JEE Main & Advanced Practice Test- 28 Free Online Test 2026
MCQ Practice Test & Solutions: JEE Main Practice Test- 28 (90 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- 28". These 90 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: 90
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The area bounded by the curve y = x3, x-axis and two ordinates x = 1 to x = 2 is equal to
If x is so small that x3 and higher power of x may be neglected, then [(1+x)3/2 - [1+(1/2)x]3] / [(1-x)1/2] may be approximated as
Area of the triangle in the Argand diagram formed by the complex numbers z, iz and z + iz is
The solution of differential equation (dy/dx)=[((1+x)y)/((y-1)x)] is
and if f is continuous at x = 1 then f(1) =The eccentricity of the ellipse 9x2 + 5y2 - 30y = 0 is
The negation of the statement ' he is rich and happy' is given by
In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): 
is discontinuous for integral values of x.
Reason(R): f(x) is not defined for integral values of x.
Assertion(A):
is discontinuous for integral values of x.Reason(R): f(x) is not defined for integral values of x.
In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): The function f (x) ≡ x5 − 5x3 − 20x + 7 attains its maximum at x = − 1 .
Reason(R) : f ′ (−1) = 0
Assertion(A): The function f (x) ≡ x5 − 5x3 − 20x + 7 attains its maximum at x = − 1 .
Reason(R) : f ′ (−1) = 0
Detailed Solution: Question 15
In the following question, a Statement-1 is given followed by a corresponding Statement-2 just below it. Read the statements carefully and mark the correct answer-
Tangents are drawn from the point (17,7) to the circle x2+y2=169.
Statement-1:
The tangents are mutually perpendicular.
Statement-2:
The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338.
Tangents are drawn from the point (17,7) to the circle x2+y2=169.
Statement-1:
The tangents are mutually perpendicular.
Statement-2:
The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2+y2=338.
In the following question, a Statement of Assertion (A) is given followed by a corresponding Reason (R) just below it. Read the Statements carefully and mark the correct answer-
Assertion(A): If the 21st and 22nd terms in the expansion of (1+x)44 are equal, then the value of x is 7/8.
Reason(R): In the expansion of (x+y)n, (r+1)th term , Tr+1 = nCr+1xn-r+1 yr+1.
Assertion(A): If the 21st and 22nd terms in the expansion of (1+x)44 are equal, then the value of x is 7/8.
Reason(R): In the expansion of (x+y)n, (r+1)th term , Tr+1 = nCr+1xn-r+1 yr+1.
The number of solutions of 2x + y = 4, x - 2 y = 2, 3x + 5y = 6 is
The mean annual salary of all employees in a company is Rs.25,000. The mean salaries of male and female employees are Rs.27,000 and Rs.17,000 respectively, the percentage of males and females employees by the company is
Everybody in a room shakes hand with every body else. The total number of hand-shakes is 66.The number of persons in the room is
If A and B are events such that P(A)>0 and P(B)≠1, then P(A̅/B̅) is equal to
If x is a positive real number different from 1, then the number 1/1 + √x, 1/1 - x , 1/1 - √x, .... are in
Let A = {a,b,c}, B = {b,c,d}, C = {a,b,d,e}, then A ∩ (B ∪ C) is
The distance of the point (2,3,4) from the plane 3x-6y+2z+11=0 is
A force of magnitude 5 units acting along the vector 2 î - 2 ĵ + k̂ displaces the point of application from (1, 2, 3) to (5, 3, 7), then the work done is
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