Question

If a, b, c are the sides of a triangle ABC, then $a^{\frac{1}{p}}+b^{\frac{1}{p}}-c^{\frac{1}{p}}$ where $p>1$, is:

• A. Always negative
• B. Always positive
• C. Always zero
• D. Positive if $1<p<2$ and negative if $p>2$

Answer 1

The correct option is B Always positive

Solution:

Since, $a,b,c$ are positive numbers.

$⟹a^{\frac{1}{p}},b^{\frac{1}{p}},c^{\frac{1}{p}}$ are positive numbers.

$⟹a^{\frac{1}{p}}+b^{\frac{1}{p}}>c^{\frac{1}{p}}$

[$\because$ Sum of any two sides of a triangle must be greater than the third side.]

$⟹a^{\frac{1}{p}}+b^{\frac{1}{p}}-c^{\frac{1}{p}}>0$

Hence, B is the correct option.