If $y=a \sin x+b \cos x$ (where a and b are constants), then $y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d}…

If $y=a \sin x+b \cos x$ (where a and b are constants), then $y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2$ is
  1. a function of $x$.
  2. a function of $x$ and $y$.
  3. a function of $y$.
  4. a constant.

Solution

$y=\mathrm{a} \sin x+\mathrm{b} \cos x$
Differentiating w.r.t. $x$, we get $\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{a} \cos x-\mathrm{b} \sin x$ $\therefore \quad$ Now, $\begin{aligned} & y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2 \\ = & (\mathrm{a} \sin x+\mathrm{b} \cos x)^2+(\mathrm{a} \cos x-\mathrm{b} \sin x)^2 \\ = & \mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x+2 \mathrm{ab} \sin x \cos x+\mathrm{a}^2 \cos ^2 x \\ & +\mathrm{b}^2 \sin ^2 x-2 \mathrm{ab} \sin x \cdot \cos x \\ = & \mathrm{a}^2\left(\sin ^2 x+\cos ^2 x\right)+\mathrm{b}^2\left(\sin ^2 x+\cos ^2 x\right) \\ = & \mathrm{a}^2+\mathrm{b}^2 \\ \therefore \quad & y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2=\mathrm{a}^2+\mathrm{b}^2 \end{aligned}$
Here, $\mathrm{a}, \mathrm{b}$ are constants. $\therefore \quad y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2$ is also a constant.

Asked in: MHT CET 2024 (02 May Shift 1)

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