Question

If ${\cos }^{-1}\left(\frac{y}{b}\right)=n\log \left(\frac{x}{n}\right)$ then $x^2{y}_2+x{y}_1=$

• A. $n^{2}y$
• B. $n^2{y}^2$
• C. $-n^{2}y$
• D. $ny$

Answer 1

The correct option is C $-n^{2}y$

$Wehave{\cos }^{-1}\frac{y}{b}=n\log \frac{x}{n}⟹y=b\cos [n\log \frac{x}{n}].........equation1Differentiating\left(1\right),weget\frac{dy}{dx}=-b\sin \left(n\log \left(\frac{x}{n}\right)\right)\frac{n}{\frac{x}{n}}\cdot \frac{1}{n}x\frac{dy}{dx}=-bn\sin \left(n\log \left(\frac{x}{n}\right)\right)Againdifferentiating,wehavex\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=-bn\cos \left(n\log \frac{x}{n}\right)\frac{n}{\frac{x}{n}}\frac{1}{n}⟹x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=-b{n}^2\cos \left(n\log \frac{x}{n}\right)=-n^{2}yHence{x}^2{y}_2+x{y}_1=-n^{2}yThereforethecorrectoptionisOPTION3$