VITEEE Maths Test - 6 - JEE MCQ

If a>0, and ax² + bx + c > 0 ∀x∈R
Then, discriminant of the quadratic equation ax² +bx+c=0 will be negative,
i.e. b²−4ac<0
and the roots will be imaginary.

We know that,

Hence, this is the required solution.

Given equation of two curves


Both equations make parabola,
So,
Area of curve,

Thus, Area 
Hence, this is the required solution.

To apply Rolle's Theorem, we need to check the following conditions:

1. Continuity: The function f(x) should be continuous on the closed interval [a, b].

2. Differentiability: The function f(x) should be differentiable on the open interval (a, b).

3. Equal values at endpoints: f(a) = f(b).

Given f(x) = log(5) + log(x³ - 3), let's check these conditions for f(x) on the interval [-1, 1]:

1. Continuity: The function is continuous on the interval [-1, 1] as long as x³ - 3 > 0, because the logarithmic function is defined for positive arguments. Let's check the values of x³ - 3 for x = -1 and x = 1:

• At x = -1, x³ - 3 = (-1)³ - 3 = -1 - 3 = -4 (which is negative).

• At x = 1, x³ - 3 = (1)³ - 3 = 1 - 3 = -2 (which is negative).

Since the arguments inside the logarithmic functions are negative at both endpoints of the interval, the function f(x) is not defined for the entire interval [-1, 1].

Therefore, Rolle's Theorem does not apply in this case as the conditions are not satisfied (specifically, f(x) is not continuous on [-1, 1]).

Thus  c can't be determined using Rolle's theorem.

If tan^-1 x = θ, then x = tanθ, 
⇒ sec² θ = 1 + tan² θ = 1 + x²
⇒ secθ = 
∴ 

⁼

We have, 
= Set of points on and inside the ellipse  so this is a convex set. 

Let P be an event that drawn pen is blue and Q be an event that drawn pen is red.
Then, PQ and QP are two disjointed cases of the given event.
Therefore,

Hence, this is the required solution.

Given: P, Q and R are the angles of a triangle.
P + Q + R = π
P + Q = π - R
Now,

Now, let

Hence, this is the required solution.

Here,

Hence, this is the required solution.

|x|² − 5|x| + 4 < 0
⇒(|x|−1) (|x| −4) < 0
If x > 0 ⇒ 1 < x < 4 ⇒ x ∈ (1,4)
x < 0 ⇒ −4 < x < −1 ⇒  x ∈ (−4,−1)

The equation of tangent at (3, 1) is

The coordinates of point P are

Then, we have to find slope of the perpendicular line through P.

Thus, 
Hence, this is the required solution.

If the scalar triple product of three non-zero vectors is zero, then the vectors are coplanar.

Vertices of the triangle are (0, 0) (-2, -2), (-4, - 8).
∴ Required area

Correct Answer is A.

Given,

must be less than zero to satisfy the given inequality.
Here, (x²−x+1)>0 (∵ D=1−4=−3<0)
where, D denotes discriminant

⇒ x < −1
Hence, x∈(−∞, −1)

Given, mid-points of sides are S(p, 0, 0), T(0, q, 0) and U(0, 0, r). Consider the diagram shown below.

Also, by mid-point theorem,

Similarly,

Therefore,

Hence, this is the required solution.

Here, it is given that

From equations, (1),(2) and (3) we get

Any point on L1 is  and on L₂ is 
The lines will intersect, when

From above results, we get

Therefore,

Hence, this is the required solution.

cos 5° - sin 25° = sin 85° - sin 25° 
= 2 cos 55° sin30°.

(∵ x > a, thereforelx - al = x - a)

Given, log₄ (x − 1) = log₂ (x−3)
⇒ log₄ (x − 1) = 2 log₄ (x−3)
⇒ log₄ (x − 1) = log₄(x − 3)²
⇒ (x−3)² = x − 1
⇒ x² + 9 − 6x = x − 1
⇒ x² − 7x + 10 = 0
⇒ (x − 2) (x − 5)=0
⇒ x = 2, or  x = 5
⇒ x = 5 but x ≠ 2
[∵ x=2 makes log (x - 3) undefined].
Hence, one solution exists.

We have,

Now,

Hence, this is the required solution.

Given constraints are 3y + x ≥ 3, x ≥ 0, y ≥ 0
Let ,
l₁: 3y+x=3
l₂: x = 0
l₃: y = 0

The corner points are A(0,1), B(3,0) So, from the graph, we can say that the area of the feasible region is unbounded. 

=1/2 x √( 2² + 14² + 10² )
=1/2 x √( 4 + 196 + 100 )
=1/2 x √( 300)
=1/2 x 10√3
=5√3

A linear programming deals with the optimization (minimization or maximization) of a linear function (objective function) of a number of variables (decision variables) subject to a number of conditions on the variables, in the form of linear equations or inequations in variables involved.
Hence, L.P.P. is a process of finding the optimum value of an objective function.
So, an objective function is a linear function (of the variable involved) whose maximum or minimum value is to be found.
Hence, the objective function of a L.P.P. is a function to be optimised.