Statement 1: We are given that \( y = \log_{10} x + \log_e x \). The derivatives of the logarithmic functions are: \[ \frac{d}{dx} (\log_{10} x) = \frac{1}{x \ln 10} \] and \[ \frac{d}{dx} (\log_e x) = \frac{1}{x \ln e} = \frac{1}{x} \] Adding these, we get: \[ \frac{dy}{dx} = \frac{1}{x \ln 10} + \frac{1}{x} = \frac{\log_{10} e}{x} + \frac{1}{x} \] Hence, Statement 1 is true.
Statement 2: The derivatives of \( \log_{10}x \) and \( \log_e x \) are incorrect in the statement. The correct derivatives are: \[ \frac{d}{dx} (\log_{10}x) = \frac{1}{x \ln 10} \] and \[ \frac{d}{dx} (\log_e x) = \frac{1}{x} \] Hence, Statement 2 is false.
The correct answer is (A) : Statement 1 is true; statement 2 is false.
Let's analyze each statement.
Statement 1: If \( y = \log_{10}x + \log_e x \), then \( \frac{dy}{dx} = \frac{\log_{10}e}{x} + \frac{1}{x} \).
We know that \( \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} = \frac{\log_a e}{x} \). Therefore,
\( \frac{d}{dx}(\log_{10} x) = \frac{1}{x \ln 10} = \frac{\log_{10} e}{x} \)
\( \frac{d}{dx}(\log_e x) = \frac{1}{x \ln e} = \frac{1}{x} \)
So, \( \frac{dy}{dx} = \frac{d}{dx}(\log_{10} x) + \frac{d}{dx}(\log_e x) = \frac{\log_{10} e}{x} + \frac{1}{x} \). Statement 1 is true.
Statement 2: \( \frac{d}{dx}(\log_{10}x) = \frac{\log x}{\log 10} \) and \( \frac{d}{dx}(\log_e x) = \frac{\log x}{\log e} \).
Let's rewrite this using natural logarithms: \( \frac{d}{dx}(\log_{10}x) = \frac{d}{dx}\left( \frac{\ln x}{\ln 10} \right) = \frac{1}{\ln 10} \cdot \frac{1}{x} = \frac{1}{x \ln 10} \), which is NOT equal to \( \frac{\ln x}{\ln 10} \).
Similarly, \( \frac{d}{dx}(\log_e x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \), which is NOT equal to \( \frac{\ln x}{\ln e} = \ln x \).
Therefore, statement 2 is false.
In conclusion, Statement 1 is true, and Statement 2 is false.
The graph between variation of resistance of a wire as a function of its diameter keeping other parameters like length and temperature constant is
While determining the coefficient of viscosity of the given liquid, a spherical steel ball sinks by a distance \( x = 0.8 \, \text{m} \). The radius of the ball is \( 2.5 \times 10^{-3} \, \text{m} \). The time taken by the ball to sink in three trials are tabulated as shown: