question number 25

# Doubt from Integration

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?

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

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

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

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

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