JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If\[f(x)=\left\{ \begin{align}   & \frac{\sin (p+1)+sin\,x}{x}\,\,\,\,,\,\,\,\,x<0 \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,q\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,x=0 \\  & \frac{\sqrt{x+{{x}^{2}}}-\sqrt{x}}{{{x}^{{}^{3}/{}_{2}}}}\,\,\,\,\,\,\,\,,\,\,\,\,x>0 \\ \end{align} \right.\] is continuous at x = 0, then the ordered pair (p,q) is equal to :                         [JEE Main 10-4-2019 Morning]

    A) \[\left( \frac{5}{2},\frac{1}{2} \right)\]          

    B)   \[\left( -\frac{3}{2},-\frac{1}{2} \right)\]

    C) \[\left( -\frac{1}{2},\frac{3}{2} \right)\]                    

    D) \[\left( -\frac{3}{2},\frac{1}{2} \right)\]

    Correct Answer: D

    Solution :

    \[RHL=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{x+{{x}^{2}}}-\sqrt{x}}{{{x}^{\frac{3}{2}}}}=\]           \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{1+x}-1}{{{x}^{{}}}}=\frac{1}{2}\]           \[LHL=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin (p+1)x+sinx}{x}=(p+1)+1=p+2\] for continuity \[LHL=RHL=f\left( 0 \right)\] \[\Rightarrow (p,q)=\left( \frac{-3}{2},\frac{1}{2} \right)\]


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