datavorous/entrance-exam-dataset huggingface dataset:
If -π<θ<π, the equation cos⁡3θ+1x2+2cos⁡2θ-1x+1-2cos⁡θ=0 has more than two roots for

Options:
A no value of θ
B one value of θ
C two value of θ
D all values of θ

The correct answer is:
no value of θ

Solution:
Given equation has more than two roots if it is an identity
⇒cos⁡3θ+1=0; 2cos⁡2θ-1=0 and 1-2cos⁡θ=0
⇒cos⁡3θ=-1⇒θ=±π3 which does not satisfy 2cos⁡2θ-1=0

Hence, no value possible

Topic: Mathematics, Quadratic Equation, JEE Main
If the equation cos4θ+sin4θ+λ=0 has real solutions for θ then λ lies in interval

Options:
A -54,-1
B -1,-12
C -12,-14
D -32,-54

The correct answer is:
-1,-12

Solution:
-λ=sin4θ+cos4θ

=sin2θ+cos2θ2-2sin2θcos2θ

=1-4sin2θcos2θ2

=1-sin22θ2

⇒λ=sin22θ2-1

λ∈-1,-12

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2020 (02 Sep Shift 2)
If the equation \( \lambda x^{2}+(2 \lambda-3) y^{2}-4 x-1=0 \) represents a circle, then its radius is

Options:
A \( \frac{\sqrt{11}}{3} \)
B \( \frac{\sqrt{13}}{3} \)
C \( \frac{\sqrt{7}}{3} \)
D \( \frac{1}{3} \)

The correct answer is:
\( \frac{\sqrt{7}}{3} \)

Solution:
Given equation is
λx2+2λ-3y2-4x-1=0
Here, a=λ, b=2λ-3
It represents a circle, if a=b
⇒ λ=2λ-3
⇒ λ=3
Also, h=0
Then, equation becomes
3x2+3y2-4x-1=0
⇒ x2+y2-43x-13=0
Here, g=-23, c=-13, f=0
∴ Radius=-232+0--13=49+13
=73

Topic: Mathematics, Circle, JEE Main
If the equation of a plane P, passing through the intersection of the planes, x+4y-z+7=0 and 3x+y+5z=8 is ax+by+6z=15 for some a,b∈R, then the distance of the point 3,2,-1 from the plane P is ……

Options:

The correct answer is:
3

Solution:
Equation of plane of passing through intersection of planes P1⇒x+4y−z+7=0 ......1 and P2⇒3x+y+5z−8=0 .....2

is P1+λP2=0

x+4y−z+7+λ3x+y+5z−8=0

3λ+1x+λ+4y+5λ−1z+7−8λ=0 .......3

ax+by+6z−15=0 ........4

On comparing 3 and 4,

a3λ+1=bλ+4=65λ−1=−157−8λ

∴ 42−48λ=−75λ+15

27λ=−27

λ=−1

∴ a−2=b3=−1

⇒ a=2, b=-3

So, equation of plane is 2x−3y+6z−15=0 ......5

Distance of plane 5 from point 3, 2, -1 is p=6−6−6−154+9+36=−2149=3.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (04 Sep Shift 1)
If the equation of plane passing through the mirror image of a point 2, 3, 1 with respect to line x+12=y-31=z+2-1 and containing the line x-23=1-y2=z+11 is αx+βy+γz=24 then α+β+γ is equal to:

Options:
A 20
B 19
C 18
D 21

The correct answer is:
19

Solution:
The given line is x+12=y-31=z+2-1

x+12=y-31=z+2-1=λ (let)

⇒x+12=λ, y-31=λ, z+2-1=λ

⇒x=2λ-1, y=λ+3, z=-λ-2

Hence, any point on the line x+12=y-31=z+2-1 is 2λ-1, λ+3, -λ-2.

Let, M be the foot of perpendicular from the point P to the line x+12=y-31=z+2-1, hence M≡2λ-1, λ+3, -λ-2.

Now, direction ratios of PM→=(2λ-3, λ, -λ-3)

Given PM→ is perpendicular to the line x+12=y-31=z+2-1, then the sum of the product of their direction ratios is zero.

22λ-3+1λ-1-λ-3=0

⇒4λ-6+λ+λ+3=0⇒λ=12

∴ M≡0, 72, -52

The mid-point of the point P and the image of P in the line x+12=y-31=z+2-1 is M.

∴ Reflection (-2, 4, -6).

Thus, the plane passes through the point (-2, 4, -6).

Also, the plane contains the line x-23=1-y2=z+11

⇒x-23=y-1-2=z+11, hence it passes through the point 2, 1, -1 and the normal to the plane is perpendicular to the line.

The equation of a plane passing through the points x1, y1, z1 and x2, y2, z2 and having normal perpendicular to a line with direction ratios <a, b, c> is x-x1y-y1z-z1abcx1-x2y1-y2z1-z2=0

Thus, the required plane is x-2y-1z+13-214-35=0

⇒(x-2)(-10+3)-(y-1)(15-4)+(z+1)(-1)=0

⇒-7x+14-11y+11-z-1=0

⇒7x+11y+z=24

Given the plane is αx+βy+γz=24

∴ α=7, β=11, γ=1

⇒α+β+γ=19.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the equation of plane passing through the mirror image of a point 2, 3, 1 with respect to line x+12=y-31=z+2-1 and containing the line x-23=1-y2=z+11 is αx+βy+γz=24 then α+β+γ is equal to:

Options:
A 20
B 19
C 18
D 21

The correct answer is:
19

Solution:
The given line is x+12=y-31=z+2-1

x+12=y-31=z+2-1=λ (let)

⇒x+12=λ, y-31=λ, z+2-1=λ

⇒x=2λ-1, y=λ+3, z=-λ-2

Hence, any point on the line x+12=y-31=z+2-1 is 2λ-1, λ+3, -λ-2.

Let, M be the foot of perpendicular from the point P to the line x+12=y-31=z+2-1, hence M≡2λ-1, λ+3, -λ-2.

Now, direction ratios of PM→=(2λ-3, λ, -λ-3)

Given PM→ is perpendicular to the line x+12=y-31=z+2-1, then the sum of the product of their direction ratios is zero.

22λ-3+1λ-1-λ-3=0

⇒4λ-6+λ+λ+3=0⇒λ=12

∴ M≡0, 72, -52

The mid-point of the point P and the image of P in the line x+12=y-31=z+2-1 is M.

∴ Reflection (-2, 4, -6).

Thus, the plane passes through the point (-2, 4, -6).

Also, the plane contains the line x-23=1-y2=z+11

⇒x-23=y-1-2=z+11, hence it passes through the point 2, 1, -1 and the normal to the plane is perpendicular to the line.

The equation of a plane passing through the points x1, y1, z1 and x2, y2, z2 and having normal perpendicular to a line with direction ratios <a, b, c> is x-x1y-y1z-z1abcx1-x2y1-y2z1-z2=0

Thus, the required plane is x-2y-1z+13-214-35=0

⇒(x-2)(-10+3)-(y-1)(15-4)+(z+1)(-1)=0

⇒-7x+14-11y+11-z-1=0

⇒7x+11y+z=24

Given the plane is αx+βy+γz=24

∴ α=7, β=11, γ=1

⇒α+β+γ=19.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (17 Mar Shift 2)
If the equation of the hypotenuse of a right-angled isosceles triangle is 3x+4y=4 and its opposite vertex is 2,2, then the equations of the perpendicular and the base are respectively

Options:
A 7x+y=16 & x-7y+12=0
B 7x-y=12 & x+7y=16
C 5x+y=12 & x-5y+8=0
D x+5y=12 & 5x-y=8

The correct answer is:
7x+y=16 & x-7y+12=0

Solution:
Let, the slope of the required line is m, then

m--341+m-34=±tan⁡45o
4m+3=±4-3m
⇒m=-7,17
⇒ equation of the lines are y-2x-2=-7 & y-2x-2=17
⇒7x+y=16 & x-7y+12=0

Topic: Mathematics, Straight Lines, JEE Main
If the equation of the locus of a point equidistant from the point $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ is $\left(a_1-b_2\right) x+\left(a_1-b_2\right) y+c=0$, then the value of 'c' is

Options:
A
$\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2-\mathrm{a}_2^2-\mathrm{b}_2^2}$

B
$\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^2$

C
$\mathrm{a}_1^2-\mathrm{a}_2^2+\mathrm{b}_1^2-\mathrm{b}_2^2$

D
$\frac{1}{2}\left(a_1^2+a_2^2+b_1^2+b_2^2\right)$

The correct answer is:
$\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^2$

Solution:
$\left(h-a_1\right)^2+\left(k-b_1\right)^2=\left(h-a_2\right)^2+\left(k-b_2\right)^2$
$\left(a_1-a_2\right) x+\left(b_1-b_2\right) y+\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)=0$
$C=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2003
If the equation of the normal to the curve y=x-ax+bx-2 at the point 1,-3 is x-4y=13 then the value of a+b is equal to ______

Options:

The correct answer is:
4

Solution:
Given curve is

y=x-ax+bx-2 ...1

At point 1,-3, we have

-3=1-a1+b1-2

⇒31+b=1-a

⇒1-a=3+3b

Differentiating 1 both sides w.r.t. x, we get

dydx=(x+b)(x-2)×(1)-(x-a)(2x+b-2)(x+b)2(x-2)2

At (1,-3) slope of normal is 14, hence

dxdy=-14

⇒dydx1,-3=-4,

⇒-4=(1+b)(-1)-(1-a)b(1+b)2(-1)2

⇒-4=(1+b)(-1)-3(b+1)b(1+b)2

⇒-4=-1-3b1+b

∵b≠-1

⇒4+4b=1+3b

⇒b=-3

So, a=7

Hence, a+b=4

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (29 Jan Shift 2)
If the equation of the parabola, whose vertex is at 5,4 and the directrix is 3x+y-29=0, is x2+ay2+bxy+cx+dy+k=0, then a+b+c+d+k is equal to

Options:
A 575
B -575
C 576
D -576

The correct answer is:
-576

Solution:
Given the equation of the parabola whose vertex is 5,4 and equation of directrix is 3x+y-29=0 is x2+ay2+bxy+cx+dy+e=0

Let focus be α,β

Let equation of ZV which is perpendicular to 3x+y-29=0 be x-3y+k=0 and it passes through point 5,4. So, k=7 and equation of ZV become x-3y+7=0

Now the intersection of 3x+y-29=0 & x-3y+7=0 will be Z=8,5

Now plotting the diagram we get,

So, foot of perpendicular from 5,4 on 3x+y-29=0 is 8,5

Now using the midpoint formula we will find the focus of parabola⇒α+82=5,β+52=4⇒α,β=2,3

∴ Focus is 2,3 and directrix is 3x+y-29=0

Applying PS2=PM2, we get

x-22+y-32=3x+y-29210

⇒x2+9y2-6xy+134x-2y-711=0

Comparing withx2+ay2+bxy+cx+dy+e=0 , we get a+b+c+d+e=9-6+134-2-711=-576

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If the equation of the plane containing the line x+2y+3z-4=0=2x+y-z+5 and perpendicular to the planer→=i^-j^+λi^+j^+k^+μ(i^-2j^+3k^) is ax+by+cz=4 then (a-b+c) is equal to

Options:
A 18
B 22
C 20
D 24

The correct answer is:
22

Solution:
Given,

The equation of the plane containing the line x+2y+3z-4=0=2x+y-z+5 and perpendicular to the planer→=i^-j^+λi^+j^+k^+μ(i^-2j^+3k^) is ax+by+cz=4,

Now let equation of required plane be,

P: x+2y+3z-4+α2x+y-z+5=0

⇒(2α+1)+α+2y+3-αz=4-5α

So, the normal vector is given by,

n1→=(2α+1)i^+(α+2)j^+(3-α)k^

And normal of the plane r→=i^-j^+λi^+j^+k^+μ(i^-2j^+3k^) will be,

n2→=i^j^k^1111-23=5j^-2j^-3k^

Now given they are perpendicular, so by perpendicular condition we get,

n1→·n2→=0

⇒52α+1-2α+2-33-α=0

⇒11α+-8=0

⇒α=811

Hence, equation of required plane will be,

P:2711x+3011y+2511z-411=0

⇒27x+30y+25z=4

Now comparing with ax+by+cz=4 we get,

a=27, b=30, c=25

Hence, a-b+c=22

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If the equation of the plane passing through 1,2,3 and situated at a maximum distance from point 2,3,4 is P=0, then the distance (in units) of P=0 from origin is

Options:
A 3
B 23
C 6
D 32

The correct answer is:
23

Solution:
Let, the point A is 1,2,3,B is 2,3,4
AB→=<1,1,1>
Required plane must be perpendicular to AB→
So, the equation of the plane is 1x-1+1y-2+1z-3=0
⇒x+y+z-6=0
Perpendicular distance (in units) of the point 0,0,0 from the plane is 0+0+0-63=23

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the equation of the plane passing through the line of intersection of the planes 2x-7y+4z-3=0, 3x-5y+4z+11=0 and the point -2, 1, 3 is ax+by+cz-7=0, then the value of 2a+b+c-7 is _________.

Options:

The correct answer is:
4

Solution:
Required plane is

p1+λp2=(2+3λ)x-(7+5λ)y+(4+4λ)z-3+11λ=0;

which is satisfied by (-2, 1, 3).

Hence, λ=16

Thus, plane is 15x-47y+28z-7=0

So, 2a+b+c-7=4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If the equation of the plane passing through the line of intersection of the planes 2x-y+z=3, 4x-3y+5z+9=0 and parallel to the line x+1-2=y+34=z-25 is ax+by+cz+6=0, then a+b+c is equal to

Options:
A 12
B 14
C 16
D 13

The correct answer is:
14

Solution:
Given,

The equation of the plane passing through the line of intersection of the planes 2x-y+z=3, 4x-3y+5z+9=0 and parallel to the line x+1-2=y+34=z-25 is ax+by+cz+6=0,

Now let, P1=2x-y+z=3

And P2=4x-3y+5z+9=0

Now using family of planes we get,

P1+λP2=0

⇒(2x-y+z-3)+λ(4x-3y+5z+9)=0

⇒(2+4λ)x-(1+3λ)y+(1+5λ)z-(3-9λ)=0

Now let,

P3=(2+4λ)x-(1+3λ)y+(1+5λ)z-(3-9λ)=0

Now given P3 is parallel to line,

x+1-2=y+34=z-25

So, using perpendicular condition as normal vector of plane will be perpendicular to given line we get,

-2(2+4λ)-4(1+3λ)+5(1+5λ)=0

⇒-3+5λ=0

⇒λ=35

Hence, equation of plane P3 will be,

P3:22x5-14y5+20z5+125=0

⇒P3=11x-7y+10z+6=0

Now on comparing with ax+by+cz+6=0, we get,

a=11, b=-7 and c=10

∴a+b+c=11-7+10=14

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 1)
If the equation of the plane passing through the point (1,1,2) and perpendicular to the line x-3y+2z-1=0=4x-y+z is Ax+By+Cz=1, then 140(C-B+A) is equal to

Options:

The correct answer is:
15

Solution:
Equation of the required plane is

ax-1+by-1+cz-2=0 ....i

Since, required plane is perpendicular to the line x-3y+2z-1=0=4x-y+z, therefore normal vector of the plane is

n→=i^j^k^1-324-11

⇒n→=-i^+7j^+11k^

So, direction ratios of normal to the plane is -1,7,11

Equation of plane is

-1(x-1)+7(y-1)+11(z-2)=0

⇒-x+7y+11z=28

⇒x-28+7y28+z2811=1

Comparing with Ax+By+Cz=1, we get

A=-128, B=728, C=1128

So,

140C-B+A=1401128-728-128

⇒140C-B+A=140×328=15

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If the equation of the plane that contains the point (-2,3,5) and is perpendicular to each of the planes 2x+4y+5z=8 and 3x-2y+3z=5 is αx+βy+γz+97=0 then α+β+γ=

Options:
A 15
B 18
C 16
D 17

The correct answer is:
15

Solution:
Given

The equation of the plane that contains the point -2,3,5 and is perpendicular to each of the planes 2x+4y+5z=8 and 3x-2y+3z=5 is αx+βy+γz+97=0

Now satisfying the point -2,3,5 in plane P:αx+βy+γz+97=0 we get,

-2α+3β+5γ+97=0 ........i

And using perpendicular condition 2x+4y+5z=8 and αx+βy+γz+97=0 we get,

2α+4β+5γ=0 ......ii

Similarly, 3α-2β+3γ=0 ........iii

Now solving equation i, ii & iii we get,

α=22, β=9 & γ=-16

Hence, α+β+γ=22+9-16=15

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
If the equation of the tangent at the point P3,4 on the parabola whose axis is the x -axis is 3x-4y+7=0, then the distance of the tangent from the focus of the parabola is

Options:

The correct answer is:
2.5

Solution:
Slope of the tangent is 34=tanâ¡Î¸
Slope of the focal chord through P3,4 is tanâ¡2Î¸=2tanâ¡Î¸1-tan2Î¸
i.e. tan2Î¸=2341-916=247
Equation of the focal chord through P is y-4=247x-3
Focus is S116,0
Distance of the focus from the tangent is
3116-40+732+42=52=2.5

Topic: Mathematics, Parabola, JEE Main
If the equation x2+4+3sinâ¡ax+b-2x=0 has atleast one real solution, where a, bâ[0, 2Ï], then one possible value of (a+b) can be equal to

Options:
A 7π2
B 5π2
C 9π2
D None of these

The correct answer is:
7π2

Solution:
x-12+3+3sinâ¡ax+b=0
x-12+3=-3sinâ¡(ax+b)
L.H.S â¥3, RHSââ[ â3,3 ] now sinâ¡(ax+b)=-1
â´x=1 sinâ¡b+a=-1

Topic: Mathematics, Trigonometric Equations, JEE Main
If the equation x2+ax+b=0 has distinct real roots and x2+ax+b=0 has only one real root, then

Options:
A b=0,a>0
B b>0,a<0
C b>0,a>0
D b<0,a>0

The correct answer is:
b=0,a>0

Solution:
Since the equation x2+ax+b=0 has distinct real roots and x2+ax+b=0 has only one real root
So, one root of the equation x2+ax+b=0 will be zero and the other root will be negative.
Hence, b=0 and a>0

Topic: Mathematics, Quadratic Equation, JEE Main
If the equation x2+bx+45=0, b∈R has conjugate complex roots and they satisfy z+1=210, then

Options:
A b2-b=30
B b2+b=72
C b2-b=42
D b2+b=12

The correct answer is:
b2-b=30

Solution:
Let z=α+iβ be one of the roots of the equation x2+bx+45=0, b∈R.

So, its other conjugate complex root will be z=α+iβ¯=α-iβ.

We know that for a quadratic equation Ax2+Bx+C=0, the sum and product of its roots are -BA & CA respectively.

So, the sum of roots of the given equation, α+iβ+α-iβ=-b1⇒2α=-b ......i.

Also, the product of roots of the given equation, α+iβα-iβ=451⇒α2+β2=45 ......ii.

Now, given that z+1=210.

⇒α±iβ+1=210

⇒α+12+β2=210

⇒α+12+β2=40 ......iii

Subtracting equation ii from equation iii, we get

α+12-α2=-5

⇒α+1+αα+1-α=-5 ∵a2-b2=a+ba-b

⇒2α+1=-5

⇒2α=-6

Comparing from equation i, we get

b=6

∴b2-b=62-6=36-6=30.

Also, b2+b=62+6=36+6=42.

Hence, option a is correct.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (08 Jan Shift 1)
If the equation x3-6x2+9x+Î»=0 has exactly one root in 1,3, then Î» belongs to the interval

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

The correct answer is:
-4,0

Solution:
Let, fx=x3-6x2+9x+λ
⇒ f ′ x =3 x 2 −12x+9=3 x 2 −4x+3
=3x-1x-3
f'x<0, ∀x∈1,3f1=4+λ, f3=λ
For fx=0 to have exactly one root in 1,3, the values of f1 and f3 should have opposite signs
⇒f1⋅f3<0
⇒λ+4λ<0
⇒λ∈-4,0

Topic: Mathematics, Application of Derivatives, JEE Main
If the equations 2x2-7x+9=0 and ax2+bx+18=0 have a common root, then a,b∈R

Options:
A a=2,b=-7
B a=-72,b=1
C a=4,b=-14
D a=4,b=-7

The correct answer is:
a=4,b=-14

Solution:
2x2-7x+9=0 has complex roots since D <0
Complex roots occur in conjugate pair if the coefficients are real.
Hence, if it is given that one root of the equations is common, both roots will be common.
So, the coefficients of the equations will be proportional
⇒2a=-7b=918
⇒a=4 and b=-14

Topic: Mathematics, Quadratic Equation, JEE Main
If the equations x 2 + 2 x + 3 = 0 and a x 2 +bx+c=0, a,b,c∈R, have a common root, then a:b:c is:

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

The correct answer is:
1:2:3

Solution:
As the roots of x 2 + 2 x + 3 = 0 are imaginary, hence both the roots are common with a x 2 +bx+c=0 .
∴a1=b2=c3.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2013 (07 Apr)
If the equations $x^2+a x+b=0$ and $x^2+b x+a=0(a \neq b)$ have a common root, then $a+b$ is equal to

Options:
A $-1$
B $1$
C $3$
D $4$

The correct answer is:
$-1$

Solution:
Let $\alpha$ be the common root, then
$$
\begin{array}{rlrl}
& \alpha^2+a \alpha+b & =0 \\
\text { and } & \alpha^2+b \alpha+a & =0 \\
\therefore & & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a}=\frac{1}{b-a} \\
\Rightarrow & & \frac{\alpha}{b-a} & =\frac{1}{b-a} \Rightarrow \alpha=1 \\
\text { Now, } \quad & \frac{\alpha^2}{a^2-b^2} & =\frac{\alpha}{b-a} \\
\Rightarrow & \frac{1}{(a+b)(a-b)} & =\frac{1}{-(a-b)} \\
\Rightarrow & a+b & =-1
\end{array}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
If the equations x2+bx-1=0 and x2+x+b=0 have a common root different from -1, then b is equal to :

Options:
A 2
B 3
C 3
D 2

The correct answer is:
3

Solution:
Let common root be x
x2+bx-1=0 ...1
x2+x+b=0 ...2
1-2, we get x=b+1b-1
Put x in Equation 1, we get
b+1b-12+b+1b-1+b=0
⇒ b+12 + b+1 b-1+bb-12=0
⇒ b2+1+2b+b2-1+bb2-2b+1=0
⇒ 2b2+2b+b3-2b2+b=0
⇒ b3+3b=0
⇒ bb2+3=0⇒b=0 or b2=-3
When b=0 then common root x= -1
But the given common root x≠-1
∴ b2= -3
b= ± 3i
b= 3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2016 (09 Apr Online)
If the equations \( x^{2}+2 x+3 \lambda=0 \) and \( 2 x^{2}+3 x+5 \lambda=0 \) have a non-zero common root, then \( \lambda \) is equal to

Options:
A \( 1 \)
B \( -1 \)
C \( 3 \)
D \( 2 \)

The correct answer is:
\( -1 \)

Solution:
We have,

x2+2x+3λ=0 ...i

2x2+3x+5λ=0 ...ii

Multiplying equation i by 2 and subtracting from ii, we get

-x-λ=0

⇒x=-λ

Putting in i, we get

λ2-2λ+3λ=0

⇒λ2+λ=0

⇒λ=0, -1.

For a non-zero common root, λ=-1.

Topic: Mathematics, Quadratic Equation, JEE Main
If the equilibrium constant for A⇌B+C is Keq(1) and that of B+C⇌P is Keq(2), the equilibrium constant for A⇌P is :

Options:
A Keq(1)/Keq(2)
B Keq(2)-Keq(1)
C Keq(1)+Keq(2)
D Keq(1)Keq(2)

The correct answer is:
Keq(1)Keq(2)

Solution:
On adding Reaction 1st and Reaction 2nd we get.

A⇌P Keq=Keq(1)·Keq (2)

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If the equilibrium constant for the reaction $2 A B \rightleftharpoons A_2+B_2$ is 49 , what is the equilibrium constant for $A B \rightleftharpoons \frac{1}{2} A_2+\frac{1}{2} B_2 ?$

Options:
A 7
B $\frac{1}{7}$
C 24.5
D 49

The correct answer is:
7

Solution:
$\begin{array}{r}2 A B \rightleftharpoons A_2+B_2 \\ K_c=\frac{\left[A_2\right]\left[B_2\right]}{[A B]^2}\end{array}$
again, $\quad \begin{aligned} A B & \rightleftharpoons \frac{1}{2} A_2+\frac{1}{2} B_2 \\ K_c{ }^{\prime} & =\frac{\left[A_2\right]^{1 / 2}\left[B_2\right]^{1 / 2}}{[A B]} \\ K_c & =\left(K_c{ }^{\prime}\right)^2\end{aligned}$
or $\quad K_c{ }^{\prime}=\sqrt{K_c}=\sqrt{49}=7$

Topic: Chemistry, Chemical Equilibrium, JEE Main
If the equilibrium constants of the following equilibria
\(\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{SO}_3 \text { and } 2 \mathrm{SO}_3 \rightleftharpoons 2 \mathrm{SO}_2+\mathrm{O}_2\)
are given by \(\mathrm{K}_1\) and \(\mathrm{K}_2\) respectively, which of the following relations is correct

Options:
A \(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)
B \(K_1=\left(\frac{1}{K_2}\right)^3\)
C \(K_2=\left(\frac{1}{K_1}\right)\)
D \(K_2=\left(K_1\right)^2\)

The correct answer is:
\(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)

Solution:
Hints: \(\mathrm{K}_1=\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}\)
\(\mathrm{K}_2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}\)
Thus \(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)

Topic: Chemistry, Chemical Equilibrium, JEE Main
If the error in the measurement of radius of a sphere is $2 \%$, then the error in the determination of volume of the sphere will be

Options:
A $2 \%$
B $4 \%$
C $6 \%$
D $8 \%$

The correct answer is:
$6 \%$

Solution:
$$
\frac{\Delta V}{V}=\frac{3 \Delta R}{R}
$$

Topic: Physics, Mathematics in Physics, JEE Main
If the error in the measurement of radius of a sphere is $2 \%$, then the error in the determination of volume of the sphere will be

Options:
A $2 \%$
B $4 \%$
C $6 \%$
D $8 \%$

The correct answer is:
$6 \%$

Solution:
$$
\frac{\Delta V}{V}=\frac{3 \Delta R}{R}
$$

Topic: Physics, Mathematics in Physics, NEET, NEET 2008 (Mains)
If the escape speed of a projectile on Earth's surface is 11.2 km s-2 and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth

Options:
A 56.63 km s-1
B 33 km s-1
C 39 km s-1
D 31.7 km s-1

The correct answer is:
31.7 km s-1

Solution:
According to the principal of conservation of energy,
Initial kinetic energy + initial potential energy,
= final kinetic energy + final potential energy
⇒ 12mv2-GMmR=12mv′2+0
⇒ 12mv′2=12mv2-GMmR ..... (i)
As consider, ve= escape velocity
12mve2=GMmR ........ (ii)
∴ From equation (i) and (ii), we get
12mv′2=12mv2-12mve2 ...... (iii)
⇒ v′2=v2-ve2
Now, ve=11.2 km s-1v=3ve .... (iv)
From equation (iii) and (iv), we get
v′2=3ve2-ve2
v′2=9ve2-ve2=8ve2=8×11.22
= 8×11.22
⇒ v′=8×11.22=8×11.2
v′=2×1.414×11.2=31.68 km s-1
∴ Speed of the body far away from the Earth,
v'=31.68 km s-1=31.7 km s-1

Topic: Physics, Gravitation, JEE Main
If the events $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events such that $\mathrm{P}(\mathrm{A})=\frac{3 x+1}{3}$ and $\mathrm{P}(\mathrm{B})=\frac{1-x}{4}$, then the set of possible values of $x$ lies in the interval :

Options:
A
$[0,1]$

B
$\left[\frac{1}{3}, \frac{2}{3}\right]$

C
$\left[-\frac{1}{3}, \frac{5}{9}\right]$

D
$\left[-\frac{7}{9}, \frac{4}{9}\right]$

The correct answer is:
$\left[-\frac{1}{3}, \frac{5}{9}\right]$

Solution:
Since events A and B are mutually exclusive
$$
\begin{aligned}
&\therefore \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1 \\
&\Rightarrow \frac{3 x+1}{3}+\frac{1-x}{4}=1 \\
&\Rightarrow 12 x+4+3-3 x=12 \\
&\Rightarrow 9 x=5 \Rightarrow x=\frac{5}{9} \\
&\therefore x \in\left[-\frac{1}{3}, \frac{5}{9}\right]
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2013 (25 Apr Online)
If the expansion in powers of $x$ of the function $\frac{1}{(1-a x)(1-b x)}$ is $a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots$, then $a_n$ is

Options:
A
$\frac{b^n-a^n}{b-a}$

B
$\frac{a^n-b^n}{b-a}$

C
$\frac{a^{n+1}-b^{n+1}}{b-a}$

D
$\frac{b^{n+1}-a^{n+1}}{b-a}$

The correct answer is:
$\frac{b^{n+1}-a^{n+1}}{b-a}$

Solution:
$(1-a x)^{-1}(1-b x)^{-1}=\left(1+a x+a^2 x^2+\ldots \ldots\right)\left(1+b x+b^2 x^2+\ldots.\right)$
$\therefore$ coefficient of $x^n=b^n+a b^{n-1}+a^2 b^{n-2}+\ldots .+a^{n-1} b+a^n=\frac{b^{n+1}-a^{n+1}}{b-a}$
$\therefore a_n=\frac{b^{n+1}-a^{n+1}}{b-a}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2006
If the expression (1+tan⁡x+tan2⁡x)(1-cot⁡x+cot2⁡x) is positive, then the complete set of values of x is

Options:
A 0,Ï2
B 0,Ï
C R- x=nπ2, n∈I
D 0,Â â

The correct answer is:
R- x=nπ2, n∈I

Solution:
(1+tan⁡x+tan2⁡x)(1+tan2x-tan⁡x)tan2⁡x>0
⇒ 1+tan2⁡x2-tan2⁡xtan2⁡x>0
Since, 1+tan2⁡x>tan2⁡x, ∀x ∈R-x=nπ2, n∈I
Hence, given expression is positive for all values of x ∈R-x=nπ2, n∈I

Topic: Mathematics, Trigonometric Equations, JEE Main
If the expression \(x^2-11 x+a\) and \(x^2-14 x+2 a\) must have a common factor and \(\mathrm{a} \neq 0\), then, the common factor is

Options:
A \((x-3)\)
B \((x-6)\)
C \((x-8)\)
D None of these

The correct answer is:
\((x-8)\)

Solution:
Here Let \(x-\alpha\) is the common factor then \(\mathrm{x}=\alpha\) is root of the corresponding equation \(\therefore \alpha^2-11 \alpha+a=0\)
\(\alpha^2-14 \alpha+2 \mathrm{a}=0\)
Subtracting \(3 \alpha-a=0 \Rightarrow \alpha=a / 3\)
Hence \(\frac{a^2}{9}-11 \frac{a}{3}+a=0, a=0\) or \(a=24\) since \(\mathrm{a} \neq 0, \mathrm{a}=24\)
\(\therefore\) the common factor of \(\left\{\begin{array}{l}x^2-11 x+24 \\ x^2-14 x+48\end{array}\right.\) is clearly \(x -8\)

Topic: Mathematics, Quadratic Equation, JEE Main