Let ƒ : [0, 2] $$ \to $$ R be a twice differentiable function such that ƒ''(x) > 0, for all x $$ \in $$ (0, 2). If $$\phi $$(x) = ƒ(x) + ƒ(2 – x), then $$\phi $$ is :

If the function f given by f(x) = x³ – 3(a – 2)x² + 3ax + 7, for some a$$ \in $$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is :

The tangent to the curve y = x² – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point :

Let f(x) = $${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$$ x $$\, \in $$ R, where a, b and d are non-zero real constants. Then :