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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 44

  • question_answer
    The values of A and B such that the function \[f(x)=\left\{ \begin{matrix}    -2\sin x,  \\    A\sin x+B,  \\    \cos x,  \\ \end{matrix} \right.\,\,\,\,\,\,\begin{matrix}    x\le -\frac{\pi }{2}  \\    -\frac{\pi }{2}<x<\frac{\pi }{2}  \\    x\ge \frac{\pi }{2}  \\ \end{matrix}\] is continuous everywhere are                     

    A) \[A=0,\,B=1\]

    B)        \[A=1,\,B=1\]

    C) \[A=-1,\,B=1\]

    D)        \[A=-1,\,B=0\]

    Correct Answer: C

    Solution :

    For continuity at all \[x\in R,\] we must have \[f\left( -\frac{\pi }{2} \right)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{-}}}{\mathop{\lim }}\,(-2\sin x)=\underset{x\to {{\left( -\frac{\pi }{2} \right)}^{+}}}{\mathop{\lim }}\,(A\sin x+B)\] \[\Rightarrow \,\,2=-A+B\]                                   ?(i) and \[f\left( \frac{\pi }{2} \right)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{-}}}{\mathop{\lim }}\,(A\sin x+B)=\underset{x\to {{\left( \frac{\pi }{2} \right)}^{+}}}{\mathop{\lim }}\,(cosx)\] \[\Rightarrow \,\,0=A+B\]                                    ??(ii) From (i) and (ii), \[A=-1\] and \[B=1\]


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