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If `f(x)=(t+3x-x^2)/(x-4),`where `t`is a parameter that has minimum and maximum, then the range of valuesof `t`is`(0,4)`(b) `(0,oo)``(-oo,4)`(d) `(4,oo)`A. `(0,4)`B. `(0,oo)`C. `-(oo,4)`D. `(4,oo)` |
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Answer» Correct Answer - 3 `f(x)=(t+3x-x^(2))/(x-4),f(x)=(x-4)(3x-2x)-(t+3x-x^(2))/(x-4)^(2)` for maximum or minimum f(X)=0 `-2x^(2)+11x-12-t-3x+x^(2)=0` `-x^(2)+8x-(12-t)=0` For one maxima and minima `64-4(12+t)gt0` `16-12-tgt0, i.e 4gtt or tlt4` |
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