
A π4
B π+2
C 2
D π+24

The correct answer is:
π+24

Solution:
f"x=2cos⁡2x=0⇒2x=π2, 3π2⇒x=π4, 3π4
Hence, required area =∫π43π4sin2xdx=∫π43π41-cos⁡2x2dx=x2-sin⁡2x4π43π4=π+24

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by fx=sinâ¡2xxâ¥0cosâ¡2xx<0 with the x-axis, x=-Ï4 and x=Ï4 is k square units. Then, the value of 4k is equal to

Options:

The correct answer is:
4

Solution:
Required area is A=∫-π40cos⁡2xdx+∫0π4sin⁡2xdx
=sin⁡2x2-π40-cos⁡2x20π4
=0--12-0-12=1 sq. units
⇒k=1
∴4k=4

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve a2y=x2(x+a) and the x-axis is

Options:
A a23Â sq.Â units
B a24Â sq.Â units
C 3a24Â sq.Â units
D a212 sq.units

The correct answer is:
a212 sq.units

Solution:
The curve is y=x2(x+a)a2 which is a cubic polynomial.

Since x2(x+a)a2=0 has a repeated root x=0,

it touches x-axis at (0, 0) and intersects at (-a, 0) .
Hence, required area =∫-a0y dx=∫-a0x2x+aa2dx=1a2x44+ax33-a0
=a212 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is

Options:
A 32 sq units
B 16 sq units
C $\frac{16}{3}$ sq units
D $\frac{32}{3}$ sq units

The correct answer is:
$\frac{32}{3}$ sq units

Solution:
We have,
$$
x=4-y^{2} \Rightarrow y^{2}=-(x-4)
$$
$\begin{aligned} \text { Required area } &=2 \int_{0}^{4} y d x=2 \int_{0}^{4} \sqrt{4-x} d x \\ &=2\left[\frac{-(4-x)^{3 / 2}}{3 / 2}\right]_{0}^{4} \\ &=\frac{-4}{3}\left[(4-4)^{3 / 2}-(4-0)^{3 / 2}\right] \\ &=\frac{-4}{3}(-8)=\frac{32}{3} \text { sq units } \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve y=12x2, x-axis and x=2 is

Options:
A 13 sq unit
B 23 sq unit
C 1 sq unit
D 43 sq unit

The correct answer is:
43 sq unit

Solution:
Required area =â«02ydx
=â«02x22dx=x3602=43 sq unit

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve y2=1-x and the lines y=xx, x=-1 and x=12 is

Options:
A 32-116 sq. units
B 32-114Â sq.Â units
C 62-115Â sq.Â units
D None of these

The correct answer is:
32-116 sq. units

Solution:
From the diagram, the area of the shaded region,
A=â«-10-1--1-xdx+â«0121-1-xdx
=-x-1-x3232-10+x+1-x3232012
=-23-1-2Ã2323+12+23Ã232-23
=23Ã232+2Ã2323-43-12
=32-43-12
=32-116 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve $y=\frac{1}{4}\left|4-x^{2}\right|$ and $y=7-|x|$ is

Options:
A 18
B 32
C 36
D 64

The correct answer is:
32

Solution:
Required Area
$\begin{aligned} \mathrm{A}=2\left[\int_{0}^{2}((7-\mathrm{x})-\right.&\left.\frac{1}{4}\left(4-\mathrm{x}^{2}\right)\right) \mathrm{dx} \\\left.+\int_{2}^{4}\left((7-\mathrm{x})-\frac{1}{4}\left(\mathrm{x}^{2}-4\right) \mathrm{dx}\right)\right] \end{aligned}$
solve this $=32$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve y=cos-1sin⁡x+π2-cos-1cos⁡x and the x -axis, where π2≤x≤π, is equal to

Options:
A π2
B π22
C π28
D π24

The correct answer is:
π24

Solution:
y=π2-sin-1sin⁡x+π2-cos-1cos⁡x
For x∈π2,π
y=π2-π-x+π2-x
y=x-π2+π2-x
y=x-π2+x-π2=2x-π

Area =12×π2×π
=π24

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The area bounded by the curve $y=\cos x$, the line joining $(-\pi / 4, \cos (-\pi / 4))$ and $(0,2)$ and the line joining $(\pi /$ $4, \cos (\pi / 4))$ and $(0,2)$ is

Options:
A $\left(\frac{4+\sqrt{2}}{8}\right) \pi-\sqrt{2}$
B $\left(\frac{4+\sqrt{2}}{8}\right) \pi+\sqrt{2}$
C $\left(\frac{4+\sqrt{2}}{4}\right) \pi-\sqrt{2}$
D $\left(\frac{4+\sqrt{2}}{4}\right) \pi+\sqrt{2}$

The correct answer is:
$\left(\frac{4+\sqrt{2}}{8}\right) \pi-\sqrt{2}$

Solution:
required area $=2\left[\frac{1}{2}\left(\frac{1}{\sqrt{2}}+2\right) \frac{\pi}{4}-\int_{0}^{\pi / 4} \cos x \mathrm{dx}\right]$
$=\left(\frac{4+\sqrt{2}}{8}\right) \pi-\sqrt{2}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve $y=\ln (x)$ and the lines $y=0, y=\ln (3)$ and $x=0$ is equal to:

Options:
A
3

B
$3 \ln (3)-2$

C
$3 \ln (3)+2$

D
2

The correct answer is:
2

Solution:
To find the point of intersection of curves
$$
\begin{aligned}
& y=\ln (x) \text { and } y=\ln (3), \text { put } \ln (x)=\ln (3) \\
\Rightarrow & \ln (x)-\ln (3)=0 \\
\Rightarrow & \ln (x)-\ln (3)=\ln (1) \\
\Rightarrow & \frac{x}{3}=1, \Rightarrow x=3
\end{aligned}
$$
$\begin{aligned} \text { Required area } & =\int_0^3 \ln (3) d x-\int_1^3 \ln (x) d x \\ = & {[x \ln (3)]_0^3-\left[x \ln (x)-x_1^3=2\right.}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2013 (09 Apr Online)
The area bounded by the curve y=sin-1⁡sin⁡x and the x-axis from x=0 to x=4π is equal to the area bounded by the curve y=cos-1⁡cos⁡x and the x-axis from x=-π to x=a, then the value of a is equal to

Options:
A π2
B 2π
C π
D 3π2

The correct answer is:
π

Solution:
Graph of y=sin-1⁡sin⁡x
Area =4×12×π×π2=π2
Graph of y=cos-1cos⁡x is
Given, π2=12×π×π+12×a×a
π22=a22⇒a=π

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

Options:
A $1 / 6$
B $2 / 3$
C $1 / 4$
D $1 / 3$

The correct answer is:
$1 / 3$

Solution:
Required area $=2\left|\int_{1 / 4}^1(\sqrt{y}-1) d y\right|_{,(\text {From the symmetry) }}$
$\text { On solving, we get required area }=\frac{1}{3} \text { sq. unit. }$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve y=x2-9 and the line y=3 is

Options:
A 86-1612-72
B 86+812-72
C 166+1612-72
D 166-1612-64

The correct answer is:
166+1612-72

Solution:
We first find the points of intersection of the curves y=x2-9 and y=3

which gives, x2-9=3 ⇒x2-9=±3

⇒x2=12 (or) x2=6 ⇒x=±12 (or) x=±6

The graphs of the two curves are shown in figure

We have to find the area of the shaded region.

Required area =2A1+2A2 (∵ The areas A1 & A2 are symmetrical about y- axis )

where A1=∫069-x2−3dx

A1=∫066-x2dx=6x-x3306 =66-663=1263=46

A2=∫633−9−x2dx+∫3123−x2−9dx

=∫63-6+x2dx+∫31212-x2dx

=-6x+x3363+12x-x33312

=-18+9--66+26)+1212-412-36-9

=-36+46+812

Hence, required area =2A1+2A2

=246+2-36+46+812

=86-72+86+1612

=166+1612-72

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (26 Jun Shift 1)
The area bounded by the curve y=x2x-12 with the x-axis is k sq. units, then the value of 60k is equal to

Options:
A 1
B 2
C 12
D 14

The correct answer is:
2

Solution:
The required area is A=∫01x2x-12dx
or A=∫01x2-x2dx=∫01x4+x2-2x3dx
=x55+x33-2x4401
=15+13-24=130=k
∴60k=2

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curve y=x with the x-axis from x=Ï to x=3.8 is Ï2-ab-Ï sq. units, then the value of b-a is equal to (where . denotes the fractional part function)

Options:

The correct answer is:
2.7

Solution:
Required Area = Area of Trapezium
=12 (Sum of parallel sides) (Distance between them)
=12Ï+3.83.8-Ï
=12Ï-3+0.83.8-Ï
=12Ï-2.23.8-Ï
=Ï2-1.13.8-Ï sq. units
âb=3.8, a=1.1

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves $y-1=\cos x, y=\sin x$ and the $\mathrm{X}$-axis between $\mathrm{x}=0$ and $\mathrm{x}=\pi$ is

Options:
A $2+\frac{\pi}{2}$
B $-\frac{\pi}{2}$
C $2-\frac{\pi}{2}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\pi}{2}$

Solution:
We have to find the area bounded by $y-1=\cos x, y$ $=\sin x$ and $x$-axis betweein $x=0$ and $x=\pi$.
Let us draw the above region
The area of bounded region
$=\operatorname{Area}(O A C)+\operatorname{Area}(A C D)$
$\begin{aligned} & =\int_0^{\pi / 2} \sin x d x+\int_{\pi / 2}^\pi(\cos x+1) d x \\ & =[-\cos x]_0^{\pi / 2}+[\sin x+x]_{\pi / 2}^\pi \\ & =[-0+1]+\left[0+\pi-1-\frac{\pi}{2}\right] \\ & =1+\pi-1-\frac{\pi}{2}=\frac{\pi}{2}\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves $y^2=8 x$ and $y=x$ is

Options:
A $\frac{128}{3}$sq. unit
B $\frac{32}{3}$sq. unit
C $\frac{64}{3}$ sq. unit
D $32$ sq. unit

The correct answer is:
$\frac{32}{3}$sq. unit

Solution:
$y^2=8 x$ and $y=x \Rightarrow x^2=8 x \Rightarrow x=0,8$
$\begin{aligned}& \therefore \text { Required area }=\int_0^8(2 \sqrt{2} \sqrt{x}-x) d x \\& =\left[\frac{4 \sqrt{2}}{3} x^{3 / 2}-\frac{x^2}{2}\right]_0^8=\frac{128}{3}-\frac{64}{2}=\frac{32}{3} s q\end{aligned}$unit.

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves $y^2-x=0$ and $y-x^2=0$ is

Options:
A $\frac{7}{3}$
B $\frac{1}{3}$
C $\frac{5}{3}$
D $1$

The correct answer is:
$\frac{1}{3}$

Solution:
$\int_0^1\left(\sqrt{x}-x^2\right) d x=\left(\frac{x^{3 / 2}}{3 / 2}-\frac{x^3}{3}\right)_0^1=\frac{1}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves y=cos⁡x and y=sin⁡2x, ∀x∈π6,π2 is equal to

Options:
A π2 sq. units
B π3 sq. units
C π6 sq. units
D 14 sq. units

The correct answer is:
14 sq. units

Solution:
A=∫π6π2sin⁡2x-cos⁡xdx=–cos⁡2x2-sin⁡xπ6π2
=cos⁡2x2+sin⁡xπ2π6
=12×12+12+12-1=14 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3 \pi}{2}$ is

Options:
A
$4 \sqrt{2}+2$

B
$4 \sqrt{2}-1$

C
$4 \sqrt{2}+1$

D
$4 \sqrt{2}-2$

The correct answer is:
$4 \sqrt{2}-2$

Solution:
$\int_0^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}(\sin x-\cos x) d x+\int_{\frac{5 \pi}{4}}^{\frac{3 \pi}{2}}(\cos x-\sin x)=4 \sqrt{2}-2$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2010
The area bounded by the curves y = ln x , y = ln x , y = ln x and y = ln x , for x∈( −1, 1 ) is

Options:

The correct answer is:
4

Solution:
The required shaded region is as shown in figure
As the graph is symmetric in all quadrants, we calculate area in one quadrant and multiply by 4.
Hence, required area =4| â« 0 1 ( lnâx ).dx |
A=4| [ xâlnâxâx ] 0 1 |
A=4| [ ( 1âlnâ1â1 )â lim xâ 0 + ( xâlnâxâx ) ] |
A=4| [ ( 0â1 )â lim xâ 0 + lnâx 1 x â0 ] |=4| [ â1â lim xâ 0 + 1 x â 1 x 2 ] |
A=4| [ â1+ lim xâ 0 + x ] |
A=+4| â1 |
A=4

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves $y=\ln x, y=\ln |x|, y=|\ln x|$ and $y=|\ln ||x|$ is

Options:
A
4 sq. units

B
6 sq. units

C
10 sq. units

D
none of these

The correct answer is:
4 sq. units

Solution:
No Solution Available

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2002
The area bounded by the curves $y=\log _e x$ and $y=\left(\log _e x\right)^2$ is

Options:
A $3-e$
B $e-3$
C $\frac{1}{2}(3-e)$
D $\frac{1}{2}(e-3)$

The correct answer is:
$3-e$

Solution:
Required area $=\int_1^e\left[\log x-(\log x)^2\right] d x$
$\begin{aligned}A & =\int_1^e \log x d x-\int_1^e(\log x)^2 d x \\& =[x \log x-x]_1^e-\left[x(\log x)^2-2 x \log x+2 x\right]_1^e \\& =[e-e-(-1)]-\left[e(1)^2-2 e+2 e-(2)\right] \\& =(1)-(e-2)=3-e .\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves y = x - 1 2 , y = x + 1 2 and y = 1 4 is

Options:
A 1 3 sq unit
B 2 3 sq unit
C 1 4 sq unit
D 1 5 sq unit

The correct answer is:
1 3 sq unit

Solution:
The curves y = x - 1 2 , y = x + 1 2 and y = 1 4 are shown as
where point of intersection are
x - 1 2 = 1 4
â x = 1 2
and x + 1 2 = 1 4 â x=- 1 2
â´ Q 1 2 , 1 4 and R - 1 2 , 1 4
â´ Required area = 2 â« 0 1 / 2 x - 1 2 - 1 4 dx
= 2 x - 1 3 3 - 1 4 x 0 1 / 2
= 2 - 1 8 Â· 3 - 1 8 - - 1 3 - 0 = 8 2 4 = 1 3 sq unit

Topic: Mathematics, Area Under Curves, JEE Main
The area bounded by the curves y=x-1+x-2 and y=3 is equal to

Options:
A 4
B 6
C 3
D 5

The correct answer is:
4

Solution:
Given:

y=fx=x-1+x-2

⇒fx=3-2x: x<11; 1≤x<22x-3; x≥2

Let us draw diagram of y=fx & y=3 we get,

Required area is the area of trapezium which is

Area =121+3×2

=4 sq. units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (06 Apr Shift 2)