A) Function of x
B) Function of y
C) Function of x and y
D) Constant
Correct Answer: D
Solution :
\[y=a\sin x+b\cos x\] Differentiating with respect to x, we get \[\frac{dy}{dx}=a\cos x-b\sin x\] Now \[{{\left( \frac{dy}{dx} \right)}^{2}}={{(a\cos x-b\sin x)}^{2}}\] \[={{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x-2ab\sin x\cos x\] and \[{{y}^{2}}={{(a\sin x+b\cos x)}^{2}}\] \[={{a}^{2}}{{\sin }^{2}}x+{{b}^{2}}{{\cos }^{2}}x+2ab\sin x\cos x\] So, \[{{\left( \frac{dy}{dx} \right)}^{2}}+{{y}^{2}}={{a}^{2}}({{\sin }^{2}}x+{{\cos }^{2}}x)+{{b}^{2}}({{\sin }^{2}}x+{{\cos }^{2}}x)\] Hence \[{{\left( \frac{dy}{dx} \right)}^{2}}+{{y}^{2}}=({{a}^{2}}+{{b}^{2}})\]= constant.
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