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Solution of x+1/x=2*1/x​

Answer»

LET,

x−1>0

=>∣x−1∣=x−1

Now,

x−1x+2<1

x−1x+2−1<0

x−1−x−2x+2<0

−3x+2<0

3x+2>0

(3)(x+2)(x+2)2>0

=> (3)(x+2)>0 as (x+2)2 is always greater than or EQUAL to 0.

=>x+2>0

=>x>−2

But we first ASSUMED the condition, x−1>0=>x>1

So, the first soltuion will be

x∈(1,∞)∩(−2,∞) which is,

x∈(1,∞) …….(1)

Now, let, x−1<0

=>∣x−1∣=1−x

=>1−xx+2<0

=>1+2xx+2>0

=>(1+2x)(x+2)(x+2)2>0

=>(1+2x)(x+2)>0

=>x∈(−∞,−2)∪(−1/2,∞)

But the assumed condition is x−1<0

The second solution will be,

x∈(−∞,1)∩(−∞,−2)∪(−1/2,∞)

Which will be (−∞,−2)∪(−1/2,1)……(2)

So the final answer will be

x∈(−∞,−2)∪(−1/2,∞)


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