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Let A(k) be the area bounded by the curves `y=x^(2)+2x-3 and y=kx+1.` ThenA. the value of k for which A(k) is least is 2B. 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` |
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Answer» Correct Answer - A::C `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 |
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