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25. f(x)= {kcosxπ−2x, if x≠π23, if x=π2      π2

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    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

     25. Given, f(x) = {πcosxπ−2x if x≠π23 if x=π2

    For continuity at x=π2

    limx→π−2f(x)=limx→π2f(x)=f(π2).

    limx→π2xcosxπ−2x=limx→π2+xcosxπ−2x=3.

    Take limx→π2xcosxx−2x=3 .

    Putting x = π2+h such that as x→π2,h→0.

    So limh→0xcos(π2+h)x−2(π2+n)=limh→0x(−sinx)−2h

    =limh→0xsinh2h

    =k2limh→0sinhh=k2.

    i e, k2=3

    ⇒ k = 6

    Similarly from limx→π+2xcosxπ−2x=3

    ⇒limh→0hcos(π2+h)π−2(π2+h)=limh→0−hsinh−2h

    =2limh→0sinhh

    =x2

    So, x2=3

    ⇒ k = 6

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