If `f(x)=int_0^x |t-1| dt`, where `0 lt= x lt= 2` then
(a) range of `f(x)` is `[0,1]` (b) `f(x)` is differentiable at `x=1` (c) `f(x)=cos^(-1)x` has two real roots (d) `f^(prime)(1/2)=1/2`

Options

A. range of `f(x)` is `[0,1]`
B. `f(x)` is differentiable at `x=1`
C. `f(x)=cos^(-1)x` has two real roots
D. `f'(1//2)=1//2`

Answer

B

Explanation

Given that `f(x)=int_(0)^(x)|t-1|dt`
`=int_(0)^(x)(1-t)dt, 0lexle1`
`=x-(x^(2))/2`
Also `f(x)=int_(0)^(1)(1-t)dt+int_(1)^(x)(t-1)dt`, where `1 le x le 2`
`=1/2+(x^(2))/2-x+1/2=(x^(2))/2-x+1`
Thus `f(x)={(x-(x^(2))/2,0lexle1),((x^(2))/2-+1,lt x le2):}`
`:.f'(x)={(1-x,0lexlt1),(x-1,1ltxlt2):}`
Thus `f(x) `is continuous as well as differentiable at `x=1`
Also, `f(x)=cos^(-1)x` has one real root. Draw the graph and verify.
For range of `f(x)`:
`f(x)=int_(0)^(x)|t-1|dt` is the value of bounded by the curve `y=|t-1|` and `x` -axis betwen the limits `t=0` and `t=x`.
Obviously minimum area is obtained when `t=0` and `t=x` coincide or `x=0`.
Maximum value of area occurs when `t=2`.
Hence `f(2)=` ara of shaded region `=1`.