If y = a sin x + b cos x,then y^2 + left( fra | vedclass

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(D) Given $y = a \sin x + b \cos x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = a \cos x - b \sin x$. Now,calculating the square of

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Constant

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(D) Given $y = a \sin x + b \cos x$. Differentiating with respect to $x$,we get: $\frac{dy}{dx} = a \cos x - b \sin x$. Now,calculating the square of the derivative: $\left( \frac{dy}{dx} \right)^2 = (a \cos x - b \sin x)^2 = a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x$. Calculating the square of $y$: $y^2 = (a \sin x + b \cos x)^2 = a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x$. Adding both expressions: $y^2 + \left( \frac{dy}{dx} \right)^2 = (a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x) + (a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x)$. Grouping the terms: $y^2 + \left( \frac{dy}{dx} \right)^2 = a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x)$. Since $\sin^2 x + \cos^2 x = 1$,we get: $y^2 + \left( \frac{dy}{dx} \right)^2 = a^2(1) + b^2(1) = a^2 + b^2$. Since $a$ and $b$ are constants,$a^2 + b^2$ is a constant. Therefore,the correct option is $D$.

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

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