Question

If y = a sin x + b cos x, then y² + \((\frac{dy}{dx})^2\) is a

• A. function of x and y
• B. function of x
• C. constant
• D. function of y

Hint:

When dealing with functions of the form \( y = a \sin x + b \cos x \), the sum of \( y^2 \) and \( \left( \frac{dy}{dx} \right)^2 \) simplifies to a constant. This is because the sum of squares of sine and cosine terms always results in 1, leading to a constant expression. This property is useful when solving problems involving trigonometric functions.

Answer 1

The Correct Option is C

The correct answer is (C) : constant.

Answer 2

The correct answer is: (C) constant.

We are given the function \( y = a \sin x + b \cos x \), and we are asked to find the value of \( y^2 + \left( \frac{dy}{dx} \right)^2 \).

Step 1: Differentiate \( y = a \sin x + b \cos x \)

The derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = a \cos x - b \sin x \] Step 2: Find \( y^2 \) and \( \left( \frac{dy}{dx} \right)^2 \)

Now, square both \( y \) and \( \frac{dy}{dx} \): \[ y^2 = (a \sin x + b \cos x)^2 = a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x \] Similarly, square \( \frac{dy}{dx} \): \[ \left( \frac{dy}{dx} \right)^2 = (a \cos x - b \sin x)^2 = a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x \] Step 3: Add \( y^2 \) and \( \left( \frac{dy}{dx} \right)^2 \)

Add the two expressions: \[ y^2 + \left( \frac{dy}{dx} \right)^2 = \left( a^2 \sin^2 x + 2ab \sin x \cos x + b^2 \cos^2 x \right) + \left( a^2 \cos^2 x - 2ab \sin x \cos x + b^2 \sin^2 x \right) \] Simplifying: \[ y^2 + \left( \frac{dy}{dx} \right)^2 = a^2 (\sin^2 x + \cos^2 x) + b^2 (\sin^2 x + \cos^2 x) \] Since \( \sin^2 x + \cos^2 x = 1 \), this becomes: \[ y^2 + \left( \frac{dy}{dx} \right)^2 = a^2 + b^2 \] Therefore, \( y^2 + \left( \frac{dy}{dx} \right)^2 \) is a constant, specifically \( a^2 + b^2 \). Thus, the correct answer is (C) constant.