JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • 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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