datavorous/entrance-exam-dataset huggingface dataset:
The system of equations $$ \begin{array}{l} \lambda x+y+3 z=0 \\ 2 x+\mu y-z=0 \\ 5 x+7 y+z=0 \end{array} $$ has infinitely many solutions in $R$. Then,

Options:
A $\lambda=2, \mu=3$
B $\lambda=1, \mu=2$
C $\lambda=1, \mu=3$
D $\lambda=3, \mu=1$

The correct answer is:
$\lambda=1, \mu=3$

Solution:
Given, system of equations
$$
\begin{array}{l}
\lambda x+y+3 z=0 \\
2 x+\mu y-z=0 \\
5 x+7 y+z=0
\end{array}
$$
System has infinitely many solutions in $R$, if
$$
\left|\begin{array}{ccc}
\lambda & 1 & 3 \\
2 & \mu & -1 \\
5 & 7 & 1
\end{array}\right|=0
$$
$\Rightarrow \quad \lambda(\mu+7)-1(2+5)+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda(\mu+7)-7+3(14-5 \mu)=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-7+42-15 \mu=0$
$\Rightarrow \quad \lambda \mu+7 \lambda-15 \mu+35=0$
By checking the options, we get option (c) $(\lambda=1, \mu=3)$ satisfies the given equation.

Topic: Mathematics, Determinants, JEE Main
The system of equations
$$
\begin{aligned}
& \alpha x+y+z=\alpha-1 \\
& x+\alpha y+z=\alpha-1 \\
& x+y+\alpha z=\alpha-1
\end{aligned}
$$
has no solution, if $\alpha$ is

Options:
A
$-2$

B
either $-2$ or 1

C
not -2

D
1

The correct answer is:
$-2$

Solution:
$$
\begin{aligned}
& \alpha x+y+z=\alpha-1 \\
& x+\alpha y+z=\alpha-1 \\
& x+y+z \alpha=\alpha-1 \\
& \Delta=\left|\begin{array}{ccc}
\alpha & 1 & 1 \\
1 & \alpha & 1 \\
1 & 1 & \alpha
\end{array}\right| \\
& =\alpha\left(\alpha^2-1\right)-1(\alpha-1)+1(1-\alpha) \\
& =\alpha(\alpha-1)(\alpha+1)-1(\alpha-1)-1(\alpha-1) \\
& \Rightarrow(\alpha-1)\left[\alpha^2+\alpha-1-1\right]=0 \\
& \Rightarrow(\alpha-1)\left[\alpha^2+\alpha-2\right]=0 \\
& {\left[\alpha^2+2 \alpha-\alpha-2\right]=0} \\
& (\alpha-1)[\alpha(\alpha+2)-1(\alpha+2)]=0 \\
& (\alpha-1)=0, \alpha+2=0 \Rightarrow \alpha=-2,1 ; \text { but } \alpha \neq 1
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2005
The system of equations
$\begin{aligned} & \alpha x+y+z=\alpha-1 \\ & x+\alpha y+z=\alpha-1 \\ & x+y+\alpha z=\alpha-1\end{aligned}$
has no solution, if $\alpha$ is

Options:
A Not - 2
B 1
C -2
D Either -2 or 1

The correct answer is:
-2

Solution:
For no solution or infinitely many solutions $\left|\begin{array}{lll}\alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha\end{array}\right|=0 \Rightarrow \alpha=1, \alpha=-2$
But for $\alpha=1$, clearly there are infinitely many solutions and when we put $\alpha=-2$ in given system of equations and adding them together L.H.S $\neq$ R.H.S. i.e., No solution.

Topic: Mathematics, Determinants, JEE Main
The system of equations \( x+k y+3 z=0,3 x+k y-2 z=0,2 x+3 y-4 z=0 \) possess a non-trivial solution over the set of rationals, then \( 2 \mathrm{k} \), is an integral element of the interval:

Options:
A \( [10,20] \)
B \( (20,30) \)
C \( [30,40] \)
D \( (40,50) \)

The correct answer is:
\( [30,40] \)

Solution:
For the given system to have a non-trivial solution, we must have
1k33k-223-4 = 0⇒ k=332

Topic: Mathematics, Determinants, JEE Main
The system of linear equations

3x-2y-kz=10

2x-4y-2z=6

x+2y-z=5 m
is inconsistent if :

Options:
A k=3, m≠45
B k=3, m=45
C k≠3, m∈R
D k≠3, m≠45

The correct answer is:
k=3, m≠45

Solution:
For inconsistent system of equation the conditions should be ∆=0 and at least one of the ∆x or ∆y or ∆z should not be equal to zero.

Now,

Δ=3-2-k2-4-212-1=0

⇒24+20-k8=0⇒k=3

And,

Δx=10-2-36-4-25 m2-1

⇒∆x=108-2-10 m+6-312+20 m

⇒∆x=84-5m

∆x≠0⇒m≠45

Δy=310-326-215 m-1

=3-6+10 m+100-310 m-6

=0

Δz=3-2102-46125m

⇒Δz=3-20m-12+210m-6+104+4

⇒Δz=-60m-36+20m-12+80

⇒Δz=-40m+32

⇒∆z=84-5m

∆z≠0⇒m≠45

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The system of linear equations
λx+2y+2z=5
2λx+3y+5z=8
4x+λy+6z=10 has

Options:
A no solution when λ=8
B a unique solution when λ=-8
C no solution when λ=2
D infinitely many solutions when λ=2

The correct answer is:
no solution when λ=2

Solution:
∆=λ222λ354λ6=λ18-5λ-212λ-20+22λ2-12=-λ2-6λ+16=λ+82-λ
∆=0 for λ=2,
∆1=5228351026

∆1=518-10-248-50+216-30

∆1=40+4-28≠0
No solution for λ=2.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The system of linear equations : $x+y+z=0,2 x$ $+y-z=0,3 x+2 y=0$ has :

Options:
A no solution
B a unique solution
C an infinitely many solution
D None of these

The correct answer is:
an infinitely many solution

Solution:
The system is homogenuous system. $\therefore$ it has either unique solution or infinite many solution depend on $|\mathrm{A}|$
$$
\begin{aligned}
\therefore|A| &=\left|\begin{array}{rrr}
1 & 1 & 1 \\
2 & 1 & -1 \\
3 & 2 & 0
\end{array}\right|=2 \times 1+1(-3)+(4-3) \\
&=2-3+1=0
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main
The system of linear equations
x+λy-z=0
λx-y-z=0
x+y-λz=0
has a non -trivial solution for

Options:
A Exactly two values of λ
B Exactly three values of λ
C Infinitely many values of λ
D Exactly one value of λ

The correct answer is:
Exactly three values of λ

Solution:
For non - trivial solution.
Δ=1λ-1λ-1-111-λ=0
1λ+1-λ-λ2+1-1λ+1=0
λ+1-λ1-λ 1+λ-λ+1=0
λ1-λ 1+λ=0
λ=0, λ=1, λ= -1
Exactly there value of λ.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2016 (03 Apr)
The system of linear equations
x+y+z=2
2x+3y+2z=5
2x+3y+a2-1z=a+1

Options:
A is inconsistent when a=3
B has a unique solution for a=3
C has infinitely many solutions for a=4
D is inconsistent when a=4

The correct answer is:
is inconsistent when a=3

Solution:
Here, D=11123223a2-1
=a2-3
Hence at |a|=3 , D=0 . So system don't have unique solution.
At a=4 , D≠0 , So, system has unique solution. Hence not inconsistent.
At |a|=3,D=0 and 2nd & 3rd equation are parallel. Hence no solution are possible.
Hence, system is inconsistent for |a|=3

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The system of linear equations $\lambda x+y+z=3$
$x-y-2 z=6$
$-x+y+z=\mu$ has

Options:
A infinite number of solutions for $\lambda \neq-1$ and all $\mu$
B infinite number of solutions for $\lambda=-1$ and $\mu=3$
C no solution for $\lambda, x-1$
D unique solution for $\lambda=-1$ and $\mu=3$

The correct answer is:
infinite number of solutions for $\lambda=-1$ and $\mu=3$

Solution:
Augmented matrix [A:B] = $\left[\begin{array}{rrrr}1 & -1 & -2 & 6 \\ -1 & 1 & 1 & \mu \\ \lambda & 1 & 1 & 3\end{array}\right]$
$\sim\left[\begin{array}{cccc}1 & -1 & -2 & 6 \\ 0 & 0 & -1 & \mu+6 \\ \lambda+1 & 0 & -1 & 9\end{array}\right]$
Applying
$$
-\left[\begin{array}{cccc}1 & -1 & -2 & R_{2}+R_{1}, R_{3} \rightarrow R_{3}+R_{1} \\ 0 & 0 & -1 & \mu+6 \\ \lambda+1 & 0 & 0 & 3-\mu\end{array}\right]
$$
Infinite number of solutions for $\lambda=-1$ and $\mu=3$

Topic: Mathematics, Determinants, JEE Main
The system shown in the adjacent figure is in equilibrium. The mass of the container with liquid is M, the density of the liquid in the container is Ï and volume of the block is V. If the container is now displaced downwards through a distance x0 and released such that block remains well inside the liquid, during subsequent motion (ignore viscosity)

Options:
A Time period of SHM of the container will be 2πM+2ρvk
B Time period of SHM of the container will be 2πM+ρvk
C The amplitude of SHM of the container will be x02
D Container will not perform SHM

The correct answer is:
Time period of SHM of the container will be 2πM+ρvk

Solution:
Mg-kx+Îx+ÏV g-a=Ma
Where Mg+ÏVg=KÎx
â M+ÏVa=-Kx
Or a-kxM+ÏV or Ï= -kM+ÏV
â T=2ÏM+ÏVK
The amplitude of SHM will be x0

Topic: Physics, Oscillations, JEE Main
The system shown in the figure is in equilibrium and all the blocks are at rest. Assume that the masses of the strings, the pulley and the spring are negligible with respect to the masses of the blocks and friction is absent. Find the acceleration of block C, just after cutting the string 1. [g=10 m s-2]

Options:
A 5 ms-2
B 3 ms-2
C 2 ms-2
D 6 ms-2

The correct answer is:
5 ms-2

Solution:
Just after cutting the string 1 the extension of the string will remain same as it was before cutting the string '1'. So, we can take string force on block is equal '2g'.
Now making F.B.D of block 'B'.
3g-T=1a .... (i)
Making F.B.D of block 'c'
T=5a .... (ii)
From (i) and (ii)
a=g2=5 m s-2

Topic: Physics, Laws of Motion, JEE Main
The system that contains the maximum number of atoms is

Options:
A $4.25 \mathrm{g}$ of $\mathrm{NH}_{3}$
B $8 \mathrm{g}$ of $\mathrm{O}_{2}$
C $2 \mathrm{g}$ of $\mathrm{H}_{2}$
D $4 \mathrm{g}$ of He

The correct answer is:
$2 \mathrm{g}$ of $\mathrm{H}_{2}$

Solution:
Number of atoms $=\frac{\text { mass }}{\text { molar mass }} \times N_{A}$ x Number of atoms in 1 mole
$\therefore$ Number of atoms in $4.25 \mathrm{g} \mathrm{NH}_{3}$
$$
\begin{array}{l}
=\frac{4.25}{17} \times N_{A} \times 4 \\
=N_{A}
\end{array}
$$
Number of atoms in $8 \mathrm{g}$
$$
\mathrm{O}_{2}=\frac{8}{32} \times \mathrm{N}_{A} \times 2=\frac{N_{\mathrm{A}}}{2}
$$
Number of atoms in $2 \mathrm{g}$
$$
\mathrm{H}_{2}=\frac{2}{2} \times \mathrm{N}_{A} \times 2=2 \mathrm{N}_{\mathrm{A}}
$$
Number of atoms in $4 \mathrm{g}$
$$
\mathrm{He}=\frac{4}{4} \times N_{A} \times 1=N_{A}
$$
Thus, $2 \mathrm{g} \mathrm{H}_{2}$ contains the maximum number of atoms among the given.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The table below gives the populations (in thousands) of ten species $(\mathrm{A}-\mathrm{J})$ in four areas $(a-d)$ consisting of the number of habitats given within brackets against each. Study the table and answer the question which follows
Which area out of a to $\mathrm{d}$ shows maximum species diversity?

Options:
A a
B $b$
C c
$
D \mathrm{d}$

The correct answer is:
\mathrm{d}$

Solution:
In areas ' $\mathrm{c}$ ' and ' $\mathrm{d}$ ' all the 10 species are present but number of habitats is lesser in area $\mathrm{d}$.

Topic: Biology, Organisms and Populations, NEET, NEET 2008 (Mains)
The table below gives the results of three titrations carried out with 0.200 M HCl to determine the molarity of a given NaOH solution using phenolphthalein as indicator. NaOH was taken in the burette and HCl was taken in a conical flask for the titrations

The actual molarity of the prepared NaOH solution was 0.220 mol dm-3. Which among the following could be the reason for the wrong value obtained in titration III?

Options:
A Number of drops of phenolphthalein added to the titration flask was more in this titration
B The concentration of HCl was wrongly used as 0.250 M for the calculation of M NaOH
C A few drops of NaOH solution were spilled outside the titration flask during titration
D A few drops of the neutralized solution from titration II were left behind in the flask

The correct answer is:
A few drops of NaOH solution were spilled outside the titration flask during titration

Solution:
OH-=nOH-VOH-=nacidVNaOH=HClVHClVNaOH=0.21M<0.220

Topic: Chemistry, Redox Reactions, JEE Main
The table gives the first four ionization energies in kJ mol-1 of four elements (the letters are not the symbols for the elements). Which element occurs in Group 13 of the periodic table?

Options:
A IE1IE2Element P5024569 IE3IE469199550
B IE1IE2Element Q5267305 IE3IE411822-
C IE1IE2Element R5841823 IE3IE4275111584
D IE1IE2Element S7961583 IE3IE432384362

The correct answer is:
IE1IE2Element R5841823 IE3IE4275111584

Solution:
After losing three electrons in element R, there is sudden huge increase in IE from 2751 to 11584 kJ/mol. It means that after three electrons, some stable configuration is achieved. This is applicable to group 13 elements with outer configuration of ns2np1 .

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The tangent and normal to the ellipse 3x2+5y2=32 at the point P2, 2 meet the x-axis at Q and R, respectively. Then the area (in sq. units) of the triangle PQR is:

Options:
A 6815
B 163
C 143
D 3415

The correct answer is:
6815

Solution:
The equation of tangent and normal to the ellipse x2a2+y2b2=1 at a point x1, y1 are respectively xx1a2+yy1b2=1 and a2xx1-b2yy1=a2-b2.

The given ellipse is 3x2+5y2=32, ⇒3x232+5y232=1.

Hence, the tangent and normal to the ellipse 3x232+5y232=1 at 2, 2 are respectively 3×2×x32+5×2×y32=1 and 32×x3×2-32×y5×2=323-325

Hence, the tangent is 3x+5y=16 ...1

And, the normal is 5x-3y=4 ...2

The point where the tangent meets the x-axis can be obtained by putting y=0, ⇒3x+0=16 ⇒x=163, thus the point Q≡163, 0.

Similarly, the point where the normal meets the x-axis, is x=45, thus the point R≡43, 0.

Tangent and normal intersect at P2, 2, and the length of perpendicular from any point on x-axis is the absolute value of its y-co-ordinate, hence the height of the triangle is 2 units.

And, hence the required area is =12QR×h

=12×163-45×2=6815 sq units.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The tangent and the normal lines at the point 3, 1 to the circle x2+y2=4 and the x -axis form a triangle. The area of this triangle (in square units) is:

Options:
A 13
B 23
C 43
D 13

The correct answer is:
23

Solution:
Equation tangent at P3,  1
x⋅3+y⋅1=4
x-intercept of this tangent is 43.
Hence, area of Δ=12⋅43⋅1=23

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola x⁡ 2 4 - y⁡ 2 5 = 1 , meets the x-axis and y-axis at A and B, respectively. Then OA2-OB2, where O is the origin, equals

Options:
A - 2 0 9
B 1 6 9
C 4
D - 4 3

The correct answer is:
- 2 0 9

Solution:
Hyperbola, x2a2-y2b2=1 ⇒x24-y25=1

∴ a2=4, b2=5

b2=a2e2-1 ⇒5=4e2-1

⇒e2=54+1=94

⇒e=32

∴ Extremity of LR in first quadrant ≡L≡ae,b2a

≡3, 52

Equation of tangent to the hyperbola x2a2-y2b2=1 at x1, y1 is x⁡x⁡1a2-y⁡y⁡1b2=1

⇒x⁡·34-y·525=1

⇒3x⁡4-y⁡2=1

To find the point A on x--axis, put y=0

⇒ x⁡ = 4 3

⇒OA=43

To find the point B on y-axis, put x=0

⇒ y⁡ = - 2

⇒OB=2

Now, OA2-OB2=169-4=-209.

Topic: Mathematics, Hyperbola, JEE Main
The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola x⁡ 2 4 - y⁡ 2 5 = 1 , meets the x-axis and y-axis at A and B, respectively. Then OA2-OB2, where O is the origin, equals

Options:
A - 2 0 9
B 1 6 9
C 4
D - 4 3

The correct answer is:
- 2 0 9

Solution:
Hyperbola, x2a2-y2b2=1 ⇒x24-y25=1

∴ a2=4, b2=5

b2=a2e2-1 ⇒5=4e2-1

⇒e2=54+1=94

⇒e=32

∴ Extremity of LR in first quadrant ≡L≡ae,b2a

≡3, 52

Equation of tangent to the hyperbola x2a2-y2b2=1 at x1, y1 is x⁡x⁡1a2-y⁡y⁡1b2=1

⇒x⁡·34-y·525=1

⇒3x⁡4-y⁡2=1

To find the point A on x--axis, put y=0

⇒ x⁡ = 4 3

⇒OA=43

To find the point B on y-axis, put x=0

⇒ y⁡ = - 2

⇒OB=2

Now, OA2-OB2=169-4=-209.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2014 (19 Apr Online)
The tangent at any point on the curve xy=4 makes intercepts on the coordinate axes as a and b. Then the value of ab is

Options:
A 8
B 16
C 32
D 64

The correct answer is:
16

Solution:
Let Ph,k be a point on xy=4
âhk=4 â¦ 1
Differentiating xy=4 w.r.t. x, we get,
xdydx+y=0âdydx=-yx
The slope of the tangent at Ph,k=-kh
So, the equation of the tangent is
kx+hy=2hk
âkx+hy=8 (from (1))
x-intercept =-8k, y-intercept =-8h
Hence, ab=64hk=644=16

Topic: Mathematics, Application of Derivatives, JEE Main
The tangent at the point (2,-2) to the curve, x2y2-2x=41-y does not pass through the point:

Options:
A -2, -7
B 8,Â 5
C -4,Â -9
D 4,13

The correct answer is:
-2, -7

Solution:
Given curve x2y2-2x=4-4y

Differentiate both sides w.r.t x,
⇒ 2xy2+2y.x2.dydx-2=-4.dydx
⇒ dydx2y.x2+4=2-2x.y2
⇒ dydx2,-2=2-2×2×42-2×4+4=1412=76
Equation of tangent, y+2=76x-2

⇒7x-6y=26, which doesn't pass through -2-7.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2017 (08 Apr Online)
The tangent drawn at any point $P$ to the parabola $y^2=4 a x$ meets the directrix at the point $K$, then the angle which $K P$ subtends at its focus is

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Here, $P\left(a t^2, 2 a t\right)$ and $S(a, 0)$.
If the tangent at $P, t y=x+a t^2$, meets the directrix
$x=-a$ at $k$, then $k=\left(-a, \frac{a t^2-a}{t}\right)$
$m_1=$ slope of $S P-\frac{2 a t}{a\left(t^2-1\right)}$
$m_2=$ slope of $S K=\frac{a\left(t^2-1\right)}{-2 a t}$
Clearly $m_1 m_2=-1, \quad \therefore \angle P S K=90^{\circ}$.

Topic: Mathematics, Parabola, JEE Main
The tangent drawn to the hyperbola x216-y29=1, at point P in the first quadrant whose abscissa is 5, meets the lines 3x-4y=0 and 3x+4y=0 at Q and R respectively. If O is the origin, then the area of triangle OQR is (in square units)

Options:
A 6
B 12
C 3
D 24

The correct answer is:
12

Solution:
Let point P be 5,Î² since it lies on given hyperbola 2516-Î²29=1âÎ²2=8116
Hence, Î²=94 (as P lies in first quadrant)
So, equation of tangent is 5x16-9y4â9=1â5x-4y=16
So, points Q and R are 8,6 and 2,-32
Hence, the area of triangle OQR is 1212+12=12 sq. units

Topic: Mathematics, Hyperbola, JEE Main
The tangent lines to the curve $y^{2}=4 a x$ at points where $x=a$, are

Options:
A parallel
B perpendicular
C inclined at $60^{\circ}$
D inclined at $30^{\circ}$

The correct answer is:
perpendicular

Solution:
The given equation of the curve is $y^{2}=4 a x$
Differentiating both sides of $(1)$ with respect to $\mathrm{x}$, we get
$$
2 y \frac{d y}{d x}=4 a ; \Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}
$$
If $\psi$ be the angle which the tangent to the curve at ( $\mathrm{x}, \mathrm{y})$ makes with the positive direction of $x$-axis then $\tan \psi=\frac{d y}{d x}$ or
$$
\tan \psi=\frac{2 \mathrm{a}}{\mathrm{y}} \quad \ldots .(3), \quad[\mathrm{using}(2)]
$$
At $x=a$, then from $(1), y^{2}=4 a \cdot a=4 a^{2} \Rightarrow y$ $=\pm 2 \mathrm{a}$

Hence, we get two points $(\mathrm{a}, 2 \mathrm{a})$ and $(\mathrm{a},-$ 2a) on the curve.
$\mathrm{At}(\mathrm{a}, 2 \mathrm{a}) \mathrm{x}=\mathrm{a}, \mathrm{y}=2 \mathrm{a}$ and let $\psi=\psi_{1}$
$$
\begin{array}{r}
\therefore \text { from (3) } \tan \psi_{1}=\frac{2 a}{2 a}=1=\tan 45^{\circ} \\
\Rightarrow \psi_{1}=45^{\circ} .
\end{array}
$$
$\mathrm{At}(\mathrm{a},-2 \mathrm{a}), \mathrm{x}=\mathrm{a}, \mathrm{y}=-2 \mathrm{a}$ and let $\psi=\psi_{2}$
$\therefore$ from (3), $\tan \psi_{2}=\frac{2 a}{-2 a}=-1=\tan 135^{\circ}$;
or $\psi_{2}=135^{\circ}$.
Hence the required angle between tangents to $(1) \operatorname{at}(\mathrm{a}, 2 \mathrm{a})$ and $(\mathrm{a},-2 \mathrm{a})=\psi_{2}-\psi_{1}=135^{\circ}-$ $45^{\circ}=90^{\circ} .$
This shows that the tangent lines to (1) at $(a, 2 a)$ and $(a,-2 a)$ are perpendicular to each other.

Topic: Mathematics, Parabola, JEE Main
The tangent to the circle $C_1: x^2+y^2-2 x-1=0$ at the point $(2,1)$ cuts off a chord of length 4 from a circle $C_2$ whose centre is $(3,-2)$. The radius of $C_2$ is

Options:
A
$\sqrt{6}$

B
2

C
$\sqrt{2}$

D
3

The correct answer is:
$\sqrt{6}$

Solution:
Here, equation of tangent on $C_1$ at $(2,1)$ is: $2 x+y-(x+2)-1=0$
Or $x+y=3$
If it cuts off the chord of the circle $C_2$ then
the equation of the chord is:
$$
x+y=3
$$
$\therefore \quad$ distance of the chord from $(3,-2)$ is :
$$
d=\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2}
$$
Also, length of the chord is $l=4$
$$
\begin{aligned}
\therefore \text { radius of } C_2=r &=\sqrt{\left(\frac{l}{2}\right)^2+d^2} \\
&=\sqrt{(2)^2+(\sqrt{2})^2}=\sqrt{6}
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The tangent to the circle x2+y2=5 at the point 1,-2 also touches the circle x2+y2-8x+6y+20=0 at the point

Options:
A 2,1
B -3,0
C -1,-1
D 3,-1

The correct answer is:
3,-1

Solution:
Equation of the tangent to x2+y2=5 at 1,-2 is x1+y-2=5.

Let, this tangent touches x2+y2-8x+6y+20=0 at h,k
Then, hx+ky-4x+h+3y+k+20=0 is same as x-2y-5=0
⇒h-41=k+3-2=-4h+3k+20-5
⇒2h+k-5=0 and h+3k=0
⇒h=3,k=-1
⇒ Required point is 3,-1

Topic: Mathematics, Circle, JEE Main
The tangent to the conic \( x^{2}-(y-1)^{2}=2 \) passes through origin and touches this conic in first quadrant at point ( \( \left.a, b\right) \), then the value of \( [9 a] \) is (Where \( [\cdot] \) denotes greatest integer function)

Options:

The correct answer is:
22

Solution:
Let equation of tangent be y=mx (As it passes through 0,0)

On solving with conic

x2-mx-12=2⇒x2-m2x2+1-2mx=2

x21-m2+2mx-3=0

D=0

⇒2m2-41-m2 -3=0

m2+31-m2=0 ⇒ 2m2=3⇒m=32,-32 (rejected, as tangent touches in first quadrant)

∴ Equation of tangent is

y=32x

For point of contact, solving with conic

x2-3x2-12=2⇒2x2-3x-22=4

2x2-3x2+2-26x=4⇒-x2+26x=6

x2-26x=-6⇒x-62=0⇒x=6

∴ a=6 ⇒ 9a=22.045

=22

Topic: Mathematics, Hyperbola, JEE Main
The tangent to the curve y=x2-5x+5, parallel to the line 2y=4x+1, also passes through the point :

Options:
A 14, 72
B 72, 14
C -18, 7
D 18, -7

The correct answer is:
18, -7

Solution:
Equation of line parallel to given line can be taken as

y=2x+c.

Now, this line touches the curve y=x2-5x+5 hence, roots of the equation

x2-5x+5=2x+c⇒x2-7x+5-c=0 must be equal.

⇒D=0⇒-72-4·1·5-c=0⇒4c=-29

Hence, equation of tangent is 4y=8x-29.

Clearly, it passes through 18, -7.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The tangent to the curve, y=xex2 passing through the point 1,e also passes through the point:

Options:
A 43, 2e
B 2, 3e
C 53, 2e
D 3, 6e

The correct answer is:
43, 2e

Solution:
Given y=xex2
⇒dydx=2x2ex2+ex2
⇒dydx at 1, e=3e
⇒ Equation of tangent is y-e=3ex-1, which passes through 43,2e.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The tangent to the ellipse x225+y216=1 at point P lying in the first quadrant meets x-axis at Q and y-axis at R. If the length QR is minimum, then the equation of this tangent is

Options:
A 2x+5y=65
B 5x+2y=65
C 2x-5=65
D 25x+y=65

The correct answer is:
2x+5y=65

Solution:
Let point P be 5cosâ¡Î¸,4sinâ¡Î¸,Î¸â0,Ï2
So, the equation of tangent is xcosâ¡Î¸5+ysinâ¡Î¸4=1
Points Q and R are 5secâ¡Î¸,0 and 0,4cosecâ¡Î¸ respectively.
Hence, QR2=25sec2â¡Î¸+16cosec2â¡Î¸=251+tan2â¡Î¸+161+cot2â¡Î¸
=41+25tan2â¡Î¸+16cot2â¡Î¸=41+5tanâ¡Î¸-4cotâ¡Î¸2+40
QRmin2=81 when tan2â¡Î¸=45âtanâ¡Î¸=25
Therefore, sinâ¡Î¸=23,cosâ¡Î¸=53
Hence, equation of tangent is 2x+5y=65

Topic: Mathematics, Ellipse, JEE Main
The tangent to the parabola $y^2=4 a x$ at the point $(a, 2 a)$ makes with $x$-axis an angle equal to

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

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

Solution:
The equation of the tangent at point $(a, 2 a)$ of the parabola $y^2=4 a x$ is $y y_1=2 a\left(x+x_1\right)$
$\Rightarrow 2 a y=2 a(x+a) \Rightarrow y=x+a$
This line makes an angle of $\pi / 4$ with the $x$-axis, as $m=\tan \theta=1$.

Topic: Mathematics, Parabola, JEE Main
The tangent to the parabola y2=4x at the point where it intersects the circle x2+y2=5 in the first quadrant, passes through the point:

Options:
A 14,34
B -13,43
C -14,12
D 34,74

The correct answer is:
34,74

Solution:
Let’s first find points of intersection of two curves.
x2+4x=5
⇒x=1, -5 ( neglect x=-5 )
Hence point of intersections are 1, 2 and 1, -2.
Tangent to the parabola at first quadrant point 1, 2
y⋅2=2x+1
⇒y=x+1 clearly point 34,74 satisfy the given equation of tangent.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The tangent to the parabola y=x2-2x+8 at P2,8 touches the circle x2+y2+18x+14y+Î»=0 at Q. The coordinates of point Q are

Options:
A -7,-12
B -9,-13
C -11,-16
D

-315,-425

The correct answer is:

-315,-425

Solution:
The equation of the tangent at P 2,8 to the parabola is

y+82=2x-x+2+8
â2x-y+4=0 ...(i)

So, the slope of the normal to the circle is -12
The equation of normal to the circle at Q will be

y+7=-12x+9
x+2y+23=0 ...(ii)
The point Q is the intersection of the tangent at P and the normal at Q

On solving the equation (i) & (ii), we get,

x+22x+4+23=0

â5x=-31âx=-315

and y=2-315+4=-425

Hence, the coordinates of the point Q are -315,-425

Topic: Mathematics, Parabola, JEE Main