Q100 / 180 · KCET 2021 · Mathematics
Consider the following statements Statement 1 : If $y=\log _{10} x+\log _{c} x$, then $\frac{d y}{d x}=\frac{\log _{10} e}{x}+\frac{1}{x}$
Statement 2 : If $\frac{d}{d x}\left(\log _{10} x\right)=\frac{\log x}{\log 10}$ and $\frac{d}{d x}\left(\log _{e} x\right)=\frac{\log x}{\log e}$
- Correct answerAStatement 1 is true, Statement 2 is false.
- BStatement $l$ is false, statement 2 is true.
- CBoth statements $l$ and 2 are true.
- DBoth statements 1 and 2 are false.
Solution
Statement 1 ,
$\begin{aligned}
y &=\log _{10} x+\log _{e} x \\
\Rightarrow \quad y &=\frac{\log _{e} x}{\log _{e} 10}+\log _{e} x
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} &=\frac{1}{x \log _{e} 10}+\frac{1}{x} \\ \Rightarrow \quad \frac{d y}{d x} &=\frac{\log _{10} e}{x}+\frac{1}{x} \\ \text { Statement 2, } \\ \frac{d}{d x} \log _{10} x &=\frac{d}{d x} \frac{\log _{e} x}{\log _{e} 10} \\ &=\frac{\log _{x}}{x \log _{e} 10} \\ \frac{d}{d x}\left(\log _{e} x\right) &=\frac{d}{d x} \frac{\log _{e} x}{\log _{e} e}=\frac{1}{x} \end{aligned}$
Hence, statement 1 is true but statement 2 is false.
Hence, the correct answer is option: 'A'.