If the function

f(x) = $$\left\{ {\matrix{ { - x} & {x < 1} \cr {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr } } \right.$$

is differentiable at x = 1, then $${a \over b}$$ is equal to :

If   f(x) is a differentiable function in the interval (0, $$\infty $$) such that f (1) = 1 and

$$\mathop {\lim }\limits_{t \to x} $$   $${{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1,$$ for each x > 0, then $$f\left( {{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$ equal to :

Let a and b respectively be the semitransverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e² − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a² − b² is equal to :

For x $$ \in $$ R, x $$ \ne $$ -1,

if (1 + x)²⁰¹⁶ + x(1 + x)²⁰¹⁵ + x²(1 + x)²⁰¹⁴ + . . . . + x²⁰¹⁶ =

$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a₁₇ is equal to :