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)`.
The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function `y=x^(3)+3x^(2)+x+a`, and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is

Options

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

Answer

C

Explanation

`f(x)=x^(3)+3x^(2)+x+a`
`f'(x)=3x^(2)+6x+1=0`
`rArr" "x=-1pm (sqrt(6))/(3).`
Hence, f(x) cuts the x-axis at
`(1)/(2)[(-1+(sqrt(6))/(3))+(-1-(sqrt(6))/(3))]=-1.`
`f(-1)=-1+3-1+a=0`
`a=-1.`