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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If \[f(x)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\frac{\pi }{3} \right)+\cos x\cos \left( x+\frac{\pi }{3} \right)\] and \[g\left( \frac{5}{4} \right)=1\], then \[(gof)(x)=\] [IIT 1996]

    A) -2

    B) -1

    C) 2

    D) 1

    Correct Answer: D

    Solution :

    • \[{f}'(x)=2\,\sin x\cos x+2\,\sin \,\left( x+\frac{\pi }{3} \right)\,\cos \,\left( x+\frac{\pi }{3} \right)\] \[-\sin x\cos \,\left( x+\frac{\pi }{3} \right)-\cos x\sin \left( x+\frac{\pi }{3} \right)\]                   
    • \[=\sin 2x+\sin \,\left( 2x+\frac{2\pi }{3} \right)-\sin \,\left( x+x+\frac{\pi }{3} \right)\]                   
    • \[=2\sin \,\left( 2x+\frac{2\pi }{3} \right)\,\cos \,\left( \frac{\pi }{3} \right)-\sin \,\left( 2x+\frac{\pi }{3} \right)=0\]                   
    • \[\Rightarrow \,\,f(x)=k,\] where k is a constant.           
    • But \[f(0)={{\sin }^{2}}0+{{\sin }^{2}}\left( \frac{\pi }{3} \right)+\cos \,\,0\,\,\cos \,\left( \frac{\pi }{3} \right)=\frac{5}{4}\]           
    • Thus \[f(x)=\frac{5}{4}\,,\,\,\forall \,x\in R.\]           
    • Therefore, \[(gof)\,\,(x)=g\,[f(x)\,]=g\,\left( \frac{5}{4} \right)=1.\]


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