1.

Let f(x)={{:(3x^2-2x+10, x lt 1),(-2,x gt 1):} The set of values of b for which f(x) has greatest value at x=1 is

Answer»

`(-6,-2)`
(2,6)
`(-6,-2)cup (2,6)`
`(-6,6)`

Solution :we have
`f(x)={{:(3x^2-2x+10, x lt 1),(-2,x gt 1):}`
Clealy `f(x) gt 0 for x lt 1 ` and`f(x)lt 0 for x gt 1`
Thus f(x) is INCREASING for all `x lt 1` and decreasing for `x gt 1`
Therfore f(x) will have grastest value at x=1 if
`underset(x to 1)lim f(x)lt f(1)=underset(x to 1)limf(x)`
`rArr underset(x to 1)lim -2x+log_2 (b^2-4)lt -1+10-7`
`rArr-2 + LOG _2(b^2-4) lt 3`
`rArrlog_2(b^2 -4)lt 5`
`b^2 -36 lt 0 and b^2 -4 gt 0 `
`rArr-6 lt b lt 6and b in (-oo,-2) cup (2,oo)`
`RARRB in (-6,-2) cup (2,6)`


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