Question

$y=\log x$ is a solution of

• A. $x{y}_2-y_1=0$
• B. $x{y}_2+y=0$
• C. $x{y}_1+y_2=0$
• D. $x{y}_2+y_1=0$

Answer 1

The correct option is D

$x{y}_2+y_1=0$

Explanation for the correct option

Given that $y=\log x$

Differentiating with respect to $x$ we get

$\frac{dy}{dx}=y_1=\frac{1}{x}$ $\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[\because \frac{d}{\mathrm{dx}}\left(\log x\right)=\frac{1}{x}\right]$

$\Rightarrow$$y_{1}x=1$

Differentiating again with respect to $x$ we get

$\Rightarrow$$\frac{d}{dx}\left(y_1\right)\times x+1\times y_1=0$ $\left[\because \frac{d}{\mathrm{dx}}\left(U·V\right)=V·\frac{\mathrm{dU}}{\mathrm{dx}}+U·\frac{\mathrm{dV}}{\mathrm{dx}}\right]$

$\Rightarrow$ $x{y}_2+y_1=0$

Thus $y=\log x$ is a solution of $x{y}_2+y_1=0$.

Hence, option(D) is the correct answer.