Let `g (x)= int_(0)^(x) f(t) dy=t` , where f is such that ` (1)/(2) le f (t) le 1` for ` tin [0,1] and 0 le f (t) le (1)/(2) ` for ` t in [1 ,2]`. then . `g (2)` satisfies the inequality

Options

A. `-(3)/(2)leg (2) lt(1)/(2)`
B. ` 0leg (2) lt2`
C. `(3)/(2) lt g (2) le 5//2`
D. `2lt g (2) lt 4`

Answer

B

Explanation

Given , g (x) `= int_(0)^(x)f(t) dt`
` rArr g (2) = int_(0)^(2) f (t) dt = int_(0)^(1)f(t)dt + int_(1)^(2)f(t)dt`
Now , `(1)/(2) le f (t) le 1 " for" t in [0,1]`
We get `int_(0)^(1)(1)/(2)dt leint_(0)^(1)f(t)dt le int_(0)^(1)1 dt`
`rArr (1)/(2)leint_(0)^(1)f (t) dt le 1` . . . (i)
Again , `0le f (t) le (1)/(2) " for " t in [1,2]` . . . (ii)
`rArr int_(1)^(2)0 dt le int_(1)^(2) f (t) dtle int _(1)^(2) dt`
`rArr 0 le int_(1)^(2) f (t) dt le (1)/(2)`
From Eqs . (i) and (ii) , we get
`(1)/(2)le int_(0)^(1)f (t) dt + int_(1)^(2) f (t) dt le (3)/(2)`
` rArr (1)/(2) le g (2) le (3)/(2)`
` rArr 0leg (2) lt 2`