2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Applications of Derivatives

JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Critical Thinking

  • question_answer
    The function \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\] increases, if [IIT 1999; Pb. CET 2001]

    A) \[0<x<\frac{\pi }{8}\]

    B) \[\frac{\pi }{4}<x<\frac{3\pi }{8}\]

    C) \[\frac{3\pi }{8}<x<\frac{5\pi }{8}\]

    D) \[\frac{5\pi }{8}<x<\frac{3\pi }{4}\]

    Correct Answer: B

    Solution :

    • \[f(x)={{\sin }^{4}}x+{{\cos }^{4}}x\]                         
    • \[={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\]                         
    • \[=1-\frac{4{{\sin }^{2}}x{{\cos }^{2}}x}{2}=1-\frac{{{\sin }^{2}}2x}{2}\]                          
    • \[=1-\frac{1}{4}(2{{\sin }^{2}}2x)\]                         
    • \[=1-\left( \frac{1-\cos 4x}{4} \right)=\frac{3}{4}+\frac{1}{4}\cos 4x\]                   
    • Hence function\[f(x)\]is increasing when \[f'(x)>0\]                   
    • \[f'(x)=-\sin 4x>0\Rightarrow \sin 4x<0\]             
    • Hence \[\pi <4x<\frac{3\pi }{2}\]or \[\frac{\pi }{4}<x<\frac{3\pi }{8}\].


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