JEE Main Maths Test- 3 Free Online Test 2026

lim_{x → 0⁡}(1+Sin(x))^Cosec(x)

Put sin(x) = t we get

lim_{t → 0⁡}(1+t)^(1⁄_t) = e.

When the correlation coefficient (r) is exactly 1, it indicates a perfect positive linear relationship between the two variables. This means:

• All data points lie precisely on a straight line.
• There is no variation among the data points.

As a result:

• The regression lines for y on x and x on y will be identical.
• These lines are described as coincident.

Therefore, the correct answer is C. Coincident.

First we consider theta as x
1/1-cosx+isinx=1/2cos^2 x/2 + 2isinx/2cosx/2
=sec(x/2)/2 {1/cosx/2 + isinx/2}
=sec(x/2)/2 (1/e^ix/2). (e^ix=cosx+isinx){eulers theorem)
=sec(x/2)/2 e^-ix/2
=sec(x/2)/2 [ cos(-x/2) + i sun(-x/2)]
=1/2secx/2.cosx/2.-1/2itanx/2
=1/2.-i(1/2tanx/2)
=Re(z)=1/2

Loga+log=log(ab)
tan(90-A)=cotA
log(tab.tan2......tan45........cot2.cot1). tan45=1
tanA.cotA=1
=log(1)=0

Given the quadratic equation in base b:

x² - (1 * b + 1)x + (2 * b + 2) = 0 with solutions x = 3 and x = 6, substitute each root to find b.

• For x = 3:

3² - (b + 1)(3) + (2b + 2) = 0.
Simplifying gives:

• 8 - b = 0 ⇒ b = 8.

• For x = 6:

6² - (b + 1)(6) + (2b + 2) = 0.
Simplifying gives:

• 32 - 4b = 0 ⇒ b = 8.

Both roots confirm b = 8. Thus, the base is 8.

The sum S can be calculated by breaking it down into two parts involving sums of squares and cubes. Using the known formulas for these sums, we compute each part separately and then combine them to find the result.

• Part 1: Calculate the sum of squares.
• Part 2: Calculate the sum of cubes.

The calculations confirm that the final value matches option C.

1. Identify Variables: Let’s denote BA = 4 units and BC = x units.

2. Area Calculation: The area A of the right-angled triangle is given by:

• A = ¹/₂ × BA × BC
• A = ¹/₂ × 4 × x = 2x.

3. Analyse the Behaviour: As x increases, the area A also increases linearly without bound because there is no constraint on the length of side BC.

4. Conclusion: Since x can be infinitely large, the area A doesn't have a maximum value—it approaches infinity as x increases.

Given that the third term of a geometric progression (G.P.) is 4, we need to find the product of the first five terms.

1. Identify the third term: In a G.P., the nth term is given by a_n = a r^n-1. Therefore, the third term is:
• a r² = 4
2. Express the product of the first five terms:
• The first five terms are: a, a r, a r², a r³, and a r⁴. Their product is:
• a · a r · a r² · a r³ · a r⁴ = a⁵ · r^1+2+3+4 = a⁵ · r¹⁰
3. Substitute a in terms of r:
• From the third term, we have a = 4/r². Substitute this into the product:
• (4/r²)⁵ · r¹⁰ = 4⁵ · (1/r¹⁰) · r¹⁰ = 4⁵

Thus, the product of the first five terms is 4⁵, which corresponds to option B.

Given the sum of an infinite G.P. is S = a / (1 - r) = 20 and the sum of their squares is S' = a² / (1 - r²) = 100:

1. From S = 20, we have:

   a = 20(1 - r)

2. Substitute a into the second equation:

   (20(1 - r))² / (1 - r²) = 100

3. Simplify the numerator and denominator:

   400(1 - r)² / ((1 - r)(1 + r)) = 100

4. Cancel (1 - r):

   400(1 - r) / (1 + r) = 100

5. Divide by 100:

   4(1 - r) / (1 + r) = 1

6. Solve for r:

   4(1 - r) = 1 + r ⟹ 4 - 4r = 1 + r ⟹ 3 = 5r ⟹ r = 3/5

7. Substitute r back into a = 20(1 - r):

   a = 20(1 - 3/5) = 20(2/5) = 8

Thus, the first term is 8.

Starting with the given equation:

9x² - 12xy cos(θ) + 4y² = 0

We move the terms involving cos(θ) to one side:

-12xy cos(θ) = -9x² - 4y²

Divide both sides by -12xy to solve for cos(θ):

cos(θ) = (9x² + 4y²) / (12xy)

Thus, the value of cos(θ) is:

(9x² + 4y²) / (12xy)

The given equation is rewritten in standard form by dividing both sides by 225:

(x + 1)²/9 + (y + 2)²/25 = 1.

This represents a vertically oriented ellipse with:

• Centre at (-1, -2)
• a² = 25
• b² = 9

The distance to the foci is calculated as:

c = √(a² - b²) = √(16) = 4.

Thus, the foci are located at:

• (-1, -2 + 4) and (-1, -2 - 4)

This simplifies to (-1, 2) and (-1, -6).

The correct answer is B.

To determine the number of diagonals in a polygon, use the formula:

Number of Diagonals = n(n^-3) / 2

For n = 15:

• Number of Diagonals = 15(15 - 3) / 2
• 15 x 12 / 2 = 90

The equation 9x² - 16y² = 144 can be rewritten as:

x²/16 - y²/9 = 1, which is the standard form of a hyperbola.

Therefore, the correct answer is B.

Given the circle x² + y² = 4 with radius 2, we need to find the locus of midpoints of chords that subtend a right angle at the centre.

• Understanding the condition:

  If a chord subtends a right angle at the centre, the endpoints form an isosceles right triangle with legs equal to the radius.

• Midpoint calculation:

  Let the midpoint be (h, k). The distance from the centre (0, 0) to this midpoint can be found using the formula involving the chord length.

• Chord length and midpoint distance:

  The chord length is 2√2, so the distance from the centre to the midpoint is √((2)² - (√2)²) = √(4 - 2) = √2.

• Locus equation:

  Therefore, all such midpoints lie on the circle x² + y² = 2.

Thus, the correct answer is option B: x² + y² = 2.

f(0) = d
OC = d ⇒    OB = d
OA = d/2
Roots of eq. are d & d/2
Product of roots d × d/2 = d

x² + x = t
t² – 5t – 6 = 0
t = 6 or t = –1
x² + x = 6             x² + x = –1
x² + x – 6 = 0       x² + x + 1 = 0
Roots are real      sum of roots = –1

⇒ ƒ'(x) > 0 ∀ x ≥ 16  ⇒ a = 16