Let f(x) be a function defined by
`f(x)=int_(1)^(x)t(t^2-3t+2)dt,x in [1,3]`
Then the range of f(x), is

Options

A. [0,2]
B. `[-1/4,4]`
C. `[-1/4,2]`
D. none of these

Answer

C

Explanation

We have `f(x)underset(1)overset(x)int t(t^2-3t +2)dx, x in [1.3]`
`f(x)=x(x^-3x+2)`
`rArr f(x)= x(x-1)(x-2)`
The chages in signs of f(x) are shown in Fig 18 m

We observe that
`f(x) lt 0 " for all " x in (1,2) " and " f(x) gt 0 " for all " x in (2,3)`
`rArr f(x) ` is decreasing on [1,2] and increasing on [2,3]
`therefore` Minimum value of f(x)
`=f(x)=underset(1)overset(2)t(t^2-3t+2)dt=[(t^4)/(4)-t^3+t^2]_1^2=1-1/4`
and
Maximum value of f(x)Max {f(1),f(3)}
Clearly f(1)=0
and `f(3)=underset(1)overset(3)f(t^2-3t+2)dt=[(t^4)/(4)t^3+t^2]_1^3=2`
`therefore` Maximum value of f(X)=2
Since f(x) is contiuous on [1.3] so it attains every value between its minimum and maximum values.
Hence range `(f)=[-1//4,2]`