Let f(x) = x² + 1/(x²) and g(x) = x - 1/x, x R - - 1,0,1. If h(x) = (f(x))/(g(x)), then the local minimum value of h(x) is

Limit, Continuity and Differentiability: JEE Main 2018 Mathematics question

JEE Main 2018 · Session, Shift 1 · Q2636

JEE Main 2018Session, Shift 1MathematicsHardLimit, Continuity and Differentiability420× since 2002

Let

f\left( x \right) = {x^2} + {1 \over {{x^2}}}

and

g\left( x \right) = x - {1 \over x}

x \in R - \left\{ { - 1,0,1} \right\}

. If

h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}

, then the local minimum value of h(x) is

JEE Main 2018 · Session, Shift 1 · Q2636

JEE Main 2018Session, Shift 1MathematicsHardLimit, Continuity and Differentiability420× since 2002

Let

f\left( x \right) = {x^2} + {1 \over {{x^2}}}

and

g\left( x \right) = x - {1 \over x}

x \in R - \left\{ { - 1,0,1} \right\}

. If

h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}

, then the local minimum value of h(x) is