UPSC NDA II 2015 Mathematics Solved Paper

Consider any equilateral triangle:
c=b=a=1 unit
Take value of p between 1 & 2 i.e.,

3

2

∴a

1

p

+b

1

p

−c

1

p

=(1)

2

3

+(1)

2

3

−(1)

2

3

=1+1−1=1>0
Take value of p greater than 2 i.e; 3
∴By considering all the options carefully; we came to a conclusion that option (b) is correct.