Question:

For $0 \leq p \leq 1$ and for any positive $a, b$ let $I(p) = (a + b)^p, J(p) = a^p + b^p$, then

Updated On: Mar 10, 2025

$I (p) > J (p)$
$I (p) \leq J (p)$
$I(p) < J(p)$ in $[0, \frac{P}{2} ]$ & $I(p) > J(p)$ in $[\frac{P}{2} , \infty )$
$I(p) < J(p)$ in $[\frac{P}{2}, \infty )$ & $J(p) > I(p)$ in $[0, \frac{P}{2}]$

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The Correct Option is B

Given that;

For any positive a,b; let $I(p)=(a+b)^b. J(p)=a^b+b^b$

For 0≤p≤1

Then let us take a=3 , b=4

then p=0.5

$i(p)=5; J(p)=7$

$\therefore J(p)>I(p)$

Now a= ⅓, b= ¼

when p=1

$I(p)= \frac{7}{12}; J(p)=\frac{7}{12}$

$\therefore I(p)=J(p)$

$I(p)$≤$J(p)$

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