Question

Can anyone giveme the step by step explanation of this question

Answer 1

Nice question

Given condition

a^2+b^2+1/a^2 +1/b^2=4

proceeding it we can write

=>a^2+b^2+1/b^2+1/a^2-4=0

=>a^2+1/a^2-2 +1/b^2+b^2-2=0

=>{a^2+1/a^2-(2×a×1/a)}+{b^2+1/b^2-(2×b×1/b)}=0

=>(a-1/a)^2+(b-1/b)^2=0

now as we know that square of any no. is positive and sum of two positive can never be zero,

so here both the terms have to be zero to satisfy the condition

hence

(a-1/a)^2=0

a-1/a=0

a=1/a

a^2=1=>a=√1

a=+1 or -1

similarly

(b-1/b)^2=1=>(b-1/b)=√1

b-1/b=+1 or -1

now putting this value of a and be in the given question we get

(a^2+b^2)^1/2=(1+1)^1/2=√2

√2 will be the required answer of this question

hence last option is correct