Let `g(x)=2e , x <= 1` and `g(x)=log(x-1) , x>1` The equation of the normal to `y=g(x)` at the point `(3,log2),` is

Let g(x)={(2e, "if", xle1),(log(x-1),"if",xgt1):} The equation of the normal to y=g(x) at the point (3,log2) , is

Let g(x)={{:(2e",",ifxle1),(log(x-1)",",ifxgt1):} .Then equation of the normal to y = g(x) at the point (3,log2) ,is

Let g(x)={(2e,if,xle1),(log(x-1),if,xgt1):} The equation of the normal to y = g(x)at the point (3,log 2), is

Let f(x)=x^(3)+x+1 and let g(x) be its inverse function then equation of the tangent to y=g(x) at x = 3 is

Let f(x)=x^(3)+x+1 and let g(x) be its inverse function then equation of the tangent to y=g(x) at x = 3 is

Let f(x)=x^(3)+x+1 and let g(x) be its inverse function then equation of the tangent to y=g(x) at x = 3 is

The equation to the normal to the circle x^(2)+y^(2)-2x-2y-2=0 at the point (3,1) on it is

The equation of the normal to the circle x^(2)+y^(2)-2 x-2 y-2=0 is at the point (3,1) on it is

Find the equations of the tangent and normal to the curve y= (log x)^2 at point x = 1/e .

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 prependicular from point Q on x-axis meets the curve y =g (x) in point R .Find equation of tangent at to y=g(x) at P .Also the value of g (1) =