JEE Main Maths Mock Test- 3 Free Online Test 2026

We are given the equation of the circle:

x² + y² - 8x + 4y + 4 = 0

Step 1: Convert to standard form
Group x and y terms:

(x² - 8x) + (y² + 4y) + 4 = 0

Complete the square:

For x² - 8x: add and subtract 16 (since (8/2)² = 16)

For y² + 4y: add and subtract 4 (since (4/2)² = 4)

So we get:

(x - 4)² + (y + 2)² = 16

Step 2: Analyze the circle
Center: (4, -2)

Radius: √16 = 4

Now check the shortest distances from the center to the x-axis and y-axis:

Distance to x-axis = |-2| = 2 → Radius = 4 → It crosses the x-axis

Distance to y-axis = |4| = 4 → Radius = 4 → It just touches the y-axis

Final Answer: B: y-axis

We are given the equation of an ellipse:

25(x + 1)² + 9(y + 2)² = 225

Step 1: Convert to standard form
Divide the entire equation by 225:

(25/225)(x + 1)² + (9/225)(y + 2)² = 1
⇒ (x + 1)² / 9 + (y + 2)² / 25 = 1

So, we now have the ellipse in standard form:

(x + 1)² / 9 + (y + 2)² / 25 = 1

Step 2: Identify the components
Center: (-1, -2)

a² = 25 ⇒ a = 5 (major axis)

b² = 9 ⇒ b = 3 (minor axis)

Since the larger denominator is under the y-term, it's a vertical ellipse.

Step 3: Use foci formula for vertical ellipse
Foci are located at:

(h, k ± c) where
c = √(a² - b²) = √(25 - 9) = √16 = 4

So, foci:

(-1, -2 ± 4) = (-1, 2) and (-1, -6)

Final Answer: D: (-1, 2), (-1, -6)

We are given a complex number in polar (Euler) form:

This is a classic De Moivre's Theorem case:

Step 1: Understand the complex number form
Let:

z = (cos(π/3) + i·sin(π/3)) = cis(π/3)

So,

z^(3/4) = (cis(π/3))^(3/4) = cis(π/3 × 3/4) = cis(π/4)

But since we are taking the fourth roots, there will be 4 distinct roots (as per complex number theory), and we must compute the product of all these roots.

Step 2: Use formula for product of all n roots
For a complex number in polar form:
z = r·cis(θ)
Then its n roots are given by:

w_k = r^(1/n) · cis((θ + 2πk)/n), for k = 0 to n − 1

Now if z = r^m · cis(mθ), then:

Product of all n roots of z = r^(m) = (modulus)^(m)

So, the product of all roots = z^m, and then take product of its n roots = r^(m) raised to the power 1 ⇒ i.e., modulus^(m)

But more generally:
The product of all n roots of a complex number is equal to
(modulus of the number)^(m/n) × (–1)^(n(n–1)/2)

Step 3: Apply to this question
Given:

z = (cis(π/3))³ = cis(π)

We are taking 4th roots ⇒ n = 4

So the 4 roots of cis(π) will be:

4th roots of (cis(π))
Let’s find the product of these 4 roots

Use:

Product of all n roots = r × (–1)^{(n(n – 1)/2)}

Here,

r = 1 (since modulus of cis(π) = 1)

n = 4

So:

Product = 1 × (–1)^(4×3/2) = (–1)⁶ = 1

Final Answer: B: 1

We are given the equation:

log(x + y) - 2xy = 0

We are asked to find y(0), i.e., the value of y when x = 0.

Step 1: Substitute x = 0
log(x + y) - 2xy = 0
⇒ log(0 + y) - 2·0·y = 0
⇒ log(y) = 0

Step 2: Solve log(y) = 0
log(y) = 0
⇒ y = 10⁰ = 1

Final Answer: A: 1

We are given:

f(x) = 2x + 3

g(x) = x² + 7

We are asked to find x such that:

f(g(x)) = 25

Step 1: Find f(g(x))
f(g(x)) = f(x² + 7)
= 2(x² + 7) + 3
= 2x² + 14 + 3
= 2x² + 17

Step 2: Solve the equation
Set f(g(x)) = 25:
2x² + 17 = 25
⇒ 2x² = 25 - 17
⇒ 2x² = 8
⇒ x² = 4
⇒ x = ±2

Final Answer: B: +2

We are asked to find the area bounded by the curve:

y = x² - 4x, the x-axis, and the vertical line x = 2

Step 1: Understand the region
We are looking for the area between the curve and the x-axis from the leftmost intersection point up to x = 2.

First, find where y = x² - 4x = 0
⇒ x(x - 4) = 0 ⇒ x = 0 and x = 4
So the parabola touches the x-axis at x = 0 and x = 4

Since we are only considering from x = 0 to x = 2, this lies between the roots, where the parabola is below the x-axis (because it's an upward-opening parabola and the vertex is at x = 2).

So, the function is negative between 0 and 2 → area is taken as –∫ from 0 to 2 to make it positive.

Step 2: Set up the integral
Area =
= – [ (x³/3) - (4x²/2) ] from 0 to 2
= – [ (8/3) - (8) - (0 - 0) ]
= – [ 8/3 - 8 ]
= – [ (8/3 - 24/3) ]
= – (–16/3)
= 16/3

Final Answer: A: (16/3) sq. units

We are given:

Assertion (A):
f(x) = log(x³) and g(x) = 3·log(x) are equal.

Reason (R):
Two functions f and g are said to be equal if their domains, ranges are equal and f(x) = g(x) ∀ x in the domain.

Step 1: Analyze the functions
We know from logarithmic properties:

log(x³) = 3·log(x) for x > 0
⇒ So, f(x) = g(x) when x > 0

Now check domains:

• Domain of f(x) = log(x³):
  x³ is defined and positive only when x > 0, so domain is x > 0
• Domain of g(x) = 3·log(x):
  log(x) is defined only for x > 0, so again domain is x > 0

So:

• Domains are equal
• f(x) = g(x) for all x in the domain
• Therefore, f and g are equal functions

Assertion is TRUE
And since the reason correctly explains why f(x) and g(x) are equal...
Final Answer: A: Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

We are given:

• A point on the ellipse: (3√3·cosθ, sinθ)
• Ellipse equation: x²/27 + y²/1 = 1
• Tangent is drawn at the point
• We need to find the value of θ (between 0 and π/2) where the sum of intercepts (x-intercept + y-intercept) of the tangent is minimum
• Step 1: Parametric form of the ellipse

Standard ellipse: x²/a² + y²/b² = 1, with:

• a² = 27 ⇒ a = 3√3
• b² = 1 ⇒ b = 1

Parametric form of point on ellipse:
x = a·cosθ = 3√3·cosθ,
y = b·sinθ = sinθ
So the point (3√3·cosθ, sinθ) lies on the ellipse, and this is the general parametric point on it.

Step 2: Tangent to the ellipse at parametric point
The equation of tangent to an ellipse at point (a·cosθ, b·sinθ) is:
(x·cosθ)/a + (y·sinθ)/b = 1
Substitute a = 3√3, b = 1:
(x·cosθ)/(3√3) + y·sinθ = 1

Step 3: Find x- and y-intercepts
To find intercepts:

• x-intercept: Put y = 0
  ⇒ (x·cosθ)/(3√3) = 1
  ⇒ x = 3√3 / cosθ
• y-intercept: Put x = 0
  ⇒ y·sinθ = 1
  ⇒ y = 1 / sinθ

Sum of intercepts = x-intercept + y-intercept
= (3√3)/cosθ + 1/sinθ
Let this sum be S(θ):
S(θ) = (3√3)/cosθ + 1/sinθ

Step 4: Minimize S(θ)
We minimize:
S(θ) = (3√3)/cosθ + 1/sinθ for 0 < θ < π/2
Try each given option:

• θ = π/6
  cos(π/6) = √3/2, sin(π/6) = 1/2
  S = (3√3)/(√3/2) + 1/(1/2) = 6 + 2 = 8
• θ = π/3
  cos(π/3) = 1/2, sin(π/3) = √3/2
  S = (3√3)/(1/2) + 1/(√3/2) = 6√3 + 2/√3 = (6√3 + 2/√3) ≈ 11.6
• θ = π/4
  cos(π/4) = sin(π/4) = √2/2
  S = (3√3)/(√2/2) + 1/(√2/2) = (6√3)/√2 + 2/√2
  = (6√3 + 2)/√2 ≈ 13.86
• θ = π/8
  cos(π/8) ≈ 0.924, sin(π/8) ≈ 0.383
  S ≈ (3√3)/0.924 + 1/0.383 ≈ 5.62 + 2.61 = 8.23

Minimum occurs at: θ = π/6, where S = 8
Final Answer: A: π/6

Statement 1: The argument of z₁z₂ is arg(z₁) + arg(z₂), not arg(z₁) ⋅ arg(z₂). False
Statement 2: For polynomials with real coefficients, imaginary roots come in conjugate pairs. True
Statement 3: Order relations exist for real numbers, not complex numbers. False
Statement 4: Cube root of unity (ω = e^{(2πi / 3)}) differs from the fourth root (e^{(2πi / 4)}). True

arg(z₁z₂) = arg(z₁) + arg(z₂)
Imaginary roots of f(x) = 0 with real coefficients occur in conjugate pairs.

Answer: B (2 and 4 only).

Consider the equation

Since  are linearly independent (assuming p⃗ \vec{p} p​ and q⃗ \vec{q} q​ are not parallel), the coefficients of each vector must be zero:
a − b = 0, b − c = 0, c − a = 0

Solving: a = b, b = c, c = a  ⇒  a = b = c.

The reason is true and explains the assertion, as linear independence forces the coefficients to be zero.

Answer: A.

We are asked to find how many 3-digit numbers between 99 and 1000 can be formed using the digits:
2, 3, 7, 0, 8, 6
(That is: 6 digits total)
Each digit can be used only once per number.

Step 1: Understand the constraint
All numbers between 99 and 1000 are 3-digit numbers
So we are forming 3-digit numbers
The first digit (hundreds place) cannot be 0 (as that would make it a 2-digit number)
Digits must be from the set: {0, 2, 3, 6, 7, 8}
No repetition of digits

Step 2: Count valid 3-digit numbers
We choose 3 digits from the 6 available, no repetition, and form a 3-digit number where the first digit ≠ 0

Let's break this into cases depending on choices for the hundreds place

Case-wise Counting:
Total valid 3-digit numbers =
Sum over all possible valid hundreds digits ≠ 0

Possible digits for the hundreds place (first digit):
{2, 3, 6, 7, 8} → 5 choices (excluding 0)

For each such choice:
Choose 1 digit for the hundreds place from 5 options (excluding 0)
Then choose 1 digit for the tens place from the remaining 5 digits
Then choose 1 digit for the units place from the remaining 4 digits

So:
Total = 5 × 5 × 4 = 100

We have, probability that A can solve the problem = P(A) = 1/2 ,
And in this way P(B) = 1/3 and P(C) = 1/4.
P(A cannot solve the problem) = 1 – P(A) = 1/2 ,
P(B cannot solve the problem) = 1 – P(B) = 1 – 1/3 = 2/3,
P(C cannot solve the problem) = 1 – P(C) = 1 – 1/4 = 3/4.
P(A, B, and C cannot solve the problem) = 1/2 x 2/3 x 3/4 = 1/4.
Therefore , P(Problem will be solve) = 1 – P(Problem is not solved by any of them)
= 1 – 1/4 = 3/4

Odd no. on the first die

1, 3, 5

multiple of 3 on the other die

3, 6

now let's see the combination of these two events happening simultaneously

as the question says

(1,3) , (1,6) , (3,3) , (3,6) , (5,6) , (5,3)

total no of favourable events = 6

total no of events throwing two dice simultaneously = 6² = 36

so probability = 6/36 = 1/6

For ax² + bx + c = 0, roots α,β, the discriminant Δ₁ = b² − 4ac.

For Ax² + Bx + C = 0, roots α + K, β + K.

The discriminant depends on the difference of roots:

Ratio: 

Answer:

We are given:

• f(x) is a second-degree polynomial, i.e., f(x) = ax² + bx + c
• f(1) = f(−1)
• a, b, c are in A.P. (Arithmetic Progression)

We are to determine the nature of f'(a), f'(b), f'(c)

Step 1: Use f(1) = f(−1)
Let f(x) = Ax² + Bx + C

Then:

• f(1) = A(1)² + B(1) + C = A + B + C
• f(−1) = A(1)² − B + C = A − B + C

Set f(1) = f(−1):

⇒ A + B + C = A − B + C
⇒ B = −B ⇒ B = 0

So the function becomes:
f(x) = Ax² + C

Step 2: Use this to find derivative
Now, f'(x) = d/dx (Ax² + C) = 2Ax

So, f'(x) = 2Ax

Step 3: a, b, c are in A.P.
Let’s say:

• a = b − d
• b = b
• c = b + d

Then:

• f'(a) = 2A(b − d)
• f'(b) = 2Ab
• f'(c) = 2A(b + d)

So, the three derivatives are:

• f'(a) = 2Ab − 2Ad
• f'(b) = 2Ab
• f'(c) = 2Ab + 2Ad

Clearly, these form an A.P.

Final Answer: A: A.P.

• A pie diagram represents the proportion of each category in a total, making it ideal for showing expenditure distribution across different heads.
• Bar diagrams compare quantities, histograms show frequency distributions, and frequency polygons are for continuous data, none of which are as suitable.

Answer: B: pie diagram.

We are given the vertices of a triangle:

• A = (2, (√3 − 1)/2)
• B = (1/2, −1/2)
• C = (2, −1/2)

We need to find the orthocentre of this triangle.

Step 1: Understand the geometry
Let’s label:

• A = (2, (√3 − 1)/2)
• B = (1/2, −1/2)
• C = (2, −1/2)

From this, we observe:
Points A and C share the same x-coordinate ⇒ AC is vertical
Points B and C share the same y-coordinate ⇒ BC is horizontal

Step 2: Slopes and altitudes
Since BC is horizontal, its perpendicular (altitude from A) will be vertical, so it passes through A and has x = 2
Since AC is vertical, its perpendicular (altitude from B) will be horizontal, so it passes through B and has y = −1/2

Step 3: Find intersection point of the two altitudes
x = 2 (from A's altitude)
y = −1/2 (from B's altitude)
⇒ Intersection point = (2, −1/2)

Final Answer: B: (2, −1/2)

For a geometric progression (G.P.),
the terms are:

Taking logarithms of the terms:

We consider the determinant of the matrix:.

Subtract row 1 from rows 2 and 3:

Factor out q − p and r − p:

Final Observation: The last matrix has identical rows (2 and 3), so its determinant is 0.
Final Answer: 
Correct option: A is correct.

Let the house be from (0, 0) to (0, 100) and the window at (d, 64).

The angle subtended at the window by the house is 90°.

Vertical distances: Top to window = 100 - 64 = 36 m, bottom to window = 64 m.

By the right-angle condition: d2 = 36 × 64 = 2304.
⇒ d √2304
⇒ d = 48

Answer: 48 meters.

The integral:

Use the identity:

Here, a = x, b = 1 − x, and 1 + x(1 − x) = 1 + x − x².

So:

Integral becomes:

Substitute u = 1 − x in the second integral:

Thus:

Correct: The answer is 0.

We are given the region:

A = { (x, y) : (y²)/2 ≤ x ≤ y + 4 }

This describes a region bounded between two curves in terms of x as a function of y.

Step 1: Interpret the region

We are integrating with respect to y, because the limits of x are in terms of y:

• Left boundary: x = y² / 2
• Right boundary: x = y + 4

So the horizontal strip at any y runs from x = y²/2 to x = y + 4.

To compute area:
Area = ∫ [from y₁ to y₂] (right - left) dy
= ∫ [y₁ to y₂] [(y + 4) − (y² / 2)] dy

Step 2: Find limits for y
Find intersection points of:
y² / 2 = y + 4
⇒ Multiply by 2: y² = 2y + 8
⇒ y² − 2y − 8 = 0
⇒ (y − 4)(y + 2) = 0
⇒ y = −2 and y = 4

Step 3: Integrate

Break it up:

Now integrate from y = −2 to y = 4:

• ∫ y² dy = (y³)/3
• ∫ y dy = (y²)/2
• ∫ 4 dy = 4y

Plug in limits:

Compute each:

1. −(1/2)[(64/3 − (−8/3))] = −(1/2)(72/3) = −(1/2)(24) = −12
2. (16/2 − 4/2) = (8 − 2) = 6
3. (4×4 − 4×(−2)) = (16 + 8) = 24

Add them:

Total area = −12 + 6 + 24 = 18

Final Answer: 18 sq. units

We are given that a line makes angles α/2, β/2, γ/2 with the positive directions of the x-, y-, and z-axes, respectively.

We are to find the value of:
cosα + cosβ + cosγ

Step 1: Use direction cosines identity
For a line making angles θ₁, θ₂, θ₃ with x, y, z axes respectively:
cos²θ₁ + cos²θ₂ + cos²θ₃ = 1

In this case, the angles made are α/2, β/2, γ/2, so:
cos²(α/2) + cos²(β/2) + cos²(γ/2) = 1
Let’s denote:

• cos(α/2) = a
• cos(β/2) = b
• cos(γ/2) = c

Then:
a² + b² + c² = 1
We are asked to find cosα + cosβ + cosγ
Now use the double angle identity:
cos(2θ) = 2cos²θ − 1

So:

• cosα = 2a² − 1
• cosβ = 2b² − 1
• cosγ = 2c² − 1

Add all:
cosα + cosβ + cosγ = (2a² − 1) + (2b² − 1) + (2c² − 1)
= 2(a² + b² + c²) − 3
= 2(1) − 3 = −1

Final Answer: −1

We are given the following conditions:

Vectors a, b, c are each perpendicular to vector sums:
a ⊥ (b + c)
b ⊥ (c + a)
c ⊥ (a + b)

Also given:
|a + b| = 6
|b + c| = 8
|c + a| = 10

We are to find |a + b + c|

Step 1: Use the dot product conditions
a · (b + c) = 0 ⇒ a · b + a · c = 0
b · (c + a) = 0 ⇒ b · c + b · a = 0
c · (a + b) = 0 ⇒ c · a + c · b = 0

Note that:

From (1): a·b = −a·c
From (2): b·c = −b·a
From (3): c·a = −c·b

Let’s add all three equations:
(a·b + a·c) + (b·c + b·a) + (c·a + c·b) = 0
Group similar terms:
a·b + b·a + a·c + c·a + b·c + c·b = 0

Since dot product is commutative: a·b = b·a, a·c = c·a, etc., so:
2(a·b + b·c + c·a) = 0
⇒ a·b + b·c + c·a = 0

Step 2: Let’s expand |a + b + c|²
We’ll use the identity:
|a + b + c|² = (a + b + c) · (a + b + c)
= a·a + b·b + c·c + 2(a·b + b·c + c·a)

We are given:
|a + b| = 6 ⇒ (a + b)·(a + b) = 36 ⇒ a·a + b·b + 2a·b = 36
|b + c| = 8 ⇒ b·b + c·c + 2b·c = 64
|c + a| = 10 ⇒ c·c + a·a + 2c·a = 100

Now add all three:
(a·a + b·b + 2a·b)
(b·b + c·c + 2b·c)
(c·c + a·a + 2c·a)
= 36 + 64 + 100 = 200

Group terms:
a·a appears twice
b·b appears twice

c·c appears twice
2(a·b + b·c + c·a)

So:
2(a·a + b·b + c·c + a·b + b·c + c·a) = 200
Divide both sides by 2:

a·a + b·b + c·c + a·b + b·c + c·a = 100
But from earlier, we have:
a·b + b·c + c·a = 0

So:
a·a + b·b + c·c = 100

Now plug into:
|a + b + c|² = a·a + b·b + c·c + 2(a·b + b·c + c·a)
= 100 + 2×0 = 100
⇒ |a + b + c| = √100 = 10

Final Answer: 10