Equation of a line which is tangent to both the curve `y=x^2+1\ a n d\ y=x^2` is `y=sqrt(2)x+1/2` (b) `y=sqrt(2)x-1/2` `y=-sqrt(2)x+1/2` (d) `y=-sqrt(2)x-1/2`

Find the equation of tangent to the curve y=sin^(-1)(2x)/(1+x^(2)) at x=sqrt(3)

The equation of tangent to the curve y=int_(x^(2))^(x^(3))(dt)/(sqrt(1+t^(2))) at x=1 is sqrt(3)x+1=y (b) sqrt(2)y+1=x sqrt(3)x+y=1(d)sqrt(2)y=x

The area bounded by the curves y=|x|-1 and y=-|x|+1 is a.1b.2c2sqrt(2)d.4

The radius of a circle,having minimum area, which touches the curve y=4-x^(2) and the lines,y=|x| is: (a) 4(sqrt(2)+1)( b) 2(sqrt(2)+1) (c) 2(sqrt(2)-1)( d) 4(sqrt(2)-1)

Aea of the region nclosed between the curves x=y^(2)-1 and x=|y|sqrt(1-y^(2)) is

If the line ax + y = c , touches both the curves x^(2) + y^(2) =1 and y^(2) = 4 sqrt2x , then |c| is equal to:

A beam of light is sent along the line x-y=1 , which after refracting from the x-axis enters the opposite side by turning through 30^0 towards the normal at the point of incidence on the x-axis. Then the equation of the refracted ray is (a) (2-sqrt(3))x-y=2+sqrt(3) (b) (2+sqrt(3))x-y=2+sqrt(3) (c) (2-sqrt(3))x+y=(2+sqrt(3)) (d) y=(2-sqrt(3))(x-1)

if y=(sqrt(x^(2)+1)+sqrt(x^(2)-1))/(sqrt(x^(2)+1)-sqrt(x^(2)-1)), then (dy)/(dx) is