For a point $$P$$ in the plane, Let $${d_1}\left( P \right)$$ and $${d_2}\left( P \right)$$ be the distance of the point $$P$$ from the lines $$x - y = 0$$ and $$x + y = 0$$ respectively. The area of the region $$R$$ consisting of all points $$P$$ lying in the first quadrant of the plane and satisfying $$2 \le {d_1}\left( P \right) + {d_2}\left( P \right) \le 4$$, is

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