Consider the equation `int_(0)^(x) (t^(2)-8t+13)dt= x sin (a//x)`
If x takes the values for which the equation has a solution, then the number of values of `a in [0, 100]` is

Options

A. `y in [-5, 7]`
B. `y in [-7, 5]`
C. `y in [5, 7]`
D. none of these

Answer

B

Explanation

Given that `underset(0)overset(x) (int) (t^(@)-8t+13) dt=x sin (a/x)`
For `R.H.S., x ne 0`
Integrating `L.H.S.`, we get
`[t^(3)/3-4t^(2)+13t]_(0)^(x)=x sin (a/x)`
or `(1/3)[x^(3)-12 x^(2)+39x]=x sin (a/x)`
or `sin (a/x)=(1/3) [x^(2)-12 x+39]" "{ :'x ne 0}`
`=(1/3) {(x-6)^(2)+3}`
`=(1/3) (x-6)^(2)+1`
But `sin (a//x) le 1`, so `sin (a//x)=1`, which is possible only for `x=6` Then we have `sin (a//6) =1`
or `a//6=2npi+pi//2 or a=12 npi+3 pi, n in Z`
Hence, `x=6, a=12 n pi+3pi, n in Z`.
For `a in [0, 100]`, there are exactly three values of `a=3pi, 15 pi`, amd `27 pi`, i.e.,
`|y- cos a| lt x`
`rArr |y+1| lt 6`
`rArr y in [-7, 5]`