JEE Main Maths Mock Test- 2 - JEE MCQ

Given vertices of the triangle:

• A(5, -2)
• B(-1, 2)
• C(1, 4)

Step 1: Find the slope of the sides of the triangle.
The slope of a line passing through two points (x₁, y₁) and (x₂, y₂) is given by: m = (y₂ - y₁) / (x₂ - x₁)

Slope of line BC: m_BC = (4 - 2) / (1 - (-1)) = 2 / 2 = 1
Slope of line AB: m_AB = (2 - (-2)) / (-1 - 5) = 4 / -6 = -2 / 3
Slope of line AC: m_AC = (4 - (-2)) / (1 - 5) = 6 / -4 = -3 / 2

Step 2: Find the slopes of the altitudes.
The altitude is perpendicular to the side of the triangle. The slope of the altitude is the negative reciprocal of the slope of the side.

The slope of the altitude from vertex A is the negative reciprocal of the slope of line BC: m_{altitude from A} = -1

The slope of the altitude from vertex B is the negative reciprocal of the slope of line AC: m_{altitude from B} = 2 / 3

Step 3: Find the equations of the altitudes.
We use the point-slope form of the line equation: y - y₁ = m(x - x₁).

Altitude from vertex A(5, -2): y - (-2) = -1(x - 5) y + 2 = -x + 5 y = -x + 3

Altitude from vertex B(-1, 2): y - 2 = (2 / 3)(x + 1) y - 2 = (2 / 3)x + (2 / 3) y = (2 / 3)x + 8 / 3

Step 4: Solve the system of equations to find the orthocenter.
To find the intersection point of the two altitudes, we solve the system of equations:

y = -x + 3

y = (2 / 3)x + 8 / 3

Set the right-hand sides equal to each other: -x + 3 = (2 / 3)x + 8 / 3 Multiply through by 3 to eliminate the fractions: -3x + 9 = 2x + 8 Simplify: -3x - 2x = 8 - 9 -5x = -1 x = 1 / 5

Substitute x = 1 / 5 into Equation 1: y = -1 / 5 + 3 = 14 / 5

Final Answer:
The orthocenter of the triangle is (1/5, 14/5).

Thus, the correct answer is A: (1/5, 14/5).

The given equation can be write as
⇒ 4k(x + 1) ² + 3(y + 2) ² = 12
on expanding wee get x² coefficient as 4k
and y² coefficient as 3
but in equation of circle x² coefficient is equal to y² coefficient
therefore 4k = 3
⇒ k = 3/4

Let equation of ellipse be x²/a² + y²/b² = 1.

If the latus rectum is PQ, then for the points P and Q:

x = ae, y²/b² = 1 - e²
⇒ y² = b²(1 - e²)
⇒ y = ±b√(1 - e²).

Hence, P is (ae, b√(1 - e²)) and Q is (ae, -b√(1 - e²)).

If the eccentric angle of the extremities is θ, then:

a cos θ = ae and b sin θ = ±b√(1 - e²).

Thus, tan θ = ±√(1 - e²)/e = ±(b/ae).

Therefore, θ = tan⁻¹(± (b/ae)).

If the argument of a complex number z is θ, then the argument of its conjugate z̅ is the negative of θ.
Therefore, arg(z̅) = -θ.

Hence the answer will be 1.

Correct Answer : b

Explanation : A = {(a, a², a³-1) (b, b², b³-1) (c, c², c³-1)}

=> {(a, a², a³) (b, b², b³) (c, c², c³)} - {(a, a², 1) (b, b², 1) (c, c², 1)} = 0

=> abc{(1, a, a²) (1, b, b²) (1, c, c²)} - {(a, a², 1) (b, b², 1) (c, c², 1)} = 0

=> abc{(a, a², 1) (b, b², 1) (c, c², 1)} - {(a, a², 1) (b, b² 1) (c, c², 1)} = 0

=> (abc-1){(a, a², 1) (b, b², 1) (c, c², 1)} = 0

abc - 1 = 0

=> abc = 1

Angle Measurement Conversion:

Degrees to Radians: Since calculus typically operates in radians, it's essential to convert degrees to radians.

Conversion Formula:

θ radians = θ° × (π/180)

cos(x°) = cos(x × π/180)

Apply the Chain Rule:

The chain rule states that if you have a composite function f(g(x)), then its derivative is f'(g(x)) · g'(x).

Differentiate cos(x × π/180):

d/dx [cos(x × π/180)] = -sin(x × π/180) × (π/180)

Simplify the Expression:

d/dx [cos(x°)] = -(π/180) × sin(x°)

Here, sin(x°) implies that the sine function takes the angle in degrees, consistent with the original function's angle measurement.

Final Answer: d/dx [cos(x°)] = -(π/180) sin(x°)

Assertion (A):
The angle between  and  is acute, which means the angle between the two vectors is less than 90°. This condition is equivalent to the dot product of the two vectors being positive. The expression for the angle condition gives:

The first vector:

The second vector:

For the vectors to form an acute angle, their dot product must be positive. The condition derived from this is that m ∉ [-2, -1/2].

Reason (R):
For any two vectors a and b:

If the angle between them is acute, a . b > 0 (positive dot product).

If the angle between them is obtuse, a . b < 0 (negative dot product).

Explanation:
The dot product of two vectors a and b is defined as:
a . b = |a| |b| cos(θ)
where θ is the angle between them.
If θ < 90°, cos(θ) is positive, so a . b > 0 (acute angle).
If θ > 90°, cos(θ) is negative, so a . b < 0 (obtuse angle).

Answer: A: Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

To find the point on the curve y = x² that is nearest to the point (3, 0), we need to minimize the distance between the point on the curve and the point (3, 0).

The distance between any point (x, y) on the curve and the point (3, 0) is given by the distance formula:

D(x) = √[(x - 3)² + (y - 0)²]

Since y = x² on the curve, substitute y = x² into the distance formula:

D(x) = √[(x - 3)² + (x² - 0)²]

This simplifies to:

D(x) = √[(x - 3)² + x⁴]

To minimize this distance, we need to minimize D(x)² (since the square root function is increasing, minimizing the square of the distance will also minimize the distance). So, we define:

f(x) = (x - 3)² + x⁴

Now, take the derivative of f(x) to find the critical points:

f'(x) = 2(x - 3) + 4x³

Set f'(x) = 0 to find the critical points:

2(x - 3) + 4x³ = 0

Simplify this:

2x - 6 + 4x³ = 0

4x³ + 2x - 6 = 0

Factor out 2:

2(2x³ + x - 3) = 0

So, we have the cubic equation:

2x³ + x - 3 = 0

This equation can be solved by trial and error or using numerical methods. One possible solution is x = 1.

Substitute x = 1 back into the equation y = x²:

y = 1² = 1

Thus, the point on the curve y = x² nearest to (3, 0) is (1, 1).

Answer: D: (1, 1)

Given:

Raise both sides to the power of 12:

Degree is the power of d²y​ / dx², which is 4.

Assertion: Using the vector triple product identity: a⃗×(b⃗×c⃗) = (a⃗⋅c⃗)b⃗ − (a⃗⋅b⃗)c⃗
Check perpendicularity:

To  so perpendicular.

To  which is not necessarily zero.

To c⃗: Similarly, not necessarily zero.

Thus, a⃗ × (b⃗ × c⃗) is perpendicular to a⃗ but not necessarily to b⃗and c⃗. The Assertion is false.

Reason: The cross product p⃗ × q⃗​ is perpendicular to both p⃗ and q⃗, and q⃗ × p⃗ = −p⃗ × q⃗ is also perpendicular to both. The Reason is true.
Answer: D (Assertion false, Reason true).

The given question asks to find the sum of the sequence:

i + i² + i³ + i⁴ + ... up to 1000 terms, where i² = -1 (the imaginary unit).

First, recall the powers of i:

i¹ = i

i² = -1

i³ = -i

i⁴ = 1

The powers of i repeat every 4 terms, i.e., the sequence of powers of i follows a periodic cycle of 4: i, -1, -i, 1.

Thus, every 4 terms the sum is:

i + (-1) + (-i) + 1 = 0

Since the pattern repeats every 4 terms, and 1000 is divisible by 4 (1000 ÷ 4 = 250), the sum of 1000 terms will be 250 full cycles of i, -1, -i, 1. Each cycle adds up to 0.

Therefore, the sum of the first 1000 terms is: 250 × 0 = 0

Answer: D: 0

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
⇒ 0.65 = P(A) + P(B) - 0.15
⇒ P(A) + P(B) = 0.80

P(Ā) + P(B̄) = 2 - (P(A) + P(B))
= 2 - 0.80
= 1.2

A leap year has 366 days (52 weeks and 2 extra days).

Extra day pairs: {Sun, Mon}, {Mon, Tue}, {Tue, Wed}, {Wed, Thu}, {Thu, Fri}, {Fri, Sat}, {Sat, Sun}.

Probability of extra days including Tuesday: 2/7.

Probability of exactly 52 Tuesdays: 1 - 2/7 = 5/7.

To solve for the product of the real roots of the equation:

t²x² + |x| + 9 = 0

We need to analyze the equation carefully.

First, note that the absolute value function (|x|) always gives a non-negative value, so:

1. The term |x| is always greater than or equal to zero.
2. The term 9 is positive (it is +9).
3. The term t²x² will be positive if x² is positive and t² is a positive constant (t is not zero).

Now, for the equation to be equal to zero, t²x² + |x| + 9 = 0, all three terms would need to cancel each other out. However, since t²x² is always positive (or zero if x = 0) and |x| and 9 are always positive, the sum of these terms can never be zero for any real x.

Thus, the equation has no real roots.

Therefore, the product of the real roots does not exist.

The correct answer is: A: does not exist.

If x, y and z respectively represent AM, GM and HM between two numbers a and b, then
y² = xz
Here x = 5, y = 4
then 16 = 5 x z
z = 16/5

N the given problem it is SD of x is 3: (or Variance of x is 9).
As Variance = (Sx)².
We know the relation : correlation coefficient (r) = Cov (x,y) / (Sx * Sy) so, 0.28 = 7.6 / (3 * Sy)
From here we get the value of SD of Y : Sy = 9.05.

2tan²x – 5sec x = 1
2 (sec²x – 1) – 5secx = 1
2sec²x – 5sec – 3 = 1
∴  cosx = 1/3

For z:10 choices (0 to 9).

For x,y with x<y,x≥1: ⁹C₂  = 36 Pairs

Total: 10⋅36 = 360

= 5 + 5 - 4

= 6

Sum of 9 terms = 9 ⋅ 15 = 135

New sum with 10 terms = 10 ⋅ 16 = 160

Added term = 160 − 135 = 25