datavorous/entrance-exam-dataset huggingface dataset:
0.5 g of an organic compound (X) with 60% carbon will produce ×10-1 g of CO2 on complete combustion.

Options:

The correct answer is:
11

Solution:
To calculate the mass percent of an element in a compound, we divide the mass of the element in 1 mole of the compound by the compound's molar mass and multiply the result by 100.

The combustion reaction is Cs+O2g→CO2g

C%=1244× Wt. of CO2 Wt. of organic compound ×100

60=1244× Wt. of CO20.5×100

Wt. of CO2=1.1

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
$\int_{0}^{5} \frac{\mathrm{d} x}{x^{2}+2 x+10}$ =

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{12}$
C $\frac{\pi}{3}$
D $\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{12}$

Solution:
$\begin{aligned} I &=\int_{0}^{5} \frac{d x}{x^{2}+2 x+10}=\int_{0}^{5} \frac{d x}{(x+1)^{2}+(3)^{2}} \\ &=\frac{1}{3}\left[\tan ^{-1} \frac{x+1}{3}\right]_{0}^{5}=\frac{1}{3}\left[\tan ^{-1} 2-\tan ^{-1} \frac{1}{3}\right] \\ &=\frac{1}{3}\left[\tan ^{-1}\left[\frac{2-\frac{1}{3}}{1+\frac{2}{3}}\right]\right]=\frac{1}{3} \tan ^{-1}(1) \\ &=\frac{1}{3} \times \frac{\pi}{4}=\frac{\pi}{12} \end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 6} \frac{\sin x}{\cos ^3 x} d x=$

Options:
A $\frac{2}{3}$
B $\frac{1}{6}$
C 2
D $\frac{1}{3}$

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

Solution:
Let $I=\int_0^{\pi / 6} \frac{\sin x}{\cos ^3 x} d x=\int_0^{\pi / 6} \tan x \sec ^2 x d x$
Put $t=\tan x \Rightarrow d t=\sec ^2 x d x$, then we have
$I=\int_0^{\frac{1}{\sqrt{3}}} t d t=\left[\frac{t^2}{2}\right]_0^{\sqrt{3}}=\frac{1}{6}$

Topic: Mathematics, Definite Integration, JEE Main
∫0∞6e3x+6e2x+11ex+6dx=

Options:
A loge3227
B loge51281
C loge25681
D loge30227

The correct answer is:
loge3227

Solution:
Let

I=∫0∞6e3x+6e2x+11ex+6dx

⇒I=∫0∞6ex+1ex+2ex+3dx

Using partial fraction, we can write

⇒I=6∫0∞12ex+1-1ex+2+12ex+3dx

⇒I=6∫0∞12e-x+1-11+2e-x+121+3e-xe-xdx

⇒I=6-12logee-x+1+12loge2e-x+1-16loge3e-x+10∞

⇒I=60--12loge2+12loge3-16loge4

⇒I=3loge2-3loge3+loge4

⇒I=5loge2-3loge3

⇒I=loge3227

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (13 Apr Shift 1)
0.08 kg air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal kg-1 °C-1 and 1 J=4.18 joule cal-1. The change in its internal energy is approximately.

Options:
A 318 J
B 298 J
C 284 J
D 142 J

The correct answer is:
284 J

Solution:
The amount of work done for the given process should be zero, as the process is an isochoric process.

Thus, the change in internal energy will be equal to the heat exchanged in the process. Mathematically,

Q=∆U

Now,

∆U=ms∆T ...1

From equation (1), it follows that

∆U=0.08×(170×4.18)×5 J≈ 284 J

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
$\int_0^{10}\left(5-\sqrt{10 x-x^2}\right) d x=$

Options:
A $50-25 \pi$
B $(100-25 \pi)$
C $\frac{1}{2}(100-25 \pi)$
D $\frac{1}{4}(100-25 \pi)$

The correct answer is:
$\frac{1}{2}(100-25 \pi)$

Solution:
$$
\begin{aligned}
& \text { Let } \\
& I=\int_0^{10} 5-\sqrt{\left.10 x-x^2\right)} d x, I=\int_0^{10}\left(5-\sqrt{(5)^2-(x-5)^2} d x\right. \\
& I=\left[5 x-\frac{x-5}{2} \sqrt{10 x-x^2}-\frac{25}{2} \sin ^{-1} \frac{x-5}{5}\right]_0^{10} \\
& I=\left(50-0-\frac{25}{2} \sin ^{-1} 1\right)-\left(0-0-\frac{25}{2} \sin ^{-1}(-1)\right) \\
& I=50-\frac{25}{2} \times \frac{\pi}{2}-\frac{25}{2} \times \frac{\pi}{2} \\
& I=50-\frac{25 \pi}{2} \Rightarrow I=\frac{1}{2}(100-25 \pi)
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{10 \pi}|\sin x| d x$ is

Options:
A
20

B
8

C
10

D
18

The correct answer is:
18

Solution:
$\int_0^{10 \pi}|\sin x| d x=10\left[\int_0^{\pi / 2} \sin x d x+\int_{\pi / 2}^\pi \sin x d x\right]$
$=10 \times[\cos x]_0^{\pi / 2}+[\cos x]_{\pi / 2}^\pi ; \quad 10[1+1]=10 \times 2=20$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2002
0.15 g of a substance dissolved in 15 g of solvent boiled at a temperature higher by 0.216oC than that of the pure solvent. The molecular weight of the substance is [Given: Molal elevation constant for the solvent is 2.16 Kkg  mol −1 ]

Options:

The correct answer is:
100

Solution:
Here it is given that w = 0.15 g,   ΔT b  = 0 .216 o C
W = 15 g,  K b  = 2.16 Kkg  mol −1
Substituting values in the expression, m=1000ÃKbÃwÎTbÃW
m =  1000×216×0.15 0.216×15 =100 g/mol

Topic: Chemistry, Solutions, JEE Main
∫020πsinx+cosx2dx is equal to:

Options:
A 10π+4
B 10π+2
C 20π-2
D 20π+2

The correct answer is:
20π+2

Solution:
Let I=∫020πsinx+cosx2dx

We know that the period of sinx+cosx is π2

i.e. I=40∫0π2sinx+cosx2dx

I=40∫0π21+sin2xdx

=40x-cos2x20π2=40π2-cosπ2+cos02

I=20π+2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (26 Jul Shift 2)
0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is_____ kcal.
(Take R=2 cal mole K-1-1 )

Options:

The correct answer is:
12

Solution:
The RMS speed of the molecule is given by vRMS=3RTM. Therefore, to double the speed the temperature must become 4 times.

Therefore, the change in the temperature is ∆T=3T0.

Now the process is isochoric (enclosed vessel) therefore, the heat required will be,

H=nCv∆T⇒H=5628×52R×3×273+127=12000 cal=12 kcal

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (24 Jun Shift 1)
0.400 g of an organic compound (X) gave 0.376 g of AgBr in Carius method for estimation of bromine. % of bromine in the compound (X) is (Given: Molar mass AgBr=188 g mol-1 Br=80 g mol-1

Options:

The correct answer is:
40

Solution:
The percentage composition of element =mass of elementmass of organic compound×100

Moles of AgBr=0.376188

Moles of Br=0.376188

Mass of Br=0.376188×80

% of Br=0.376×80188×(0.400)×100

= 40%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
0.765 g of an acid gives 0.535 g of CO2 and 0.138 g of H2O. Then, the ratio of percentage of carbon and hydrogen is

Options:
A 19:2
B 18:17
C 20:17
D 1:7

The correct answer is:
19:2

Solution:
% for CO2

∵ 0.765 g of acid gives CO2=0.535 g

∴100 g of acid gives CO2=0.535×1000.765

% of CO2=70.00

% for carbon

∵ 44 g of CO2 gives, C=12 g

∴ 70 g of CO2 gives, C=12×7044

=19.09 g

∴ % of carbon =19.00

Similarly,

(i) for hydrogen

% of H2O

⇒ 0.138×1000.765

=18.03=18.00

% of Hydrogen

⇒ 2×1818=2.00

Hence ratio of carbon and hydrogen is 19:2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
$\int_{0}^{a} \sqrt{\frac{x}{a-x}} d x=$

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

The correct answer is:
$\left(\frac{\pi}{2}\right) a$

Solution:
(C)

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^a f(x) d x=$

Options:
A $\int_0^a f(a+x) d x$
B $\int_0^a f(2 a+x) d x$
C $\int_0^a f(x-a) d x$
D $\int_0^a f(a-x) d x$

The correct answer is:
$\int_0^a f(a-x) d x$

Solution:
It is a fundamental property.

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^a x^2\left(a^2-x^2\right)^{3 / 2} d x=$

Options:
A $\frac{\pi a^6}{32}$
B $\frac{2 a^5}{15}$
C $\frac{a^6}{32}$
D None of these

The correct answer is:
$\frac{\pi a^6}{32}$

Solution:
$\begin{aligned} & I=\int_0^a x^2\left(a^2-x^2\right)^{3 / 2} d x \\ & \text { Put } x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta \\ & I=\int_0^{\pi / 2} a^2 \sin ^2 \theta \cdot a^3 \cos ^3 \theta \cdot a \cos \theta d \theta \\ & =a^6 \int_0^{\pi / 2} \sin ^2 \theta \cos ^4 \theta d \theta=a^6 \frac{\Gamma \frac{3}{2} \cdot \Gamma \frac{5}{2}}{2 \cdot \Gamma \frac{8}{2}} \\ & =a^6 \frac{\frac{1}{2} \cdot \sqrt{\pi} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi}}{2 \cdot 3 \cdot 2 \cdot 1}=\frac{\pi a^6}{32} .\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^a x^2 \sin x^3 d x$ equals

Options:
A $\left(1-\cos a^3\right)$
B $3\left(1-\cos ^3\right)$
C $-\frac{1}{3}\left(1-\cos ^3\right)$
D $\frac{1}{3}\left(1-\cos a^3\right)$

The correct answer is:
$\frac{1}{3}\left(1-\cos a^3\right)$

Solution:
$\begin{aligned} & I=\int_0^a x^2 \sin x^3 d x ; \text { Put } x^3=t \Rightarrow x^2 d x=\frac{d t}{3} \\ & \therefore I=\frac{1}{3} \int_0^{a^3} \sin t d t=-\frac{1}{3}[\cos t]_0^{a^3}=-\frac{1}{3}\left[\cos a^3-1\right] \\ & =\frac{1}{3}\left[\cos a^3 -1\right] \\ & \end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^a x^4 \sqrt{a^2-x^2} d x=$

Options:
A $\frac{\pi}{32}$
B $\frac{\pi}{32} a^6$
C $\frac{\pi}{16} a^6$
D $\frac{\pi}{8} a^6$

The correct answer is:
$\frac{\pi}{32} a^6$

Solution:
Put $x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta$
Now $\int_0^3 x^4 \sqrt{a^2-x^2} d x=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos \theta \cos \theta d \theta$
$=a^6 \int_0^{\pi / 2} \sin ^4 \theta \cos ^2 \theta d \theta=a^6 \frac{\Gamma\left(\frac{5}{2}\right) \Gamma\left(\frac{3}{2}\right)}{2 \Gamma 4}=\frac{\pi}{32} a^6$,(Using gamma function).

Topic: Mathematics, Definite Integration, JEE Main
$\left(\int_0^a x d x\right) \leq(a+4), \quad$ then

Options:
A $0 \leq a \leq 4$
B $-2 \leq a \leq 4$
C $-2 \leq a \leq 0$
D $a \leq-2$ or $a \geq 4$

The correct answer is:
$-2 \leq a \leq 4$

Solution:
$\begin{aligned} & \int_0^a x d x \leq a+4 \Rightarrow \frac{a^2}{2} \leq a+4 \\ & \Rightarrow a^2 \leq 2 a+8 \quad \Rightarrow a^2-2 a-8 \leq 0 \\ & \Rightarrow(a-4)(a+2) \leq 0 \Rightarrow-2 \leq a \leq 4\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{b-c} f^{\prime}(x+a) d x=$

Options:
A $f^{\prime}(a)-f^{\prime}(b)$
B $f^{\prime}(b-c+a)-f^{\prime}(a)$
C $f^{\prime}(b+c-a)+f^{\prime}(a)$
D None of these

The correct answer is:
$f^{\prime}(b-c+a)-f^{\prime}(a)$

Solution:
$\int_0^{b-c} f^{\prime}(x+a) d x=\left[f^{\prime}(x+a)\right]_0^{b-c}=f^{\prime}(b-c+a)-f^{\prime}(a)$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^\pi[\cot x] d x,[\bullet]$ denotes the greatest integer function, is equal to

Options:
A
$\frac{\pi}{2}$

B
1

C
$-1$

D
$-\frac{\pi}{2}$

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

Solution:
$$
\begin{aligned}
& \text { Let } \mathrm{I}=\int_0^\pi[\cot x] d x \\
& =\int_0^\pi[\cot (\pi-x)] d x=\int_0^\pi[-\cot x] d x
\end{aligned}
$$
Adding (1) and (2)
$$
\begin{aligned}
& 2 I=\int_0^\pi[\cot x] d x+\int_0^\pi[-\cot x] d x==\int_0^\pi(-1) d x \quad[\because[x]+[-x]=-1 \text { if } x \notin Z \\
& =[-x]_0^\pi=-\pi \\
& \therefore I=-\frac{\pi}{2}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2009
∫0∞e-xsin6xdx=

Options:
A 2485
B 124285
C 136529
D 144629

The correct answer is:
144629

Solution:
The formula for the integration,

Im=∫0∞e-xsinmxdx, where m>2

⇒Im=m(m-1)1+m2Im-2

Now,

I6=∫0∞e-xsin6xdx

⇒I6=6(5)1+62I4

⇒I6=3037I4

Similarly,

I4=∫0∞e-xsin4xdx

⇒I4=4(3)1+42I2

⇒I4=1217I2

Similarly,

I6=∫0∞e-xsin6xdx

⇒I2=2(1)1+22I0

⇒I2=25I0

And

I0=∫0∞e-xsin0xdx

⇒I0=-e-x0∞=1

The value of I6 is

I6=3037×1217×25×1=144629.

Topic: Mathematics, Definite Integration, JEE Main
$\left|\begin{array}{ccc}0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0\end{array}\right|=$

Options:
A 0
B $(p-q)(q-r)(r-p)$
C pqr
D $3 p q r$

The correct answer is:
0

Solution:
Since determinant of a skew-symmetric matrix of odd order is zero.

Topic: Mathematics, Determinants, JEE Main
$\int_0^\pi \frac{x \tan x}{\sec x+\tan x} d x=$

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

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

Solution:
$\begin{aligned} & I=\int_0^\pi \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x \\ & \qquad \because \text { using } \\ & \qquad \int_0^a f(x) d x=\int_0^a f(a-x) d x\end{aligned}$
$=\int_0^\pi \frac{-(\pi-x) \tan (x)}{-\sec x-\tan x} d x$
Adding Eqs. (i) and (ii)
$$
2 I=\int_0^\pi \frac{\pi \tan x}{\sec x+\tan x} d x=\pi \int_0^\pi \frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}} d x
$$
$\begin{aligned} &=\pi \int_0^\pi \frac{\sin x}{1+\sin x} d x=\pi \int_0^\pi \frac{\sin x+1-1}{1+\sin x} d x \\ &=\pi\left[\int_0^\pi \frac{(1+\sin x)}{1+\sin x} d x-\int_0^\pi \frac{1}{1+\sin x} d x\right] \\ &=\pi\left[\int_0^\pi 1 d x-\int_0^\pi \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x\right] \\ &=\pi\left[[x]_0^\pi-\int_0^\pi \frac{1-\sin x}{\cos ^2 x} d x\right]=\pi[(\pi-0)- \\ & \int_0^\pi\left(\sec ^2 x-\sec x \tan x\right) d x=\pi\left[\pi-(\tan x-\sec x)_0^\pi\right] \\ &=\pi[\pi-[\tan \pi-\sec \pi-\tan 0+\sec 0]] \\ &=\pi[\pi-\{0+1-0+1\}] \\ & \quad 2 I=\pi[\pi-2] \Rightarrow I=\frac{\pi}{2}[\pi-2]\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\infty} \frac{x^2 d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=$

Options:
A $\frac{\pi}{2(a-b)}$
B $\frac{\pi}{2(b-a)}$
C $\frac{\pi}{(a+b)}$
D $\frac{\pi}{2(a+b)}$

The correct answer is:
$\frac{\pi}{2(a+b)}$

Solution:
$\begin{aligned} & \int_0^{\infty} \frac{x^2 d x}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=\int_0^{\infty} \frac{\left(x^2+a^2\right)-a^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ & \int_0^{\infty} \frac{1}{x^2+b^2} d x-a^2 \int_0^{\infty} \frac{1}{\left(x^2+a^2\right)\left(x^2+b^2\right)} d x \\ = & {\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}\right]_0^{\infty}-\frac{a^2}{\left(a^2-b^2\right)} \int_0^{\infty}\left(\frac{1}{x^2+b^2}-\frac{1}{x^2+a^2}\right) d x } \\ = & \frac{1}{b} \cdot \frac{\pi}{2}-\frac{a^2}{\left(a^2-b^2\right)}\left[\frac{1}{b} \tan ^{-1} \frac{x}{b}-\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]_0^{\infty}=\frac{\pi}{2(a+b)}\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^\pi x f(\sin x) d x$ is equal to

Options:
A
$\pi \int_0^\pi f(\cos x) d x$

B
$\pi \int_0^\pi f(\sin x) d x$

C
$\frac{\pi}{2} \int_0^{\pi / 2} f(\sin x) d x$

D
$\pi \int_0^{\pi / 2} f(\cos x) d x$

The correct answer is:
$\pi \int_0^{\pi / 2} f(\cos x) d x$

Solution:
$I=\int_0^\pi x f(\sin x) d x=\int_0^\pi(\pi-x) f(\sin x) d x$
$=\pi \int_0^\pi f(\sin x) d x-1$
$2 I=\pi \int_0^\pi f(\sin x) d x$
$I=\frac{\pi}{2} \int_0^\pi f(\sin x) d x=\pi \int_0^{\pi / 2} f(\sin x) d x$
$=\pi \int_0^{\pi / 2} f(\cos x) d x$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2006
$\int_0^\pi x f(\sin x) d x=$

Options:
A $\pi \int_0^\pi f(\sin x) d x$
B $\frac{\pi}{2} \int_0^\pi f(\sin x) d x$
C $\frac{\pi}{2} \int_0^{\pi / 2} f(\sin x) d x$
D None of these

The correct answer is:
$\frac{\pi}{2} \int_0^\pi f(\sin x) d x$

Solution:
$\begin{aligned} & \int_0^\pi x f(\sin x) d x=\frac{\pi}{2} \int_0^\pi f(\sin x) d x \\ & \text {Since } \int_0^a x f(x) d x=\frac{1}{2} a \int_0^{a} f(x) d x \text {, if } f(a-x)=f(x) .\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\left|\begin{array}{lll}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=$

Options:
A 0
B $a^3+b^3+c^3-3 a b c$
C $3 a b c$
D $(a+b+c)^3$

The correct answer is:
0

Solution:
$\left|\begin{array}{ccc}1 & a & a^2-b c \\ 1 & b & b^2-a c \\ 1 & c & c^2-a b\end{array}\right|=\left|\begin{array}{ccc}0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & c^2-a b\end{array}\right|$ by $\left\{\begin{array}{l}R_1 \rightarrow R_1-R_2 \\ R_2 \rightarrow R_2-R_3\end{array}\right.$
$(a-b)(b-c)\left|\begin{array}{lll}0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & c^2-a b\end{array}\right|=0 \quad, \quad\left\{R_1=R_2\right\}$

Topic: Mathematics, Determinants, JEE Main
$\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$ is equal to

Options:
A $\frac{3\pi}{2}$
B $\frac{\pi}{2}$
C $0$
D $-1$

The correct answer is:
$0$

Solution:
$I=\int_{-1}^1 \frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}} d x$
Let $f(x)=\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}$
Replacing $x$ by $-x$, we get
$\begin{aligned}
& f(-x)=\frac{\sqrt{1-x+x^2}-\sqrt{1+x+x^2}}{\sqrt{1-x+x^2}+\sqrt{1+x+x^2}} \\
& =\frac{\sqrt{1+x+x^2}-\sqrt{1-x+x^2}}{\sqrt{1+x+x^2}+\sqrt{1-x+x^2}}=-f(x)
\end{aligned}$
So, $f(x)$ is an odd function.
$\therefore \quad \int_{-1}^1 f(x) d x=0$

Topic: Mathematics, Definite Integration, JEE Main
$\int_{-1}^1 \sin ^{11} x d x$ is equal to

Options:
A $\frac{10}{11}, \frac{8}{9}, \frac{6}{7}, \frac{4}{5}, \frac{2}{3}$
B $\frac{10}{11}, \frac{8}{9}, \frac{6}{7}, \frac{4}{5}, \frac{2}{3}, \frac{\pi}{2}$
C 1
D 0

The correct answer is:
0

Solution:
Since $f(x)=\sin ^{11} x$ is an odd function, therefore $\int_{-1}^1 \sin ^{11} x d x=0$

Topic: Mathematics, Definite Integration, JEE Main
1, 1-Dichloropropane on hydrolysis gives:

Options:
A propanone
B propanal
C ethanal
D 1, 1-Propanediol

The correct answer is:
propanal

Solution:
It gives propanal as follows

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$1+\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\ldots \ldots \infty$ is equal to

Options:
A $3$
B $6$
C $9$
D $12$

The correct answer is:
$12$

Solution:
It is an arithmetico-geometric series
$S_{\infty}=\frac{a}{1-r}+\frac{d r}{(1-r)^2}$ $=\frac{1}{1-\frac{1}{2}}+\frac{2}{\left(1-\frac{1}{2}\right)^2}=\frac{2}{\frac{1}{2}}+\frac{2}{\frac{1}{4}}$
$=4+8=12$.

Topic: Mathematics, Sequences and Series, JEE Main
1.

2.

3.

4.

The dipeptides which may be obtained from the aminoacids glycine and alanine are

Options:
A only 1
B only 2
C both 1 and 2
D all of them

The correct answer is:
all of them

Solution:

Topic: Chemistry, Biomolecules, JEE Main
1 , 2 di-bromo cyclohexane on dehydro halogenation gives

Options:
A

B

C

D None of these

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
$\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=$

Options:
A $x-\log \left(1-2 e^{-x}\right)+\mathrm{c}$
B $\log \left(1-2 e^{-x}\right)+\mathrm{c}$
C $x+\log \left(1-2 e^{-x}\right)+\mathrm{c}$
D $x+2\log \left(1-2 e^{-x}\right)+\mathrm{c}$

The correct answer is:
$x+2\log \left(1-2 e^{-x}\right)+\mathrm{c}$

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

Topic: Mathematics, Indefinite Integration, JEE Main
1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198 °C. The percentage of dissociation of the acid is (Nearest integer) [Given : Density of acetic acid is 1.02 g mL-1 Molar mass of acetic acid is 60 gmol-1
KfH2O=1.85 K kg mol-1]

Options:

The correct answer is:
5

Solution:
M=d×V=1.02×1.2=1.224gm

Moles of acetic acid =0.0204 moles in 2 L

So molality =0.0102 mol/kg

Now ΔTf=i×Kf×m

i=1+α for acetic acid

0.0198=1+α×1.85×0.0102

α=0.04928

≅5%

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (29 Jun Shift 1)
$1-2 \sin ^2\left(\frac{\pi}{4}+\theta\right)=$

Options:
A $\cos 2 \theta$
B $-\cos 2 \theta$
C $\sin 2 \theta$
D $-\sin 2 \theta$

The correct answer is:
$-\sin 2 \theta$

Solution:
$1-2 \sin ^2\left(\frac{\pi}{4}+\theta\right)=\cos \left(\frac{\pi}{2}+2 \theta\right)=-\sin 2 \theta$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
$\int \frac{3-x^2}{1-2 x+x^2} e^x d x=e^x f(x)+c \Rightarrow f(x)$

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

The correct answer is:
$\frac{1+x}{1-x}$

Solution:
Let $\begin{aligned} I & =\int \frac{3-x^2}{1-2 x+x^2} e^x d x \\ & =\int \frac{3-x^2}{(1-x)^2} e^x d x \\ & =\int\left(\frac{2}{(1-x)^2}+\frac{1+x}{1-x}\right) e^x d x\end{aligned}$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& e^x\left(\frac{1+x}{1-x}\right)+c=e^x f(x)+c \\
& \Rightarrow \quad f(x)=\frac{1+x}{1-x} \\
&
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
1, 2-dibromoethane when heated with alcoholic potash gives

Options:
A Ethane
B Acetylene
C Ethylene
D Methane

The correct answer is:
Acetylene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
132-1+152-1+172-1+…+1(201)2-1 is equal to

Options:
A 101404
B 25101
C 101408
D 99400

The correct answer is:
25101

Solution:
Given S=132-1+152-1+172-1+…+1(201)2-1

Tn=1(2n+1)2-1=14(n)(n+1)

Tn=14(n+1)-nn(n+1)=141n-1n+1

∴ S=∑n=1100Tn=∑n=1100141n-1n+1

∴ S=141-12+12-13+...+1100-1101

⇒S=141-1101=14100101=25101

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (18 Mar Shift 1)
$1^3-2^3+3^3-4^3+\ldots .+9^3=$

Options:
A
425

B
-425

C
475

D
-475

The correct answer is:
425

Solution:

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2002
$\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x=$

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

The correct answer is:
$2 \pi$

Solution:
$\begin{aligned} I &=\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\tan ^{-1}\left(\frac{x^{2}+1}{x}\right)\right] d x \\ &=\int_{-1}^{3}\left[\tan ^{-1}\left(\frac{x}{x^{2}+1}\right)+\cot ^{-1}\left(\frac{x}{x^{2}+1}\right)\right] d x \\ &=\int_{-1}^{3} \frac{\pi}{2} d x=\frac{\pi}{2}[x]_{-1}^{3}=\frac{4 \pi}{2} \\ &=2 \pi \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
1, 3-butadiene reacts with ethylene to form

Options:
A Benzene
B Cyclohexane
C Cyclohexene
D 2, 3 dimethyl butane

The correct answer is:
Cyclohexene

Solution:
Cyclohexene
the addition of ethene to 1,3 -butadiene to give cyclohexene.

Topic: Chemistry, Hydrocarbons, JEE Main
∫18+2x-x2dx=

Options:
A 13sin-1⁡x-13+c
B sin-1⁡x+13+c
C 13sin-1⁡x+13+c
D sin-1⁡x-13+c

The correct answer is:
sin-1⁡x-13+c

Solution:
Let I=â«dx8+2x-x2
â I=â«dx9+2x-x2-1
â I=â«dx32-x-12
â´ I=sin-1â¡x-13+c

Topic: Mathematics, Indefinite Integration, JEE Main
1×10-5 M AgNO3 is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is _______×10-8 S m-1.

[Given : Ksp(AgBr)=4.9×10-13 at 298 K

λAg+0=6×10-3Sm2 mol-1

λBr-0=8×10-3Sm2 mol-1

λNO3-0=7×10-3Sm2 mol-1

Options:

The correct answer is:
13039

Solution:
The concentrations of the ions can be calculated as follows,

Ag+=10-5M

NO3-=10-5M

Br-=KspAg+=4.9×10-8M

Λm=k1000×M=Specific conductance1000×Molarity

For Ag+

6×10-3=kAg+1000×10-5

kAg+=6×10-5Sm-1

⇒6000×10-8Sm-1

for Br-

8×10-3=kBr-1000×4.9×10-8

kBr-=39.2×10-8Sm-1

for NO3-

7×10-3=kNO3-1000×10-5

kNO3-=7×10-5Sm-1

=7000×10-8Sm-1

Conductivity of solution

⇒(6000+7000+39.2)×10-8

⇒13039.2×10-8 S m-1

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (01 Feb Shift 2)
1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42×10-3bar

The molar mass of the biopolymer is -×104 g mol-1. (Round off to the Nearest Integer)
[Use: R=0.083 L bar mol-1 K-1

Options:

The correct answer is:
15

Solution:
π=CRT; π= osmotic pressure
C= molarity
T= Temperature of solution let the molar mass be

Molar mass of polymer = M gm/mol

2.42×10-3 bar==1.46 gM gm/mol0.1ℓ×0.083ℓ.barmol.K×300 K

⇒ M=15.02×104 g/mol

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (27 Jul Shift 1)
1.56Ã105 J of heat is conducted through is 2 m2 wall of 12 cm thick in one hour. The temperature difference between the two sides of the wall is 20 â. What is the thermal conductivity of the material of the wall (in W m-1 K-1)?

Options:

The correct answer is:
0.13

Solution:
dQdt=KA∆Tx
1.56×1053600=K×2×2012×10-2
K=1.56×105×12×10-23600×2×10
=1.5612=0.13

Topic: Physics, Thermal Properties of Matter, JEE Main
1.84âg mixture of CaCO 3 and MgCO 3 when heated to constant weight of 0.96âg as residue. Find the mass of CaCO 3 (in grams) in the mixture. M CaCO 3 â=â100 g/mole,â M MgCO 3 â=â84 g/mole

Options:

The correct answer is:
1.00

Solution:
MgCO 3 + CaCO 3 â¶ MgO + CaO + 2 CO 2 g 1.8 4 - x g x g 0.8 8 g Moles { 1 Â· 8 4 - x 8 4 x 1 0 0 0 Â· 8 8 4 4
-------------------------------------------------------------------------------------
Apply POAC to C
1.8 4 - x 8 4 Ã 1 + x 1 0 0 Ã 1 = 0.8 8 4 4 Ã 1 = 0.0 2
1 8 4 - 1 0 0 x + 8 4 x = 1 6 8
â´ 1 6 = 1 6 x
â´ x = 1 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main