VITEEE: Subject Wise and Full Length VITEEE PCME Mock Test - 3 Free Online

D = 2008! 2009! 2010!
R₃ R₃ - R₂; R₂ R₂ - R₁
D = 2008! 2009! 2010!
D = (2008!)³ (2009)² (2010) (2)
∴ - 4 = (2009)² (2010) (2) - 4
= 2[[(2008 + 1]² (2010) - 2]
= 2[(2008)² (2010) + 2(2008) (2010) + 2010 - 2]
= 2[(2008)² (2010) + 2(2008) (2010) + 2008]
Which is divisible by 2008.

A vector coplanar with 2i + j + k and i - j + k is of the form

This vector will be orthogoal to 5i + 2j + 6k if 

⇒ 
So a is of the form α(3j - k)
Thus, a required unit vector is 

We have the p.d.f. of a random variable X as:
f(x) = 3(1 - 2x²), 0 < x < 1
= 0, otherwise
∴





= 3 x (179/2592)
= 179/864

Using characteristic equation = 0
⇒ -20 - λ + λ² + 21 = 0
⇒ λ² - λ + 1 = 0
⇒ A² - A + I = 0
⇒ A³ - A² + A = 0
⇒ A³ + I = 0
⇒ A³ = - I
So, A⁴⁸² = (A³)¹⁶⁰⋅A² = (−I)¹⁶⁰⋅A² = I⋅(A − I) = A − I
A⁷⁰⁰ = (A³)²³³ ⋅ A = (−I)²³³ ⋅ A = −A
A³⁴⁵ = (A³)¹¹⁵ = (−I)¹¹⁵ = −I. Thus, A⁴⁸² = A − I, A⁷⁰0 = −A , A³⁴⁵ = −I
Answer: Correct (C)

Given: Parabola: y² = 4ax
Check when line y = c is tangent to the parabola.

Substitute y = c into the parabola:
⇒ c² = 4ax ⇒ x = c² / (4a)
So, line y = c intersects parabola at one point ⇒ it is a tangent if there's only one point of intersection.
That’s already true here, so y = c is a tangent if the point lies on the parabola.

Point of contact:
(x, y) = (c²/4a, c)
Find derivative at this point:
From y² = 4ax ⇒ dy/dx = 2a / y
At point (c²/4a, c): dy/dx = 2a / c

Since line y = c has slope 0, it can be tangent only if 2a/c = 0 ⇒ not possible unless a = 0
Hence, y = c is tangent only when it touches at exactly one point and satisfies tangency condition.

From standard result:
Line y = c is tangent to y² = 4ax when c = 2√a
Final Answer: D: None of these

Answer: A → 10

Use the Leibniz rule for the n-th derivative of a product: (uv)⁽ⁿ⁾ = Σ_k=0ⁿ C(n,k) u^(k) v^(n-k).

Let u = x and v = sin x. Then u⁽⁰⁾ = x, u⁽¹⁾ = 1 and u^(k) = 0 for all k ≥ 2, so only the terms with k = 0 and k = 1 contribute.

Therefore y⁽ⁿ⁾ = x·sin⁽ⁿ⁾(x) + n·sin^(n-1)(x), and evaluating at x = 0 gives y⁽ⁿ⁾(0) = n·sin^(n-1)(0) because the first term vanishes.

For n = 10, y⁽¹⁰⁾(0) = 10·sin⁽⁹⁾(0). Derivatives of sin x cycle every 4, so sin⁽⁹⁾(x) = cos x, hence sin⁽⁹⁾(0) = cos(0) = 1.

Thus y⁽¹⁰⁾(0) = 10, so option A is correct.

Take A = a₁ + a₂ and B = a₄·eᵃ⁵ · a₄·eᵃ⁶

There are 3 constants, namely A, a₃, and B. Then, we have:

y = A·sin(x + a₃) − B·eˣ

Differentiate step-by-step:

⇒ dy/dx = A·cos(x + a₃) − B·eˣ

⇒ d²y/dx² = −A·sin(x + a₃) − B·eˣ

Now add y:

⇒ d²y/dx² + y = −2B·eˣ

Differentiate both sides:

⇒ d³y/dx³ + dy/dx = −2B·eˣ

From earlier:

dy/dx = A·cos(x + a₃) − B·eˣ

So:

d³y/dx³ + dy/dx = d²y/dx² + y

Rewriting:

d³y/dx³ − d²y/dx² + dy/dx − y = 0

Final Result: This is a differential equation of order 3 and degree 1.

If A is a non-singular matrix, then
(1) (adj A)^-1 =
(2) adj (kA) = k^n-1. adj A
(3) adj (adj A) = |A|^n-2A
(4) adj (A^-1) =

g(x) = f(15 - |x - 10|)
= 15 - |15 - |x - 10| - 10| = 15 - |5 - |x - 10||

Non-differentiable at points where slopes change: x = 5,10,15
Answer: Correct (A).

Given:



Integrating, we get


Since it passes through (0, 1), so c = 0.
Thus, we get the equation of the curve as: y(x + 1) = 1.

Put f(x) = t
f′(x) dx = dt

= 

= 

Thus 

Answer: Correct (B).

Given, the length of latus rectum = (1/3) × length of major axis 

⇒ 6b² = 2a²
⇒ 3a²(1 − e²) = a²,  (∵ b² = a²(1 − e²))
⇒ 3e² = 2
⇒ e =  eccentricity is always positive of an ellipse. 

Let the equation of asymptotes be
2x² + 5xy + 2y² + 4x + 5y + λ = 0  ...(1)
The above equation represents a pair of straight lines.
Therefore, abc + 2fgh − at² − bg² − ch² = 0
Here, a = 2, b = 2, h = 5/2, g = 2, f = 5/2
and c = λ,

⇒ λ = 2
On putting the value of λ in equation (1) we get
2x² + 5xy + 2y² + 4x + 5y + 2 = 0
Which is the equation of the asymptotes.

Given that A = {1, 2, 3}.

A relation R is defined as R = {(1,2), (2,1)}.

It is seen that {(1,1), (2,2), (3,3)} ∉ R.

Therefore, R is not reflexive.

Now, (1,2) ∈ R and (2,1) ∈ R, so R is symmetric.

Now, (1,2) ∈ R and (2,1) ∈ R, but (1,1) ∉ R. So, R is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Given the inequalities:

x² + 6x − 27 > 0

⇒ x² + 9x − 3x − 27 > 0

⇒ x(x + 9) − 3(x + 9) > 0

⇒ (x − 3)(x + 9) > 0

⇒ x > 3 or x < −9 …(1)

For −x² + 3x + 4 > 0

⇒ x² − 3x − 4 < 0

⇒ x² − 4x + x − 4 < 0

⇒ x(x − 4) + 1(x − 4) < 0

⇒ (x − 4)(x + 1) < 0

⇒ −1 < x < 4 …(2)

From (1) and (2), the solution is (3, 4).

Let the equation of the line be y = mx + c

x-intercept: -c/m
y-intercept: c

Arithmetic mean of the reciprocals of the intercepts:

(-m/c + 1/c) / 2 = 1/4

Solving, we get:
2(1 - m) = c

Equation of the line:

y = mx + 2(1 - m)
Rearranging:
(y - 2) - m(x - 2) = 0

Thus, the line always passes through (2,2).

Constraints:
5x + 4y ≥ 20, x ≤ 6, y ≤ 4, x ≥ 0, y ≥ 0.

Intersection points:
5x + 4y = 20, x = 0: y = 5 (not feasible, y ≤ 4).
y = 4: 5x + 16 = 20 ⟹ x = 4/5. Point: (4/5, 4).
x = 6: 30 + 4y = 20 ⟹ y = –5/2 (not feasible).
y = 0: 5x = 20 ⟹ x = 4. Point: (4, 0).
x = 6, y = 4: 5·6 + 4·4 = 46 > 20.
Feasible. Vertices: (4, 0), (6, 0), (6, 4), (4/5, 4).
Forms a quadrilateral. Answer: Correct (D).

Since the projections of  on the axes are given, we can represent as:

The magnitude of  is given by:

= 13

Substitute y = x + 2
a( x+ 2)² + b(x + 2) + c = ay² + by + c = 0
Roots are y = α, β so:
x = y − 2 = α −2 , β − 2

Major axis = 6 = 2a
⇒ a = 3
Given e = 1/2, we use the formula b² = a²(1 − e²)
⇒ b = 3√3/2

Thus, the required equation is:

Multiply by 36:

4(x−7)² + 16y² = 36

⇒ 3x² + 4y² − 42x + 120 = 0

Let a, b ∈ A, then (a, b) ∈ R ⇒ (b, a) ∈ R⁻¹.
Since R = R⁻¹, we get (b, a) ∈ R.
This confirms that R is symmetric.

Since the asymptotes 3x + 4y = 2 and 4x - 3y + 5 = 0 are perpendicular to each other, the hyperbola is a rectangular hyperbola. The eccentricity of a rectangular hyperbola is √2.

We are given that:

sin²θ + sin²ϕ = 1/2 .....(1)
cos²θ + cos²ϕ = 3/2 .....(2)

After squaring and adding both equations, we get:

(sin²2θ + cos²2θ) + (sin²2ϕ + cos²2ϕ) + 2(sin²θ sin²ϕ + cos²θ cos²ϕ) = 1/4 + 9/4

⇒ cos²θ cos²ϕ + sin²θ sin²ϕ = 1/4 (∵ sin²A + cos²A = 1)

Now using the identity:
cos(A − B) = cosAcosB + sinAsinB

We get:

⇒ cos(2θ − 2ϕ) = 1/4
⇒ cos²(θ − ϕ) = 5/8

Thus, the final answer is B) 5/8.

Given that,


∴I₁ > I₂ and I₄ > I₃

For continuity at x = 5


We know that

⇒ b(5 − π) + 3 = a(5 − π) + 1
⇒ 5 − b + 3 = 5 − a + 1
⇒ b(5 − π) + 3 = a(5 − π) + 1
⇒ 5 − b + 3 = 5 − a + 1
⇒ 3 − 1 = a(5 − π) − b(5 − π)
⇒ 3 − 1 = 5a − 5b
⇒ (a − b)(5 − π) = 2
⇒ a − b = (2 / (5 − π))

We know that Im(z) = (1 / 2i) (z - z̅).

Thus,

z₁Im(z̅₂z₃) + z₂Im(z̅₃z₁) + z₃Im(z̅₁z₂)

= (1 / 2i) [(z₁ (z̅₂z₃ - z₂z̅₃)) + (z₂ (z̅₃z₁ - z₃z̅₁)) + (z₃ (z̅₁z₂ - z₁z̅₂))]

= (1 / 2i) (z₁z̅₂z₃ - z₁z₂z̅₃ + z₂z̅₃z₁ - z₂z₃z̅₁ + z₃z̅₁z₂ - z₃z₁z̅₂)

= 0

Thus, the given expression is independent of z₁, z₂, and z₃.

Correct Answer: D) Independent of z₁, z₂, and z₃.

We know that

Since a > 1/√3​, the principal value may include π.
Therefore