datavorous/entrance-exam-dataset huggingface dataset:
$\int_0^1 \frac{e^{-x}}{1+e^{-x}} d x=$

Options:
A $\log \left(\frac{1+e}{e}\right)-\frac{1}{e}+1$
B $\log \left(\frac{1+e}{2 e}\right)-\frac{1}{e}+1$
C $\log \left(\frac{1+e}{2 e}\right)+\frac{1}{e}-1$
D None of these

The correct answer is:
$\log \left(\frac{1+e}{2 e}\right)-\frac{1}{e}+1$

Solution:
Put $1+e^{-x}=t \Rightarrow-e^{-x} d x=d t$, then we have
$\begin{aligned} I & =\int_2^{1+\frac{1}{e}} \frac{(t-1)(-d t)}{t}=\int_2^{1+\frac{1}{e}}\left(\frac{1}{t}-1\right) d t \\ & =\left[\log _e t-t\right]_2^{1+\frac{1}{e}}=\log _e\left(1+\frac{1}{e}\right)-\left(1+\frac{1}{e}\right)-\log _e 2+2 \\ & =\log _\theta\left(\frac{e+1}{2 e}\right)-\frac{1}{e}+1\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=$

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

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

Solution:
Put $t=\tan ^{-1} x \Rightarrow d t=\frac{1}{1+x^2} d x$, then
$\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x=\int_0^{\pi / 4} t d t=\left[\frac{t^2}{2}\right]_0^{z / 4}=\frac{\pi^2}{32}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^1(1-x)^9 d x=$

Options:
A 1
B $\frac{1}{10}$
C $\frac{11}{10}$
D 2

The correct answer is:
$\frac{1}{10}$

Solution:
Required value $= \left[\frac{-(1-x)^{10}}{10}\right]_0^1=\frac{1}{10}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

Options:
A $\frac{\pi^2}{2}$
B $\frac{\pi^2}{3}$
C $\pi^2$
D $\frac{\pi^2}{4}$

The correct answer is:
$\frac{\pi^2}{4}$

Solution:
$=\int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$
On adding Eqs. (i) and (ii), we get
$2 I=\int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta$
Let $\cos \theta=t \Rightarrow-\sin \theta d \theta=d t$
$\begin{aligned}
\therefore \quad 2 I & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t \\
& =2 \pi \int_0^1 \frac{1}{1+t^2} d t \\
& =2 \pi\left[\tan ^{-1} t\right]_0^1=2 \pi \cdot \frac{\pi}{4}
\end{aligned}$
$\Rightarrow \quad I=\frac{\pi^2}{4}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^1 \frac{d}{d x}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right] d x$ is equal to

Options:
A 0
B $\pi$
C $\pi / 2$
D $\pi / 4$

The correct answer is:
$\pi / 2$

Solution:
$I=\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)\right]_0^1=\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}$

Topic: Mathematics, Definite Integration, JEE Main
0.1 M KMnO4 is used for following titration. What volume of the solution in mL will be required to react with 0.158 g of Na2S2O3?
Not balanced: S2O32-+MnO4-+H2O→MnO2(s⁡)+SO42-+OH-

Options:
A 26.7 mL
B 50 mL
C 65 mL
D 75 mL

The correct answer is:
26.7 mL

Solution:
S+22O32-+MnO4-+H2OâMnO2(sâ¡)+S+6O42-+OH-
'n' factor of Na2S2O3=8
'n' factor of KMnO4=3
m.eq of Na2S2O3= m.eq of KMnO4
0.158 158 ×8×1000=0.1×V×3
V=26.7 mL

Topic: Chemistry, Redox Reactions, JEE Main
0.1âMâNaOH is titrated with 0.1 M, 20 ml HA till the end point, K a HA â=â 6Ã10 â6 and degree of dissociation of HA is small as compared to unity. What is the pH of the resulting solution at the end point?

Options:
A 6.23
B 9.22
C 7.21
D 8.95

The correct answer is:
8.95

Solution:
NaOH2â¡+HA2â¡âNaA+H2Oâ¡
At end point â¡0.1Ã20=2
âµ 20 mLâNaOH is required for the complete neutralization of HA
NaA is a salt of strong base and weak acid
Thus, will undergo hydrolysis and solution will becomes basic
C =  NaA  =  2 20 + 20  = 0.05 M
And pK a  = −log 6 ×  10 −6  = 5.2
pH at the end point = 7 +  1 2 pK a  + log C
7 +  1 2   5.2 + log 0.05  = 8.95

Topic: Chemistry, Ionic Equilibrium, JEE Main
0.01MKMnO4 solution was added to 20.0 mL of 0.05M Mohr 's salt solution through a burette. The initial reading of 50 mL burette is zero. The volume of KMnO4 solution left in the burette after the end point is....... mL. (nearest integer)

Options:

The correct answer is:
30

Solution:
Fe2++MnO4-⟶Fe3++Mn2+

vf=1 vf=5

vf=Valency factor

Milli equivalents of mohr's salt = milli equivalents of KMnO4

1×0.05×20=5 0.01×Vml

Vml =20ml

Volume of KMnO4 left = 50-20=30ml

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2022 (28 Jun Shift 2)
$\int_0^\pi \frac{1}{1+\sin x} d x$ is equal to

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

The correct answer is:
2

Solution:
$\begin{aligned} & \text { Let } I=\int_0^\pi \frac{1}{1+\sin x} d x \\ & =\int_0^\pi \frac{1}{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}} d x \\ & =\int_0^\pi \frac{\sec ^2 \frac{x}{2}}{\left(1+\tan \frac{x}{2}\right)^2} d x \\ & \text { Put } \tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \\ & \therefore \quad I=\int_0^{\infty} \frac{2 d t}{(1+t)^2}=\left[-\frac{2}{1+t}\right]_0^{\infty}=2 \\ & \end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

Options:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{2}$
D $\pi$

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

Solution:
Let
$I=\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$
Put $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$
$\therefore \quad I=\int_{\pi / 2}^0 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)(-\sin \theta) d \theta$
$=\int_0^{\pi / 2} \sin \left(2 \tan ^{-1} \tan \left(\frac{\pi}{2}-\frac{\theta}{2}\right)\right) \sin \theta d \theta$
$=\int_0^{\pi / 2} \sin (\pi-\theta) \sin \theta d \theta=\int_0^{\pi / 2} \sin ^2 \theta d \theta$
$=\int_0^{\pi / 2}\left[\frac{1-\cos 2 \theta}{2}\right] d \theta=\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\pi / 2}$
$=\frac{1}{2}\left[\frac{\pi}{2}-0\right\rceil=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\infty} \frac{\log \left(1+x^2\right)}{1+x^2} d x=$

Options:
A $\pi \log \frac{1}{2}$
B $\pi \log 2$
C $2 \pi \log \frac{1}{2}$
D $2 \pi \log 2$

The correct answer is:
$\pi \log 2$

Solution:
$\begin{aligned} & \text { Let } I=\int_0^x \frac{\log \left(1+x^2\right)}{1+x^2} d x \\ & \text { Put } x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta\end{aligned}$
$\begin{gathered}\therefore \quad 1=\int_0^{\pi / 2} \log (\sec \theta)^2 d \theta=2 \int_0^{\pi / 2} \log \operatorname{(sed} \theta) \theta \\ =-2 \int_0^{\pi / 2} \log \cos \theta d \theta=-2 \cdot \frac{\pi}{2} \log \frac{1}{2}=-\pi \log \frac{1}{2}=\pi \log 2\end{gathered}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to

Options:
A 0
B $\frac{\pi}{2}$
C $\frac{\pi}{4}$
D $\pi$

The correct answer is:
0

Solution:
Let $I=\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cos x} d x$
On putting $x=\left(\frac{\pi}{2}-x\right)$ in Eq. (i), we get
$$
\begin{aligned}
I &=\int_{0}^{\pi / 2} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1-\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x \\
&=\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1-\sin x \cos x} d x \\
&=-\int_{0}^{\pi / 2}\left(\frac{\sin x-\cos x}{1-\sin x \cos x}\right) d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=\int_{0}^{\pi / 2} 0 d x=0
$$
$\Rightarrow \quad I=0$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \frac{d x}{2+\cos x}=$

Options:
A $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
B $\sqrt{3} \tan ^{-1}(\sqrt{3})$
C $\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
D $2 \sqrt{3} \tan ^{-1}(\sqrt{3})$

The correct answer is:
$\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Solution:
$\begin{aligned} & I=\int_0^{\pi / 2} \frac{d x}{2+\cos x} \\ &=\int_0^{\pi / 2} \frac{d x}{2 \sin ^2 \frac{x}{2}+2 \cos ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}} \\ &=\int_0^{\pi / 2} \frac{d x}{\sin ^2 \frac{x}{2}+3 \cos ^2 \frac{x}{2}}=\int_0^{\pi / 2} \frac{\sec ^2 \frac{x}{2}}{3+\tan ^2 \frac{x}{2}} d x \\ & \text { Put } t=\tan \frac{x}{2} \Rightarrow d t=\frac{1}{2} \sec ^2 \frac{x}{2} d x \text {, then }\end{aligned}$
$I=2 \int_0^1 \frac{d t}{3+t^2}=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Topic: Mathematics, Definite Integration, JEE Main
$\int_{0}^{\pi / 2} \frac{2^{\sin x}}{2^{\sin x}+2^{\cos x}} d x$ equals

Options:
A 2
B $\pi$
C $\pi / 4$
D $\pi / 2$

The correct answer is:
$\pi / 4$

Solution:
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin \mathrm{x}}}{2^{\sin x}+2^{\cos \mathrm{x}} \mathrm{dx}}$
$\mathrm{I}=\int_{0}^{\pi / 2} \frac{2^{\sin (\pi / 2-\mathrm{x})}}{2^{\sin (\pi / 2-\mathrm{x})}+2^{\cos (\pi / 2-\mathrm{x})} \mathrm{dx}}$
$=\int \frac{2^{\cos \mathrm{x}}}{2^{\cos \mathrm{x}}+2^{\sin \mathrm{x}}} \mathrm{dx} \Rightarrow 21=\int_{0}^{\pi / 2} \mathrm{~d} \mathrm{x}=\frac{\pi}{2} \Rightarrow 1=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, JEE Main
∫022x2-3x+x-12dx, where t is the greatest integer function, is equal to

Options:
A 76
B 1912
C 3112
D 32

The correct answer is:
1912

Solution:
∫022x2-3x+x-12dx

=∫022x2-3xdx+∫02x-12dx

Let I1=∫022x2-3xdx=∫0323x-2x2dx+∫3222x2-3xdx

=3x22-2x33032+2x33-3x22322

=278-94+83-6-94+278=94-103=1912

Let I2=∫02x-12dx

Assume x-12=t

i.e. I2=∫-1232tdt

=∫-120-1dt+∫010·dt+∫1321·dt

=-t-120+t132=-12+32-1=0

Hence, ∫022x2-3x+x-12dx=1912

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (27 Jul Shift 2)
$\int_0^{2 a} f(x) d x=$

Options:
A $2 \int_0^a f(x) d x$
B 0
C $\int_0^a f(x) d x+\int_0^a f(2 a-x) d x$
D $\int_0^a f(x) d x+\int_0^{2 a} f(2 a-x) d x$

The correct answer is:
$\int_0^a f(x) d x+\int_0^a f(2 a-x) d x$

Solution:
It is a fundamental property.

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x d x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x}=$

Options:
A 0
B $\pi$
C $\pi / 2$
D $\pi / 4$

The correct answer is:
$\pi / 4$

Solution:
Let $I=\int_0^{\pi / 2} \frac{\sin ^{3 / 2} x d x}{\cos ^{3 / 2} x+\sin ^{3 / 2} x}$ ...(i)
$=\int_0^{\pi / 2} \frac{\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)}{\cos ^{3 / 2}\left(\frac{\pi}{2}-x\right)+\sin ^{3 / 2}\left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\pi / 2} \frac{\cos ^{3 / 2} x d x}{\sin ^{3 / 2} x+\cos ^{3 / 2} x}$ ...(ii)
Adding (i) and (ii), we get $I=\frac{1}{2} \int_0^{\pi / 2} 1 d x=\frac{1}{2}[x]_0^{\pi / 2}=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \cos ^2 x d x=$

Options:
A $1-\frac{\pi}{4}$
B $1+\frac{\pi}{4}$
C $\frac{\pi}{4}$
D $\frac{\pi}{2}$

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

Solution:
Using gamma function,
$\int_0^{\pi / 2} \cos ^2 x d x=\frac{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2 \Gamma(2)}=\frac{\frac{1}{2} \Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{2 \cdot 1 \cdot \Gamma(1)}=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \log _e(\sin 2 x) d x$

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

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

Solution:
$I=\int_0^{\pi / 2} \log _e(\sin 2 x) d x$
Let
$$
2 x=t
$$
For lower limit at $x=0, t=0$
and upper limit at $x=\pi / 2, t=\pi$ and $d x=\frac{1}{2} d t$
So, $I=\frac{1}{2} \int_0^\pi \log _e \sin (t) d t=\int_0^{\pi / 2} \log _e \sin (t) d t$
$$
=-\frac{\pi}{2} \log _e 2
$$

Topic: Mathematics, Definite Integration, JEE Main
0.002 molal aqueous solution of an ionic compound with moleciular formula Co(NH 3 ) 5 NO 2 Cl freezes at -0.00744∘C . How many moles of ions does 3 moles of the salt produce on being dissolved in water?
[Given: K f of water =1.86 K kg  mol −1 ]

Options:

The correct answer is:
6

Solution:
(ÎT f )â=âiâÃâ k f âÃâm
(ÎT f )â=âiâÃâ k f âÃâMolality
0.00744â=âiâÃâ1.86âÃâ0.002
iâ2
âiâ=â1â+âÎ±(nâ1)
â2â=â1â+â1(nâ1)
⇒n−1 = 1
â´ân=2
â´ number of ions produced from 3 moles of compound = 2Ã3=6

Topic: Chemistry, Solutions, JEE Main
â«0Ï2secâ¡ xnsecâ¡ xn+cosecâ¡ xndx=

Options:
A π2
B π3
C π4
D π6

The correct answer is:
π4

Solution:
∫0π2sec⁡ xnsec⁡ xn+cos⁡ ec xndx ..... (i)
⇒ I=∫0π2sec⁡ π2-xnsec⁡ π2-x+ cosce⁡π2-xnndx
⇒ I=∫0π2cosce⁡ xncosce⁡xn + sec⁡ xndx ..... (ii)
Adding equation (i) and (ii)
2I=∫0π2dx
⇒ 2I=x0π2
⇒ π4

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \sin ^{2 m} x d x=$

Options:
A $\frac{2 m!}{\left(2^m \cdot m!\right)^3} \cdot \frac{\pi}{2}$
B $\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \cdot \frac{\pi}{2}$
C $\frac{2 m!}{2^m \cdot(m!)^4} \cdot \frac{\pi}{2}$
D None of these

The correct answer is:
$\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \cdot \frac{\pi}{2}$

Solution:
Here the power is even, so from formula
$\begin{aligned} & \int_0^{\pi / 2} \sin ^{2 m} x d x=\frac{(2 m-1)}{2 m} \cdot \frac{(2 m-3)}{(2 m-2)} \cdots \cdots \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} \\ & =\frac{2 m \cdot(2 m-1)(2 m-2) \ldots \cdot 3 \cdot 2 \cdot 1 \cdot \frac{\pi}{2}}{[2 m \cdot(2 m-2)(2 m-4) \ldots . .4 \cdot 2]^2}\end{aligned}$
Multiplying the numerator and the denominator by $2 m(2 m-2) \ldots . .4 .2$ $=\frac{(2 m)!}{\left[2^m \cdot m(m-1)(m-2) \ldots . .2 .1\right]^2} \frac{\pi}{2}$
$=\frac{(2 m)!}{\left(2^m \cdot m!\right)^2} \frac{\pi}{2}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \sin ^2 x \cos ^3 x d x=$

Options:
A 0
B $\frac{2}{15}$
C $\frac{4}{15}$
D None of these

The correct answer is:
$\frac{2}{15}$

Solution:
Using gamma function,
$\int_0^{\pi / 2} \sin ^2 x \cos ^3 x d x=\frac{\Gamma\left(\frac{3}{2}\right) \Gamma 2}{2 \Gamma\left(\frac{7}{2}\right)}=\frac{2}{15}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{2 \pi} \sin ^6 x \cos ^5 x d x$ is equal to

Options:
A $2 \pi$
B $\pi / 2$
C $0$
D $-\pi$

The correct answer is:
$0$

Solution:
Let $\begin{aligned} I & =\int_0^{2 \pi} \sin ^6 x \cos ^5 x d x \\ I & =2 \int_0^\pi \sin ^6 x \cos ^5 x d x\end{aligned}$
$[\because f(2 \pi-x)=f(x)]$
Let $f(x)=\sin ^6 x \cos ^5 x$
$[\because f(\pi-x)=-f(x)]$
$\begin{aligned} f(\pi-x) & =\sin ^6(\pi-x) \cdot \cos ^5(\pi-x) \\ & =\sin ^6 x \cdot\left(-\cos ^5 x\right) \\ & =-\sin ^6 x \cdot \cos ^5 x \\ & =-f(x) \\ \therefore \quad I & =0\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 2} \sin ^8 x \cos ^2 x d x$ is equal to

Options:
A $\frac{\pi}{512}$
B $\frac{3 \pi}{512}$
C $\frac{5 \pi}{512}$
D $\frac{7 \pi}{512}$

The correct answer is:
$\frac{7 \pi}{512}$

Solution:
$\begin{aligned} & \int_0^{\pi / 2} \sin ^8 x \cdot \cos ^2 x d x=\frac{\frac{8+1}{2} ! \frac{2+1}{2} !}{2 \frac{8+2+2}{2} !} \\ &=\frac{\frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} \cdot \frac{1}{2} \cdot \pi}{2 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=\frac{7 \pi}{512}\end{aligned}$
[OBJECT END

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{2 \pi}(\sin x+|\sin x|) d x=$

Options:
A 0
B 8
C 4
D 1

The correct answer is:
4

Solution:
$\begin{aligned} & \int_0^{2 \pi}(\sin x+|\sin x|) d x= \\ & =\int_0^\pi(\sin x+\sin x) d x+\int_0^\pi(\sin x-\sin x) d x \\ & =\int_0^\pi 2 \sin x d x+0 \\ & =[-2 \cos x]_0^\pi=4\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^2 \frac{x^3 d x}{\left(x^2+1\right)^{\frac{3}{2}}}=$

Options:
A $(\sqrt{2}-1)^2$
B $\frac{(\sqrt{2}-1)^2}{\sqrt{2}}$
C $\frac{\sqrt{2}-1}{\sqrt{2}}$
D None of these

The correct answer is:
None of these

Solution:
$\begin{aligned} & \text { Put } t=x^2+1 \Rightarrow d t=2 x d x \\ & \qquad \int_0^2 \frac{x^3}{\left(x^2+1\right)^{3 / 2}} d x=\frac{1}{2} \int_1^5 \frac{(t-1)}{t^{3 / 2}} d t=\frac{1}{2} \int_1^5\left[t^{-1 / 2}-t^{-3 / 2}\right] d t \\ & =\frac{1}{2}\left[2 \sqrt{t}+2 \frac{1}{\sqrt{t}}\right]_1^5=\frac{1}{2}\left[2 \sqrt{5}+\frac{2}{\sqrt{5}}-2-2\right] \\ & =\left[\sqrt{5}+\frac{1}{\sqrt{5}}-2\right]=\frac{6-2 \sqrt{5}}{\sqrt{5}}\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\sqrt{2}}\left[x^2\right] d x$ is

Options:
A
$2-\sqrt{2}$

B
$2+\sqrt{2}$

C
$\sqrt{2}-1$

D
$\sqrt{2}-2$

The correct answer is:
$\sqrt{2}-1$

Solution:
$\int_1^0\left[x^2\right] d x+\int_1^{\sqrt{2}}\left[x^2\right] d x=0+\int_1^{\sqrt{2}} d x=\sqrt{2}-1$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2002
∫032+x2dx=

Options:
A limn→∞1n2n+12+22+…+3n2n2
B limn→∞1n3n+12+22+…+6n2n2
C limn→∞1n6n+12+22+…+9n2n2
D limn→∞1n3n+12+22+…+3n2n2

The correct answer is:
limn→∞1n6n+12+22+…+9n2n2

Solution:
Let 2+x2=f(x)

⇒∫032+x2dx=∫03fxdx

Now, ∫03fxdx=limn→∞1n∑r=13nfrn

=limn→∞1n∑r=13n2+rn2

=limn→∞1n2×3n+12+22+32+…+3n2n2

=limn→∞1n6n+12+22+32+…+3n2n2

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^3|2-x| d x \quad$ equals

Options:
A $2 / 7$
B $5 / 2$
C $3 / 2$
D $-3 / 2$

The correct answer is:
$5 / 2$

Solution:
$I=\int_0^3|2-x| d x=\int_0^2(2-x) d x+\int_2^3-(2-x) d x$
$\begin{gathered}=\int_0^2(2-x) d x-\int_2^3(2-x) d x=\left[2 x-\frac{x^2}{2}\right]_0^2-\left[2 x-\frac{x^2}{2}\right]_2^3 \\ \left.\Rightarrow I=[4-2]-6-\frac{9}{2}-(4-2)\right]=2-\left[4-\frac{9}{2}\right]=\frac{5}{2} .\end{gathered}$

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

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

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

Solution:
(d) Let
$$
\begin{gathered}
I=\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x \\
\Rightarrow \quad I=\int_0^{\alpha / 3} \frac{f\left(\frac{\alpha}{3}-x\right)}{f\left(\frac{\alpha}{3}-x\right)+f\left(\frac{\alpha-3\left(\frac{\alpha}{3}-x\right)}{3}\right)} d x \\
\left.\Rightarrow \quad I=\int_0^{\alpha / 3} \frac{f\left(\frac{\mathrm{i})}{3} f(x) d x=\int_0^\alpha f(a-x)\right.}{f\left(\frac{\alpha-3 x}{3}\right)+f(x)} d x\right]
\end{gathered}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 2 I=\int_0^{\alpha / 3} 1 d x \\
& \Rightarrow \quad 2 I=[x]_0^{\alpha / 3} \\
& \Rightarrow \quad 2 I=\frac{\alpha}{3} \\
& \Rightarrow \quad I=\frac{\alpha}{6} \text {. } \\
&
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^3\left|x^2-3 x+2\right| d x=$

Options:
A $\frac{3}{2}$
B $\frac{1}{6}$
C $\frac{11}{6}$
D $\frac{11}{2}$

The correct answer is:
$\frac{11}{6}$

Solution:
Let $I=\int_0^3\left|x^2-3 x+2\right| d x$
$\because \quad x^2-3 x+2=(x-2)(x-1)$
So, $\quad x^2-3 x+2 < 0$ for $x \in(1,2)$
and $x^2-3 x+2 \geq 0$ for $x \in R-(1,2)$
So,
$$
\begin{aligned}
& I= \int_0^1\left(x^2-3 x+2\right) d x-\int_1^2\left(x^2-3 x+2\right) d x \\
&+\int_2^3\left(x^2-3 x+2\right) d x \\
&=\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_0^1-\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_1^2+ \\
& \qquad\left[\frac{x^3}{3}-\frac{3 x^2}{2}+2 x\right]_2^3 \\
&=\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left[\left(\frac{8}{3}-\frac{12}{2}+4\right)-\left(\frac{1}{3}-\frac{3}{2}+2\right)\right] \\
& \quad+\left[\left(\frac{27}{3}-\frac{27}{2}+6\right)-\left(\frac{8}{3}-\frac{12}{2}+4\right)\right]=\frac{11}{6}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^3 \frac{3 x+1}{x^2+9} d x$ is equal to :

Options:
A $\log (2 \sqrt{2})+\frac{\pi}{12}$
B $\log (2 \sqrt{2})+\frac{\pi}{2}$
C $\log (2 \sqrt{2})+\frac{\pi}{6}$
D $\log \left(2(\sqrt{2})+\frac{\pi}{3}\right.$

The correct answer is:
$\log (2 \sqrt{2})+\frac{\pi}{12}$

Solution:
$=\frac{3}{2}\left[\log \left(x^2+9\right)\right]_0^3+\frac{1}{3}\left[\tan ^{-1} \frac{x}{3}\right]_0^3$
$=\frac{3}{2}[\log 18-\log 9]+\frac{1}{3}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]$
$=\frac{3}{2}[\log 2]+\frac{\pi}{12}$
$=\log (2 \sqrt{2})+\frac{\pi}{12}$

Topic: Mathematics, Definite Integration, JEE Main
$\int_0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x$ is equal to

Options:
A $1 / 6$
B $1 / 3$
C $1 / 12$
D $1 / 9$

The correct answer is:
$1 / 6$

Solution:
Divide $\mathrm{Nr} \& \mathrm{Dr}$ by $\cos \mathrm{x}$
$\int_0^{\pi / 4} \frac{\tan ^2 x \sec ^2 x d x}{\left(1+\tan ^3 x\right)^2} d x$
Let $1+\tan ^3 \mathrm{x}=\mathrm{t}$
$\begin{aligned}
& \tan ^2 x \sec ^2 x d x=\frac{d t}{3} \\
& \frac{1}{3} \int_1^2 \frac{\mathrm{dt}}{\mathrm{t}^2}=\frac{1}{6}
\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (06 Apr Shift 1)
0.4 g mixture of NaOH,Na2CO3 and some inert impurities was first titrated with N10HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is (Rounded-off to the nearest integer)

Options:

The correct answer is:
4

Solution:
Upto first end point gm equi. of NaOH+Na2CO3=HCl

x+y×1=110×17.5

x+y=1.75 ....1

Upto second end point

NaOH+Na2CO3≡HCl

x+y×2=110×19

x+2 y=1.9 ....2

y=0.15

%Na2CO3=0.15×10-3×1060.4×100

=3.975%

=4%

Hence answer is 4

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (25 Feb Shift 1)
0.4 g of an organic compound on Kjeldahl’s analysis gave enough ammonia to just neutralize 20 ml of 1N  H 2 SO 4 . What is the percentage of nitrogen in the compound?

Options:

The correct answer is:
70

Solution:
milli equivalents of NH 3 = milli equivalents of H 2 SO 4
milli moles of NH 3 ×1=1 N×20 ml
milli moles of NH 3  = 20 milli moles
∴ mass of N= 2 0 ×14 100 0
=0.28
∴ % of N= 0.28 0.4 ×100=70%

Topic: Chemistry, Amines, JEE Main
$\int_0^{\pi / 4}[\sqrt{\tan x}+\sqrt{\cot x}] d x$ is equal to

Options:
A $\frac{\pi}{\sqrt{2}}$
B $\frac{\pi}{2}$
C $\frac{3 \pi}{\sqrt{2}}$
D $\pi$

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

Solution:
$$
\begin{aligned}
& I=\int_0^{\pi / 4} \sqrt{\tan x}+\sqrt{\cot x} d x \\
&=\int_0^{\pi / 4}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) d x \\
&=\int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}} d x \\
&=\int_0^{\pi / 4} \frac{\sqrt{2}(\sin x+\cos x)}{\sqrt{2 \sin x \cos x}} d x \\
&=\sqrt{2} \int_0^{\pi / 4} \frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}} d x
\end{aligned}
$$
Let $\quad \sin x-\cos x=t$
$$
\Rightarrow \quad(\cos x+\sin x) d x=d t
$$
$$
\begin{aligned}
\therefore I & =\sqrt{2} \int_0^{\pi / 4} \frac{d t}{\sqrt{1-t^2}} \\
& =\sqrt{2}\left[\sin ^1 t\right]_0^{\pi / 4} \\
& =\sqrt{2}\left[\sin ^1(\sin x-\cos x)\right]_0^{\pi / 4} \\
& =\sqrt{2}\left[\sin ^1\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)-\sin ^1\left(\sin \infty^{\circ}-\cos \infty^{\circ}\right)\right] \\
& =\sqrt{2}\left[\sin ^1 0-\sin ^1(-1)\right] \\
& =\sqrt{2}\left[0+\frac{\pi}{2}\right]=\sqrt{2} \frac{\pi}{2}=\frac{\pi}{\sqrt{2}}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main
0.05 cm thick coating of silver is deposited on a plate of 0.05 m2 area. The number of silver atoms deposited on plate are _______×1023. (At mass Ag =108, d=7.9 g cm-3 Round off to the nearest integer.

Options:

The correct answer is:
11

Solution:
Volume of silver coating =0.05×0.05×10000=25 cm3

Given, Density =7.9 g cm-3

Density = Mass/ Volume

Mass of silver deposited =25×7.9 g

Moles of silver atoms = mass / molar mass =25×7.9108

Number of silver atoms =25×7.9108×6.023×1023

=11.01×1023

Ans. 11

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
∫05cosπx-x2dx, where t denotes greatest integer less than or equal to t, is equal to

Options:
A 0
B 2
C -3
D 4

The correct answer is:
0

Solution:
I=∫05cosx2π-πxdx

∵x∈0,5 ∴x∈0,52⇒x2∈0,1,2

I=∫02cos-πxdx+∫24cosπ-πxdx+∫45cos2π-πxdx

⇒I=sinπxπ02+-sinπxπ24+sinπxπ45=0

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (29 Jun Shift 1)
0.5F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper which can be deposited will be

Options:
A 63.5 g
B 31.75 g
C 15.80 g
D unpredictable

The correct answer is:
15.80 g

Solution:
Cu2++2e-⟶Cu ...i

According to Faraday's first law, when an electric current is passed through an electrolyte, amount of substance deposited is directly proportional to the quantity of electric charge passed through the electrolyte.

∵1 mol of 63.5 g of Cu is obtained by passing 2F of electricity.

Mass of Cu obtained by passing 1F of electricity =63.52g

∴ Mass of Cu obtained by passing 0.5F of electricity
=63.52×0.5=15.80 g

Topic: Chemistry, Electrochemistry, JEE Main