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 bounded by `f(x)=(x^(3))/(3)-x^(2)+a` and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

Options

A. `1//2`
B. 2
C. 1
D. `2//3`

Answer

D

Explanation

`f(x)=(x^(3))/(3)-x^(2)+a`
`f'(x)=x^(2)-2x=x(x-2)lt0" "` (note that f(x) is monotonic in (0,2))
Hence, for the minimum and f(x) must cross the x-axis at
`(0+2)/(2)=1`.
`"Hence, "f(1)=(1)/(3)-1+a=0`
`rArr" "a=(2)/(3).`