datavorous/entrance-exam-dataset huggingface dataset:
The area of the region (in square units) above the x-axis bounded by the curve yâ¡=tanx, 0â¤xâ¡â¤Ï2 and the tangent to the curve at xâ¡ = Ï 4 is

Options:
A 1 2 log 2 - 1 2
B 1 2 1 + log 2
C 1 2 1 - log 2
D 1 2 log 2 + 1 2

The correct answer is:
1 2 log 2 - 1 2

Solution:
Required Area
=∫0π4tanxdx-Area of triangle formed by tangent, x=π4 and x-axis
=log⁡secπ4-log⁡sec0-12×1π4-x0
=log⁡2-14 (where x0=π4-12 is x-interept of tangent)
=12log⁡2-14

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region (in the square unit) bounded by the curve $x^2=4 y$, line $x=2$ and $x$-axis is

Options:
A $1$
B $\frac{2}{3}$
C $\frac{4}{3}$
D $\frac{8}{3}$

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

Solution:
Given $x^2=4 y$
$\text { Shaded area }=\int_0^2 \frac{x^2}{4} d x=\frac{1}{12}\left[x^3\right]_0^2=\frac{8}{12}=\frac{2}{3}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region \( \left\{(\mathrm{x}, \mathrm{y}): \mathrm{xy} \leq 8,1 \leq \mathrm{y} \leq \mathrm{x}^{2}\right\} \) is

Options:
A \( 16 \log _{e} 2-\frac{14}{3} \)
B \( 8 \log _{e} 2-\frac{7}{3} \)
C \( 8 \log _{e} 2-\frac{14}{3} \)
D \( 16 \log _{e} 2-6 \)

The correct answer is:
\( 16 \log _{e} 2-\frac{14}{3} \)

Solution:
To draw the inequality, let us draw the equation
xy=8 and y=1 and y=x2
For point of intersection
(i) xy=8 and y=1⇒A(8,1)
(ii) xy=8 and y=x2 ⇒x3=8 ⇒x=2 ⇒B(2, 4)
(iii) y=x2 and y=1⇒C1, 1 and D 1, 1
Now xy≤8⇒ region which contains origin
y≥1⇒ region above line y=1
y≤x2⇒ region outside the parabola
Now required area
Method I:
Using x-axis:
A=∫12x2-1⋅dx+∫288x-1⋅dx
A=x33-x12+8lnx-x28
A=-143+16ln2
Method II:
Using y-axis:
A=∫148y-y⋅dy
A=8lny-23y3214⇒A=-143+16ln2
Note: The question should include bounded area term as in 2nd quadrant there exist a area which satisfy the inequality and is unbounded.

Topic: Mathematics, Area Under Curves, JEE Main
The area of the region: R=x,y:5x2≤y≤2x2+9 is

Options:
A 93 square units
B 123 square units
C 113 square units
D 63 square units

The correct answer is:
123 square units

Solution:
Required area=2∫032x2+9-5x2dx

=29x-x303

=293-33=123 square units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The area of the region S=(x,y):3x2≤4y≤6x+24 is______.

Options:

The correct answer is:
27

Solution:
We have,

S=(x,y):3x2≤4y≤6x+24

Then,

y=3x24

4y=6x+24

Solving both equations, we get -2,3 and 4,12.

Required area

A=∫−24 6x+244−3x24dx

⇒A=14∫−24 6x+24−3x2dx

⇒A=34x2+6x−x34−24

⇒A=12+24−16-3+12-2

⇒A=27 sq.units

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The area of the region S=x,y:y2≤8x,y≥2x,x≥1 is

Options:
A 526
B 1926
C 1326
D 1126

The correct answer is:
1126

Solution:
Plotting the region for the given inequalities y2≤8x, y≥2x, x≤1, we get

Required area will be ∫1422x-2xdx

=22x323/214-2x2214

=4238-1-2316-1

=2823-1522=1126

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The area of the region x,y:x-1≤y≤5-x2 is equal to

Options:
A 52sin-135-12
B 5π4-32
C 3π4+32
D 5π4-12

The correct answer is:
5π4-12

Solution:
Plotting region for x-1≤y≤5-x2, we get,

The intersection points of the curves x2+y2=5 and y=x-1 will be -1,2 & 2,1

Here chord AB subtends a right angle at the centre of the circle.
So, required area = area of ∆ABC+ area of segment of circle on chord AB

=12AC·BC+ [area of quarter circle - area of ∆AOB]

=12×2×22+90°360°×π52-12×52

=5π4-12

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The area of the region x,y: x2≤y≤8-x2,y≤7 is

Options:
A 27
B 18
C 20
D 21

The correct answer is:
20

Solution:
The given curves are x2≤y, y≤8-x2 and y≤7

The point of intersection of the curves x2≤y and y≤8-x2 is obtained by,

⇒x2=8-x2

⇒x=±2 and y=4.

Hence the points are 2,4,-2,4.

The points of intersection of the curves y≤8-x2 and y≤7 is obtained by,

⇒8-x2=7

⇒x=±1 and y=7

Hence the points are 1,7,-1,7.

The required graph is

The required area is symmetrical about y-axis.

Hence required area is A=2∫04ydy+∫478-ydy

A=2y323204-8-y 323247

=43432-032-8-732-8-432

=438-0-1+8

=20 sq units.

Hence, the required area is 20 sq units.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The area of the region x,y:x2≤y≤x2-4,y≥1 is

Options:
A 43(42-1)
B 43(42+1)
C 34(42+1)
D 34(42-1)

The correct answer is:
43(42-1)

Solution:
The required diagram is:

The shaded region is symmetric about y-axis.

Required area =2∫12ydy+∫244-ydy

=2y3/23212-2(4-y)3/2324

=43(42-1)

Hence this is the correct option.

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The area of the smaller part of the circle x2+y2=2 cut off by the line x=1 is

Options:
A π2 sq. units
B Ï2-1 sq. units
C Ï2+1 sq. units
D Ï2-12 sq. units

The correct answer is:
π2-1 sq. units

Solution:
x2+y2=2 and x=1
y2=1⇒y=±1
Required area =2×∫122-x2dx
=2x22-x2+sin-1x212
=212×0+π2-12+π4
=2π4-12=π2-1 sq. units

Topic: Mathematics, Area Under Curves, JEE Main
The area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43x-4=0 is equal to

Options:
A 132-123+8π
B 132-123+6π
C 134-123+8π
D 134-123+6π

The correct answer is:
134-123+8π

Solution:
Plotting the area of the smaller region enclosed by the curves y2=8x+4 and x2+y2+43x-4=0 we get,

Now finding point of intersection of x2+y2+43x-4=0 and

y2=8x+4

We get, 0,2 and 0,-2

Both are symmetric about x-axis

So, area will be =2∫0216-y2-23-y2-48dy

=212y16-y2+16sin-1y4-23y-y324+12y02

=138π+4-123

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2022 (27 Jul Shift 1)
The area of the smaller segment cut off from the circle $x^2+y^2=9$ by $x=1$ is

Options:
A $\frac{1}{2}\left(9 \sec ^{-1} 3-\sqrt{8}\right)$
B $9 \sec ^{-1}(3)-\sqrt{8}$
C $\sqrt{8}-9 \sec ^{-1}(3)$
D None of these

The correct answer is:
$9 \sec ^{-1}(3)-\sqrt{8}$

Solution:
Area of smaller part
$\begin{aligned}& \qquad I=2 \int_1^3 \sqrt{9-x^2} d x \\
& \text { Area of smaller part } \\& =2 \cdot \frac{1}{2}\left[x \sqrt{9-x^2}+9 \sin ^{-1} \frac{x}{3}\right]_1^3=\left[9 \frac{\pi}{2}-\sqrt{8}-9 \sin ^{-1}\left(\frac{1}{3}\right)\right]\end{aligned}$
$\begin{aligned} & =\left[9\left(\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{3}\right)\right)-\sqrt{8}\right]=\left[9 \cos ^{-1}\left(\frac{1}{3}\right)-\sqrt{8}\right] \\ & =\left[9 \sec ^{-1}(3)-\sqrt{8}\right] .\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the triangle formed by the coordinate axes and a tangent to the curve $x y=a^2$ at the point $\left(x_1, y_1\right)$ on it is

Options:
A $\frac{a^2 x_1}{y_1}$
B $\frac{a^2 y_1}{x_1}$
C $2 a^2$
D $4 a^2$

The correct answer is:
$2 a^2$

Solution:
$\begin{aligned} & \text { Since } y=\frac{a^2}{x}, \therefore \frac{d y}{d x}=-\frac{a^2}{x^2} \\ & \therefore \text { At }\left(x_1, y_1\right), \frac{d y}{d x}=\frac{-a^2}{x_1^2}\end{aligned}$
Thus tangent to the curve will be $\quad y-y_1=\frac{-a^2}{x_1^2}\left(x-x_1\right)$
$\begin{aligned}& \Rightarrow y x_1^2-y_1 x_1^2=-a^2 x+a^2 x_1 \\ & \Rightarrow \quad a^2 x+x_1^2 y=x_1\left(x_1 y_1+a^2\right)=2 a^2 x_1, \quad\left(x_1 y_1=a^2\right)\end{aligned}$
This meets the $x$-axis where $y=0$
$\therefore a^2 x=2 a^2 x_1, \quad \therefore x=2 x_1$
$\therefore$ Point on the $x$-axis is $\left(2 x_1, 0\right)$
Again tangent meets the $y$-axis where $x=0$
$\boxtimes x_1^2 y=2 a^2 x_1, \quad \therefore y=\frac{2 a^2}{x_1}$
So point on the $y$-axis is $\left(0, \frac{2 a^2}{x_1}\right)$
Required area
$=\frac{1}{2}\left(2 x_1\right)\left(\frac{2 a^2}{x_1}\right)=2 a^2$

Topic: Mathematics, Application of Derivatives, JEE Main
The area of the triangle formed by the tangent and normal at the point \((1, \sqrt{3})\) on the circle \(\mathrm{x}^2\) \(+y^2=4\) and the \(x\)-axis is

Options:
A 3 sq. units
B \(2 \sqrt{3}\) sq. units
C \(3 \sqrt{2}\) sq. units
D 4 sq. units

The correct answer is:
\(2 \sqrt{3}\) sq. units

Solution:
The tangent on \(x^2+y^2=4\) at \((1, \sqrt{3})\) is \([x+\sqrt{3} y=4]\) and equation of normal at \((1, \sqrt{3})\) is \(\mathrm{y}=\mathrm{x} \sqrt{3}\)
\(\begin{aligned}
& \text { Required area }=\int_0^1 x \sqrt{3} \mathrm{dx}+\int_1^4 \frac{4-x}{\sqrt{3}} \mathrm{dx} \\
& =\sqrt{3}\left[\frac{\mathrm{x}^2}{2}\right]_0^1+\frac{1}{\sqrt{3}}\left[4 \mathrm{x}-\frac{\mathrm{x}^2}{2}\right]_1^4 \\
& =\sqrt{3} \times \frac{1}{2}+\frac{1}{\sqrt{3}}\left[4(4-1)-\frac{1}{2}(16-1)\right] \\
& =\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{3}}\left[12-\frac{15}{2}\right]=\frac{\sqrt{3}}{2}+\frac{9}{2 \sqrt{3}} \\
& =\frac{\sqrt{3}}{2}+\frac{3 \sqrt{3}}{2}=2 \sqrt{3} \text { sq. units }
\end{aligned}\)

Topic: Mathematics, Area Under Curves, JEE Main
The area of the triangle formed by the tangent to the hyperbola $x y=a^2$ and co-ordinate axes is

Options:
A $a^2$
B $2 a^2$
C $3 a^2$
D $4 a^2$

The correct answer is:
$2 a^2$

Solution:
Given $x y=a^2$ or $y=\frac{a^2}{x}$
There are two points on the curve $(a, a),(-a,-a)$
The equation of the line at $(a, a)$ is,
$y-a=\left(\frac{d y}{d x}\right)_{(a, a)}(x-a)=\left(\frac{-a^2}{x^2}\right)_{(a, a)}(x-a)$
$y-a=-(x-a)$ therefore, equation of the tangent at $(a, a)$ is $x+y=2 a$. The interception of line $x+y=2 a$ with $x$-axis is $2 a$ and with $y$-axis is $2 a$.
$\therefore$ Required area $=\frac{1}{2} \times 2 a \times 2 a=2 a^2$

Topic: Mathematics, Area Under Curves, JEE Main
The area of the triangle whose vertices are complex numbers $z, i z, z+i z$ in the Argand diagram is

Options:
A
$2|z|^2$

B
$1 / 2|z|^2$

C
$4|z|^2$

D
$|z|^2$

The correct answer is:
$1 / 2|z|^2$

Solution:
Vertices of triangle in complex form is $z, i z, z+i z$
In cartesian form vertices are $(x, y),(-y, x)$ and $(x-y, x+y)$
$$
\begin{aligned}
& \therefore \text { Area of triangle }=\frac{1}{2}\left|\begin{array}{ccc}
x & y & 1 \\
-y & x & 1 \\
x-y & x+y & 1
\end{array}\right| \\
& =\frac{1}{2}[x(x-x-y)-y(-y-x+y)+1 \\
& \left.\left(-y x-y^2-x^2+x y\right)\right]
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{2}\left[-x y+x y-y^2-x^2\right]=\frac{1}{2}\left(x^2+y^2\right) \\
& (\because \text { Area can not be negative }) \\
& =\frac{1}{2}|z|^2 \quad\left(\because z=x+i y,|z|^2=x^2+y^2\right)
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2012 (12 May Online)
The area of the triangle with vertices P(z), Q(iz) and R(z+iz) is

Options:
A 1
B 12|z|2
C 12
D 12|z+iz|2

The correct answer is:
12|z|2

Solution:
From the above diagram, it is clear that z, iz are perpendicular to each other.

Here, base =z, height =iz

So, area of the triangle formed by the vertices i.e.,

A=12×base×height

⇒A=12×z×iz

⇒A=12z2

Hence, the required area is 12z2.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The area of triangle formed by the lines joining the vertex of the parabola, $x^2=8 y$, to the extremities of its latus rectum is

Options:
A
2

B
8

C
1

D
4

The correct answer is:
8

Solution:
Given parabola is $\mathrm{x}^2=8 \mathrm{y}$ $\Rightarrow 4 a=8 \Rightarrow a=2$
To find: Area of $\triangle A B C$
$$
\begin{aligned}
& A=(-2 a, a)=(-4,2) \\
& B=(2 a, a)=(4,2) \\
& \mathrm{C}=(0,0)
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \text { Area }=\frac{1}{2}\left|\begin{array}{ccc}
-4 & 2 & 1 \\
4 & 2 & 1 \\
0 & 0 & 1
\end{array}\right| \\
& =\frac{1}{2}[-4(2)-2(4)+1(0)] \\
& =\frac{-16}{2}=-8 \approx 8 \text { sq. unit } \\
& (\because \text { area cannot be negative })
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2012 (12 May Online)
The area of triangle formed by the lines x-y=0,x+y=0 and any tangent to the hyperbola x2-y2=16 is equal to

Options:
A 2 sq. units
B 4 sq. units
C 8 sq. units
D 16 sq. units

The correct answer is:
16 sq. units

Solution:
Equation of the tangent to the hyperbola x2-y2=16 in parametric form is
4xsec⁡ϕ-4ytan⁡ϕ=16
⇒xsec⁡ϕ-ytan⁡ϕ=4
Solving with y=x and y=-x, we get,
A4sec⁡ϕ+tan⁡ϕ,4sec⁡ϕ+tan⁡ϕ,
B4sec⁡ϕ-tan⁡ϕ,4tan⁡ϕ-sec⁡ϕ
∴ Area bounded by ∆AOB
=1216tan2⁡ϕ-sec2⁡ϕ-16sec2⁡ϕ-tan2⁡ϕ
=12-16-16=16 sq. units

Topic: Mathematics, Hyperbola, JEE Main
The area under the curve $y=|\cos x-\sin x|$, $0 \leq x \leq \frac{\pi}{2}$, and above $x$-axis is :

Options:
A
$2 \sqrt{2}$

B
$2 \sqrt{2}-2$

C
$2 \sqrt{2}+2$

D
0

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

Solution:
$$
\text { } y=|\cos x-\sin x|
$$
$$
\begin{aligned}
\text { Required area } & =2 \int_0^{\pi / 4}(\cos x-\sin x) d x \\
& =2\left[\sin x+\cos x 0_0^{\pi / 4}\right] \\
& =2\left[\frac{2}{\sqrt{2}}-1\right]=(2 \sqrt{2}-2) \text { sq. units }
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2013 (23 Apr Online)
The arenium ion which is not involved in the bromination of Aniline is

Options:
A
B
C
D

The correct answer is:

Solution:
Arenium ion: A carbocation formed by electrophilic attack on an aromatic ring, it is a reactive intermediate in electrophilic aromatic substitution. The formation of an arenium ion takes place by electrophilic attack on benzene.

As the delocalisation of electrons stops at an sp3 hybridized carbon atom, the arenium ion is not aromatic in nature.

Since -N¨H2 group is o/p directing hence arenium ion will not be formed by attack at meta position i.e.

Hence, the answer is C.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The arithmetic mean and standard deviation of a data of nine numbers are 13 and 5 respectively. If 3 is included as the 10th item of the data, then the variance of the data of ten number is

Options:
A 23.5
B 21.5
C 31.5
D 27

The correct answer is:
31.5

Solution:
$\because \frac{\sum_{i=1}^9 x_i}{9}=13 \Rightarrow \sum_{i=1}^9 x_i=117$.
$\begin{array}{rlrl} & & & \\ \text { and } & & \sigma^2=25=\frac{\sum_{i=1}^9 x_i^2}{9}-(13)^2 \\ \Rightarrow & \quad \sum_{i=1}^9 x_i^2=9[25+169]\end{array}$
Now, after including $10^{\text {th }}$ item as ' 3 '
$$
\begin{gathered}
\text { New mean } \bar{y}=\frac{\sum_{i=1}^9 x_i+3}{10}=\frac{117+3}{10}=12 \\
\text { and new variance }=\frac{\left(\sum_{i=1}^9 x_i^2+9\right)}{10}-(12)^2 \\
=\frac{(169+25+1) 9}{10}-144=\frac{1755-1440}{10}=31.5
\end{gathered}
$$

Topic: Mathematics, Statistics, JEE Main
The arithmetic mean and the geometric mean of two distinct 2 -digit numbers $x$ and $y$ are two integers one of which can be obtained by reversing the digits of the other (in base 10 representation). Then $x+y$ equals-

Options:
A 82
B 116
C 130
D 148

The correct answer is:
130

Solution:
$\frac{x+y}{2}=10 a+b, \sqrt{x y}=10 b+a \quad(a, b \in N)$
$x y=(10 b+a)^{2}$
$(x-y)^{2}=4(11 a+11 b)(9 a-9 b)$
$=4 \cdot 11 \cdot(a+b) \cdot 9(a-b)$
$\Rightarrow a+b=11, a-b=1$
$a=6, b=5$
$\left((x-y)^{2}\right.$ is perfect square of an integer)
$x+y=130$

Topic: Mathematics, Sequences and Series, JEE Main
The arithmetic mean of a set of 50 numbers is 38. If two numbers of the set, namely 60 and 40 are discarded, the arithmetic mean of the remaining set of numbers is

Options:
A 38.5
B 37.5
C 36.5
D 36

The correct answer is:
37.5

Solution:
Given Î£xi50=38,âÎ£xi=1900
New value of Î£xi=1900-60-40=1800,ân=48
New mean =180048=37.5

Topic: Mathematics, Statistics, JEE Main
The arithmetic mean of a set of observation is X¯. If each observation is divided by α and increased by 10, then the mean of the new series is

Options:
A X¯α
B X¯+10α
C X¯+10αα
D αX¯+10

The correct answer is:
X¯+10αα

Solution:
Let x1, x2,…,xn be n observations.

Then, X¯=1nΣxi

Let yi=xiα+10

Then. 1n∑i=1nyi=1α1nΣxi+1n10n

⇒X¯new=1αX¯+10=X¯+10αα

Hence option (c) is correct.

Topic: Mathematics, Statistics, JEE Main
The arithmetic mean of first $n$ natural number

Options:
A $\frac{n-1}{2}$
B $\frac{n+1}{2}$
C $\frac{n}{2}$
D $n$

The correct answer is:
$\frac{n+1}{2}$

Solution:
A.M. $=\frac{1+2+3+\ldots \ldots+n}{n}=\frac{\frac{1}{2} n(n+1)}{n}=\frac{n+1}{2}$

Topic: Mathematics, Sequences and Series, JEE Main
The arithmetic mean of marks in Mathematics for four divisions A, B, C and D were 80, 75, 70 and 72 respectively. Their standard deviations were $12,6,8$ and and 10 respectively. Then, division
has more uniformity.

Options:
A D
B B
C C
D A

The correct answer is:
B

Solution:
$$
\begin{aligned}
& \text { We know that C.V. }=\frac{\text { Standard Deviation }}{\text { Mean }} \\
& \therefore(\text { C.V. })_A=\frac{12}{18}=0.15 \text { and }(\text { C.V. })_B=\frac{6}{75}=0.08 \\
& (\text { C.V. })_C=\frac{8}{70}=0.11 \text { and }(\text { C.V. })_D=\frac{10}{72}=0.14
\end{aligned}
$$
C.V. is the least for division B.

Topic: Mathematics, Statistics, JEE Main
The arithmetic mean of the numbers 2sin⁡2o,4sin⁡4o,6sin⁡6o,…….178sin⁡178o,180sin⁡180o is

Options:
A sin⁡1o
B cot⁡1o
C tan⁡1o
D cos⁡1o

The correct answer is:
cot⁡1o

Solution:
Total number of given numbers =90
Sum of numbers,
S=2sinâ¡2o+4sinâ¡4o+.â¦â¦..+178sinâ¡178o+180sinâ¡180o
S=178sinâ¡2o+176sinâ¡4o+.â¦â¦..+2sinâ¡178o
Adding both the equation, we get,
2S=180sinâ¡2o+sinâ¡4o+.â¦â¦..+sinâ¡178o+0
S=90sinâ¡2o+sinâ¡4o+.â¦â¦..+sinâ¡178o
â Mean =S90=sinâ¡2o+sinâ¡4o+.â¦â¦+sinâ¡178o
=sinâ¡89osinâ¡1osinâ¡2o+178o2
=cosâ¡1osinâ¡1oÃsinâ¡90o=cotâ¡1o

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The arithmetic mean of two numbers is 1834 and the positive square root of their product is 15. The larger of the two numbers is

Options:
A 24
B 25
C 20
D 30

The correct answer is:
30

Solution:
Let a and b be the positive numbers.
Then a+b2=1834⇒a+b=752
and ab=15⇒ab=225
so, a-b2=a+b2-4ab=7522-4×225
=752+30752-30=1352×152
⇒a-b=±152×3=±452
Case1 →a+b=752,a-b=452
⇒a=30,b=452
Case2 →a+b=752,a-b=-452
⇒a=152,b=30
So, larger of the two numbers is 30

Topic: Mathematics, Sequences and Series, JEE Main
The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 2 and the harmonic mean is one-fifth of the greater of a and b, such that α=a+b and β=a-b, then the value of α+β2 is equal to

Options:
A 96
B 234
C 74
D 84

The correct answer is:
74

Solution:
A=G+2
H=b5
(let b>a )
G2=AH⇒ab=a+b2b5⇒10a=a+b⇒b=9a
A=a+b2=5a,G=9a.a=3a
⇒5a=3a+2⇒a=1⇒b=9
⇒α=10 and β=8
⇒α+β2=74

Topic: Mathematics, Sequences and Series, JEE Main
The arithmetic mean of two positive numbers a and b exceeds their geometric mean by 32 and the geometric mean exceeds their harmonic mean by 65. If a+b=Î± and a-b=Î², then the value of 10Î²Î± is equal to

Options:

The correct answer is:
6

Solution:
A-G=32 and G-H=65
Also, we know that G2=AH
⇒G2=32+GG-65
⇒G2=G2-65G+32G-95
⇒32-65G=95⇒G=95×103=6
G=6⇒A=152
⇒a+b2=152 and ab=6
⇒a+b=15 and ab=36
a-b=a+b2-4ab=225-4×36=9
α=a+b=15, β=a-b=9
⇒10βα=10×915=6

Topic: Mathematics, Sequences and Series, JEE Main
The arm PQ of a rectangular conductor is moving from x=0 to x=2b outwards and then inwards from x=2b to x=0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x=0 to x=b. Identify the graph showing the variation of different quantities with distance:

Options:
A A—Flux, B—Power dissipated, C—EMF
B A—Power dissipated, B—Flux, C—EMF
C A—Flux, B—EMF, C—Power dissipated
D A—EMF, B—Power dissipated, C—Flux

The correct answer is:
A—Flux, B—EMF, C—Power dissipated

Solution:
As rod moves in field area increases upto x=b, then field is absent and again flux is generated on return journey from x=b to x=0. Thus plot A for flux.
⇒e=-dϕdt⇒ curve B for emf
⇒ Power dissipated =vi⇒ curve C for power dissipated

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (20 Jul Shift 1)
The arrangement $A B C A B C A B C \ldots .$. is referred as

Options:
A Octahedral close packing
B Hexagonal close packing
C Tetragonal close packing
D Cubic close packing

The correct answer is:
Hexagonal close packing

Solution:
The system $A B C A B C \ldots \ldots$ is also referred to as face-centred cubic or $f c c$.

Topic: Chemistry, Solid State, JEE Main
The arrangement is released from rest. The minimum angle Î¸ at which block A starts sliding downwards is

Options:
A tan⁡θ=0
B tan⁡θ=0.5
C tan⁡θ=16
D tan⁡θ=13

The correct answer is:
tan⁡θ=16

Solution:
For sliding 3 mgsin⁡θ=0.5 mgcos⁡θ
tan⁡θ=16

Topic: Physics, Laws of Motion, JEE Main
The arrangement of X- ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X- is 250 pm , the radius of A+ is

Options:
A 104 pm
B 125 pm
C 183 pm
D 57 pm

The correct answer is:
104 pm

Solution:
rA+rx-=0.414=rA+250
rA+=0.414Ã250 pm=104 pm

Topic: Chemistry, Solid State, JEE Main
The Arrhenius plots of two reactions, I and II are shown graphically :
The graph suggests that

Options:
A $E_{1}>E_{\|}$and $A_{1}>A_{11}$
B $E_{11}^{\prime}>E_{1}$ and $A_{\|}>A^{\prime \prime}$
C $E_{1}^{\prime \prime}>E_{\|}$and $A_{\|}^{\prime \prime}>A$
D $E_{\|}^{\prime}>E_{1}^{\prime \prime}$ and $A_{1}^{n}>A_{1}$

The correct answer is:
$E_{1}>E_{\|}$and $A_{1}>A_{11}$

Solution:
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{Ea}}{\mathrm{R}} \times \frac{1}{\mathrm{~T}}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The artificial sweetener that contains chlorine that has the look and taste of sugar and is the stable temperature for cooking?

Options:
A Aspartame
B Saccharin
C Sucralose
D Alitame

The correct answer is:
Sucralose

Solution:
The artificial sweetener containing chlorine that has the appearance and taste as that of sugar and is stable at cooking temperature is sucralose.
Sucralose is derived from sugar through a multi-step patented manufacturing process that selectively substitutes three atoms of chlorine for three hydroxyl groups on the sugar molecule. This change produces a sweetener that has no calories, yet is 600 times sweeter than sucrose.
Sucralose tastes like sugar. Its chemical formula is $\mathrm{C}_{12} \mathrm{H}_{19} \mathrm{Cl}_3 \mathrm{O}_8$.
Hence, option 3 is correct.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The artificial sweetener that has the highest sweetness value in comparison to cane sugar is:

Options:
A Sucralose
B Aspartame
C Saccharin
D Alitame

The correct answer is:
Alitame

Solution:
Natural sweeteners, e.g., sucrose add to calorie intake and therefore many people prefer to use artificial sweeteners. Ortho-sulphobenzimide, also called saccharin, is the first popular artificial sweetening agent. It is about 550 times as sweet as cane sugar. Aspartame is 100 times as sweet as cane sugar. Sucralose is 600 times as sweet as cane sugar. Alitame is an artificial sweetener that has 1000 times the sweetness value of cane sugar.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2016 (09 Apr Online)
The artificial sweetener that has the highest sweetness value when compared to cane sugar is:

Options:
A Aspartame
B Alitame
C Sucralose
D Saccharin

The correct answer is:
Alitame

Solution:
The correct option is $\mathbf{2}$ Alitame
Alitame is an artificial sweetener that has the highest sweetness value in comparison to cane sugar.
It is about 2000 times sweeter than cane sugar. It is also stable at higher temperatures.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The ascending acidity order of the following H atoms is

Options:
A C < D < B < A
B A < B < C < D
C A < B < D < C
D D < C < B < A

The correct answer is:
C < D < B < A

Solution:
Alkynes are acidic because they can release hydrogen atoms to form alkylide ions, because the hydrogen is attached more electronegative sp hybridised carbon. After that alkenes are more acidc than alkanes due to the presence of sp2 carbon atom in alkenes. Among the two alkanes acidic nature order is decided by the inductive effect. Methyl groups are electron releasing groups, hence they decrease the acidic nature. Hence, the overall order is

Stability of conjugate base ∝ acidic strength

C<D<B<A

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)