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

Options

A. 0
B. -1
C. 1
D. 2

Answer

A

Explanation

`f(x)= sin x + cos x`
`"or "(df(x))/(dx)=cos x- sin x`
`"If "(df(x))/(dx)=0," then " cos x= sin x rArr x=(pi)/(4) and x=(5pi)/(4)`
(considering any two of consecutive points of extremum).
For minimum area bounded by `y=f(x) and y=a,` between
`x=(pi)/(4) and x=(5pi)/(4),` graphs of g(x) must cut y =a at
`c=((pi)/(4)+(5pi)/(4))/(2)=(3pi)/(4).`
`a=f((3pi)/(4))`
`rArr" "a=sin ((3pi)/(4))+cos ((3pi)/(4))=0.`