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JEE Main & Advanced Practice Test- 15 Free Online Test 2026


MCQ Practice Test & Solutions: JEE Main Practice Test- 15 (75 Questions)

You can prepare effectively for JEE JEE Main & Advanced Mock Test Series 2026 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "JEE Main Practice Test- 15". These 75 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 180 minutes
  • - Number of Questions: 75

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JEE Main Practice Test- 15 - Question 1

The equation to the directrix of a parabola if the two extremities of its latus rectum are (2, 4) and (6, 4) and the parabola passes through the point (8, 1) is

Detailed Solution: Question 1

focus is (4, 4) & D can be y = 6 or y = 2

where ‘O’ is origin and S is the focus and D is directrix

JEE Main Practice Test- 15 - Question 2


then the maximum value of  Δ is

Detailed Solution: Question 2

Apply R3 → R3 → R1, we get

= (3cos θ – sin θ)2 So, maximum value of Δ equals 10.

JEE Main Practice Test- 15 - Question 3

The number of solution(s) of the equation 


(where z = x + iy, x, y ∈ R, i2 = –1 and x ≠ 2)

Detailed Solution: Question 3



⇒ x = – √2
∴ z = – √2
Hence only one z will satisfy above equation.

JEE Main Practice Test- 15 - Question 4

Two circles of radii r1 and r2 are both touching the coordinate axes and intersecting each other orthogonally. The value of r1/r2 (where r1 > r2) equals

Detailed Solution: Question 4

Circle is (x – r)2 + (y – r)2 = r2
⇒ x2 + y2 – 2xr – 2yr + r2 = 0
Hence the circles are x2 + y2 – 2xr1 – 2yr1 + r12 = 0    ......(1)

x2 + y2 – 2xr2 – 2yr2 + r22 = 0 .....(2)
As (1) and (2) are orthogonal so
2r1r2 + 2r1r2 = r12 + r22


JEE Main Practice Test- 15 - Question 5

Let X and Y be two matrices satisfying this relations

then Tr.(X) – Tr.(Y) equals
[Note : Tr.(P) denotes trace of matrix P.]

JEE Main Practice Test- 15 - Question 6

The differential equation dx/dy = 3y/2x represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity can be

Detailed Solution: Question 6

2x dx – 3y dy = 0 gives, on integration, 
 
The solution represents a family of hyperbolas given by

whose eccentricity


and eccentricity

it gives a pair of lines which are the asymptotes of the hyperbolas.

JEE Main Practice Test- 15 - Question 7

Line L, perpendicular to the line with equation y = 3x – 5, contains the point (1, 4). The x-intercept of L, is

Detailed Solution: Question 7

put y = 0, x = 13

JEE Main Practice Test- 15 - Question 8

Let A = [aij] (1 ≤ i , j ≤ 3) be a 3 × 3 matrix and B = [bij] (1 ≤ i , j ≤ 3) be a 3 × 3 matrix such that


If det. A = 4, then the value of det. B is

Detailed Solution: Question 8

B = AAT.
Hence, det.
B = |AAT| = |A| |AT| = |A|2 = 42 = 16.

JEE Main Practice Test- 15 - Question 9

Number of words that can be formed using all the letters of the word GARGEE if no two alike letters are together, is

Detailed Solution: Question 9

Total – n(A ∪ B)


Set A represents number of ways when G's are together
Set B represents number of ways when E's are together

Aliter: GG EE A R
Number of words when

Number of words when G's are separated but E's are together = 3! × 4C2 = 36
∴ Number of ways when no two alike letters are together = 120 – 36 = 84

JEE Main Practice Test- 15 - Question 10

If acute angle between the line  and xy plane is α and acute angle between the planes x + 2y = 0 and 2x + y = 0 is β then (cos2α + sin2β) equals

Detailed Solution: Question 10

JEE Main Practice Test- 15 - Question 11

If area of pentagon PQRST be 7, where P(–1, –1), Q(2, 0), R(3, 1), S(2, 2) and T(–1, t), t > 0, then the value of t is

Detailed Solution: Question 11


Area of pentagon PQRST = 7
⇒ ar.(trapezium PQST) + ar.(ΔQRS) = 7

⇒ t = 1

JEE Main Practice Test- 15 - Question 12

The sum of all value of λ for which the lines 2x + y + 1 = 0; 3x + 2λy + 4 = 0; x + y - 3λ = 0 are concurrent, is

Detailed Solution: Question 12


(3λ + 1)(2λ + 1) + 3λ(2λ - 4) = 0
⇒ 6λ2 + 5λ + 1 + 6λ2 - 12λ = 0
⇒ 12λ2 - 7λ + 1 = 0
⇒ (3λ – 1)(4λ – 1) = 0

⇒ Sum = 7/12 

JEE Main Practice Test- 15 - Question 13

If A and B are two independent events such that P(A' ∩ B') = 2/15, P(A ∩ B') = 1/6 then P(B) =

JEE Main Practice Test- 15 - Question 14

A hyperbola has centre at origin and one focus at (6, 8). If its two directrices are 3x + 4y + 10 = 0 and 3x + 4y –10 = 0 and eccentricity is e,then the value of 4e2/5 is equal to

Detailed Solution: Question 14

Distance between centre and focus = ae = 10
Distance between directrices = 2a/e = 4

JEE Main Practice Test- 15 - Question 15

Number of integral values of 'k' for which the chord of the circle x2 + y2 = 125 passing through P(8, k) gets bisected at P (8, k) and has integral slope is

Detailed Solution: Question 15

The slope of the chord is 

⇒ k = ± 1, ± 2, ± 4, ± 8 but (8, k) must also lie inside the circle x2 + y2 = 125

⇒ 64 + k2 – 125 < 0
⇒ k2 < 61
⇒ k can be equal to ± 1, ± 2, ± 4
⇒ 6 values

JEE Main Practice Test- 15 - Question 16

Locus of the feet of perpendiculars drawn from points (1, 2) and (3, 4) on a variable tangent to the conic | z - (1+ 2i) | - | z - (3+ 4i) | = 2 is

Detailed Solution: Question 16


represents a hyperbola with foci (1, 2) and (3, 4) and length of transverse axis = 2.
∴ 2a = 2 ⇒ a = 1
∵ Feet of perpendiculars from foci on any tangent lie on auxilliary circle of the hyperbola.
∴ Locus will be auxilliary circle.
∴ Centre = mid point of foci = (2, 3)

and radius = semi transverse axis = 1
∴ Equation of auxilliary circle is |z - (2 + 3i) |= 1

JEE Main Practice Test- 15 - Question 17

Number of numbers greater than a million and divisible by 5 which can be formed by using only the digits 1, 2,1, 2, 0, 5 and 2 is :

Detailed Solution: Question 17

JEE Main Practice Test- 15 - Question 18

The dual of statement (p v ~ q) ∧ (~ p) is

Detailed Solution: Question 18

In dual statement v replace by ∧ and ∧ replace by v so answer is (p ∧ ~ q) v (~ p).

JEE Main Practice Test- 15 - Question 19

A normal is drawn to the parabola y2 = 9x at the point P(4, 6), S being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

Detailed Solution: Question 19

Required intercept will be equal to the perpendicular distance from the focus on the tangent at P.
Tangent at P,


⇒ 12y = 9x + 36
⇒ 9x - 12y + 36 = 0

JEE Main Practice Test- 15 - Question 20

Consider ellipse  Let C is centre of the ellipse and P is a variable point lying on the ellipse. If the angle between CP and tangent at P is minimum, then P may be

Detailed Solution: Question 20


and C (0, 0)


∴ angle is minimum, when θ = 45°

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 21

(Instruction to attempt numerical value (integer) type question: If your answer is 100 write 100 only. Do not write 100.0)

If then the value of a5 + b4 must be


Detailed Solution: Question 21

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 22

The value of x satisfying the equation

then k is


Detailed Solution: Question 22

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 23

The sum of squares of all integral values of a for which the quadratic expression (x−a)(x−10)+1 can be factored as a product (x+α)(x+β) of two factors and α, β ∈ I must be equal to


Detailed Solution: Question 23

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 24

If f(x) = tan-1 (sin x + cos x)3 is an increasing function, then the value of x in (0, 2π) is x ∈  Then the value of a + 10b + 100c + 1000d must be


Detailed Solution: Question 24

*Answer can only contain numeric values
JEE Main Practice Test- 15 - Question 25

The sum of two digit even numbers which do not end with zero is


Detailed Solution: Question 25

Required sum
= (12 + 14 + 16 + …. + 98) – (20 + 30 + 40 + …. + 90)
44/2 (12 + 98) – 8/2 (20 + 90)
= 22(110) – 4(110)
= (110) (18)
= 1980

JEE Main Practice Test- 15 - Question 26

A non-conducting rod AB of length l has a total charge q. The rod is rotated about an axis passing through its center of mass with a constant angular velocity ω as shown in the figure. The magnetic moment of the rod is

Detailed Solution: Question 26

JEE Main Practice Test- 15 - Question 27

The distance between two parallel plates of a capacitor is a. A conductor of thickness b(b < a) is inserted between the plates as shown in the figure. The variation of effective capacitance between the plates of the capacitor as a function of the distance (x) is best represented by

Detailed Solution: Question 27

JEE Main Practice Test- 15 - Question 28

A solid sphere of radius R, and dielectric constant ‘k’ has spherical cavity of radius R/4. A point charge q1 is placed in the cavity. Another charge q2 is placed outside the sphere at a distance of r from q1. Then Coulombic force of interaction between them is found to be ‘F1’. When the same charges are separated by same distance in vacuum then the force of interaction between them is found to be F2 then

Detailed Solution: Question 28

Coulombic force between them remains same.

Charge flown from battery = CV
Work done = CV2
Heat produced ΔH = ΔU + ΔW

JEE Main Practice Test- 15 - Question 29

Energy stored in the capacitor in it’s steady state is

Detailed Solution: Question 29

Potential across capacitor is zero, hence energy stored is zero.

JEE Main Practice Test- 15 - Question 30

A point charge of 0.1C is placed on the circumference of a non-conducting ring of radius 1m which is rotating about an axis passing from centre and perpendicular to the plane of ring with a constant angular acceleration of 1 rad/sec2. If ring starts from rest at t = 0, the magnetic field at the centre of the ring at t = 10 sec, is

Detailed Solution: Question 30

ω = 0 + 1 × 10 = 10 rad/sec2
∴ v = rω = 1 × 10 = 10 m/s

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