datavorous/entrance-exam-dataset huggingface dataset:
If in the given figure $\overrightarrow{O A}=\mathbf{a}, \overrightarrow{O B}=\mathbf{b}$ and $A P: P B=m: n$, then $\overrightarrow{O P}=$

Options:
A $\frac{m \mathbf{a}+n \mathbf{b}}{m+n}$
B $\frac{n \mathbf{a}+m \mathbf{b}}{m+n}$
C $m \mathbf{a}-n \mathbf{b}$
D $\frac{m \mathbf{a}-n \mathbf{b}}{m-n}$

The correct answer is:
$\frac{m \mathbf{a}+n \mathbf{b}}{m+n}$

Solution:
$\begin{aligned} \therefore & \frac{A P}{P B}=\frac{n}{m} \\ \therefore & \frac{P B}{A P}=\frac{m}{n} \\ \therefore & \frac{P B+A P}{A P}=\frac{m+n}{n} \\ \therefore & A P=\frac{n}{m+n} \times(p B+A P)\end{aligned}$
$\begin{aligned} & =\frac{n}{m+n}(\overrightarrow{O B}-O \vec{A}) \\ & =\frac{n}{m+n}(b-a) \\ & =\frac{n b-n a}{m+n}\end{aligned}$
$\begin{aligned} \therefore O P & =O A+A P \\ & =a+\frac{n b-n a}{m+n} \\ & =\frac{m a+m a+n b-n a}{m+n} \\ & =\frac{m a+n b}{m+n} \end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If \( I_{n}=\int_{0}^{\frac{\pi}{2}} x^{n} \sin x d x \), where \( n>1 \), then

Options:
A \( I_{n}+n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n} \)
B \( I_{n}+n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n-1} \)
C \( I_{n}-n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n} \)
D \( I_{n}-n(n-1) I_{n-2}=-n\left(\frac{\pi}{2}\right)^{n} \)

The correct answer is:
\( I_{n}+n(n-1) I_{n-2}=n\left(\frac{\pi}{2}\right)^{n-1} \)

Solution:
We have,

In=∫0π2xnsinxdx

⇒In=-xncosx0π2+∫0π2nxn-1cosxdx

⇒In=n∫0π2xn-1cosxdx

⇒In=nxn-1sinx0π2-∫0π2n-1xn-2sinxdx

⇒In=nπ2n-1-n-1In-2

Hence,

In+nn-1In-2=nπ2n-1

Topic: Mathematics, Definite Integration, JEE Main
If \( \int \frac{3 x+1}{(x-3)(x-5)} d x=\int \frac{-5}{(x-3)} d x+\int \frac{B}{(x-5)} d x \), then the value of \( B \) is

Options:
A \( 3 \)
B \( 4 \)
C \( 6 \)
D \( 8 \)

The correct answer is:
\( 8 \)

Solution:
Given,

∫3x+1x-3x-5dx=∫-5x-3dx+∫Bx-5dx

So, let

3x+1x-3x-5=−5x−3+Bx−5

⇒3x+1=−5x-5+Bx-3
Putting x=5 in the above equation, we get
35+1=0+B5-3
⇒16=2B

or B=8.

Topic: Mathematics, Indefinite Integration, JEE Main
If \( \int_{0}^{x^{2}(1+x)} f(t) \mathrm{d} t=x \), then the value of \( f(2) \) is

Options:
A \( 1 / 2 \)
B \( 1 / 3 \)
C \( 1 / 4 \)
D \( 1 / 5 \)

The correct answer is:
\( 1 / 5 \)

Solution:
Differentiating both sides w.r.t. x, we get

fx21+x×2x+3x2=1 (Use Newton-Leibniz Rule for differentiation)

At x=1⇒f2=15

Topic: Mathematics, Definite Integration, JEE Main
If \( \int_{0}^{10 \pi+\alpha}|\sin x| d x=k-\cos \alpha \), where \( 0 < \alpha < \pi \), then \( k= \)

Options:
A \( 101 \)
B \( 100 \)
C \( 201 \)
D none of these

The correct answer is:
\( 201 \)

Solution:
Let I=∫0100π+αsin x dx

=∫0100πsin x dx+∫100π100π+αsin x dx

=100∫0πsin x dx+∫0αsin x dx

∵ Period of sin x is π & 0<α<π

=100 -cosx0π+-cosx0α

=100 1+1+-cos α+1=201-cos α

∴ k=201

Topic: Mathematics, Definite Integration, JEE Main
If intensity of incident radiation in a photocell is increased, the stopping potential

Options:
A first increases and then decreases.
B remains unchanged.
C decreases.
D increases.

The correct answer is:
remains unchanged.

Solution:
The stopping potential does not depend on the intensity of incident light.

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2020 (12 Oct Shift 1)
If $\omega$ is a complex cube root of unity, then $(1+\omega)\left(1+\omega^2\right) \quad\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots$ to $2 n$ factors $=$

Options:
A $0$
B $1$
C $-1$
D None of these

The correct answer is:
$1$

Solution:
$(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots \ldots$ upto $2 n$ factors $=\left(-\omega^2\right)(-\omega)(1+\omega)\left(1+\omega^2\right) \ldots .$. upto $2 n$ factors
$=1.1 .1 . . .$. upto $n$ factors $=1$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a complex cube root of unity, then $(1-\omega)\left(1-\omega^2\right)\left(1-\omega^4\right)\left(1-\omega^8\right)=$

Options:
A $0$
B $1$
C $-1$
D $9$

The correct answer is:
$9$

Solution:
$\begin{aligned} & (1-\omega)\left(1-\omega^2\right)\left(1-\omega^4\right)\left(1-\omega^8\right) \\ & =(1-\omega)\left(1-\omega^2\right)(1-\omega)\left(1-\omega^2\right)=(1-\omega)^2\left(1-\omega^2\right)^2 \\ & =(-3 \omega)\left(-3 \omega^2\right)=9 \omega^3=9\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a complex cube root of unity, then $225+\left(3 \omega+8 \omega^2\right)^2+\left(3 \omega^2+8 \omega\right)^2$ is equal to:

Options:
A 72
B 192
C 200
D 248

The correct answer is:
248

Solution:
$225+\left(3 \omega+8 \omega^2\right)^2+\left(3 \omega^2+8 \omega\right)^2$
$=225+9 \omega^2+64 \omega^4+48 \omega^3+9 \omega^4+64 \omega^2+48 \omega^3$
$=225+9 \omega^2+64 \omega+48+9 \omega+64 \omega^2+48$
$=225+73\left(\omega^2+\omega\right)+96$
$=225-73+96=225+23=248$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a complex cube root of unity, then $\sin \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\}$ is equal to

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

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

Solution:
Since, $\omega$ is a cube root of unity.
$$
\begin{aligned}
\therefore \quad \sin & \left\{\left(\omega^{10}+\omega^{23}\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left\{\left(\omega+\omega^2\right) \pi-\frac{\pi}{4}\right\} \\
& =\sin \left(-\pi-\frac{\pi}{4}\right) \quad\left(\because 1+\omega+\omega^2=0\right) \\
& =-\sin \left(\pi+\frac{\pi}{4}\right)=\sin \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a complex cube root of unity, then $(x-y)(x \omega-y)\left(x \omega^2-y\right)=$

Options:
A $x^2+y^2$
B $x^2-y^2$
C $x^3-y^3$
D $x^3+y^3$

The correct answer is:
$x^3-y^3$

Solution:
$\begin{aligned} & (x-y)(x \omega-y)\left(x \omega^2-y\right) \\ & =\left(x^2 \omega-x y-x y \omega+y^2\right)\left(x \omega^2-y\right) \\ & =x^3-x^2 y\left(1+\omega+\omega^2\right)+x y^2\left(1+\omega+\omega^2\right)-y^3 \\ & =x^3-y^3 \quad\left(1+\omega+\omega^2=0\right)\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a cube root of unity, then $(1+\omega)^3-\left(1+\omega^2\right)^3=$

Options:
A $0$
B $\omega$
C $\omega^2$
D None of these

The correct answer is:
$0$

Solution:
$\begin{aligned} & (1+\omega)^3-\left(1+\omega^2\right)^3=\left(-\omega^2\right)^3-(-\omega)^3 \\ & =-\omega^6+\omega^3=-\omega^3 \omega^3+\omega^3=-1+1=0\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a cube root of unity, then the value of $\left(1-\omega+\omega^2\right)^5+\left(1+\omega-\omega^2\right)^5=$

Options:
A $16$
B $32$
C $48$
D $-32$

The correct answer is:
$32$

Solution:
$\begin{aligned} & \left(1-\omega+\omega^2\right)^5+\left(1+\omega-\omega^2\right)^5 \\ & =(-2 \omega)^5+\left(-2 \omega^2\right)^5=-32 \omega^3 \omega^2-32\left(\omega^3\right)^3 \omega \\ & =-32\left(\omega^2+\omega\right)=-32(-1)=32\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is a cube root of unity, then $\left|\begin{array}{ccc}x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega\end{array}\right|=$

Options:
A $x^3+1$
B $x^3+\omega$
C $x^3+\omega^2$
D $x^3$

The correct answer is:
$x^3$

Solution:
$\Delta=\left|\begin{array}{ccc}x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega\end{array}\right|$
$=\left|\begin{array}{ccc}x+1+\omega+\omega^2 & \omega & \omega^2 \\ x+1+\omega+\omega^2 & x+\omega^2 & 1 \\ x+1+\omega+\omega^2 & 1 & x+\omega\end{array}\right|$, $\left(C_1 \rightarrow C_1+C_2+C_3\right)$
$=x\left|\begin{array}{ccc}1 & \omega & \omega^2 \\ 1 & x+\omega^2 & 1 \\ 1 & 1 & x+\omega\end{array}\right|$ ( $\left.1+\omega+\omega^2=0\right)$
$=x\left[1\left\{\left(x+\omega^2\right)(x+\omega)-1\right\}+\omega\{1-(x+\omega)\} \quad+\omega^2\left\{1-\left(x+\omega^2\right)\right\}\right]$
$=x\left(x^2+\omega x+\omega^2 x+\omega^3-1+\omega-\omega x-\omega^2 \quad+\omega^2-\omega^2 x-\omega^4\right)$
$=x^3, \left(\omega^3=1\right)$.

Topic: Mathematics, Determinants, JEE Main
If $\alpha$ is a non-real root of $x^6=1$, then $\frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}$ is equal to

Options:
A $\alpha^2$
B 0
C $-\alpha^2$
D $\alpha$

The correct answer is:
$-\alpha^2$

Solution:
Given that
$\begin{aligned}
& \Rightarrow(x-1)\left(x^5+x^4+x^3+x^2+x+1\right)=0 \\
& \Rightarrow \quad x^5+x^4+x^3+x^2+x+1=0 \\
&
\end{aligned}$
$[\because$ roots are non-real]
Since $\alpha$ is a root of the equation (i)
$\begin{array}{ll}
\therefore & \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0 \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\left(\alpha^4+\alpha^2\right) \\
\Rightarrow & \alpha^5+\alpha^3+\alpha+1=-\alpha^2\left(\alpha^2+1\right) \\
& \frac{\alpha^5+\alpha^3+\alpha+1}{\alpha^2+1}=-\alpha^2
\end{array}$

Topic: Mathematics, Complex Number, JEE Main
If Î± is a root of the equation 4x2+2x-1=0 and fx=4x3-3x+1, then 2fÎ±+Î± is equal to

Options:
A -1
B 0
C 1
D 2

The correct answer is:
1

Solution:
4α2+2α-1=0
2fα+α=24α3-3α+1+α
=2α4α2+1-2α
=2α1-2α+1-2α
=2α-2α2+1-2α
=2-2α2-α+1
=-4α2-2α+2
=-1+2=1

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha$ is a root of $z^2-z+1=0$, then $\left(\alpha^{2014}+\frac{1}{\alpha^{2014}}\right)+\left(\alpha^{2015}+\frac{1}{\alpha^{2015}}\right)^2$
$$
\begin{aligned}
& +\left(\alpha^{2016}+\frac{1}{\alpha^{2016}}\right)^3+\left(\alpha^{2017}+\frac{1}{\alpha^{2017}}\right)^4+ \\
& \left(\alpha^{2018}+\frac{1}{\alpha^{2018}}\right)^5=
\end{aligned}
$$

Options:
A 8
B 5
C 3
D -5

The correct answer is:
8

Solution:
$z^2-z+1=0$, then $z=-\omega$
$$
\begin{aligned}
& \text { So, }\left|(-\omega)^{2014}+\frac{1}{(-\omega)^{2014}}\right|+\left[(-\omega)^{2015}+\frac{1}{(-\omega)^{2015}}\right]^2 \\
& +\left[(-\omega)^{2016}+\frac{1}{(-\omega)^{2016}}\right]^3+\left[(-\omega)^{2017}+\frac{1}{(-\omega)^{2017}}\right]^4 \\
& +\left[(-\omega)^{2018}+\frac{1}{(-\omega)^{2018}}\right]^5 \\
& {\left[\begin{array}{r}
{\left[-\omega-\frac{1}{\omega}\right]+\left[\omega^2+\frac{1}{\omega^2}\right]^2+8+\left[-\omega+\frac{1}{-\omega}\right]^4+\left[\omega^2+\frac{1}{\omega^2}\right]^5} \\
=\left[+\omega+\omega^2\right]+\left[\omega^2+\omega\right]^2+8+\left[-\omega-\omega^2\right]^4+\left[\omega^2+\omega\right]^5 \\
=-1+1+8+1-1=8
\end{array}\right.}
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
If $\omega$ is an $n^{\text {th }}$ root of unity, other than unity, then the value of $1+\omega+\omega^2+\ldots+\omega^{n-1}$ is

Options:
A $0$
B $1$
C $-1$
D None of these

The correct answer is:
$0$

Solution:
$1+\omega+\omega^2+\ldots . .+\omega^{n-1}=\left(\frac{\omega^n-1}{\omega-1}\right)=0,\left(\omega^n=1\right)$

Topic: Mathematics, Complex Number, JEE Main
If β is one of the angles between the normals to the ellipse x2+3y2=9 at the points 3cosθ,3sinθ and -3sinθ,3cosθ; θ∈0,π2; then 2cotβsin2θ is equal to :

Options:
A 13
B 34
C 23
D 2

The correct answer is:
23

Solution:
Given equation of ellipse is x29+y23=1

Normal at 3cosθ,3sinθ is

3secθ.x-3cosecθy=6

⇒Slope m1=3secθ3cosecθ=3tanθ

Normal at −3sinθ,3cosθ

-3cosecθ.x-3secθ.y=6

⇒Slope m2=-3cosecθ3secθ=-3cotθ

Angle between normal is β.

⇒tanβ=m1-m21+m1m2=3tanθ+3cotθ1-3=32sinθcosθ

⇒tanβ=3sin2θ⇒2cotβsin2θ=23

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2018 (15 Apr)
If $\beta$ is one of the angles between the normals to the ellipse, $x^2+3 y^2=9$ at the points $(3 \cos \theta, \sqrt{3} \sin \theta)$ and $(-3 \sin \theta, \sqrt{3} \cos \theta) ; \in\left(0, \frac{\pi}{2}\right) ;$ then $\frac{2 \cot \beta}{\sin 2 \theta}$ is equal to

Options:
A
$\sqrt{2}$

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

C
$\frac{1}{\sqrt{3}}$

D
$\frac{\sqrt{3}}{4}$

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

Solution:
Since, $x^2+3 y^2=9$
$$
\begin{aligned}
&\Rightarrow 2 x+6 y \frac{d y}{d x}=0 \\
&\Rightarrow \frac{d y}{d x}=\frac{-x}{3 y}
\end{aligned}
$$
Slope of normal is $-\frac{d x}{d y}=\frac{3 y}{x}$
$$
\begin{gathered}
\Rightarrow\left(-\frac{d y}{d x}\right)(3 \cos \theta, \sqrt{3} \sin \theta) \\
=\frac{3 \sqrt{3} \sin \theta}{3 \cos \theta}=\sqrt{3} \tan \theta=m_{\mathrm{l}} \\
\&\left(-\frac{d x}{d y}\right)(-3 \sin \theta, \sqrt{3} \cos \theta) \\
=\frac{3 \sqrt{3} \cos \theta}{-3 \sin \theta}=-\sqrt{3} \cot \theta=m_2
\end{gathered}
$$
As, $\beta$ is the anagle between the normals to the given ellipse then
$$
\begin{aligned}
&\tan \beta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
&=\left|\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1-3 \tan \theta \cot \theta}\right| \\
&=\left|\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1-3}\right|
\end{aligned}
$$
So, $\tan \beta=\frac{\sqrt{3}}{2}|\tan \theta+\cot \theta|$
$$
\begin{aligned}
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{2}\left|\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right| \\
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{2}\left|\frac{1}{\sin \theta \cos \theta}\right| \\
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{\sin 2 \theta} \Rightarrow \frac{2 \cot \beta}{\sin 2 \theta}=\frac{2}{\sqrt{3}}
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If $\theta$ is the acute angle between the curves $x^2+y^2=2020 \sqrt{2}$ and $x^2-y^2=2020$, then $\frac{\sin \theta+\cos \theta}{\tan \theta}$ is equal to

Options:
A $\sqrt{2}$
B $\frac{3+\sqrt{3}}{2}$
C $\frac{3+\sqrt{3}}{4}$
D $\frac{3+\sqrt{3}}{6}$

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

Solution:
Adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
& 2 x^2=2020(\sqrt{2}+1) \\
& \therefore \quad x^2=1010(\sqrt{2}+1), y^2=1010(\sqrt{2}-1) \\
& \therefore \quad x^2 y^2=(1010)^2\left\{(\sqrt{2})^2-(1)^2\right\} \\
& =(1010)^2 \\
& \therefore \quad x y= \pm 1010 \\
& \because \quad x^2+y^2=2020 \sqrt{2}, x^2-y^2=2020 \\
& \Rightarrow \quad 2 x+2 y y^{\prime}=0 \text { and } 2 x-2 y y^{\prime}=0 \\
& \Rightarrow \quad y^{\prime}=\frac{-x}{y}=m_1 \text { (Let) } \\
& \Rightarrow \quad y^{\prime}=\frac{x}{y}=m_2 \text { (let) } \\
& \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{-\frac{x}{y}-\left(+\frac{x}{y}\right)}{1+\frac{x}{y}\left(\frac{-x}{y}\right)}\right| \\
& =\left|\frac{-2 x}{y} \times \frac{y^2}{y^2-x^2}\right| \\
& =\left|\frac{-2 x y}{y^2-x^2}\right|=\left|\frac{-2( \pm 1010)}{-2020}\right|=| \pm 1|=1 \\
&
\end{aligned}
$$
$\begin{aligned} \therefore \quad \theta & =\tan ^{-1}(1)=\pi / 4 \\ \therefore \frac{\sin \theta+\cos \theta}{\tan \theta} & =\frac{\sin (\pi / 4)+\cos (\pi / 4)}{\tan (\pi / 4)} \\ & =\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1}=\frac{2}{\sqrt{2}}=\sqrt{2}\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Main
If Î¸ is the angle between the pair of tangents drawn to the ellipse 3x2+2y2=5 from the point 1,2, then the value of tan2Î¸ is equal to

Options:

The correct answer is:
28.8

Solution:
The equation of the tangent to 3x2+2y2=5 in slope form is y=mx±53m2+52
If it passes through 1,2, then

2-m2=53m2+52
⇒2m23+4m-32=0
⇒m1+m2=-423 & m1m2=-3223
⇒m1+m2=-6 & m1m2=-94
tan⁡θ=m1+m22-4m1m21+m1m2
tan2θ=36+9-542=45×1625=1445=28.8

Topic: Mathematics, Ellipse, JEE Main
If $\theta$ is the angle between the vectors $\mathbf{a}$ and $\mathbf{b}$, then $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a} \cdot \mathbf{b}|}$ equal to

Options:
A $\tan \theta$
B $-\tan \theta$
C $\cot \theta$
D $-\cot \theta$

The correct answer is:
$\tan \theta$

Solution:
Angle between the given vectors $\mathbf{a}$ and $\mathbf{b}$ is $\theta$. We know that, $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a} \cdot \mathbf{b}|}=\frac{|\mathbf{a} \| \mathbf{b}| \sin \theta}{|\mathbf{a} \| \mathbf{b}| \cos \theta}=\tan \theta$.

Topic: Mathematics, Vector Algebra, JEE Main
If $\omega$ is the complex cube root of unity, the
$$
\left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2=
$$

Options:
A -1
B 0
C 4
D -4

The correct answer is:
-4

Solution:
$\begin{aligned} & \left(3+5 \omega+3 \omega^2\right)^2+\left(3+3 \omega+5 \omega^2\right)^2 \\ & =\left(3+3 \omega+3 \omega^2+2 \omega\right)^2+\left(3+3 \omega+3 \omega^2+2 \omega^2\right)^2 \\ & =\left[3\left(1+\omega+\omega^2\right)+2 \omega\right]^2+\left[3\left(1+\omega+\omega^2\right)+2 \omega^2\right]^2 \\ & =(3(0)+2 \omega)^2+\left[3(0)+2 \omega^2\right]^2=4 \omega^2+4 \omega^4 \\ & =4 \omega^2\left(1+\omega^2\right)=4 \omega^2(-\omega) \\ & =-4 \omega^2=-4\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\alpha$ is the degree of dissociation of $\mathrm{Na}_2 \mathrm{SO}_4$, the vant Hoff's factor (i) used for calculating the molecular mass is

Options:
A
$1+\alpha$

B
$1-\alpha$

C
$1+2 \alpha$

D
$1-2 \alpha$

The correct answer is:
$1+2 \alpha$

Solution:
Total moles $=1+2 \alpha$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2005
If $\alpha$ is the degree of ionization, $C$ the concentration of a weak electrolyte and $K_a$ the acid ionization constant, then the correct relationship between $\alpha, C$ and $K_{\mathrm{s}}$ is

Options:
A $\alpha^2=\sqrt{\frac{K_a}{C}}$
B $\alpha^2=\sqrt{\frac{C}{K_s}}$
C $\alpha=\sqrt{\frac{K_a}{C}}$
D $\alpha=\sqrt{\frac{C}{K_a}}$

The correct answer is:
$\alpha=\sqrt{\frac{K_a}{C}}$

Solution:
According to the Ostwald's dilution formula $\alpha^2=\frac{K(1-\alpha)}{C}$. But for weak electrolytes $\alpha$ is very small. So that $(1-\alpha)$ can be neglected. So that, $\quad \alpha=\sqrt{\frac{K_a}{C}}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
If ω is the non-real cube root of unity, then the number of ordered pairs of integers a,b, such that aω+b=1, is equal to

Options:

The correct answer is:
6

Solution:
We have, aÏ+b2=1
âaÏ+baÏ-+b=1
âa2+abÏ+Ï-+b2=1
âa2-ab+b2=1
âa-b2+ab=1â¦..i
As, 1+Ï+Ï2=0
When a-b2=0 and ab=1 then 1,1;-1,-1
When a-b2=1 and ab=0 then 0,1;1,0;0,-1;-1,0
Hence, 0,1;1,0;0,-1;-1,0;1,1;-1,-1 i.e., 6 ordered pairs.

Topic: Mathematics, Complex Number, JEE Main
If Î± is the only real root of x3+bx2+cx+1=0 b<c, then the value of Î± is (where, . represents the greatest integer function)

Options:

The correct answer is:
1

Solution:
Let, fx=x3+bx2+cx+1
fx=0 has only one real root so it is an increasing function
Now, f0=1,f-1=-1+b-c+1
f-1=b-c<0
using mean value theorem, real root must lie between -1,0
so, Î±=-1
Î±=1

Topic: Mathematics, Application of Derivatives, JEE Main
If α, β is the orthocenter of the triangle ABC with vertices A3, –7, B–1, 2 and C4, 5, then 9α-6β+60 is equal to

Options:
A 25
B 35
C 30
D 40

The correct answer is:
25

Solution:
The required diagram will be:

Equation of AD:y+7=-53x-3

⇒3y+21=-5x+15

⇒5x+3y+6=0 ......i

Equation of BE:y-2=-112x+1

⇒12y-24=–x-1

⇒x=23-12y....(ii)

by ii 115-60y+3y+6=0

⇒57y=121

y=12157, x=23-12×12157

∴ 9α-6β+60=9×23-108×12157-6×12157+60

=207-242+60=25

Hence this is the required option.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (15 Apr Shift 1)
If α is the positive root of the equation, px=x2−x−2=0, then limx→α+1−cospxx+α−4is equal to

Options:
A 32
B 32
C 12
D 12

The correct answer is:
32

Solution:
Given, px=0

⇒x2-x-2=0

⇒x-2x+1=0

⇒x=2,-1 ∴ α=2

Now,limx→2+1−cosx2−x−2x−2

=limx→2+2sin2x2−x−22x−2

=limx→2+2·sinx2-x-22x-2

=limx→2+2·sinx2-x-22x-2 (because x>2)

=limx→2+ 2sinx2-x-22x2-x-22·2·x2-x-2x-2

=limx→2+12·x-2x+1x-2

=32.

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If Î» is the remainder when 22021 is divided by 17, then the value of Î» must be equal to

Options:
A 3
B 7
C 13
D 15

The correct answer is:
15

Solution:
22021=2.22020=2.24505=216505=217-1505
=217505-505C117504+505C217503............+505C504171-505C5051
=217k-1
=17⋅2k-2=17⋅2k-17+15
=172k-1+15

Topic: Mathematics, Binomial Theorem, JEE Main
If λ is the wavelength of hydrogen atom from the transition n =3 to n=1, then what is the wavelength for doubly ionised lithium ion for same transition?

Options:
A Î»3
B 3Î»
C λ9
D 9 Î»

The correct answer is:
λ9

Solution:
For wavelength
1λ=RZ21n12-1n22
Here, transition is same
So, λ∝1Z2
λHλLi=ZLi2ZH2=32 12=9
λLi=λH 9=λ9

Topic: Physics, Atomic Physics, JEE Main
If isobutane and n-butane are present in a gas, then how much oxygen should be required for complete combustion of $5 \mathrm{~kg}$ of this gas

Options:
A $17.9 \mathrm{~kg}$
B $9 \mathrm{~kg}$
C $27 \mathrm{~kg}$
D $1.8 \mathrm{~kg}$

The correct answer is:
$17.9 \mathrm{~kg}$

Solution:
Isobutane and $n$-butane $\left[\mathrm{C}_4 \mathrm{H}_{10}\right]$ have same molecular formula; $\mathrm{C}_4 \mathrm{H}_{10}+\frac{13}{2} \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}$
For $58 \mathrm{gm}$ of $C_4 H_{10} 208 \mathrm{gm} O_2$ is required then for $5 \mathrm{~kg}$ of $C_4 H_{10} \quad O_2=\frac{5 \times 208}{58}=17.9 \mathrm{~kg}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
If it is assumed that ${ }_{92}^{235} U$ decays only by emitting $\alpha$ and $\beta$ particles, the possible product of the decay is :

Options:
A ${ }_{89}^{225} A c$
B ${ }_{89}^{227} A c$
C ${ }_{89}^{230} A c$
D ${ }_{89}^{231} A c$

The correct answer is:
${ }_{89}^{227} A c$

Solution:
New mass no. $=235-2 \times 4=227$
New at. no. $=92-2 \times 2+1=92-4+1=89$

Topic: Chemistry, Structure of Atom, JEE Main
If it takes 5 minutes to fill a 15 litre bucket from a water tap of diameter 2π cm then the Reynolds number for the flow is (density of water = 103 kg/m3 and viscosity of water=10-3 Pa.s ) close to:

Options:
A 5500
B 550
C 1100
D 11,000

The correct answer is:
5500

Solution:
Reynolds number
R=ρvDη
D = Diameter of litre
Also rate of flow = volumetime=A v
Vt= π D24×v ⇒v=4VπD2t
∴ R= ρ Dη×4 Vπ D2 t
=4 ρ Vπ η D t
=4×103×15×10-3π×10-3×2×5×60 π×102
=10000π ≈5500

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2015 (10 Apr Online)
If IUPAC name of an element is "Unununnium" then the element belongs to nth group of periodic table. The value of n is _______.

Options:

The correct answer is:
11

Solution:
Unununnium is the IUPAC name given to the element Roentgenium and the atomic number 111. In the periodic table in the 7th period it is located and placed in the 11th group. Roentgenium is a radioactive element and its isotopes are unstable in nature.

111 belongs to 11th group.

Hence the value of n is 11.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If iz3+z2-z+i=0, then |z| is equal to

Options:

The correct answer is:
1

Solution:
Given, iz3+z2−z+i=0
z3+1iz2-zi+1=0
⇒ z3-iz2+iz+1=0
⇒ z2z-i+iz-i=0
⇒ z-iz2+i=0
⇒z=i or z2=−i
Hence, |z|=1

Topic: Mathematics, Complex Number, JEE Main
If ${ }^{\mu_j}$ represents refractive index when a light ray goes from medium $i$ to medium $j$, then the product $2 \mu_1 \times{ }_3 \mu_2 \times{ }_4 \mu_3$ is equal to

Options:
A $3^{\mu_1}$
B $3 \mu_2$
C $\frac{1}{{ }_1 \mu_4}$
D ${ }_4 \mu_2$

The correct answer is:
$\frac{1}{{ }_1 \mu_4}$

Solution:
${ }_2 \mu_1 \times{ }_3 \mu_2 \times{ }_4 \mu_3=\frac{\mu_1}{\mu_2} \times \frac{\mu_2}{\mu_3} \times \frac{\mu_3}{\mu_4}=\frac{\mu_1}{\mu_4}={ }_4 \mu_1=\frac{1}{{ }_1 \mu_4}$

Topic: Physics, Ray Optics, JEE Main
If ∑k=0100kk+1Ck100=a⋅2100+bc, where a,b,c∈N, then the least value of a+b+c100 is equal to

Options:

The correct answer is:
2.01

Solution:
Let S=∑k=0100kk+1C100k=∑k=0100k+1-1k+1C100k
=∑k=0100C100k-∑k=01001k+1C100k
=2100-1101∑k=0100101k+1C100k
=2100-1101∑k=0100C101k+1
=2100-11012101-1
= 101⋅ 2 100 − 2 101 +1 101 = 99⋅ 2 100 +1 101
So, a=99, b=1, c=101
Hence, a+b+c100least=2.01

Topic: Mathematics, Binomial Theorem, JEE Main
If Î£k=1â1k+2k+kk+2=a+bc, where a, b, câN and a, b, câ[1, 15], then a+ b+c is equal to

Options:

The correct answer is:
11

Solution:
Here, general term Tk=1(k+2)k+kk+2
Rationalising the expression, we get,
â T k = k + 2 k - k k + 2 k k + 2 2 - k 2 k + 2
âTk=(k+2)kâkk+22k(k+2)=12(1kâ1k+2)
Now, T1=12(11â13),
T2=12(12â14),
T 3 = 1 2 1 3 - 1 5 and so on
Tk=12(1kâ1k+2)
Adding all the terms, we get,
T1+T2+...Tk=12(1+12â1k+1â1k+2)
Now, if kââ
âk=1âTk=12(1+12â0â0)
= 1 + 2 2 2
1 + 2 8 = 2 + 4 16
As, câ[1,15]âc=8, a=1, b=2
â any other solution is not possible
âa+b+c=11

Topic: Mathematics, Sequences and Series, JEE Main