BITSAT Mathematics Test - 5 - JEE MCQ

sin2 25° = sin² 65°
= sin² 25° + sin² (90° - 25°)
= sin² 25° + cos² 25° = 1

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

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

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

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

= log(√2+1).

= log (√2 + 1)

Required length = 

Let 
Therefore,

Hence, this is the required solution.

Since, are coplanar,

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

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)

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

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.

Consider the given expression:

Differentiate both sides with respect to x.

Hence, this is the required solution.

Consider the graph of the above function.

Therefore,
the function is not differentiable at x = -3, 0, 3.
Hence, this is the required solution.

Consider the given expression.

Hence, this is the required solution.

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.

We can write the given expression as

Multiplying numerator and denominator by the conjugate of

Given:

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

Given:


Hence,

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

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

Multiplying and dividing (1)by (–√3−√2), we get

Now, multiplying and dividing (2) by (√3 + √2), we get

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

As we know if direction cosines of a line are (l,m,n) then l²+m²+n²=1
Since direction cosines of line are 

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,