Find `A+B if A = [(2,5),(3,6)] and B = [ (7,9),(6,5)]`

Answer

`(A) rarr (p,r); (B) rarr (s); (C) rarr (q); (D) rarr (s)`

Explanation

`(A) rarr (p,r), (B) rarr (s), (C) rarr (q), (D) rarr (s)`
On comparing, we get
`{4f(-1)-3}a^(2) + {4f(1)-3} a+f(2)=0`
`{4f(-1)-3}b^(2) + {4f(1)-3} b+f(2)=0`
and `{4f(-1)-3}c^(2) + {4f(1)-3}C+f(2)=0`
It is clear that a, b, c are the roots of
`{4f(-1)-3}x^(2) + {4f(1)-3}x+f(2)=0` then
`4f(-1)-3=0, 4f(1)-3=0,f(2)=0`
`rArr f(-1)=3/4,f(1)=3/4,f(2)=0` Let `f(x) = (x-2) (ax+b)`
Now, `f(-1) = 3/4 rArr (-3) (-a+b) =3/4 rArr a-b = 1/4 `
`f(1) = 3/4 rArr (-1) (-a+b) =3/4 rArr a+b = -3/4 `
`therefore a= -1/4 , b = -1/2`
`rArr f(x) = 1/4 (4-x^(2))`
Graph of `y = f(x)`

(A) x- coordinates of the point intersection of `y = f(x) ` with
ther X- axis are - 2 and 2.
(B) Area `=3/2 int_(-2)^(2) 1/4 (4-x^(2)) dx = 3/4 int _(0)^(2) (4-x^(2)) dx`
`= 3/4 [4x-x^(3)/3]_(0)^(2)= 3/4xx 16/3 = 4`
(C) Maximum value of `f(x)` is 1.
(D) Length of intercept on the X_ axis is 4.