datavorous/entrance-exam-dataset huggingface dataset:
If an electron moves from rest from a point at which potential is 50 volt to another point at which potential is 70 volt, then its kinetic energy in the final state will be

Options:
A $3.2 \times 10^{-10} \mathrm{~J}$
B $3.2 \times 10^{-18} \mathrm{~J}$
C $1 \mathrm{~N}$
D $1$ dyne

The correct answer is:
$3.2 \times 10^{-18} \mathrm{~J}$

Solution:
K.E. $=q_0\left(V_A-V_B\right)=$ $1.6 \times 10^{-19}(70-50)=3.2 \times 10^{-18} \mathrm{~J}$

Topic: Physics, Electrostatics, JEE Main
If an electron of hydrogen atom moves from fourth excited state to ground state in Lymen series find the total number of spectral lines?

Options:

The correct answer is:
10

Solution:
Total number of spectral lines =nn-12=55-12=10

Topic: Chemistry, Structure of Atom, JEE Main
If an ellipse with its axes as coordinate axes, $2 \mathrm{a}$ and $2 \mathrm{~b}$ as the lengths of its major and minor axes respectively passes through the points $(2,2)$ and $(3,1)$, then $3 a^2+5 b^2=$

Options:
A 32
B 8
C 64
D 16

The correct answer is:
64

Solution:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{4}$ ...(i)
$\Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1$ ...(ii)
Equation (ii) - (i),
$\begin{aligned} & \Rightarrow \frac{8}{a^2}=\frac{3}{4} \Rightarrow a^2=\frac{32}{3} \\ & \Rightarrow \quad 3 a^2=32 \\ & \Rightarrow \quad \frac{1}{b^2}=1-\frac{9}{a^2}=1-\frac{27}{32}=\frac{5}{32} \\ & \Rightarrow \quad b^2=\frac{32}{5} \Rightarrow 5 b^2=32 \\ & \therefore \quad 3 a^2+5 b^2=64 .\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of β will be:

Options:
A 3.5
B 0.5
C 0.875
D 7

The correct answer is:
7

Solution:
Iε=IC+IB

⇒ΔIε=ΔIC+ΔIB

4 mA=3.5 mA+ΔIB

⇒ΔIB=0.5 mA

⇒β=ΔICΔIB

β=3.50.5

⇒β=7

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (24 Feb Shift 1)
If an endothermic reaction is non- spantaneous at freezing point of water and becomes feasible at its boiling point, then

Options:
A
$\Delta \mathrm{H}$ is - ve, $\Delta \mathrm{S}$ is + ve

B
$\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are $+$ ve

C
$\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are $-$ ve

D
$\Delta \mathrm{H}$ is $+$ ve, $\Delta \mathrm{S}$ is $-\mathrm{ve}$

The correct answer is:
$\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ both are $+$ ve

Solution:
$\Delta \mathrm{H}$ +ve at low temperature and $\Delta \mathrm{S}$ +ve at low temperature shows that reaction is non spontaneous
At high temperature (boiling point) becomes feasible

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2002
If an equation of a tangent to the curve, $y-\cos (x+f),-1-1 \leq x \leq 1+\pi$, is $x+2 y=k$ then $k$ is equal to :

Options:
A
1

B
2

C
$\frac{\pi}{4}$

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

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

Solution:
Let $y=\cos (x+y)$
$$
\Rightarrow \frac{d y}{d x}=-\sin (x+y)\left(1+\frac{d y}{d x}\right)
$$
Now, given equation of tangent is
$$
\begin{aligned}
&x+2 y=k \\
&\Rightarrow \text { Slope }=\frac{-1}{2}
\end{aligned}
$$
So, $\frac{d y}{d x}=\frac{-1}{2}$ put this value in (1), we get
$$
\begin{aligned}
&\frac{-1}{2}=-\sin (x+y)\left(1-\frac{1}{2}\right) \\
&\Rightarrow \sin (x+y)=1 \\
&\Rightarrow x+y=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2}-x
\end{aligned}
$$
Now $\frac{\pi}{2}-x=\cos (x+y)$
$$
\Rightarrow x=\frac{\pi}{2} \text { and } y=0
$$
Thus $x+2 y=k \quad \Rightarrow \frac{\pi}{2}=k$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (25 Apr Online)
If an extra electron is added to the hypothetical molecule C2, this extra electron will occupy the molecular orbital:-

Options:
A π2P*
B π2P
C σ2P*
D σ2P

The correct answer is:
σ2P

Solution:
Configuration of C2
σ1 s2σ1 s*2σ2 s2σ2 s*2π2px2=π2py2σ2pz
an extra electron added to the σ2p of the above configuration.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
If an integer is chosen at random from first 100 positive integers, then the probability that the chosen number is a multiple of 4 or 6 , is

Options:
A $\frac{41}{100}$
B $\frac{33}{100}$
C $\frac{1}{10}$
D None of these

The correct answer is:
$\frac{33}{100}$

Solution:
Let $A$ be the event to be multiple of 4 and $B$ be the event to be multiple of 6
So,$P(A)=\frac{25}{100}, \quad P(B)=\frac{16}{100}$
Thus required probability is
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow P(A \cup B)=\frac{25}{100}$ $+\frac{16}{100}-\frac{8}{100}=\frac{33}{100}$

Topic: Mathematics, Probability, JEE Main
If an interference experiment is performed using two wavelengths close to each other, two interference patterns corresponding to the two wavelengths are obtained on the screen. The fringe system remains distinct up to a point on the screen where the nth order maximum of one wavelength falls on the nth order minimum of the other wavelength. Suppose in Young's double sit experiment, sodium light composed of two wavelengths Î»1 and Î»2 close to each other ( with Î»2 greater than Î»1 ) is used. The order n, up to which the fringes can be seen on the screen, is given by

Options:
A n=λ2λ2-λ1
B n=λ1λ2-λ1
C n=λ22λ2-λ1
D n=λ12λ2-λ1

The correct answer is:
n=λ22λ2-λ1

Solution:
If an interference experiment is performed using two wavelengths close to each other, two interference patterns corresponding to the two wavelengths are obtained on the screen. The fringe system remains distinct upto a point on the screen where the nth order maximum of one wavelength, say Î»1=5890 AË falls on the nth order minimum of the order wavelength Î»2=5895 AË . Thus, interference pattern can be seen upto a distance yn from the centre of the screen if
yn=nÎ»1Dd; nth maximumâ¦â¦â¦..(i)
=n-12Î»2Dd;nth maximumâ¦â¦â¦(ii)
Or nÎ»1=n-12Î»2 or 2nÎ»1=2n-1Î»2
This gives,
n=Î»22Î»2-Î»1 , which is choice n=Î»22Î»2-Î»1

Topic: Physics, Wave Optics, JEE Main
If an iron (III) complex with the formula \(\left[\mathrm{Fe}\left(\mathrm{NH}_3\right)_x(\mathrm{CN})_y\right]^{-}\)has no electron in its \(e_g\) orbital, then the value of \(x+y\) is

Options:
A 4
B 5
C 6
D 3

The correct answer is:
6

Solution:
Complex is
$\begin{aligned} & x=2 \\ & y=4 \\ & \text { so } x+y=6\end{aligned}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (04 Apr Shift 2)
If an observer is walking away from the plane mirror with $6 \mathrm{~m} / \mathrm{sec}$. Then the velocity of the image with respect to observer will be

Options:
A $6 \mathrm{~m} / \mathrm{sec}$
B $-6 m /$ sec
C $12 \mathrm{~m} / \mathrm{sec}$
D $3 m / \sec$

The correct answer is:
$12 \mathrm{~m} / \mathrm{sec}$

Solution:
The velocity of the image with respect to the observer will be $12 \mathrm{~m} / \mathrm{s}$. This is because the velocity of the image with respect to the mirror is the same in magnitude as the velocity of the object (the observer) with respect to the mirror, but in the opposite direction. So, if an observer is walking away from the plane mirror with a speed of 6 $\mathrm{m} / \mathrm{s}$, the velocity of the image with respect to the observer will be the sum of the observer's velocity and the image's velocity, which is $6 \mathrm{~m} / \mathrm{s}+6 \mathrm{~m} / \mathrm{s}=12 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Ray Optics, JEE Main
If an unbiased dice is rolled thrice, then the probability of getting a greater number in the $i^{\text {th }}$ roll than the number obtained in the $(i-1)^{\text {th }}$ roll, $i=2,3$, is equal to

Options:
A $3 / 54$
B $2 / 54$
C $1 / 54$
D $5 / 54$

The correct answer is:
$5 / 54$

Solution:
Favourable cases $={ }^6 \mathrm{C}_3$
Total out comes $=6^3$
Probability of getting greater number than previous
$\text {one }=\frac{{ }^6 \mathrm{C}_3}{\mathrm{r}^3}=\frac{20}{216}=\frac{5}{54}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (09 Apr Shift 2)
If an unbiased die, marked with -2,-1,0,1,2,3 on its faces is thrown five times, then the probability that the product of the outcomes is positive, is :

Options:
A 8812592
B 5212592
C 4402592
D 27288

The correct answer is:
5212592

Solution:
The unbiased die is thrown five times such that the product of the outcomes is positive, then either all outcomes are positive or any two are negative.

Either all outcomes are positive or any two are negative.

Now,

p=p (positive) =36=12

q=p (negative) =26=13

Required probability

=C55125+C25132123+C45134121

=5212592

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If an uncharged sphere of metal is placed in the uniform electric field produced by two large oppositely charged plates, then the lines of force will look like

Options:
A
B
C
D

The correct answer is:

Solution:
Due to induction charges will appear on the surface of sphere as shown in diagram. Electric field lines never enters any conducting surface and are perpendicular at the surface.

Topic: Physics, Electrostatics, JEE Main
If $\alpha$ and $\beta$ are imaginary cube roots of units then value of $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$ is

Options:
A $1$
B $0$
C $-1$
D $2$

The correct answer is:
$0$

Solution:
Let $\alpha=\omega, \beta=\omega^2$
Now $\alpha^4+\beta^{28}+\frac{1}{\alpha \beta}$
$\begin{aligned} & =\omega^4+\left(\omega^2\right)^{28}+\frac{1}{\omega \omega^2} \\ & =\omega^4+\omega^{56}+\frac{1}{\omega^3} \\ & =\omega+\omega^2+\frac{1}{1} \\ & \left.=0 \quad \quad \quad \text { [as } \omega^3=1\right]\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\alpha$ and $\beta$ are imaginary cube roots of unity, then $\alpha^4+\beta^4+\frac{1}{\alpha \beta}=$

Options:
A $3$
B $0$
C $1$
D $2$

The correct answer is:
$0$

Solution:
Complex cube root of unity are $1, \omega, \omega^2$
Let $\alpha=\omega, \beta=\omega^2$; Then $\alpha^4+\beta^4+\alpha^{-1} \beta^{-1}$ $=\omega^4+\left(\omega^2\right)^4+\left(\omega^{-1}\right)\left(\omega^2\right)^{-1}=\omega+\omega^2+1=0$

Topic: Mathematics, Complex Number, JEE Main
If $\alpha, \beta$ and $\gamma$ are length of the altitudes of a $\triangle A B C$ with area $\Delta$, then $\frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$ is equal to

Options:
A $\sin ^2 A+\sin ^2 B+\sin ^2 C$
B $\cos ^2 A+\cos ^2 B+\cos ^2 C$
C $\tan ^2 A+\tan ^2 B+\tan ^2 C$
D $\cot ^2 A+\cot ^2 B+\cot ^2 C$

The correct answer is:
$\sin ^2 A+\sin ^2 B+\sin ^2 C$

Solution:
$\because$ Area of triangle $=\frac{1}{2}$ base $\times$ altitude
$\therefore \Delta=\frac{1}{2} a \alpha=\frac{1}{2} b \beta=\frac{1}{2} c \gamma$
$\Rightarrow \alpha=\frac{2 \Delta}{a}, \beta=\frac{2 \Delta}{b}$ and $\gamma=\frac{2 \Delta}{c}$
$\therefore \frac{\Delta^2}{R^2}\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac{1}{\gamma^2}\right)$
$=\frac{\Delta^2}{R^2}\left(\frac{a^2}{4 \Delta^2}+\frac{b^2}{4 \Delta^2}+\frac{c^2}{4 \Delta^2}\right)$
$\begin{aligned} & =\frac{1}{4 R^2}\left(4 R^2 \sin ^2 A+4 R^2 \sin ^2 B+4 R^2 \sin ^2 C\right) \\ & \qquad\left[\because \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2 R}\right] \\ & =\sin ^2 A+\sin ^2 B+\sin ^2 C\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\alpha$ and $\beta$ are roots of the equation, $x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ for some $k$, and $\alpha^2+\beta^2=66$, then $\alpha^3+\beta^3$ is equal to:

Options:
A
$248 \sqrt{2}$

B
$280 \sqrt{2}$

C
$-32 \sqrt{2}$

D
$-280 \sqrt{2}$

The correct answer is:
$-280 \sqrt{2}$

Solution:
$x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ or, $x^2-4 \sqrt{2} k x+2 k^4-1=0$ $\alpha+\beta=4 \sqrt{2} k$ and $\alpha \cdot \beta=2 k^4-1$ Squaring both sides, we get
$$
\begin{aligned}
&(\alpha+\beta)^2=(4 \sqrt{2} k)^2 \\
&\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=32 k^2 \\
&66+2 \alpha \beta=32 k^2 \\
&66+2\left(2 k^4-1\right)=32 k^2 \\
&66+4 k^4-2=32 k^2 \\
&\Rightarrow 4 k^4-32 k^2+64=0 \\
&\text { or, } k^4-8 k^2+16=0 \\
&\Rightarrow\left(k^2\right)^2-8 k^2+16=0 \\
&\Rightarrow\left(k^2-4\right)\left(k^2-4\right)=0 \\
&\Rightarrow k^2=4, k^2=4 \\
&\Rightarrow k=\pm 2
\end{aligned}
$$
Now, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)$
$$
\therefore \alpha^3+\beta^3=(4 \sqrt{2} k)\left[66-\left(2 k^4-1\right)\right]
$$
Putting $k=-2,(k=+2$ cannot be taken because it does not satisfy the above equation)
$\therefore \alpha^3+\beta^3=(4 \sqrt{2}(-2))\left[66-2(-2)^4-1\right]$
$\alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)=$
$(-8 \sqrt{2})(35)$
$\therefore \alpha^3+\beta^3=-280 \sqrt{2}$

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha$ and $\beta$ are roots of the equation, $x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ for some $k$, and $\alpha^2+\beta^2=66$, then $\alpha^3+\beta^3$ is equal to:

Options:
A
$248 \sqrt{2}$

B
$280 \sqrt{2}$

C
$-32 \sqrt{2}$

D
$-280 \sqrt{2}$

The correct answer is:
$-280 \sqrt{2}$

Solution:
$x^2-4 \sqrt{2} k x+2 e^{4 \ln k}-1=0$ or, $x^2-4 \sqrt{2} k x+2 k^4-1=0$ $\alpha+\beta=4 \sqrt{2} k$ and $\alpha \cdot \beta=2 k^4-1$ Squaring both sides, we get
$$
\begin{aligned}
&(\alpha+\beta)^2=(4 \sqrt{2} k)^2 \\
&\Rightarrow \alpha^2+\beta^2+2 \alpha \beta=32 k^2 \\
&66+2 \alpha \beta=32 k^2 \\
&66+2\left(2 k^4-1\right)=32 k^2 \\
&66+4 k^4-2=32 k^2 \\
&\Rightarrow 4 k^4-32 k^2+64=0 \\
&\text { or, } k^4-8 k^2+16=0 \\
&\Rightarrow\left(k^2\right)^2-8 k^2+16=0 \\
&\Rightarrow\left(k^2-4\right)\left(k^2-4\right)=0 \\
&\Rightarrow k^2=4, k^2=4 \\
&\Rightarrow k=\pm 2
\end{aligned}
$$
Now, $\alpha^3+\beta^3=(\alpha+\beta)\left(\alpha^2+\beta^2-\alpha \beta\right)$
$$
\therefore \alpha^3+\beta^3=(4 \sqrt{2} k)\left[66-\left(2 k^4-1\right)\right]
$$
Putting $k=-2,(k=+2$ cannot be taken because it does not satisfy the above equation)
$\therefore \alpha^3+\beta^3=(4 \sqrt{2}(-2))\left[66-2(-2)^4-1\right]$
$\alpha^3+\beta^3=(-8 \sqrt{2})(66-32+1)=$
$(-8 \sqrt{2})(35)$
$\therefore \alpha^3+\beta^3=-280 \sqrt{2}$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2014 (11 Apr Online)
If Î± and Î² are roots of the equation x2+5x-6=0 then the value of tan-1â¡Î±-tan-1â¡Î² is

Options:
A π2
B 0
C π
D π4

The correct answer is:
π2

Solution:
x2+5x-6=0
x2+5x-6=0
x2+6x-x-6=0
xx+6-1x+6=0
x-1x+6=0
x=1, nbsp;xâ -6 (Since modulus can not be giving negative values)
â´ x=1
â´ x=Â± 1
Î±=1,Î²=-1
â´ tan-1â¡Î±-tan-1â¡Î²=tan-1â¡1-tan-1â¡-1
=Ï4--Ï4
=Ï2 = π 2

Topic: Mathematics, Quadratic Equation, JEE Main
If $\alpha$ and $\beta$ are roots of the equation $x^2+p x+\frac{3 p}{4}=0$, such that $|\alpha-\beta|=\sqrt{10}$, then $p$ belongs to the set :

Options:
A
$\{2,-5\}$

B
$\{-3,2\}$

C
$\{-2,5\}$

D
$\{3,-5\}$

The correct answer is:
$\{-2,5\}$

Solution:
Given quadratic eqn. is
$$
x^2+p x+\frac{3 p}{4}=0
$$
So, $\alpha+\beta=-p, \alpha \beta=\frac{3 p}{4}$
Now, given $|\alpha-\beta|=\sqrt{10}$
$$
\begin{aligned}
& \Rightarrow \alpha-\beta=\pm \sqrt{10} \\
& \Rightarrow(\alpha-\beta)^2=10 \Rightarrow \alpha^2+\beta^2-2 \alpha \beta=10 \\
& \Rightarrow(\alpha+\beta)^2-4 \alpha \beta=10 \\
& \Rightarrow p^2-4 \times \frac{3 p}{4}=10 \Rightarrow p^2-3 p-10=0 \\
& \Rightarrow p=-2,5 \Rightarrow p \in\{-2,5\}
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2013 (22 Apr Online)
If $\alpha$ and $\beta$ are roots of $x^{2}-x+1=0,$ then the value of $\alpha^{2013}+\beta^{2013}$ is

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

The correct answer is:
-2

Solution:
Given equation is $x^{2}-x+1=0$
$$
\begin{aligned}
x &=\frac{1 \pm \sqrt{1-4}}{2} \quad\left(\because x=\frac{b \pm \sqrt{b^{2}-4 a c}}{2 a}\right) \\
&=\frac{1 \pm i \sqrt{3}}{2}=\frac{1+i \sqrt{3}}{2}, \frac{1-i \sqrt{3}}{2} \\
\Rightarrow-x &=\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2} \\
\Rightarrow+x &=\omega,-\omega^{2}
\end{aligned}
$$
Since, $(\alpha, \beta)$ are the roots of given equation. Then, $\alpha=-\omega$ and $\beta=-\omega^{2}$
$$
\begin{aligned}
\therefore \alpha^{2013}+\beta^{2013} &=-(\omega)^{2013}+\left(-\omega^{2}\right)^{2013} \\
&=-\omega^{2013}-\omega^{4026} \\
&=-\left(\omega^{3}\right)^{671}-\left(\omega^{3}\right)^{1342} \\
&=-(1)^{671}-(1)^{1342} \\
&=-1-1=-2
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main
If α and β are such that the function f(x) defined by fx=αx2-β, for |x|<1-1|x|, for |x|≥1 is differentiable everywhere then the ordered pair (α,β)=

Options:
A -12,-32
B 12,-32
C 12,32
D -12,32

The correct answer is:
12,32

Solution:
It is given that the function is differentiable at all points,
By the first principle,

LHDat x=1=RHDat x=1

limx→1-f(x)-f(1)x-1=limx→1+f(x)-f(1)x-1

limx→1-αx2-β-α+βx-1=limx→1+-1x+1x-1

limx→1-αx2-1x-1=limx→1+x-1xx-1

limx→1-α(x+1)(x-1)x-1=limx→1+1x

limx→1-αx+1=limx→1+1x

α(1+1)=1

α=1/2

Every differentiable function is also continuous,

RHL=LHL

limx→1-fx=limx→1+fx

12(1)2-β=-11

β=12+1

β=3/2

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $\alpha$ and $\beta$ are the complex cube roots of unity, then $\alpha^3+\beta^3+\alpha^{-2} \times \beta^{-2}$ is equal to

Options:
A $1$
B $-3$
C $3$
D $0$

The correct answer is:
$3$

Solution:
$\begin{aligned} & \alpha^3+\beta^3+\alpha^{-2} \cdot \beta^{-2}=\alpha^3+\beta^3+\frac{1}{\alpha^2 \beta^2} \\ & =\omega^3+\omega^6+\frac{1}{\omega^2 \omega^4}=1+1+1=3\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $\alpha$ and $\beta$ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is

Options:
A $\frac{\cos \alpha+\cos \beta}{\cos (\alpha-\beta)}$
B $\frac{\sin (\alpha-\beta)}{\sin \alpha-\sin \beta}$
C $\frac{\cos \alpha-\cos \beta}{\cos \cos (\alpha-\beta)}$
D $\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}$

The correct answer is:
$\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}$

Solution:
Let $\mathrm{a}, \mathrm{b}$ are the length of semi-major axis and semi-minor axis respectively, then $(a \cos \alpha, b \sin \alpha),(a \cos \beta, b \sin \beta),(a e, 0)$ are collinear.
$\frac{b(\sin \alpha-\sin \beta)}{a(\cos \beta-\cos \cos \alpha)}=\frac{b \sin \alpha-0}{a \cos \alpha-a e}$
$\begin{aligned} & (\cos \alpha-e)(\sin \beta-\sin \alpha)=\sin \alpha(\cos \beta-\cos \alpha) \end{aligned}$
$\begin{aligned} & e=\frac{\cos \alpha(\sin \beta-\sin \alpha)-\sin \alpha(\cos \beta-\cos \alpha)}{\sin \beta-\sin \alpha} \\ & e=\frac{\sin \sin (\alpha-\beta)}{\sin \alpha-\sin \beta} \\ & e=\frac{\sin (\alpha-\beta)(\sin \alpha+\sin \beta)}{\sin ^2 \alpha-\sin ^2 \beta}\end{aligned}$
$\therefore e=\frac{\sin \alpha+\sin \beta}{\sin (\alpha+\beta)}$ $\left[\because \sin \sin (A+B) \cdot \sin \sin (A-B)=\sin ^2 A-\sin ^2 B\right]$

Topic: Mathematics, Ellipse, JEE Main
If Î± and Î² are the roots of 4x2-16x+t=0, ât>0 such that 1<Î±<2<Î²<3, then the number of integral values of t are

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
fx=4x2-16x+t
Opening upward parabola
f1>0, f2<0, f3>0
f1>0â4-16+t>0ât>12
f2<0â16-32+t<0ât<16
f3>0â36-48+t>0ât>12
â12<t<16ât=13, 14, 15
Hence, the number of integral values t=13,14,15 is 3.

Topic: Mathematics, Quadratic Equation, JEE Main
If α and β are the roots of the equation 1 513-472719-319419119413-472x2-5x+5-3=-4, then the value of 2-α2-β is

Options:
A 51
B -12
C 12
D -7

The correct answer is:
-12

Solution:
Let, A=1-437 then A-1=719-319419119
Since, AA-1=I=A-1A
Hence, 13-47719-319419119=I=719-31941911913-47
⇒15x2-5x+5-3=-4
⇒x2+5x+5-15=-4
⇒x2-5x-6=0⇒x=6,-1
Hence, 2-α2-β=-4×3=-12

Topic: Mathematics, Matrices, JEE Main
If Î± and Î² are the roots of the equation 2x2+4x-5=0, then the equation whose roots are 12Î±-3 and 12Î²-3 is

Options:
A x2+10x-11=0
B 11x2+10x+1=0
C x2+10x+11=0
D 11x2-10x+1=0

The correct answer is:
11x2+10x+1=0

Solution:
Substituting y=12Î±-3, we get,

2Î±=1y+3âÎ±=121y+3
Since, Î± is a root of the given equation
â2121y+32+4121y+3-5=0
â21+3y4y22+41+3y2y-5=0
â1+3y2+4y1+3y-5Ã2y2=0
â1+9y2+6y+4y+12y2-10y2=0
â11y2+10y+1=0
Hence, the required equation is 11x2+10x+1=0

Topic: Mathematics, Quadratic Equation, JEE Main
If α and β are the roots of the equation 2x2x+1=1, then β is equal to :

Options:
A 2αα+1
B -2αα+1
C 2αα-1
D 2α2

The correct answer is:
-2αα+1

Solution:
Given equation is

2x2x+1=1⇒4x2+2x-1=0 .........(1)

roots of equation 1

are α and β

∴α+β=-12

⇒β=-12-α ........(2)

and

4α2+2α-1=0⇒α2=14-α2 .........(3)

now

-2αα+1=-2α2-2a

=-214-α2-2α=-12-α=β

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If α and β are the roots of the equation, 7x2-3x-2=0, then the value ofα1-α2+β1-β2 is equal to:

Options:
A 2732
B 124
C 38
D 2716

The correct answer is:
2716

Solution:
Given α and β are the roots of the equation,7x2-3x-2=0, then

α+β=37 ; αβ=-27

Now

α1-α2+β1-β2=(α+β)-αβ(α+β)1-α21-β2=(α+β)-αβ(α+β)1+(αβ)2-α2+β2

⇒(α+β)-αβ(α+β)1+(αβ)2-(α+β)2+2αβ=37+27371+272-372-227=2716

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If α and β are the roots of the equation 8x2-3x+27=0, then the value of α2β13+β2α13 is

Options:
A 13
B 14
C 15
D 16

The correct answer is:
14

Solution:
α+β=38,αβ=278
α2β13+β2α13=α313+β313αβ13
α+βαβ13=3827813=3832=14

Topic: Mathematics, Quadratic Equation, JEE Main
If α and β are the roots of the equation 375 x2-25x-2=0, then limn→∞ ∑r=1nαr+limn→∞ ∑r=1nβr is equal to:

Options:
A 112
B 21346
C 7116
D 29358

The correct answer is:
112

Solution:
For given quadratic 375x2-25x-2=0, its roots are,

α,β=25±252+2×4×3752×375⇒α<1,β<1

Also, α+β=25375, αβ=-2375

Now limn→∞∑r=1nαr+limn→∞∑r=1nβr

=α1-α+β1-β a+ar+ar2+..... infinite  terms=a1-r    if  r<1

=α+β-2αβ1-α+β+αβ=25375+43751-25375-2375=29348=112

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2019 (12 Apr Shift 1)
If α, β and γ are the roots of the equation px3+qx2+r=0, then the value of the determinant αββγγαβγγααβγααββγ is

Options:
A pq
B qr
C 0
D pr

The correct answer is:
0

Solution:
The roots of the equation px3+qx2+r=0 are α,β,γ
i.e. αβ+βγ+γα=0
∴αββγγαβγγααβγααββγ

Applying C1→C1+C2+C3

αβ+βγ+γαβγγααβ+βγ+γαγααβαβ+βγ+γααββγ=0

Topic: Mathematics, Determinants, JEE Main
If Î± and Î² are the roots of the equation x2-2x+3=0, then the sum of roots of the equation having roots as Î±3-3Î±2+5Î±-2 and Î²3-Î²2+Î²+5 is

Options:
A 1
B 3
C 5
D 7

The correct answer is:
3

Solution:
Since, Î±2-2Î±+3=0
âÎ±3-3Î±2+5Î±-2=Î±Î±2-2Î±+3-Î±2-2Î±+3+1=1
And Î²2-2Î²+3=0
âÎ²3-Î²2+Î²+5=Î²Î²2-2Î²+3+Î²2-2Î²+3+2=2
So, the required equation is
x2-2+1x+2.1=0
âx2-3x+2=0
Sum of roots =3

Topic: Mathematics, Quadratic Equation, JEE Main
If α and β are the roots of the equation x2+px+2=0 and 1α and 1β are the roots of the equation 2x2+2qx+1=0, then α-1αβ-1βα+1ββ+1α is equal to :

Options:
A 949+q2
B 949-q2
C 949+p2
D 949-p2

The correct answer is:
949-p2

Solution:
α, β are roots of x2+px+2=0

⇒α+β=-p, αβ=2

1α,1β are roots of 2x2+2qx+1=0

∴ α, β are roots of x2+2qx+2=0

⇒α+β=-2q, αβ=2

∴p=2q ...(i)

⇒α-1αβ-1βα+1ββ+1α=αβ-αβ-βα+1αβ·αβ+1αβ+2

⇒2-α+β2-2αβαβ+12·2+12+2

⇒52-p2-42·92=949-p2 or, 949-4q2

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (03 Sep Shift 1)