BITSAT Mathematics Test - 10 - JEE MCQ

Given:

Given: 
α, β, γ are the roots of the 2x³ - 3x² + 6x + 1 = 0

Given
Objective function
Maximize, Z = X₁ + 2X₂
Constraints

Non neagative constarints
X₁, X₂ ≥ 0
The above equations can be written as,

Plot the above equations on X₁ - X₂ graph and find out the solution space.

Now, find out the value of the objective function at every extreme point of solution space.

Since the value of the objective function is maximum at A. There A(0, 2) is the optimal solution.

It is given that,

Taking LHS,

On rationalizing the denominator,

We know that, i² = -1

Now according to the question,
⇒ i = 1, which is only possible if,
⇒ i^x = 1 only when x = 4n, where n is any positive integer.

Arrange the data in order : 31, 34, 34, 37, 38, 41, 43, 44. Find the middle value (s) : 37 and 38.
Since there are two middle values find their mean : 37 + 38 = 75.
75/2 = 37.5
Therefore, the median of this data set is 37.5.

Given-
100 cards are numbered from 1 to 100.
Total outcomes = 100
Perfect square numbers = 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Favorable outcomes = 10
P(getting a card with a perfect square number) 

⇒ P(getting a card with a perfect square number) = 10/100
⇒ P(getting a card with a perfect square number) = 1/10

Circle is given as x² + y² = ax + by
or, x² + y² - ax - by = 0               ...(i)
Then line is given as cx - by + b² = 0

Substituting the value of y from (ii) in (i), we get

If it is a perfect square, then c - a = 0 or c = a

Given: Equation of curve is x² + y² - 2x - 4y + 1 and tangent to the given curve is parallel to Y - axis.
Let point of the contact be (x₁, y₁).
As we know that, if a tangent of any curve is parallel to Y- axis then dx/dy = 0
Now by differentiating equation of curve x² + y² - 2x - 4y + 1 = 0 with respect to y we get, 

∵ (x₁, 2) is the point of contact i.e x = x₁, y = 2  will satisfy the equation x² + y² - 2x - 4y + 1 = 0

So, the required points are: (-1, 2) and (3, 2)

Given:



Max. of cos 4θ = 1

Minimum of 4θ = -1

We have, y = tan⁻¹(sec x − tan x)

[sin²A + cos²A = 1, sin²A = 2sin A cos A, cos²A = cos²A − sin²A]

On differentiating 2^x + 2^y = 2^x+y w.r.t. x, we get

Using chain rule 
we get,

So, |cos x − sin (x)| can be written as cos x − sinx

Given,
y = (1 + x)(1 + x²)(1 + x⁴)…(1 + x²ⁿ)  ......(1)
Taking the logarithm on both side of above equation (1), we get:

Differentiating equation (2) with respect to x, we get:


From equation (1), we have:
At x = 0, y = 1.
Put x = 0 in equation (3), we get:

  (differentiating both sides w.r.t. x)

 (again differentiating both sides w.r.t. x)

  (using the equation (i)).

Given,
y = log_e (sin(x²))
On differentiating w.r.t. x, we get

On replacing x by 1/x, we get

On multiplying Eq. (1) by 5 and Eq. (2) by 3 and then on subtracting, we get

Let y = x^x, then by taking log on both sides, we have:-
log y = x.logx
Differentiating both sides w.r.t. x, we have,

endpoints of double ordinate are
A(a cos θ,− b sin θ) B(a cos θ,− b sin θ)
the point which divides AB in 1 : 2 ratio

Let the equation of the required ellipse be 
But the ellipse passes through the point (2, 1)


Hence, equation is

Equation of normal to the ellipse at the point (0,3) is

Which is the equation of y-axis

We have,

Comparing with the standard equation of ellipse 

a = 8, b = 10
Hence,

Given point is (4√3, 5)
Focal distance = b ± ey

Given equation of tangent is  and equation of tangent at the point (a cos ϕ, b sin ϕ) on the ellipse is 
Comparing these 2 equation we get

Any point on the ellipse  can be taken as P(a cosθ, b sinθ)
The equation of tangent at point P is 

The equation of line perpendicular to tangent is

Since, the equation (i) passes through the focus (a_e,0), then we get, 


Thus, the equation (i) becomes 

Now, the equation of line joining centre and point of contact (a cos θ, b sin θ) is

Solving the equation (ii) & (iii), we get the point of intersection Q whose abscissa is a/e.
Hence, Q lies on the corresponding directrix x = a/e.

Centre of the ellipse will be 
Major axis = 6 = 2a
⇒ a = 3

Thus required equation is 
Putting the values of a and b, we get

Let S₁ = 3x² + 5y² − 32
and S₂ = 25x² + 9y² − 450

For the point (3, 5)
S₁ = 3(3)² + 5(5)² − 32 = 120 > 0
and S₂ = 25(3)² + 9(5)² − 450
= 225 + 225 − 450 = 0
∴ Point (3, 5) lies outside the first ellipse and for the second ellipse lies on the ellipse.
Hence, two tangents for first ellipse and one tangent for the second ellipse can be drawn
So total number of tangents are = 3

Let, the equation of tangent which is perpendicular to the line 3x + 4y = 7 is

Since, it is a tangent to the ellipse 
Applying the condition of tangency c² = a²m² + b², we get, 

∴ Equations are 4x − 3y = ±√65