Equation(s) of the straight line(s), inclined at `30^0` to the x-axis such that the length of its (each of their) line segment(s) between the coordinate axes is 10 units, is (are) (a) `x+sqrt(3)y+5sqrt(3)=0` (b)`x-sqrt(3)y+5sqrt(3)=0` (c)`x+sqrt(3)y-5sqrt(3)=0` (d)`x-sqrt(3)y-5sqrt(3)=0`

The equation of the lines passing through the point (1,0) and at a distance (sqrt(3))/2 from the origin is (a) sqrt(3)x+y-sqrt(3) =0 (b) x+sqrt(3)y-sqrt(3)=0 (c) sqrt(3)x-y-sqrt(3)=0 (d) x-sqrt(3)y-sqrt(3)=0

Solve sqrt(3)x^(2) - sqrt(2)x + 3sqrt(3)=0

The equation of a straight line on which the length of perpendicular from the origin is four units and the line makes an angle of 120^0 with the x-axis is (A) xsqrt(3)+y+8=0 (B) xsqrt(3)-y=8 (C) xsqrt(3)-y=8 (D) x-sqrt(3)y+8=0

Solve : sqrt(3)x^(2)+5x+2sqrt(3)=0

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)

Find the equation of the straight line which passes through the origin and makes angle 60^0 with the line x+sqrt(3)y+sqrt(3)=0 .

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)

If the area enclosed between the curves y=kx^2 and x=ky^2 , where kgt0 , is 1 square unit. Then k is: (a) 1/sqrt(3) (b) sqrt(3)/2 (c) 2/sqrt(3) (d) sqrt(3)

Solve, using the method of substitution sqrt( 2)x - sqrt(3) y =1, sqrt(3) x - sqrt( 8) y = 0

The values of parameter a for which the point of minimum of the function f(x)=1+a^2x-x^3 satisfies the inequality (x^2+x+2)/(x^2+5x+6)<0a r e (2sqrt(3),3sqrt(3)) (b) -3sqrt(3),-2sqrt(3)) (-2sqrt(3),3sqrt(3)) (d) (-2sqrt(2),2sqrt(3))