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Show that ((x+b)(x+c))/((b-a)(c-a))+((x+...

Show that `((x+b)(x+c))/((b-a)(c-a))+((x+c)(x+a))/((c-b)(a-b))+((x+a)(x+b))/((a-c)(b-c))=1` is an identity.

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If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2) . "Proof : " A=int_(a)^(c)(f(c)-f(x))dx+int_(c)^(b)(f(c))dx =f(c)(c-a)-int_(a)^(c) (f(x))dx+int_(a)^(b)(f(x))dx-f(c)(b-c) rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b)(f(x))dx-int_(a)^(c)(f(x))dx Differentiating w.r.t. c, we get (dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0) For maxima and minima , (dA)/(dc)=0 rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0) Hence, c=(a+b)/(2) "Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0 Hence, A is minimum when c=(a+b)/(2) . If the area enclosed by f(x)= sin x + cos x, y=a between two consecutive points of extremum is minimum, then the value of a is