JEE Main Maths Mock Test- 1 Free Online Test 2026

In the given question:

Assertion (A):
The coefficient of x^r in the expansion of (1+x)²⁰ is:

Reason (R):
The expression for  is:

Both statements are true:

• The assertion is correct because ​ is indeed the coefficient of xrx^rxr in the expansion.
• The reason is also correct because  is a well-known property of binomial coefficients.

However, the reason is not the explanation for the assertion. The assertion stands alone, and the reason is simply a property of the binomial coefficient.

Therefore, the correct answer is:

B: Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

To determine the truth of the assertion and reason, let's analyze the matrix.

A matrix is non-singular (or invertible) if its determinant is non-zero.

For the matrix

the determinant is:

det = (1 × 5) − (3 × 4) = 5 − 12 = −7

Since the determinant is non-zero (−7), the matrix is non-singular, and hence invertible.

• The assertion says the inverse does not exist, which is false because the matrix is invertible.
• The reason says the matrix is non-singular, which is true.

Conclusion:
The assertion is false, but the reason is true.
Thus, the correct answer is:
D: Assertion is false but Reason is true.

The equation of the line is:

3x - 4y = λ

The equation of the circle is:

x² + y² - 4x - 8y - 5 = 0

Rewriting the equation of the circle by completing the square:

(x - 2)² + (y - 4)² = 25

This means the center of the circle is (2, 4) and the radius is 5.

To find the value of λ when the line is tangent to the circle, we use the formula for the distance from a point to a line. The line equation 3x - 4y = λ can be written as:

3x - 4y - λ = 0

The distance from the center of the circle (2, 4) to the line is given by:

Substituting A = 3, B = -4, C = -λ, and the center (2, 4):

For the line to be tangent to the circle, the distance must be equal to the radius (5), so:

Multiplying both sides by 5:

|-10 - λ| = 25

This gives two possible solutions:
-10 - λ = 25, so λ = -35
-10 - λ = -25, so λ = 15

Thus, λ can be -35 or 15.
The correct answer is: A: -35, 15

To find the area between the curves y² = 4x and x² = 4y:

Step 1: Find the points of intersection

• The first curve, y² = 4x, is a parabola that opens to the right.
• The second curve, x² = 4y, is a parabola that opens upwards.

To find where they intersect, we substitute x = y² / 4 (from y² = 4x) into x² = 4y, which gives:

(y⁴) / 16 = 4y, which simplifies to y⁴ = 64y.

Solving for y, we get the points of intersection at y = 0 and y = 4.

Step 2: Set up the integral

We now need to find the area between the curves from y = 0 to y = 4.

• For the first curve, x = y² / 4.
• For the second curve, x = 2√y.

The area is given by the integral of the difference between the two x-values:

Step 3: Solve the integrals

Integrating 2√y from 0 to 4 gives

Integrating y² / 4 from 0 to 4 gives

Step 4: Subtract the two results

Now, subtract the second result from the first:

A = 64/3 - 16/3 = 48/3 = 16.

Thus, the area is 16/3 square units.

The correct answer is B: 16/3.

We have

S=2(x²+y²)+7x−7y+5=0

=x²+y²+7/2x−7/2y+5/2=0

Length of tangent from (0,0) =√S1

=√5/2

Hence, the required answer is √5/2.

Step 1:
Calculate the distance between points A and B

Step 2:

Step 3:

AB² + CA² = (√40)² + (√160)² = 40 + 160 = 200
BC² = (√360)² = 360

Since AB² + CA² ≠ BC², it is not a right-angled triangle.
Since AB ≠ BC ≠ CA, it is not an equilateral or isosceles triangle.

Check if the points are collinear:
If the points are collinear, the area of the triangle formed by them is zero.
The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:
1/2 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|.
Area = 1/2 |3(2 - 16) + 5(16 - 4) + (-1)(4 - (-2))|
= 1/2 |3(-18) + 5(12) - 1(6)|
= 1/2 |-54 + 60 - 6|
= 1/2 |0| = 0

Since the area is zero, the points are collinear.
The points A, B, and C are collinear.

y = e^cx
dy/dx = c. e^cx
y' = cy

The given function is:

f(x) = 2 cos(1/3)(x - π)

To find the fundamental period of this function, we need to examine the cosine function.

The general form of a cosine function is:

f(x) = A cos(B(x - C)) + D

Where:
A is the amplitude
B affects the period
C is the phase shift
D is the vertical shift

For a cosine function, the period is given by the formula:

Period = (2π) / |B|

In this case, the coefficient B is 1/3, so the period is:

Period = (2π) / |1/3| = 6π

Thus, the fundamental period of the function f(x) = 2 cos(1/3)(x - π) is: A: 6π.

Consider the planes 3x - 6y - 2z = 15 and 2x + y - 2z = 5.
Statement-1: The parametric equations of the line of intersection are x = 3 + 14t, y = 1 + 2t, z = 15t. To find the line of intersection, compute the direction vector via the cross product of the normal vectors (3, -6, -2) and (2, 1, -2):

The direction vector is (14, 2, 15).
Now, find a point on the line by setting t = 0: Solve 3x - 6y = 15 and 2x + y = 5.
From the second, y = 5 - 2x.
Substitute into the first: 3x - 6(5 - 2x) = 15 → 3x - 30 + 12x = 15 → 15x = 45 → x = 3. Then, y = 5 - 6 = -1. Point: (3, -1, 0). Parametric equations: x = 3 + 14t, y = -1 + 2t, z = 15t.

Statement-1 claims y = -1 + 2t, which is incorrect (should be y = -1 + 2t).
Thus, Statement-1 is false. Statement-2: The vector 14i + 2j + 15k is parallel to the line.
This is true, as it matches the direction vector.
Answer: Statement-1 is false, Statement-2 is true.

The system of equation has a non-zero solution

Clearly, m₁m₂ = - 1. 
Hence, the two tangents arc mutually perpendicular. 
Statement 1 is true. 
Now, the locus of the point of intersection of two mutually perpendicular tangents to the circle x² + y² = r² is the director circle i.e. the circle x² +y² = 2r². 
For the given circle r = 13. .. 
Its director circle is x² + y² = 338. 
Hence, statement 2 is true and  a cogect explanation of statement as the point (17, 7) lies on the director circle of the circle (i). 

ⁿC₁₂ = ⁿC₈ 

Using symmetry property, we know that

ⁿC₁₂ = ⁿC_n-12 

Therefore, ⁿC_n-12 = ⁿC₈ 

Hence, n - 12 = 8

n = 20

Given line is 2x + 3y − 4 = 0 given parabola is 
y² = 4x

l = 2, m = 3, n = −4, a = 4.

Answer (B): (-2, -3)

Total outcomes = 36
Doublet are 6 (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)
Probability of getting doublet = 6/36
= 1/6

Covariance is a quantitative measure of the extent to which the deviation of one variable from its mean matches the deviation of the other from its mean. It is a mathematical relationship that is defined as:
Cov(X,Y) = E[(X − E[X])(Y − E[Y])]
Correlation between two random variables, ρ(X,Y) is the covariance of the two variables normalized by the variance of each variable. This normalization cancels the units out and normalizes the measure so that it is always in the range [0, 1]:

Answer: Correct (A: 0)

Given: Here,AB is the diagonal of square.
The vertices of a square A 
Let the mid−point of AB be EThen coordinates of E are 
Therefore equation of other diagonal is

Given y = f(x)
differentiating w.r.t x 
y' = f'(x) which is the slope of the tangent
Hence the slope of the normal is - 1/f'(x) = 3pi/4 = -1
therefore f'(x) = 1
Hence f'(3) = 1

f (x) = a cos (bx + c) + d ..(i)

As we know, -1 ≤ cos θ ≤ 1

For minimum, cos (bx + c) = −1

From (i), f (x) = −a + d = (d − a)

For maximum, cos (bx + c) = 1

From (i), f (x) = a + d = (d + a)

Range of f (x) = [d − a, d + a]

Alternatively,

-1 ≤ cos(bx + c) ≤1

-a ≤ a cos(bx + c) ≤ a

-a + d ≤ a cos(bx + c) + d ≤ a + d

Range of f(x) = [d – a, a + d]

OP² = x² + y²
y = e^x, y' = e^x,

Final Answer:
So, the minimum value is 0.6, and the expression (2m - n)/5 becomes:
m/n = 14/25, where m = 14 and n = 25.
The minimum value of x² + y² is 0.6.

We are given that the function f(x) is non-constant and thrice differentiable, defined on (-∞, ∞), with the following properties:

f(x) = f(6 - x) (i.e., the function is symmetric about x = 3).

f'(0) = 0, f'(2) = 0, and f'(5) = 0 (the first derivative is zero at these points).

We are asked to find the minimum number of roots of the equation:

(f''(x))² + f'(x) f'''(x) = 0

in the interval x ∈ [0, 6], and then find the sum of the digits of this minimum number of roots.

Step 1: Analyze the symmetry
Since f(x) = f(6 - x), this implies that f'(x) = -f'(6 - x), and hence the function is symmetric about x = 3. This symmetry will help us analyze the behavior of f(x) and its derivatives in the given interval [0, 6].

Step 2: Analyze the equation
We are given the equation:

(f''(x))² + f'(x) f'''(x) = 0

This can be factored as:

f'(x) f'''(x) = -(f''(x))²

We know that f'(0) = 0, f'(2) = 0, and f'(5) = 0, so at these points, the first derivative f'(x) is zero, which means that f'(x) is a factor of the equation.

We need to determine the roots of this equation in the interval [0, 6].

Step 3: Solve for the minimum number of roots
By analyzing the equation and considering the given conditions, we observe that the minimum number of roots occurs at the points where the symmetry of the function holds, which corresponds to three roots. These roots are symmetric about x = 3 and correspond to the points where f'(x) and f''(x) vanish.

Step 4: Calculate the sum of the digits of n
Since the minimum number of roots is 3, the sum of the digits of n is: 3

Thus, the sum of the digits of n is 3.

x²(y + z)y²(z + x)z²(x + y) = a³b³c³ = x³y³z³
⇒ (x + y) (y + z) (z + x) = xyz
⇒ x²(y + z) + y²(z + y) + z²(x + y) + xyz = 0
⇒ a³ + b³ + c³ + abc = 0

Given Parametric Equations:
The parametric equations are:
x = t² - t + 1 — (1)
y = t² + t + 1 — (2)

Step 1: Eliminate the parameter t
To eliminate t, let's first subtract equation (1) from equation (2):

y - x = (t² + t + 1) - (t² - t + 1) y - x = 2t — (Equation 3)

Next, add equation (1) and equation (2):

x + y = (t² - t + 1) + (t² + t + 1) x + y = 2t² + 2 — (Equation 4)

Step 2: Eliminate t between equations (3) and (4)
We have two equations:

y - x = 2t (Equation 3)

x + y = 2t² + 2 (Equation 4)

From Equation (3), solve for t:

t = (y - x)/2

Substitute this value of t into Equation (4):

x + y = 2((y - x)/2)² + 2

Simplify:

x + y = (y - x)² / 2 + 2

Multiply both sides by 2 to eliminate the fraction:

2(x + y) = (y - x)² + 4

Now, move the constant term to the other side:

(y - x)² = 2(x + y - 2)

This is the equation of the parabola. The axis of symmetry is the line where x - y = 0.

Step 3: Find the Vertex
From the equation above, we observe that the vertex lies at the intersection of the axis and the tangent at the vertex.

The axis is x - y = 0, and the tangent at the vertex is x + y - 2 = 0.

By solving the system of equations:

x - y = 0 (axis of symmetry)

x + y - 2 = 0 (tangent at the vertex)

Add these two equations:

(x - y) + (x + y - 2) = 0 2x - 2 = 0 x = 1

Substitute x = 1 into the equation x - y = 0:

1 - y = 0 → y = 1

Thus, the vertex of the parabola is (1, 1).

Step 4: Calculate 2a + 4b
The coordinates of the vertex are (a, b) = (1, 1).

Now, calculate 2a + 4b:

2a + 4b = 2(1) + 4(1) = 2 + 4 = 6

Final Answer: The value of 2a + 4b is 6.

Step 1: Substitute values into f(A, B) and g(A, B)

f(A, B) for A = 18° and B = 36°:

g(A, B) for A = 18° and B = 36°:

Since both the numerator and denominator are identical, the value of g(18°, 36°) simplifies to 1.

Step 2: Compute h(18°, 36°)

Now that we know g(18°, 36°) = 1, we have:

So, the final answer is the value of f(18°, 36°), which requires calculating the sine and cosine of the angles 18° and 36°. Using a calculator:

• sin(18°) ≈ 0.3090
• cos(36°) ≈ 0.8090
• cos(18°) ≈ 0.9511
• sin(36°) ≈ 0.5878

Substitute these values into the expression for f(18°, 36°):

Thus, the value of h(18°, 36°) is approximately 1.55.