• # question_answer Define F(x) as the product of two real functions ${{f}_{1}}(x)=x,x\in R,$and ${{f}_{2}}(x)=\left\{ \begin{matrix} \sin \frac{1}{x}, & if\,x\ne 0\, \\ 0, & if\,x=0 \\ \end{matrix} \right.$as follows: $F(x)=\left\{ \begin{matrix} {{f}_{1}}(x).{{f}_{2}}(x), & if\,x\ne 0\, \\ 0, & if\,x=0 \\ \end{matrix} \right.$ Statement -1 : F(x) is continuous on R. Statement - 2 : ${{f}_{1}}(x)$ and ${{f}_{2}}(x)$ are continuous on R     AIEEE  Solved  Paper (Held On 11 May  2011) . 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

Correct Answer: B

Solution :

$f(x)=\left\{ \begin{matrix} x\sin (1/x), & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.$ at$x=0$ $LHL=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\left\{ -h\sin \left( -\frac{1}{h} \right) \right\}$ $=0\times a$ finite quantity between - 1 and 1 $RHL=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,h\,\sin \frac{1}{h}$ $=0$ $f(0)=0$ $\therefore$$f(x)$is continuous on R. ${{f}_{2}}(x)$ is not continuous at x = 0

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