If `x=1+1/(3+1/(2+1/(3+1/(2oo))))` a `sqrt(5/2)` b. `sqrt(3/2)` c. `sqrt(7/3)` d. `sqrt(5/3)`

The shortest distance between the lines (x-3)/3=(y-8)/(-1)=(z-3)/1a n d(x+3)/(-3)=(y+7)/2=(z-6)/4 is a. sqrt(30) b. 2sqrt(30) c. 5sqrt(30) d. 3sqrt(30)

Values (s)(-i)^(1/3) is/are (sqrt(3)-i)/2 b. (sqrt(3)+i)/2 c. (-sqrt(3)-i)/2 d. (-sqrt(3)+i)/2

int_(-1)^(1/2)(e^x(2-x^2)dx)/((1-x)sqrt(1-x^2))i se q u a lto (sqrt(e))/2(sqrt(3)+1) (b) (sqrt(3e))/2 sqrt(3e) (d) sqrt(e/3)

The domain of f(x)=sin^(-1)[2x^2-3],w h e r e[dot] denotes the greatest integer function, is (a) (-sqrt(3/2),sqrt(3/2)) (b) (-sqrt(3/2),-1)uu(-sqrt(5/2),sqrt(5/2)) (c) (-sqrt(5/2),sqrt(5/2)) (d) (-sqrt(5/2),-1)uu(1,sqrt(5/2))

If tangents P Q and P R are drawn from a point on the circle x^2+y^2=25 to the ellipse (x^2)/16+(y^2)/(b^2)=1,(b (a) (sqrt(5))/4 (b) (sqrt(7))/4 (c) (sqrt(7))/2 (d) (sqrt(5))/3

Solve sqrt(y+1)+sqrt(2y-5)=3 .

The abscissas of point Pa n dQ on the curve y=e^x+e^(-x) such that tangents at Pa n dQ make 60^0 with the x-axis are. 1n((sqrt(3)+sqrt(7))/7)a n d1n((sqrt(3)+sqrt(5))/2) 1n((sqrt(3)+sqrt(7))/2) (c) 1n((sqrt(7)-sqrt(3))/2) +-1n((sqrt(3)+sqrt(7))/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)

If 3sin^(-1)((2x)/(1+x^2))-4cos^(-1)((1-x^2)/(1+x^2))+2tan^(-1)((2x)/(1-x^2))=pi/3, w h e r e|x|<1, then x is equal to (a) 1/(sqrt(3)) (b) -1/(sqrt(3)) (c) sqrt(3) (d) -(sqrt(3))/4