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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    Let \[2{{\sin }^{2}}x+3\sin x-2>0\] and \[{{x}^{2}}-x-2<0\] (x is measured in radians). Then x lies in the interval            [IIT 1994]

    A) \[\left( \frac{\pi }{6},\ \frac{5\pi }{6} \right)\]

    B) \[\left( -1,\ \frac{5\pi }{6} \right)\]

    C) \[(-1,\ 2)\]

    D) \[\left( \frac{\pi }{6},\ 2 \right)\]

    Correct Answer: D

    Solution :

    • \[2\,{{\sin }^{2}}x+3\,\sin x-2>0\]           
    • \[2\,{{\sin }^{2}}x+4\,\sin x-\sin x-2>0\]           
    • \[2\,\sin x\,(\sin x+2)-1\,(\sin x+2)>0\]           
    • \[\,(\sin x+2)\,(2\sin x-1)>0\]           
    • \[2\,\sin x-1>0\,\,\Rightarrow \,\,\sin x>1/2\]           
    • \[x>\pi /6\,\,\Rightarrow \,\,x\in \,\,(\pi /6,\,\,\infty )\]            .....(i)           
    • and \[{{x}^{2}}-x-2<0\,\Rightarrow \,{{x}^{2}}-2x+x-2<0\]           
    • \[x\,(x-2)+1\,(x-2)<0\]           
    • \[(x+1)\,(x-2)<0\,\Rightarrow \,\,x\in (-1,\,\,2)\]                   .....(ii)           
    • From (i) and (ii), \[x\in (\pi /6,\,\,2)\].


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