Question

Define $f(x)$ as the product of two real functions $f_1(x) = x, x \in$ R, and $f_2(x) =$ \begin{cases} sin \frac{1}{x} , & \text{If x\ne 0} \\\ 0, & \text{If x = 0} \end{cases} as follows : f(x) = \begin{cases} f_1(x).f_2(x) , & \text{If x \ne 0} \\\ \quad 0, & \text{If x = 0} \end{cases} $f(x)$ is continuous on R. $f_1(x)$ and $f_2(x)$ are continuous on R.

• A. Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is true, Statement-2 is false
• D. Statement-1 is false, Statement-2 is true

Answer 1

Answer: (B)

Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1

Answer 2

F(x) = \begin{cases} x \,sin(1/ x) , & \text{ x \ne 0} \\\ \quad 0, & \text{ x = 0} \end{cases} at $x = 0$ LHL = \displaystyle\lim_{h\to0^{+}}\left\\{-h\,sin \left(-\frac{1}{h}\right)\right\\} $= 0 ? a$ finite quantity between - 1 and 1 $RHL = \displaystyle\lim_{h\to0^{+}} h sin \frac{1}{h}$ $= 0$ $f\left(0\right) = 0$ $\therefore f\left(x\right)$ is continuous on R. $f_{2}\left(x\right)$ is not continuous at $x = 0$