Doubt from Integration

question number 25

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sorry, it is question number 26, not 25 ...

Try it by cauchy integral inequality didnt tried but looking like that only (problem is seen in fitjee aits).

can you try it?


question number 12 , can you also try this problem , @Sneha_2021

Already done in forum asked by azimuddin seikh

No it was asked by jagdish sir
assume c=\dfrac{a}{1-a}
I_1=\int_0^{a} 1^3dx + \int_a^{1} c^3dx
After diffrenciation c=\dfrac{1}{4}
Ans =5

  • credit stackexchange seen 6 to 8 month before
    @arush_2019 sir can share post of jagdish sir

can you explain , why did you took c=a/(1-a)

Actully
i=\int_0^{a} 1dx +\int_a^{1} -cdx=0 given in problem
\int_0^{1} f(x)dx=0...... given i used this only

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i thought the function would be symmetric about x=1/2, is this correct?

@Sneha_2021 can you solve the first one , actually, I was unable to proceed after Cauchy integral inequality...

@sneha i have doubt if we take f(x)=sin2pix then ans is coming zero

Will share at night only busy
@Achyut_2020 it will follow because of range given in problem is |f(x)| \le 1 but the problem is to get max value of cube

  • the given condition is of chebhyshevs polynomial

Getting 4 but too long mthd
@Hari_Shankar sir any shortcut??

can you share your solution?

Done too roughly just assumed a biquadratic
Nd find all coefficent by Leibniz nd manipulation
Is ans correct i will try some other approach.
Ans??

sorry, don't have the answer with me

#26
f(1) =0
f'(1) =f''(1) =f"'(1) =0
hence
f(x) = (x-1)^{4}
now killed the problem

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For 26 diff until you don't get derivative non zero at 1