1.The triangle formed by the tangent to the curve
f(x)=x^2+bx-b at the point(1,1) & the coordinate axes lies in first quadrant . if it's area is 2 then b=....

options are
1.) -1 2.) 3 3.) -3 4.) 1


2.area bounded by the parabola y =4x^2, y=x^2/9 and the line y=2 is...

options r in terms of root2/3

Q.N. 1 ---> b = (-3)
Q.N. 2 ---> area = 20(root2)/3

if you want the detailed solutions then just let me know...

the answers are correct.i want the detailed solution plz..

Q1.

we have,
area of the triangle formed by ref. axes and the tangent to the parabola at (1,1) = 2
=> area of the two smaller triangles + area of the sq. = 2
=> [(1/2)*1*tan(theta)] + [1*1] + [(1/2)*{1/tan(theta)}*1] = 2
=> [(1/2)*1*tan(theta)]+ [(1/2)*{1/tan(theta)}*1] = 1
=> tan(theta) + 1/tan(theta) = 2
=> sq. of tan(theta) - 2*tan(theta) + 1 = 0
=> tan(theta) = 1
But, tan(theta) is negative of the slope of the curve (or slope of the tangent to the curve) at (1,1)
therefore,
tan(theta) = 1 = -(dy/dx)(at x=1,y=1) = -(2x+b) = -2-b
=> b = -3.

Q.2


required area = shaded region in the above pic
= 2*[{integral of 4*x^2 from x=0 to x=a} +
{2*(b-a)} - {integral of (x^2)/9 from x=0 to x=b}]
= 20(root2)/3