Mathematics Test - 44

y = Ae^2x + Be^-2x 

dy/dx = 2Ae^2x – 2Be^-2x

d²y/dx² = 4Ae^2x + 4Be^-2x

= 4*y

d²y/dx² – 4y = 0

If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by:

Var (X) = E[(X - μ)]², where μ = E(X)

For a discrete random variable X, the variance of X is obtained as follows:

where the sum is taken all over the values of x for which pX(x) > 0. So the variance of X is the weighted average of the squared deviations from the mean μ, where the weights are given by the probability function pX(x) of X.

Important Points

• The standard deviation of X is defined as the square root of the variance.
• The variance cannot be negative, because it is an average of squared quantities.

• Var (X) is often denoted as σ².

Hence, If X is a random variable with mean μ, then the variance of X, denoted by Var (X) is given by Var (X) = E[(X - μ)]²

Step 1: Separate Variables

We are given:

dy/dx = (9y * log x) / (5x * log y)

Now rearrange the terms to separate x and y:

(log y)/y dy = (9 log x)/(5x) dx

Step 2: Integrate Both Sides

Left side:

∫ (log y)/y dy

Let u = log y → du = (1/y) dy

So, the integral becomes: ∫ u du = (u²)/2 = (log y)² / 2

Right side:

∫ (9 log x)/(5x) dx

Take constant 9/5 out: (9/5) ∫ (log x)/x dx

Let v = log x → dv = (1/x) dx

So, the integral becomes: ∫ v dv = (v²)/2 = (log x)² / 2

Thus the right side = (9/5) * (log x)² / 2

Step 3: Combine Results

So we get:

(log y)² / 2 = (9/5) * (log x)² / 2 + C

Multiply both sides by 2 to simplify:

(log y)² = (9/5) * (log x)² + K (where K = 2C)

Step 4: Analyze the Solution

We want a particular form from the options.

If K = 0, then:

(log y)² = (9/5) * (log x)²

This can be rearranged to:

(log y)² - (9/5)(log x)² = 0

Final Answer: Option B

For any scalar k and an n × n square matrix A, the determinant satisfies |kA| = kⁿ|A|. Here n = 3 and k = 2, so |2A| = 2³ × 4 = 8 × 4 = 32. This formula arises from the multilinearity of the determinant function with respect to the rows (or columns) scaled by k.

The equation of tangent to the curve y = f(x), at point (x, y), is

Y – y = dy/dx * (X – x) …..(1)

Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))

Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))

Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,

(x – y/(dy/dx))* (y – x/(dy/dx)) = 4

Or, (y – x/(dy/dx))² – 4dy/dx = 0

Or, x²(dy/dx)² – 2(xy – 2)dy/dx + y² = 0

Solving for dy/dx we get,

dy/dx = [(xy – 2) ± √(1 – xy)]/ x²

(dy/dx)= (dx/dy)-1

So, d²y/dx² = – (dx/dy)^-2 d/dx(dx/dy)

= – (dy/dx)²(d²x/dy²)(dy/dx)

⇒ d²y/dx² + (dy/dx)³ d²x/dy² = 0