2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2006

  • question_answer
    If \[f:R\to R\] given by \[f(x)=\left\{ \begin{matrix}    2\,\cos x, & if & x\le -\frac{\pi }{2}  \\    a\,\sin \,x+b, & if & -\frac{\pi }{2}<x<\frac{\pi }{2}  \\    1+{{\cos }^{2}}x, & if & x\ge \frac{\pi }{2}  \\ \end{matrix} \right.\] is a continuous function on R, then (a, b) is equal to

    A)  \[(1/2,\,1/2)\]

    B)  \[(0,-1)\]

    C)  \[(0,2)\]

    D)  \[(1,0)\]

    Correct Answer: A

    Solution :

    For \[x=-\frac{\pi }{2},\,f\left( -\frac{\pi }{2} \right)=0\] \[LHL=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -\frac{\pi }{2}-h \right)\] \[\underset{h\to 0}{\mathop{\lim }}\,\,\,2\cos \,\left( -\frac{\pi }{2}-h \right)=0\] \[RHL=\underset{x\to \frac{{{\pi }^{+}}}{2}}{\mathop{\lim }}\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f\left( -\frac{\pi }{2}+h \right)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,a\,\sin \,\left( -\frac{\pi }{2}+h \right)+b\] \[=-a+b\] Since, function is continuous. \[\therefore \] \[RHL=LHL\] \[\Rightarrow \] \[-a+b=0\] \[\Rightarrow \] \[a=b\] From the given options only  ie, \[\left( \frac{1}{2},\frac{1}{2} \right)\] satisfies this condition.


You need to login to perform this action.
You will be redirected in 3 sec spinner