2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    The value of \[\underset{x\to -\pi }{\mathop{\lim }}\,\frac{\int\limits_{-0}^{\sin x}{{{\sin }^{-1}}t\,dt}}{{{(x+\pi )}^{2}}}\] is equal to

    A)  \[\frac{-1}{2}\]

    B)  \[0\]

    C)  \[2\sqrt{2}\]

    D)  \[2\]

    Correct Answer: C

    Solution :

    \[=\underset{x\to -\pi }{\mathop{\lim }}\,\frac{\left( {{\sin }^{-1}}(\sin x) \right)\cos x}{2(x+\pi )}\left( \frac{0}{0}\text{form} \right)\] \[=\underset{x\to -\pi }{\mathop{\lim }}\,\frac{-(x+\pi )}{2(x+\pi )}\cos x=\frac{1}{2}\]


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