Let A(k) be the area bounded by the curves `y=x^(2)+2x-3 and y=kx+1.` Then

Options

A. the value of k for which A(k) is least is 2
B. the value of k for which A(k) is least is `3//2`
C. least value of A(k) is `32//3`
D. least value of A(k) is `64//3`

Answer

A, C

Explanation

`x_(1) and x_(2)` are the roots of the equation
`x^(2)+2x-3=kx+1`
`"or "x^(2)+(2-k)x-4=0`
`{:(x_(1)+x_(2)=k-2),(x_(1)x_(2)=-4):}}`
`A=overset(x_(2))underset(x_(1))int[(kx+1)-(x^(2)+2x-3)]dx`
`=[(k-2)(x^(2))/(2)-(x^(3))/(3)+4x]_(x_(1))^(x_(2))`
`=[(k-2)(x_(2)^(2)-x_(1)^(2))/(2)-(1)/(3)(x_(2)^(3)-x_(1)^(3))+4(x_(2)-x_(1))]`
`=(x_(2)-x_(1))[((k-2)^(2))/(2)-(1)/(3)((x_(2)-x_(1))^(2)-x_(1)x_(2))+4]`
`=sqrt((x_(2)+x_(1))^(2)-4x_(1)x_(2))[((k-2)^(2))/(2)-(1)/(3)((k-2)^(2)+4)+4]`
`=(sqrt((k-2)^(2)+16))/(6)[(1)/(6)(k-2)^(2)+(8)/(3)]`
`=([(k-2)^(2)+16]^(3//2))/(6)`
which is least when `k=2 and A_("least")=32//3` sq. units