Mathematics Test - 41

Given the differential equation:

y'' + 9y = 0

We need to find which function satisfies this equation.

Step 1: Test each option by calculating y'' + 9y

Option A: y = 5 tan 3x

• y = 5 tan 3x

• y' = 5 * 3 sec² 3x = 15 sec² 3x

• y'' = 15 * derivative of sec² 3x = 15 * 2 sec² 3x * tan 3x * 3 = 90 sec² 3x tan 3x (using chain rule)

Now, y'' + 9y = 90 sec² 3x tan 3x + 9 * 5 tan 3x = tan 3x (90 sec² 3x + 45) which is not zero for all x.

Option A does not satisfy the equation.

Option B: y = 5 cos 3x

• First derivative: y' = 5 * (-3 sin 3x) = -15 sin 3x

• Second derivative: y'' = -15 * 3 cos 3x = -45 cos 3x

Now, compute y'' + 9y:

y'' + 9y = -45 cos 3x + 9 * 5 cos 3x = -45 cos 3x + 45 cos 3x = 0

So, Option B satisfies the differential equation.

Option C: y = cos 3x

• y' = -3 sin 3x

• y'' = -9 cos 3x

Compute y'' + 9y:

-9 cos 3x + 9 cos 3x = 0

Option C satisfies the equation.

Option D: y = 6 cos 3x

• y' = -18 sin 3x

• y'' = -54 cos 3x

Compute y'' + 9y:

-54 cos 3x + 9 * 6 cos 3x = -54 cos 3x + 54 cos 3x = 0

Option D satisfies the equation.

Final answer:

Options B, C, and D satisfy the differential equation y'' + 9y = 0.

Among the given choices,

B: y = 5 cos 3x
C: y = cos 3x
D: y = 6 cos 3x

are solutions.

Only Option A is not a solution.

Consider the function y = 7x²

Differentiating w.r.t x, we get

dy/dx = 14x

⇒ dy/dx -14x = 0

Hence, the function y = 7x² is a solution for the differential equation dy/dx - 14x = 0

To find the smallest value of the polynomial

f(x) = x³ - 18x² + 96 on the interval [0, 9], follow these steps:

Step 1: Find the derivative

To locate critical points, we differentiate the function:

f'(x) = 3x² - 36x

Step 2: Set the derivative equal to zero

3x² - 36x = 0

Factor the expression:

3x(x - 12) = 0

This gives two critical points:

x = 0 and x = 12

Step 3: Check which points lie in the interval [0, 9]

Since x = 12 is outside the interval, we only consider:

• x = 0

• x = 9 (right endpoint of the interval)

Step 4: Evaluate f(x) at x = 0 and x = 9

• f(0) = 0³ - 18(0)² + 96 = 96

• f(9) = 9³ - 18(9)² + 96
  = 729 - 1458 + 96
  = -633

Step 5: Conclusion

The smallest value of the function on [0, 9] is:

-633 at x = 9

Solving eqns. (i) and (ii), we get points of intersection (2, 2√3) and (2, -2√3) Substituting these values of x in eq. (ii). Since both curves are symmetrical about r-axis.

Hence the required area 

Let, y = sin(x + α)/sin(x + β)

Then,

dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin²(x + β)

= sin(x+β – x-α)/sin²(x + β)

Or sin(β – α)/sin²(x + β)

So, for minimum or maximum value of x we have,

dy/dx = 0

Or sin(β – α)/sin²(x + β) = 0

Or sin(β – α) = 0 ……….(1)

Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).

Therefore, y has neither a maximum or minimum value.

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)

Differentiating both sides of (1) with respect to x we

f’(x) = 3x² – 24x + 45

3x² – 24x + 45 = 0

Or x² – 8x + 15 = 0

Or (x – 3)(x – 5) = 0

Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5

Therefore, f’(x) = 0 for x = 3 and x = 5.

If h be a positive quantity, however small, then,

f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.

f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.

Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.

So, f(x) has maximum at 3.

Putting, x = 3 in (1)

Thus, its maximum value is,

f(3) = 3³ – 12*3² + 45*3 + 8 = 62.

Poisson distribution shows the number of times an event is likely to occur within a specified time. It is used only for independent events that occur at a constant rate within a given interval of time.