Factorise the following : `25a^2–4b^2+28bc–49c^2`

Answer

`atoq;btor,s;ctop;dtop`

Explanation

a. `I_(1)=int_(pi//6)^(pi//3) sec^(2) thetaf(2sin theta) d theta`
Applying property `int_(a)^(b)f(a+b-x)dx=int_(a)^(b)f(x)dx`
`I_(1)=int_(pi//6)^(pi//3)sec^(2)((pi)/2-theta)f(2sin2((pi)/2-theta))d theta`
`=int_(pi//6)^(pi//3)cosec^(3) theta f(2sin 2 theta) d theta=I_(2)`
b. `f(x+1)=f(x+3)` or `f(x)=f(x+2)`
Thus, `f(x)` is periodic with period 2.
Then `int_(a)^(a+b)f(x)dx` is independent of `a` for which `b` is multiple of 2.
Thus, `b=2,4,6`...........
c. Let `I=int_(1)^(4)(tan^(-1)[x^(2)])/(tan^(-1)[x^(2)]+tan^(-1)[25+x^(2)-10x])`.............1
Appling `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`, we get
`I=int_(1)^(4)(tan[(5-x)^(2)])/(tan^(-1)[(5-x)^(2)]+tan^(-1)[x^(2)])dx`.............2
Adding equatiions 1 and 2 we get
`2I=int_(1)^(4)dx` or `2I=3` or `I=3//2`
d. Let` y=sqrt(x+sqrt(x+sqrt(x+)))..............=sqrt(x+y)`
or `y^(2)-y-x=0`
or `y=(1+-sqrt(1+4x))/(2.1)=(1+sqrt(1+4x))/2( :' ygt1)`
`:.I=int_(0)^(2)(1+sqrt(1+4x))/2 dx`
`=[x/2+((1+4x)^(3//2))/(3/2xx2xx4)]_(0)^(2)`
`=[(1+27/12)=(0+1/12)]`
`=1+26/12=19/6`
`:. [I]=3`
`[I]=3`