Easy Algebra Question - 2

sir,i did i like this............
\frac{(x^{4}+\frac{1}{x^{4}}+1)}{(x^{2}+\frac{1}{x^{2}}+1)}
in this expresion as we can see the optimum is attained x=1...
thats 3/3=1..coz..when x>1...we can see numerator is bigeer than denominator.....hence k will bcom big and big.......
wen x<1.......agin numerator >denominator...as now the 1/x^4 factor plays the role.......we need not bother as x<0 as function is even....if we have proved for +ve part, the same holds good for the negative also

K=\frac{x^{4}+ \frac{1}{x^{4}}+ 1}{x^{2} + \frac{1}{x^{2}} + 1 }

numerator= \left( x^{2}+ \frac{1}{x^{2}}\right) ^{2} -2 + 1

numerator= \left( x^{2}+ \frac{1}{x^{2}}\right) ^{2} -1

using a² -b² = (a+b)(a-b)

numerator= ( x^{2}+ \frac{1}{x^{2}} -1 )(x^{2} + \frac{1}{x^{2}} + 1)

Thus:

K=\frac{ \left( x^{2} + \frac{1}{x^{2}} + 1\right) \left( x^{2} + \frac{1}{x^{2}} - 1 \right) }{x^{2}+ \frac{1}{x^{2}}+ 1}

Thus\; K = ( x^{2}+ \frac{1}{x^{2}} -1 )\; \; \;

we\; can\; cancel\; the\; term\; x^{2} + \frac{1}{x^{2}} + 1\; since\; it\; can\; never\; be\; zero

also \; x^{2}\; can\; also\; never\; be\; zero\; since\; x<0\;

since we can multiply by x² thus\; x^{4}-x^{2}(k+1)+1 =0

then\; use\; discriminant \geq 0\; and\; neglect\; -ve\; value\; of k\; , is\; it\; correct\;??