BITSAT Mathematics Test - 9 - JEE MCQ

Maximize z = 60X₁ + 50X₂
Subject to

Graphical representation:

The feasible point are (0, 0), (20, 0), (0, 20), and (10,15) Max(0,0) = 0
Max(20, 0) = 60 × 20 ⇒ 1200
Max(0, 20) = 50 × 20 ⇒ 1000
Max(10,15) = (60 × 10) + (50 × 15) ⇒ 1350
∴ the given LPP has unique optimal solution.

The given integral is,

Now by differentiating the above equation we get,

Therefore, equation (1) will become,

Now by substituting log⁡(sin⁡ x) = t in the above equation we get,

Given linear differential equation is:

∴ Complete solution is,

The given equation of the plane is 3x + 4y − 5z − 60 = 0, it can be written in the form

which meets the coordinate axes at the points A(20, 0, 0), 8(0, 15, 0) and (0, 0, −12).
The coordinates of the origin are (0, 0, 0).
Therefore, volume of the tetrahedron OABC is

Given: 3x + 2y + 1 = 0

⇒ 2y = −3x−1

Compare this with the slope-intercept form to find the slope of the line: y = mx + c
So, m = −3/2
Now the slope of the given line is m = −3/2, since the given line is parallel to the required line
we know that the slope of the parallel is the same, i.e. under the condition of parallel lines m₁ = m₂
Now the slope of the required line is also equals to m = −3/2
Equation of line:

Solving y² = 6(x−1) and y² = 3x we get 6(x − 1) = 3x ⇒ x = 2

So, y=±√6 y² = 6(x − 1)

Now, A and B are independent events. 

In a rectangular hyperbola:
⇒ a = b
Eccentricity:

Simplifying, we have:

Canceling the common terms in the numerator and denominator, we have:
⇒ e = √2

We know that, volume of sphere 
Surface area of sphere S = 4πr², where r is the radius of the sphere.
So, rate of change of volume of sphere,

Rate of change of surface area of sphere,

From equation (1) and equation (2),

Therefore, the rate of change of volume of a sphere with respect to its surface area is 2 cm³/cm².

Given: 

Obviously α = −i  (where i² = √−1)
taking αⁿ common from R₂ and  common from R₃ 

Since, |PQ| = |PS| = 2
Shaded part represents the external part of circle having centre (−1,0) and radius 2.
As we know equation of circle having centre z₀ and radius r, is |z − z₀| = r
|z − (−1 + 0i)| > 2
⇒ |z + 1| > 2
Also, argument of z+1 with respect to positive direction of x-axis is π/4

and argument of z+1 in anticlockwise direction is 

From Equations. (1) and (2), we get

x² + y² + z² = (a + b)² + ω²(a + bω)² + (aω² + bω)²
= a² + b² + 2ab + a²ω² + b²ω⁴ + 2abω³ + a²ω⁴ + b²ω² + 2abω³
= a²(1 + ω + ω²) + b²(1 + ω + ω²) + 6ab   [∵  ω⁴ =ω]
= 6ab     [∵  1 + ω + ω² = 0]

Given,
|z| ≥ 3
Now,

We have,

Here, coefficient of x is negative and y is positive, therefore it lies in the second quadrant

If m & n are co-prime then only solution is z = 1.

Given that

Let O,G and C be the orthocenter, centroid and circumcentre respectively, then

|z − (2 + 2i)| = 1 represents a circleof radius 1 and centre C(2 + 2i).
Minimum value of OP (where, P is any point on the circle and O is the origin) i.e 
Minimum value of |z| = OC − radius = 2√2 − 1

f(x) is discontinuous at x = 0.

Given function is

The graph of the function, is

We know that, a function is not differentiable at a point where its graph has a sharp corner.
And, from the graph we can easily conclude that f(x) is non-derivable at x = −2, −1, 0,1, 2
Hence, the set S = {−2, −1, 0,1, 2}.

At x = 0, 1, 2, 3 f changes its definition.
∴ At x = 0 LHL = −1, RHL = 0 ⇒ f is discontinuous at
x = 0
x = 1 LHL = 1, RHL = 2 ⇒ f is discontinuous at x = 1
x = 2 LHL = RHL = f(2) = 4 ⇒ f is continuous at x = 2
x = 3 LHL = 5, f(3) = 6 ⇒ f is discontinuous at x = 1
∴ Points of discontinuity are 0, 1, 3.

Given that f is continuos at π/2

For f(x) to be continuous at x = 0

As we know that domain of the sin⁻¹x is [−1,1] and domain of [x] is x ∈ R
So domain of function f(x) is x ∈ [−1,1].
Now  [sin⁻¹x] is discontinuous at points where sin⁻¹x = −1,0,1
⇒ x = −sin1, 0, sin1
And [x] is discontinuous at points x = 0,1
Now, lets check for x = 0
f(0⁺) = [0⁺] − [0⁺] = 0 − 0 = 0
f(0⁻) = [0⁻] − [0⁻] = −1−(−1) = 0
Hence, f(x) is continuous at x = 0
Therefore, points of discontinuity are −sin1, sin1.

When x ∉ I, both the functions [x] and cos
∴ f(x) is continuous on all non-integral points.

∴ f is continuous at all integral points.
Thus, f is continuous everywhere.

Given that,
Function f : R → R, f(0) = 1
f(2x) − f(x) = x    ...(1)
Now, replacing x by x/2 in (1), we get

Replacing x by x/2 in (2) and so on and repeatedly adding, we get


So, formula for sum of infinite  can be used.

A real valued function  f is said to be continuous if it is continuous at every point in the domain of f i.e. 
∀a∈ domain of function 

f(x) = x² − a, x > 2
b − x, x ≤ 2
f(x) is continuous at x = 2