JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    \[\underset{x\to 0}{\mathop{\lim }}\,\frac{x+2\sin x}{\sqrt{{{x}^{2}}-2\sin x+1}-\sqrt{{{\sin }^{2}}x-x+1}}\]is : [JEE Main 12-4-2019 Afternoon]

    A) 3                                 

    B) 2

    C) 6                                 

    D) 1

    Correct Answer: B

    Solution :

    Rationalize \[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( x+2\sin x \right)\left( \sqrt{{{x}^{2}}+2\sin x+1}+\sqrt{{{\sin }^{2}}x-x+1} \right)}{{{x}^{2}}+2\sin x+1-{{\sin }^{2}}x+x-1}\]\[\underset{x\to 0}{\mathop{\lim }}\,\frac{\left( x+2\sin x \right)\left( 2 \right)}{{{x}^{2}}+2\sin x-{{\sin }^{2}}x+x}\] \[\frac{0}{0}\]form using L' hospital \[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\frac{\left( 1+2\cos x \right)\times 2}{2x+2\cos x-2\sin x\cos x+1}\]\[\Rightarrow \frac{2\times 3}{\left( 2+1 \right)}=2\]


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