JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
    \[\underset{x\to \pi /4}{\mathop{\lim }}\,\frac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\frac{\pi }{4} \right)}\] is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) \[4\sqrt{2}\]                                          

    B) \[8\sqrt{2}\]

    C) 4                                 

    D)   8

    Correct Answer: D

    Solution :

    \[\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{{{\cot }^{3}}x-\tan x}{\cos \left( x+\frac{\pi }{4} \right)}\] \[=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{1-{{\tan }^{4}}x}{\cos \left( x+\frac{\pi }{4} \right)}=2\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{1-{{\tan }^{2}}x}{\cos \left( x+\frac{\pi }{4} \right)}\] \[=2\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\frac{1}{\sqrt{2}}(\cos \,x-\sin x)}.\frac{1}{{{\cos }^{2}}x}\] \[=4\sqrt{2}\,\,\,\underset{x\to \frac{\pi }{4}}{\mathop{\lim }}\,(\cos \,x+\sin x)=8\]


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