The abscissa of a point on the curve `x y=(a+x)^2,` the normal which cuts off numerically equal intercepts from the coordinate axes, is `-1/(sqrt(2))` (b) `sqrt(2)a` (c) `a/(sqrt(2))` (d) `-sqrt(2)a`

find the abscissa of the point on the curve xy=(c-x)^2 the normal at which cuts off numerically equal intercepts from the axes of co–ordinates.

The abscissa of the point on the curve ay^(2)=x^(3), the normal at which cuts off equal intercepts from the coordinate axes is

The abscissa of the point on the curve sqrt(xy)=a+x the tangent at which cuts off equal intercepts from the coordinate axes is -(a)/(sqrt(2))( b) a/sqrt(2)(c)-a sqrt(2)(d)a sqrt(2)

A(1/sqrt(2), 1/sqrt(2)) is a point on the circle x^2 + y^2 = 1 and B is another point on the circle such that AB = pi/2 units. Then coordinates of B can be : (A) (1/sqrt(2), - 1/sqrt(2)) (B) (-1/sqrt(2), 1/sqrt(2)) (C) (-1/sqrt(2), - 1/sqrt(2)) (D) none of these

int_0^(sqrt(2)) [x^2] dx is equal to (A) 2-sqrt(2) (B) 2+sqrt(2) (C) sqrt(2)-1 (D) sqrt(2)-2

The length of the chord of the parabola y^(2)=x which is bisected at the point (2,1) is (a) 2sqrt(3)( b) 4sqrt(3)(c)3sqrt(2) (d) 2sqrt(5)

The area of the figure bounded by the curves y=cosx and y=sinx and the ordinates x=0 and x=pi/4 is (A) sqrt(2)-1 (B) sqrt(2)+1 (C) 1/sqrt(2)(sqrt(2)-1) (D) 1/sqrt(2)

Radius of the circle that passes through the origin and touches the parabola y^(2)=4ax at the point (a,2a) is (a) (5)/(sqrt(2))a (b) 2sqrt(2)a (c) sqrt((5)/(2))a (d) (3)/(sqrt(2))a