Saved Bookmarks
| 1. |
Match the following lists: |
|
Answer» `alphax^(3)+betax` is an odd function `I=0+2int_(0)^(2)gamma dx=2xx2gamma=4gamma` b. `I=1/2 int_(0)^(1)2sin alpha x sin beta x dx` `=1/2int_(0)^(1)(cos(alpha-beta)x-cos(alpha+beta)x)dx` `=1/2[(sin(alpha-beta)x)/(alpha-beta)-(sin (alpha+beta)x)/(alpha+beta)]_(0)^(1)` `=1/2[(sin (alpha-beta))/(alpha-beta)-(sin (alpha+beta))/(alpha+beta)]`...........1 Also `2 alpha=tan alpha` and `2beta=tan beta` `:. 2 (alpha-beta)=tan alpha-tan beta ` and `2(alpha+beta)=tan alpha+tan beta` `2(alpha-beta)=(sin(alpha-beta))/(cos alpha cos beta)` and `2(alpha+beta)=(sin (alpha +beta))/(cos alpha cos beta)` substituting these values, we get C.`f(x+alpha)+f(x)=0` or `(x+2alpha)+f(x)+alpha)=0` or `f(x+2alpha)=f(x)` Thus, `f(x)` is periodic with periodic `2 alpha`. Hence `int_(beta)^(beta+2gamma alpha) f(x)dx=gamma int_(0)^(2alpha) f(x)dx` d. Let `I=int_(0)^(alpha)[SINX]dx, alpha epsilon[(2beta+1)pi,(2beta+2)pi],beta epsilon N`, ( wher [.] denotes the greatest INTEGER function) `=int_(0)^(2betapi) [sinx]dx+int_(2betapi)^((2beta+1)pi) [sinx]dx+int_((2beta+1)pi)^(alpha) [sinx] dx` `=betaint_(0)^(2PI) [sinx]dx+0+int_((2beta+1)pi)^(alpha) (-1)dx` `=-betapi+(2beta+1)pi-alpha` `=(beta+1)pi-alpha` Thus, `gamma int_(0)^(alpha) [sinx]dx` depends on `alpha beta` and `gamma`. |
|