1.

y= f(x) is a polynomial function passing through point (0, 1) and which increases in the intervals (1, 2) and (3, oo) and decreases in the intervals (oo,1) and (2, 3). If f(1) = -8, then the value of f(2) is

Answer»

-3
-6
-20
-7

Solution :From the given data, we can CONCLUDE that `(dy)/(dx)=0 " at " x=1,2,3.`
Hence, `f'(x)=a(x-1)(x-2)(x-3), a gt 0`
`implies f(x)=int a(x^(3)-6x^(2)+11x-6)dx`
`=a((x^(4))/(4)-2x^(3)+(11x^(2))/(2)-6x)+C`
Also, `f(0)=1 implies c=1`
` :. f(x)=a((x^(4))/(4)-2x^(3)+(11x^(2))/(2)-6x)+1 "(1)" `
So, graph is symmetrical about line `x=2` and range is `[f(1), OO) or [f(3),oo).`
`f(1)=8`
`implies a(-(9)/(4))+1=-8`
`impliesa=4`
`:. f(2)= -7`


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