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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 |
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Answer» `(-6,-2)` `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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