Search any question & find its solution
Question:
Answered & Verified by Expert
If $y=a \sin x+b \cos x$, then $y^2+\left(\frac{d y}{d x}\right)^2$ is a
Options:
Solution:
2629 Upvotes
Verified Answer
The correct answer is:
constant
Given, $y=a \sin x+b \cos x$
$\begin{aligned} & \frac{d y}{d y}=a \cos x-b \sin x \\ & \begin{aligned} y^2+\left(\frac{d y}{d x}\right)^2=(a \sin x+b \cos x)^2 \\ +(a \cos x-b \sin x)^2\end{aligned}\end{aligned}$
$\begin{aligned} & =a^2 \sin ^2 x+b^2 \cos ^2 x+2 a b \sin x \\ & \quad \cdot \cos x+a^2 \cos ^2 x+b^2 \sin ^2 x-2 a b \sin x \cdot \cos x \\ & =a^2\left(\sin ^2 x+\cos ^2 x\right)+b^2\left(\sin ^2 x+\cos ^2 x\right) \\ & =a^2+b^2 \quad\left[\because \sin ^2 x+\cos ^2 x=1\right] \\ & =\text { constant }\end{aligned}$
$\begin{aligned} & \frac{d y}{d y}=a \cos x-b \sin x \\ & \begin{aligned} y^2+\left(\frac{d y}{d x}\right)^2=(a \sin x+b \cos x)^2 \\ +(a \cos x-b \sin x)^2\end{aligned}\end{aligned}$
$\begin{aligned} & =a^2 \sin ^2 x+b^2 \cos ^2 x+2 a b \sin x \\ & \quad \cdot \cos x+a^2 \cos ^2 x+b^2 \sin ^2 x-2 a b \sin x \cdot \cos x \\ & =a^2\left(\sin ^2 x+\cos ^2 x\right)+b^2\left(\sin ^2 x+\cos ^2 x\right) \\ & =a^2+b^2 \quad\left[\because \sin ^2 x+\cos ^2 x=1\right] \\ & =\text { constant }\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.