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.