Most Important Selected Qs for JEE AdvancedMathematicsDefinite Integration
g(n)= ₀^ n^2+n+1 e^ x / 2-[x / 2] ( x 2 - [ x 2 ] ) d(x-[x]) ; n N then g(n)
Options
- Ahas minimum value as 1 4 + e
- Bhas maximum value as 3- e
- Chas minimum value as 3 4 - e 4
- Dnone of these
Correct answer
D. none of these
Step-by-step solution
g(n)= ₀^ n^2+n+1 e^ x / 2] x 2 d x = (n^2+n+1 ) ₀^1 e^ (x / 2) x 2 d x= (n^2+n+1 ) ₀^1 e^ x / 2 ( x 2 ) d x=n^2+n+1 [4-2 e^ 1 / 2 ] So, minimum value is 12-6 e