If `int_0^1 e^(x^2)(x-alpha)dx=0`, then (a)`alphalt2` (b)`alphalt0` `(c)"" 0ltalphalt1` (d) `alpha="0`

Options

A. `1 lt alpha lt 2`
B. `alpha lt 0`
C. `0 lt alpha lt 1`
D. `alpha=0`

Answer

C

Explanation

We have `int_(0)^(1) e^(x^(2))(x-alpha)dx=0`
or `int_(0)^(1)e^(x^(2))xdx=int_(0)^(1)e^(x^(2)) alpha dx`
or `1/2 int_(0)^(1) e^(t) dt=alpha int_(0)^(1) e^(x^(2))dx`, where `t=x^(2)`
or`1/2 (e-1)=alpha int_(0)^(1)e^(x^(2))dx`…………….1
Since `e^(x^(2))` is an increasing function for `0lexle1`.
`1le e^(x^(2))le e` when `0 le x le 1`
or `1(1-0)le int_(0)^(1)e^(x^(2))dxle(1-0)`
or `1 le int_(0)^(1) e^(x^(2))dxle e`................2
From equation 1 and 2 we find that L.H.S of equation 1 is positive and `int_(0)^(1)e^(x^(2))dx` lies between 1 and `e`. Therefore `alpha` is a positive real number.
Now, from equation 1 `alpha=(1/2(e-1))/(int_(0)^(1)e^(x^(2))dx)`.................3
The denominator of equation 3 is greater than unity and the numerator lies between 0 and 1. Threfore, `0 lt alpha lt 1`.