datavorous/entrance-exam-dataset huggingface dataset:
Let S=z=x+iy:2z-3i4z+2i is a real number }. Then which of the following is NOT correct?

Options:
A y+x2+y2≠-14
B (x,y)=0,-12

C x=0
D y∈-∞,-12∪-12,∞

The correct answer is:
(x,y)=0,-12

Solution:
We have,

Z=2z-3i4z+2i

⇒Z=2x+i2y-34x+i4y+2

⇒Z=2x+i2y-34x+i4y+2×4x-i4y+24x-i4y+2

⇒Z=8x2+4y+22y-3+i4x2y-3-8xy+4x16x2+4y+22

Since, Z is purely real, so 4x2y-3-8xy+4x16x2+4y+22=0

⇒-12x+8xy-8xy+4x=0

⇒x=0

And,

4x+i4y+2≠0

⇒i4y+2≠0

⇒y≠-12

Hence this is the required option.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let S=z∈ℂ:z2+z¯=0. Then ∑z∈SRez+Imz is equal to _______.

Options:

The correct answer is:
0

Solution:
Given,

S=z∈C:z2+z¯=0

Let z=x+iy

z2=x2-y2+2ixy

z¯=x-iy

Now putting the value in z2+z¯ we get,

z2+z¯=x2-y2+x+i2xy-y=0

Now on comparing both side we get,

⇒x2+x-y2=0 and 2xy-y=0

⇒y=0 or x=12

Now if y=0; x=0,-1

If x=12 then y=32,-32

So, ∑z∈SRez+Imz=0-1+12+12+0+0+32-32=0

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of A and adding 2 to each element of B. Then the sum of the mean and variance of the elements of C is

Options:
A 40
B 32
C 38
D 36

The correct answer is:
38

Solution:
Let the elements in set A be x1,x2,x3,x4,x5

Now given mean is=5

Now subtracting 3 from each term we get,

New mean as x1-3,x2-3…,x5-3=5-3=2

So, sum of elements will be 2×5=10

Also given variance,

VarX=12

Now we know that subtracting 3 from each term will not change the variance,

So, Varx1-3,x2-3,…x5-3=12

⇒∑xi-325-4=12

⇒∑xi-32=80

Now let elements in set B be y1,y2…y5

Given mean y1,y2…y5=8

Now adding each element by 2 we get,

New mean y1+2,y2+2,…y5+2=10

So, sum of elements will be 10×5=50

Also given Vary1,y2…y5=20

Similarly new variance,

Vary1+2,y2+2…y5+2=20

⇒∑yi+225-100=20

⇒∑yi+22=120×5

Now finding, combined mean we get,

∑i=15xi-3+∑yi+210=10+5010=6

And combined variance =∑xi-32+∑yi+2210-62

=80+120×510-36=32

Hence, the sum of combined mean and variance will be 32+6=38

Topic: Mathematics, Statistics, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let∫sin⁡2xln⁡(cos⁡x)dx=⁡f(x)cos2⁡x+C, (where, C is the constant of integration) and f0=12. If fπ3 is equal to 1a+ln⁡b, then the value of a+b is

Options:

The correct answer is:
4

Solution:
Given integral is
I=∫2sin⁡xcos⁡xln⁡⁡(cos⁡x)dx
Let, cos⁡x=t
⇒-sin⁡xdx=dt
⇒I=-2∫tln⁡tdt
Using Integration by parts
I=-2t22ln⁡t-∫t221tdt
I=-t2ln⁡t+t22+C
I=cos2⁡x12-ln⁡cos⁡x+C
fx=12-ln⁡⁡cos⁡x
fπ3=12-ln⁡12
=12+ln⁡2
⇒a=b=2
∴a+b=4

Topic: Mathematics, Indefinite Integration, JEE Main
Let slope of the tangent line to a curve at any point Px,y be given by xy2+yx. If the curve intersects the line x+2y=4 at x=-2, then the value of y, for which the point 3,y lies on the curve, is :

Options:
A -43
B 1835
C -1819
D -1811

The correct answer is:
-1819

Solution:
Given dydx=xy2+yx

xdy-ydxy2=xdx

-dxy=xdx

-xy=x22+c

∵ Curve intersects the line x+2y=4 at x=-2⇒ point of intersection is -2,3.

∴ Curve passes through -2,3.

⇒23=2+c⇒c=-43

⇒-xy=x22-43

Now put 3,y we get,

⇒-3y=196

⇒y=-1819

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Let T be a circle with diameter AB and centre O. Let l be the tangent to T at B. For each point M on T different from A, consider the tangent t at M and let interest l at P. Draw a line parallel to AB through P intersecting OM at Q. The locus of Q as M varies over T is

Options:
A an arc of a circle
B a parabola
C an arc of an ellipse
D a branch of a hyperbola

The correct answer is:
a parabola

Solution:
Let the equation of circle
x2+y2=r2

Let Mrcosθ,rsinθ

Equation of tangents at
Mrcosθ,rsinθ is rxcosθ+rysinθ=r2

⇒xcosθ+ysinθ=r

Tangent is passes through P.

∴xcosθ+ysinθ=r

⇒y=r1-cosθsinθ=rtanθ2

PQ is parallel to AB.

∴k=rtanθ2

OQ is normal of circle.

∴ Slope of normal =kh=tanθ

⇒kh=2tanθ21-tan2θ2

⇒kh=2kr1-k2r2 ∵tanθ2=kr

⇒2hr=r2-k2⇒k2=r22hr

Locus of Q is y2=-2rx-r2

which is parabola.

Topic: Mathematics, Circle, JEE Main
Let T be the tangent to the ellipse E:x2+4y2=5 at the point P(1,1). If the area of the region bounded by the tangent T, ellipse E, lines x=1 and x=5 is α5+β+γcos-115, then |α+β+γ| is equal to______.

Options:

The correct answer is:
1.25

Solution:
We have,

x2+4y2=5

⇒x25+y254=1

Tangent at P1,1 to the ellipse is

x+4y=5
Required Area

=∫155-x4-5-x22dx

=5x4-x28-x45-x2-54sin-1x515

=554-58-54·π2-54+18+12+54sin-115

=554-54-54·π2+54π2-cos-115

=554-54-5π8+5π8-54cos-115

=545-54-54cos-115

So,

α=54, β=-54 & γ=-54

α+β+γ=1.25

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let [t] denote the greatest integer function. If
∫02.4x2dx=α+β2+γ3+δ5, then α+β+γ+δ is equal to

Options:

The correct answer is:
6

Solution:
Given that ∫02.4x2dx=α+β2+γ3+δ5, then α+β+γ+δ

We know that for 1≤x<2â, x=1ââââââ⇒∫02.4x2dx=∫010dx+∫121dx+∫232dx+∫323dx+∫254dx+∫52.45dx

=0+(2-1)+2(3-2)+3(2-3)+4(5-2)+52.4-5

=9-2-3-5

This is in the form of α+β2+γ3+δ5

α=9, β=-1, γ=-1, δ=-1

⇒α+β+γ+δ=9-1-1-1=6

Hence this is the required answer.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let $[\mathrm{t}]$ denote the greatest integer less than or equal to $\mathrm{t}$. Let $f:[0, \infty) \rightarrow \mathbf{R}$ be a function defined by $f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0,8]$ at which $f$ is not continuous. Then $\sum_{\mathrm{a} \in \mathrm{S}} \mathrm{a}$ is equal to _______

Options:

The correct answer is:
17

Solution:
$\left[\frac{x}{2}+3\right]$ is discontinuous at $x=2,4,6,8$
$\sqrt{\mathrm{x}}$ is discontinuous at $\mathrm{x}=1,4$ $\mathrm{F}(\mathrm{x})$ is discontinuous at $\mathrm{x}=1,2,6,8$
$\sum \mathrm{a}=1+2+6+8=17$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let [t] denote the greatest integer less than or equal to t.
Let f(x)=x-[x], g(x)=1-x+[x], and h(x)=min{f(x),g(x)},x∈[-2,2]. Then h is :

Options:
A continuous in [-2,2] but not differentiable at more than four points in (-2,2)
B Continous in [-2,2] but not differentiable at exactly three poionts in (-2,2)
C not continuous at exactly four points in [-2,2]
D not continuous at exactly three points in [-2,2]

The correct answer is:
continuous in [-2,2] but not differentiable at more than four points in (-2,2)

Solution:
f(x)=x-[x]={x}

g(x)=1-x+[x]=1-{x}

Now, Graph of min {f(x),g(x)}

Clearly graph is continuous in [-2,2] but non differentiable at 7 points (i.e. greater than 4) in (-2,2)

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let t denote the greatest integer less than or equal to t. Then the value of ∫122x-3xdx is

Options:

The correct answer is:
1

Solution:
∫123x-3x-xdx

=∫123x-xdx

=∫12x-3xdx

=∫12xdx-∫123xdx

= x2212-3∫01/33xdx

=42-12-9x2201/3

=32-92(1/3)2-02

=1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Let t denote the greatest integer ≤t and limx→0⁡x4x=A. Then the function, fx=x2sin⁡πx is discontinuous, when x is equal to:

Options:
A A+1
B A+5
C A+21
D A

The correct answer is:
A+1

Solution:
logx→0⁡4x-4x=A⇒logx→O⁡4-x4x=A

⇒4-0

⇒A=4

Hence, for 5, fx is discontinuous as it is an non-integral value.

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let t denote the greatest integer ≤t. If λ ε R−0, 1, limx→01−x+xλ−x+x=L, then L is equal to

Options:
A 1
B 2
C 12
D 0

The correct answer is:
2

Solution:
limx→01−x+xλ−x+x=L

Let's find left-hand and right-hand limit.

For left-hand limit

x=0-h, x→0, h→0, h>0

LHL=limh→01−-h+-hλ−-h+-h

⇒ LHL=limh→01+h--hλ+h+-h

Since h>0 ⇒-h=-1

⇒ LHL=limh→01+h+hλ+h-1=1λ-1

For right-hand limit

x=0+h, x→0, h→0, h>0

RHL=limh→01−h+hλ−h+h

Since h>0 ⇒h=0

⇒ RHL=limh→01+h+hλ+h+0=1λ

For existence of limit LHL=RHL

⇒1λ-1=1λ

Cross multiply and square both side.

⇒λ2=λ-12 ⇒λ2=λ2+1-2λ

⇒λ=12

Limit L=112=2

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Let t denote the greatest integer ≤t. if the constant term in the expansion of 3x2-12x57 is α then α is equal to _____

Options:

The correct answer is:
1275

Solution:
The given expansion is 3x2-12x57.

The general term in the binomial expansion of x+an is given by Tr+1=Crnxn-rar.

⇒Tr+1=Cr73x27-r-12x5r

=Cr7-1r37-r2rx14-2r-5r

Now for term independent of x⇒14-2r-5r=0

⇒r=2

Coefficient of x0 is C27-12×37-222

=7×62×3522.

⇒α=51034

⇒α=1275

Hence this is the required answer.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let [t] denote the greatest integer ≤t. Then 2π∫π65π68cosecx-5cotxdx is equal to _______

Options:

The correct answer is:
14

Solution:
Let,

I=∫π65π68cosecx-5cotxdx

⇒I=8 ∫π65π6cosecxdx-5∫π65π6cotxdx

⇒I=8I1-5I2, where I1=8 ∫π65π6cosecxdx & I2=5∫π65π6cotxdx

Now solving,

I1=∫π65π6cosecxdx

⇒I1=∫π65π61dx=2π3

As when x∈π6,5π6, cosecx∈[1,2),

So, cosecx=1

Now solving,

I2=∫π65π6cot xdx

⇒I2=∫π6π41dx+∫π4π20dx+∫π23π4-1dx+∫3π45π6-2dx

⇒I2=π4-π6-3π4-π2-25π6-3π4

⇒I2=-π3

Required value will be,

2πI=2π8×2π3+5×π3=14

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let [t] denote the greatest integer ≤t. Then the value of 8·∫-121([2x]+|x|)dx is

Options:

The correct answer is:
5

Solution:
Let I=∫-121([2x]+|x|)dx

2x will be discontinuous at x = 0, 12 ; x ∈ -12, 1

We know that x = x, x ≥ 0 ; -x , x <0.

=∫-120(-1-x)dx+∫01/2(0+x)dx+∫121(1+x)dx

We know that ∫xndx = xn+1n+1 + c

=-x-x22-120+x2201/2+x+x22121

=(0)-+12-18+18+1+12-12+18=58

⇒8∫-121([2x]+|x|)dx=5.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1-Tn=10, then the value of n is :

Options:
A 10
B 8
C 5
D 7

The correct answer is:
5

Solution:
Number of triangles =C3n
Given:C3n+1-C3n=10
⇒n2-n-20=0

⇒n-5n+4=0
⇒n=5, ∵n≠0

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2013 (07 Apr)
Let $T_r$ be the rth term of an A.P. whose first term is a and common difference is $d$. If for some positive integers $m, n, m \neq n, T_m=\frac{1}{n}$ and $T_n=\frac{1}{m}$, then $a-d$ equals

Options:
A
0

B
1

C
$\frac{1}{\mathrm{mn}}$

D
$\frac{1}{m}+\frac{1}{n}$

The correct answer is:
0

Solution:
$$
T_m=\frac{1}{n}=a+(m-1) d
$$
and $T_n=\frac{1}{m}=a+(n-1) d$
from (1) and (2) we get $a=\frac{1}{m n}, \quad d=\frac{1}{m n}$ Hence $\mathrm{a}-\mathrm{d}=0$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2004
Let Tr denote the rth term in the expansion of 2x+14xn. If the ratio T3:T2=7:1 and the sum of the coefficients of the second and third term is 36, then the value of x is

Options:
A 12
B -12
C -14
D -13

The correct answer is:
-13

Solution:
It is given that T3T2=7⇒nC22xn-214x2nC12xn-114x1=7
⇒n-122-3x=7
Also, nC1+ nC2=36
⇒ n+nn-12=36
⇒  n2+n-72=0⇒  n=8,  -9
n=8 ⇒ 72 2-3x=7
⇒2-3x-1=1=20
⇒-3x-1=0⇒x=-13

Topic: Mathematics, Binomial Theorem, JEE Main
Let tan-1 y=tan-1 x+tan-12x1-x2, where x<13,Then a value of y is

Options:
A 3x+x31+3x2
B 3x-x31-3x2
C 3x+x31-3x2
D 3x-x31+3x2

The correct answer is:
3x-x31-3x2

Solution:
Given tan-1 y=tan-1 x+tan-12x1-x2, x<13
=tan-1 x+2tan-1 x
=3tan-1 x
⇒tan-1 y=tan-13x-x31-3x2
∴ y=3x-x31-3x2

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2015 (04 Apr)
Let tangents PQ and PR are drawn from the point P-2,4 to the parabola y2=4x. If S is the focus of the parabola y2=4x, then the value (in units) of RS+SQ is equal to

Options:

The correct answer is:
22

Solution:
Let, Q=t12,2t1 & R=t22,2t2, then P=t1t2,t1+t2=-2,4
⇒t1t2=-2 & t1+t2=4
Now, QS=1+t12, RS=1+t22
⇒QS+RS=2+t12+t22
=2+t1+t22-2t1t2
=2+42-2-2
=2+16+4=22 units

Topic: Mathematics, Parabola, JEE Main
Let tangents \( P Q \) and \( P R \) are drawn from the point \( P(-2,4) \) to the parabola \( y^{2}=4 x . \) If \( S \) is the focus of the parabola \( y^{2}=4 x \), then the value (in units) of \( R S+S Q \) is equal to

Options:

The correct answer is:
22

Solution:
Let, Q=t12,2t1 & R=t22,2t2,

then P=t1t2,t1+t2=-2,4
⇒t1t2=-2 & t1+t2=4
Now, QS=1+t12, RS=1+t22
⇒QS+RS=2+t12+t22
=2+t1+t22-2t1t2
=2+42-2-2
=2+16+4=22 units

Topic: Mathematics, Parabola, JEE Main
Let the abscissae of the two points P and Q on a circle be the roots of x2-4x-6=0 and the ordinates of P and Q be the roots of y2+2y-7=0. If PQ is a diameter of the circle x2+y2+2ax+2by+c=0, then the value of a+b-c is

Options:
A 12
B 13
C 14
D 16

The correct answer is:
12

Solution:
Given that the roots of x2-4x-6=0 are the abscissa of the end of diameter

i.e. x1+x2=4, x1x2=-6

and roots of y2+2y-7=0 are ordinate of the end of diameter

i.e. y1+y2=-2, y1y2=-7

Now, equation of the circle will be

x-x1x-x2+y-y1y-y2=0

i.e. x2-x1+x2x+x1x2+y2-y1+y2y+y1y2=0

⇒x2+y2-4x+2y-13=0

∴a=-2, b=1, c=-13

⇒a+b-c=-2+1+13=12

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Let the acute angle bisector of the two planes x-2y-2z+1=0 and 2x-3y-6z+1=0 be the plane P. Then which of the following points lies on P ?

Options:
A (0, 2, -4)
B -2, 0, -12
C (4, 0, -2)
D 3, 1, -12

The correct answer is:
-2, 0, -12

Solution:
Angle bisectors of the two given planes are

x-2y-2z+11+4+4=±2x-3y-6z+14+9+36

⇒x-5y+4z+4=0 and 13x-23y-32z+10=0

Let θ be the angle between the planes x-5y+4z+4=0 and x-2y-2z+1=0

So, cosθ=1+10-81+4+4·1+25+16=142⇒tanθ=41>1

⇒θ>45°

Then acute angle bisector is plane P: 13x-23y-32z+10=0

Point -2, 0, -12 lies on the the plane P

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let the area enclosed by the lines x+y=2, y=0, x=0 and the curve f(x)=minx2+34,1+[x] where [x] denotes the greatest integer ≤x, be A. Then the value of 12A is

Options:

The correct answer is:
17

Solution:
Given,

The lines x+y=2, y=0, x=0 and the curve f(x)=minx2+34,1+[x] where [x] denotes the greatest integer ≤x,

Now plotting the diagram of given function we get,

Now from above diagram, the area enclosed is given by,

A=∫012x2+34dx+1212+32×1=512+1

⇒ 12A=17

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Let the area of the region enclosed by the curve $y=\min \{\sin x, \cos x\}$ and the $x$ axis between $x=-\pi$ to $x=\pi$ be $A$. Then $A^2$ is equal to ___________

Options:

The correct answer is:
16

Solution:
$\begin{aligned} & y=\min \{\sin x, \cos x\} \\ & x-\text { axis } \quad x-\pi \quad x=\pi\end{aligned}$
$\begin{aligned} & \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} \\ & \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} \\ & =(\cos x+\sin x)_{-3 \pi / 4}^{-\pi} \\ & =(-1+0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) \\ & =-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ & \int_{\pi / 4}^{\pi / 2} \cos x d x=(\sin x)_{\pi / 4}^{\pi / 2}=1-\frac{1}{\sqrt{2}} \\ & A=4 \\ & A^2=16\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let the area of the triangle with vertices A1,α,Bα,0 and C0,α be 4 sq. units. If the points α,-α,-α,α and α2,β are collinear, then β is equal to

Options:
A 64
B -8
C -64
D 512

The correct answer is:
-64

Solution:
Given, the area of the triangle with vertices A1,α,Bα,0 and C0,α be 4 sq. units

Now applying formula for area we get,

12α011α10α1=±4

⇒α=±8

Now given points will be 8,-8,-8,8,64,β

OR -8,8,8,-8,64,β

Given above points are collinear, so their slope will be equal, so slope=8+8-8-8=-1

⇒β-864+8=-1

⇒β=-64

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x+y=3. If R and r be the radius of circumcircle and incircle respectively of ΔABC, then (R+r) is equal to :

Options:
A 92
B 72
C 22
D 32

The correct answer is:
92

Solution:
Distance from origin to line x+y=3 is inradius.

r=OM=32

& sin30°=12=rR⇒R=62

∴r+R=92

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $C_2$ be a circle having centre at $(-1,0)$ and radius 2 . If the line of the common chord of $\mathrm{C}_1$ and $\mathrm{C}_2$ intersects the $y$-axis at the point $\mathrm{P}$, then the square of the distance of $\mathrm{P}$ from the centre of $\mathrm{C}_1$ is :

Options:
A 2
B 1
C 4
D 6

The correct answer is:
2

Solution:
$\begin{aligned} & S_1: x^2+y^2-2 x-2 y+1=0 \\ & S_2: x^2+y^2+2 x-3=0 \\ & \text { Common chord }=S_1-S_2=0 \\ & -4 x-2 y+4=0 \\ & 2 x+y=2 \Rightarrow P(0,2) \\ & d_{(c, p)}^2=(1-0)^2+(2-1)^2=2\end{aligned}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let the circle S:36x2+36y2-108x+120y+C=0 be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x-2y=4 and 2x-y=5 lies inside the circle S, then:

Options:
A 259<C<133
B 100<C<165
C 81<C<156
D 100<C<156

The correct answer is:
100<C<156

Solution:
The given circle is S:36x2+36y2-108x+120y+C=0

⇒x2+y2-3x+103y+C36=0

We know that the centre and radius of a circle x2+y2+2gx+2fy+c=0 is -g, -f and g2+f2-c respectively.

Thus, centre e≡-g,-f≡32, -106 and radius =r=94+10036-C36

Now, this circle neither intersects nor touches the co-ordinate axes, hence, the radius of the circle is less than the absolute value of the co-ordinate with smaller value.

⇒r<32

94+10036-C36<32

⇒ 94+10036-C36<94

⇒ C>100 …1

Now, point of intersection of x-2y=4 and 2x-y=5 is (2, -1), which lies inside the circle S.

The point x1, y1 lies inside a circle x2+y2+2gx+2fy+c=0, if x12+y12+2gx1+2fy1+c<0

⇒ 22+-12-32+103-1+C36<0

⇒ 4+1-6-103+C36<0

⇒ C<156 …2

From 1 and 2, we have 100<C<156

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let the circumcentre of âABC is S-1,0 and the midpoints of the sides AB and AC are E1,-2 and F-2,-1 respectively, then the coordinates of A are

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

The correct answer is:
0,3

Solution:
Slope of SF is 1
⇒ Slope of AB is 1
Now, the slope of ES is -1
⇒ Slope of AC is -1
Equation of AB is x+y+3=0
Equation of AC is x-y-3=0
Hence, point A is 0,3

Topic: Mathematics, Straight Lines, JEE Main
Let the circumcentre of ÎABC is S-1,0 and the midpoints of sides AB and AC are E(1,-2) and F-2,-1 respectively, then the equation of the circumcircle of ÎABC is

Options:
A x+12+y2=5
B x+12+y2=10
C x+12+y2=15
D x+12+y2=1

The correct answer is:
x+12+y2=10

Solution:
Slope of SF is 1
⇒ Slope of AB is -1
Now, the slope of ES is -1
⇒ Slope of AC is 1
Equation of AB is x+y+3=0
Equation of AC is x-y-3=0
So, point A is 0,3
SA=-1-02+0+32=10
Hence, the equation of circumcircle is x+12+y2=10

Topic: Mathematics, Straight Lines, JEE Main
Let the circumcentre of a triangle with vertices Aa,3,Bb,5 and Ca,b,ab>0 be P1,1. If the line AP intersects the line BC at the point Qk1,k2, then k1+k2 is equal to

Options:
A 2
B 47
C 27
D 4

The correct answer is:
47

Solution:
On plotting the graph we get,

Now by diagram mAC⟶∞ and mPD=0 as PD is perpendicular bisector, so Da+a2,b+32

≡Da,b+32

Now since mPD=0, so b+32-1=0

⇒b+3-2=0⇒b=-1

Now similarly finding the coordinates of Eb+a2,5+b2=a-12,2 as b=-1

Now we know that product of perpendicular slopes is mCB·mEP=-1

⇒5-bb-a=2-1a-12-1=-1

⇒6-1-a=2a-3=-1

⇒12=1+aa-3

⇒12=a2-3a+a-3

⇒a2-2a-15=0

⇒a-5a+3=0

So, a=5 or a=-3

Now given ab>0

So, a-1>0⇒a<0

So, a=-3

Now equation of line AP by two point form will be, {where A-3,3 & P1,1}

y-1=3-1-3-1x-1

⇒-2y+2=x-1

⇒ x+2y=3 ⋯1

Now equation of line BC {where B-1,5 and C-3,-1} will be,

y-5=62x+1

⇒y-5=3x+3

⇒y=3x+8 ⋯2

Now solving 1 and 2 we get,

x+23x+8=3

⇒7x+16=3⇒7x=-13

⇒x=-137

And y=3-137+8=-39+567

So, y=177

So the value of k1,k2 will be x+y=-13+177=47 where k1 is x & k2 is y

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let the coefficients of the middle terms in the expansion of 16+βx4,1-3βx2 and 1-β2x6,β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50-2dβ2 is equal to _____ .

Options:

The correct answer is:
57

Solution:
Let the coefficients of the middle terms in the expansion of 16+βx4,1-3βx2 and 1-β2x6,β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50-2dβ2 is equal to _____ .

Given,

Coefficients of middle terms of given expansions 16+βx4, 1-3βx2 and1-β2x6 are C2416β2, C12-3β & C36-β23 respectively,

Now it is given that middle terms are forming an A.P

So, 2C12-3β = C36-β23+C2416β2

⇒2.2-3β=β2-5β32

⇒-24=2β-5β2

⇒5β2-2β-24=0

⇒5β2-12β+10β-24=0

⇒β5β-12+25β-12=0

⇒β+25β-12=0

So, β=125 and β=-2 {which is not possible as β>0}

Now d=C12-3β-C2416β2=-6β-β2

Now putting the value of β & d in 50-2dβ2 we get,

50-2dβ2=50-2-6β-β2β2=50+12β+2=57

Topic: Mathematics, Binomial Theorem, JEE Main
Let the coefficients of the middle terms in the expansion of 16+βx4,1-3βx2 and 1-β2x6,β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50-2dβ2 is equal to _____ .

Options:

The correct answer is:
57

Solution:
Let the coefficients of the middle terms in the expansion of 16+βx4,1-3βx2 and 1-β2x6,β>0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50-2dβ2 is equal to _____ .

Given,

Coefficients of middle terms of given expansions 16+βx4, 1-3βx2 and1-β2x6 are C2416β2, C12-3β & C36-β23 respectively,

Now it is given that middle terms are forming an A.P

So, 2C12-3β = C36-β23+C2416β2

⇒2.2-3β=β2-5β32

⇒-24=2β-5β2

⇒5β2-2β-24=0

⇒5β2-12β+10β-24=0

⇒β5β-12+25β-12=0

⇒β+25β-12=0

So, β=125 and β=-2 {which is not possible as β>0}

Now d=C12-3β-C2416β2=-6β-β2

Now putting the value of β & d in 50-2dβ2 we get,

50-2dβ2=50-2-6β-β2β2=50+12β+2=57

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let the coefficients of third, fourth and fifth terms in the expansion of x+ax2n, x≠0, be in the ratio 12:8:3. Then the term independent of x in the expansion, is equal to _______.

Options:

The correct answer is:
4

Solution:
The general term in the expansion of a+bn is Tr+1=Crnan-rbr.

Hence, the general term in the expansion of x+ax2n is Tr+1=Crn(x)n-rax2r

⇒Tr+1=Crnarxn-3r

Given the coefficients of third, fourth and fifth terms are in the ratio 12:8:3

⇒C2na2:C3na3:C4na4=12:8:3

⇒C2na2C3na3=128 and C3na3C4na4=83

Using CrnCr+1n=r+1n-r,

⇒3n-2a=32 and 4n-3a=83

On dividing the two equations, we get 3n-34n-2=916

⇒4n-12=3n-6

⇒n=6 and a=12.

For term independent of x⇒n=3r

⇒r=2.

∴ The coefficient isC26122=154.

Hence, the nearest integer is 4.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Let the coefficients of three consecutive terms in the binomial expansion of (1+2x)n be in the ratio 2: 5: 8. Then the coefficient of the term, which is in the middle of these three terms, is

Options:

The correct answer is:
1120

Solution:
Given,

The coefficients of three consecutive terms in the binomial expansion of (1+2x)n be in the ratio 2: 5: 8.

Now rth term in expansion of (1+2x)n is given by, tr+1=Crn(2x)r

Now let Tr, Tr+1, Tr+2 are in the ratio 2:5:8

⇒TrTr+1=Cr-1n(2)r-1Crn(2)r=25

⇒n!(r-1)!(n-r+1)!n!(2)r!(n-r)!=25

⇒rn-r+1=45⇒5r=4n-4r+4

⇒9r=4(n+1) .......1

Now taking other ratio Tr+1Tr+2=Crn(2)rCr+1n(2)r+1=58

⇒n!r!(n-r)!n!(r+1)!(n-r-1)!=54⇒r+1n-r=54

⇒4r+4=5n-5r

⇒5n-4=9r .........2

From equation 1 & 2 we get,

4n+4=5n-4⇒n=8 and r=4

So, coefficient of middle term will be

C4824=16×8×7×6×54×3×2×1=16×70=1120

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (29 Jan Shift 1)