Question

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

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

Answer 1

Answer: (B)

$-n^{2}y$

Given $co{s}^{-1}\left(\frac{y}{b}\right)=log{\left(\frac{x}{n}\right)}^n$

$\left(\frac{y}{b}\right)=cos\left\{nlog\left(\frac{x}{n}\right)\right\}$

$y=bcos\left\{nlog\left(\frac{x}{n}\right)\right\}$

Differentiate with respect to $x$ we get

$\Rightarrow y_1=-bsin\left\{nlog\left(\frac{x}{n}\right)\right\}\times \frac{n}{x}$

$\Rightarrow x{y}_1=-bnsin\left\{nlog\left(\frac{x}{n}\right)\right\}$

Again differentiate with respect to $x$ we get

$\Rightarrow x{y}_2+y_1=-bncos\left\{nlog\left(\frac{x}{n}\right)\right\}\times \frac{n}{x}$

$\Rightarrow x^2{y}_2+x{y}_1=-b{n}^{2}cos\left\{nlog\left(\frac{x}{n}\right)\right\}$

Since $y=bcos\left\{nlog\left(\frac{x}{n}\right)\right\}$, we get

$x^2{y}_2+x{y}_1=-n^{2}y$