The minimum area of circle which touches the parabolas `y=x^2+1` and `y^2=x-1` is `(9pi)/(16)s qdotu n i t` (b) `(9pi)/(32)s qdotu n i t` `(9pi)/8s qdotu n i t` (d) `(9pi)/4s qdotu n i t`

Find the maximum area of circle which touches the parabolas y=x^2+1 and y=x^2-1 (i) ((9pi)/16) sq.unit (ii) ((9pi)/32) sq.unit (iii) ((9pi)/8) sq.unit (iv) ((9pi)/4) sq.unit

The area bounded by the curve y^2=8x\ a n d\ x^2=8y is (16)/3s qdotu n i t s b. 3/(16)s qdotu n i t s c. (14)/3s qdotu n i t s d. 3/(14)s qdotu n i t s

The area bounded by the curve a^2y=x^2(x+a) and the x-axis is (a^2)/3s qdotu n i t s (b) (a^2)/4s qdotu n i t s (3a^2)/4s qdotu n i t s (d) (a^2)/(12)s qdotu n i t s

The area of a circle whose radius is the diagonal of a square whose area is 4 sq. units is 16pi s qdotu n i t s (b) 4pi s qdotu n i t s (c) 6pi s qdotu n i t s (d) 8pi s qdotu n i t s

Area bounded by y=sqrt(5-x^2)a n dy=|x-1| is: (A) (5pi-2)/3 s qdotu n i t s (B) (5pi-2)/4 s qdotu n i t s (C) (5pi)/4 s qdotu n i t s (D) none of these

What is the area of the shaded region? (FIGURE) 32-4 pi s qdotu n i t s (b) 32-8 pi s qdotu n i t s (c) 16-4 pi s qdotu n i t s (d) 16-8 pi s qdotu n i t s

The area inside the parabola 5x^2-y=0 but outside the parabola 2x^2-y+9=0 is 12sqrt(3)s qdotu n i t s 6sqrt(3)s qdotu n i t s 8sqrt(3)s qdotu n i t s (d) 4sqrt(3)s qdotu n i t s

Consider two curves C_1: y^2=4[sqrt(y)]x a n dC_2: x^2=4[sqrt(x)]y , where [.] denotes the greatest integer function. Then the area of region enclosed by these two curves within the square formed by the lines x=1,y=1,x=4,y=4 is 8/3s qdotu n i t s (b) (10)/3s qdotu n i t s (11)/3s qdotu n i t s (d) (11)/4s qdotu n i t s

The line 2x+3y=12 meets the x-axis at A and the y-axis at B . The line through (5, 5) perpendicular to A B meets the x-axis, y-axis & the line A B at C , D , E respectively. If O is the origin, then the area of the OCEB is (20)/3s qdotu n i t (b) (23)/3s qdotu n i t (26)/3s qdotu n i t (d) (5sqrt(52))/9s qdotu n i t

If f(x)=sinx ,AAx in [0,pi/2],f(x)+f(pi-x)=2,AAx in (pi/2,pi)a n df(x)=f(2pi-x),AAx in (pi,2pi), then the area enclosed by y=f(x) and the x-axis is pis qdotu n i t s (b) 2pis qdotu n i t s 2s qdotu n i t s (d) 4s qdotu n i t s