Let `g(x)=int_(0)^(x)f(t) dt`, where f is such that `(1)/(2)lef(t)le1 "for" t in [0,1] and , 0 le f(t) le(1)/(2) "for" t in [1,2]` then, g(2) satisfiwes the inquqality,

Options

A. `-(3)/(2) le g (2) lt (1)/(2)`
B. `0 leg(2) le2`
C. `(3)/(2) le g (2) le (5)/(2)`
D. `2 lt g(2) lt4`

Answer

B

Explanation

We have,
`g(x)=underset(0)overset(x)intf(t)`
`rArr g(2) =overset(2)underset(0)int f(t)dt=overset(1)underset(0)int f(t)dt+overset(2)underset(1)int f(t)dt`
We know that
`m le f(x) le M "for" x in [a,b]`
`rArr m(b-a) le underset(a)overset(b) int f(x)dx le M(b-a)`
`:. (1)/(2) le f(t) le 1 "for" t in [0,1] and ,0 le f(t) le (1)/(2) "for" t in [1,2]`
` and (1)/(2) (1-0) leunderset(0)overset(1)intf(t)dt le 1(2-1)`
`rArr (1)/(2) le underset(0)overset(1)intf(t)dt le 1 and, 0 le underset(1)overset(2)int f(t)dt le(1)/(2)`
`rArr (1)/(2) le underset(0)overset(1)intf(t)dt + underset(1)overset(2)int f(t)dtle(3)/(2)`
`rArr (2)/(2)n le g(2) (3)/(2) rArr 0 le g(2) lt2`