Let $$f:\left( {0,\infty } \right) \to R$$ be given by $$f\left( x \right) $$= $$\int\limits_{{1 \over x}}^x {{{{e^{ - \left( {t + {1 \over t}} \right)}}} \over t}} dt$$. Then

The slope of the tangent to the curve $${\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}$$ at the point $$(1, 3)$$ is

The value of $$\int\limits_0^1 {4{x^3}\left\{ {{{{d^2}} \over {d{x^2}}}{{\left( {1 - {x^2}} \right)}^5}} \right\}dx} $$ is

From a point $$P\left( {\lambda ,\lambda ,\lambda } \right),$$ perpendicular $$PQ$$ and $$PR$$ are drawn respectively on the lines $$y=x, z=1$$ and $$y=-x, z=-1.$$ If $$P$$ is such that $$\angle QPR$$ is a right angle, then the possible value(s) of $$\lambda $$ is/(are)