Q92 / 155 · WBJEE 2018 · Mathematics
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
- A$l(p)>J(p)$
- Correct answerB$l(p) \leq J(p)$
- C$l(p) < J{(} p)$ in $\left[0 ,\frac{p}{2}\right]$ and $l(p)>J(p)$ in $\left[\frac{p}{2}, \infty\right)$
- D$l(p) < J(p)$ in $\left[\frac{p}{2}, \infty\right)$ and $J(p) < l(p)$ in $\left[0, \frac{p}{2}\right]$
Solution
Here, let $p=\frac{1}{m}$
then $\left(a^{p}+b^{p}\right)^{1 / p}=\left(a^{1 / m}+b^{1 / m}\right)^{m}$
$=a+b+k, k \geq 0$
$\therefore \quad a^{p}+b^{p} \geq(a+b)^{p} \geq(a+b)$
$\Rightarrow \quad J(p) \geq I(p)$
Hence, the correct answer is option: 'B'.