Q176 / 270 · NDA 2015 (Phase 2) · Mathematics

If $a, b, c$ are the sides of a triangle $A B C$, then $a^{\frac{1}{p}}+b^{\frac{1}{p}}-c^{\frac{1}{p}}$ where $\mathrm{p} \gt 1$, is

  • A
    always negative
  • Correct answer
    B
    always positive
  • C
    always zero
  • D
    positive if $1 \lt \mathrm{p} \lt 2$ and negative if $\mathrm{p} \gt 2$

Solution

Consider any equilateral triangle: $\mathrm{c}=\mathrm{b}=\mathrm{a}=1 \text { unit }$ Take value of p between $1 \& 2$ i.e., $\frac{3}{2}$ $\begin{aligned} & \therefore \mathrm{a}^{1 / p}+\mathrm{b}^{1 / p}-\mathrm{c}^{1 / p}=(1)^{2 / 3}+(1)^{2 / 3}-(1)^{2 / 3} \\ & =1+1-1=1 \gt 0 \end{aligned}$ Take value of $p$ greater than 2 i.e; 3 . $\therefore \mathrm{a}^{1 / p}+\mathrm{b}^{1 / p}-\mathrm{c}^{1 / p}=(1)^{1 / 3}+(1)^{1 / 3}-(1)^{1 / 3}=1 \gt 0 .$ $\therefore$ By considering all the options carefully; we came to a conclusion that option (2) is correct.

Hence, the correct answer is option: 'B'.

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