If line `y-x-1+lambda=0` is equally inclined to axes and equidistant from the point `(1,-2)` and `(3,4)` ,the `lambda` is

Answer

`atop,q;btop,q,r,ctoq,s;dtos

Explanation

a. `I=int_(-2)^(2)(alphax^(3)+betax+gamma)dx`
`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`.