SRMJEEE Maths Mock Test - 8 - JEE MCQ

Given, and

Let the circumcentre of the triangle be the origin.
Orthocentre isand the centroid is 

Consider the given expression:

Differentiate both sides with respect to x.

Hence, this is the required solution.

Given, 

In order to rationalise we will first obtain the rationalising factor of

Therefore, the rationalising factor is

Now multiplying and dividing the given expression by we get

Let edge of a cube be 1 units. The diagonals of a cube are OA and BC. So, DR's of diagonals OA are (1, 1, 1) and BC are (0, -1, 1, 1), i.e., (-1, 1, 1)

Now, angle between diagonals,

We know that the intercept form of the equation of a plane is
where x₁, y₁ and z₁ are the intercepts made by the plane on the axes.
Therefore,

Therefore, the equation of the plane is

Hence, this is the required solution.

Total number of functions = 7⁷
Number of onto functions = 7!
Therefore,
required probability =
Hence, this is the required solution.

Consider the given expression.

Hence, this is the required solution.

cos 2 = cos 114° 35’ 30” is surely negative.

Let, y = log (Hog xl) and z = log x,
then 

The given equations are
5x−y=3 ...(i)
y²−6x²=25 ...(ii)
From (i), we have
y = 5x - 3
Substituting this value of y in (ii), we get
(5x−3)² −6x² = 25
⇒19x² − 30x − 16 = 0
⇒19x2−38x+8x−16=0
⇒19x (x − 2) + 8(x − 2)=0
⇒(19x + 8)(x − 2)=0

And substituting these values in (i), we get

If A(x₁, y₁, z₁)  & B(x₂, y₂, z₂) and P divides AB internally in ratio m₁:m₂ then

Required length = 

Let t = sinθ

hence we have that 2t2 + 3t + 1 = 0 ⇒ 2t2 + 2t + t + 1 = 0 ⇒ 2t(t+1) + (t+1) = 0 ⇒ (t+1)(2t+1) = 0 ⇒ t = −1 and t = −12

Hence we have that sinθ = −1 and sinθ = −12.

Solving these we get Remember that θ belongs to [0,2π] hence we have that sinθ = −1 ⇒θ = 3π/2 and sinθ = −12 ⇒ θ = 11π/6 and θ = 7π/6

∴ The solutions are θ1 = 3π/2, θ2 = 11π/6, θ3 = 7π/6
 

Given:
P(x) = x2 + xQ'(1) + Q'(2)
Q(x) = x2 + xP'(2) + P'(3)

Now,
P'(x) = 2x + Q'(1)
P''(x) = 2
Q'(x) = 2x + P'(2)
Q''(x) = 2

Now, at x = 1,
P'(1) = 2 + Q'(1)
Q'(1) = 2 + P'(2)
⇒ P'(1) = 4 + P'(2)
At x = 2,
P'(2) = 4 + Q'(1)
Q'(2) = 4 + P'(2)
⇒ Q'2 = 8 + Q'(1)

Consider the given expression:

Since is positive when 0 < x < 2, and negative when 2< x < 4.
Therefore,

Therefore,

Hence, this is the required solution.

Here, area between 0 and k is A₁, and between k and 1 is A₂. Therefore,

Given:

= log (√2 + 1)

The given product contains the factor cos 90° = 0.

We can write the given expression as

Multiplying numerator and denominator by the conjugate of

Let P be a position vector, say 
Let 
∵C is the midpoint of 

= log(√2+1).

The two curves must in (0,0) and (1,1).
∴ Required area lies above the curve y = x₂ and below x = y² is

Given, 
In order to rationalise we will first obtain the rationalising factor of 

Therefore, the rationalising factor is

Now multiplying and dividing the given expression by (i), we have