JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    One root of the equation \[\cos x-x+\frac{1}{2}=0\]lies in the interval [Kurukshetra CEE 1996]

    A) \[\left[ 0,\,\frac{\pi }{2} \right]\]

    B) \[\left[ -\frac{\pi }{2},\,0 \right]\]

    C) \[\left[ \frac{\pi }{2},\,\pi  \right]\]

    D) \[\left[ \pi ,\frac{3\pi }{2} \right]\]

    Correct Answer: A

    Solution :

    \[f(x)=\cos x-x+\frac{1}{2}\], \[f(0)=\frac{3}{2}>0\] \[f\left( \frac{\pi }{2} \right)=0-\frac{\pi }{2}+\frac{1}{2}=\frac{1-\pi }{2}<0\], \[\left( \because \ \pi =\frac{22}{7}\text{ nearly} \right)\] \ One root lies in the interval\[\left[ 0,\,\frac{\pi }{2} \right]\].


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