If `|f(x_1) - f(x_2)| le (x_1 - x_2)^(2), AA x_1 , x_2 in R`. Find the equation of tangent to the curve `y = f(x)` at the point `(1, 2)`.

If |f(x_(1))-f(x_(2))|<=(x_(1)-x_(2))^(2), for all x_(1),x_(2)in R, then the equation of tangent to the curve y=f(x) at the point (1,2) is a.y=2 b.y=3 c.x=2 d.x=3

if |f(x_(1))-f(x_(2))|<=(x_(1)-x_(2))^(2) Find the equation of gent to the curve y=f(x) at the point (1,2).

A polynomial function f(x) satisfies the condition f(x+1)=f(x)+2x+1. Find f(x) if f(0)=1 Findalso the equations of the pair of tangents from the origin on the curve y=f(x) and compute the areaenclosed by the curve and the pair of tangents.

If f(x)+2f(1-x)=x^(2)+1,AA x in R, then the range of f is :

Let f (x) = [{:(1-x,,, 0 le x le 1),(0,,, 1 lt x le 2 and g (x) = int_(0)^(x) f (t)dt.),( (2-x)^(2),,, 2 lt x le 3):} Let the tangent to the curve y = g( x) at point P whose abscissa is 5/2 cuts x-axis in point Q. Let the pependiculat from point Q on x-axis meets the curve y =g (x) in point R. If 'theta' be the angle between tangents to the curve y =g (x) at point P and R, then tan theta equals to :

Let f(x) = x(2-x), 0 le x le 2 . If the definition of f(x) is extended over the set R-[0,2] by f (x+1)= f(x) , then f is a

f(x)+2f(1-x)=x^(2)+2,AA x in R then f(x)=

Let f(x)=x^(2)-x+1, AA x ge (1)/(2) , then the solution of the equation f(x)=f^(-1)(x) is

Determine a differentiable function y=f(x) which satisfies f'(x)=[f(x)]^(2) and f(0)=-(1)/(2). Find also the equation of the tangent at the point where the curve crosses the y-axis.