MCQ
If the function $ f (x) =\frac{{t + 3x - {x^2}}}{{x - 4}}$ , where $'t'$ is a parameter has a minimum and a maximum then the range of values of $'t'$ is
- A$(0, 4)$
- B$(0, \infty )$
- ✓$(- \infty , 4)$
- D$(4, \infty )$
✓
Answer
Correct option: C.
$(- \infty , 4)$
c
$f (x) =\frac{{t + 3x - {x^2}}}{{x - 4}}$ ;$f ' (x) = \frac{{(x - 4)(3 - 2x) - (t + 3x - {x^2})}}{{{{(x - 4)}^2}}}$
$f (x) =\frac{{t + 3x - {x^2}}}{{x - 4}}$ ;$f ' (x) = \frac{{(x - 4)(3 - 2x) - (t + 3x - {x^2})}}{{{{(x - 4)}^2}}}$
for maximum or minimum, $f ' (x) = 0$
$- 2x^2 + 11x - 12 - t - 3x + x^2 = 0$
$- x^2 + 8x - (12 + t) = 0$
for one $M$ and $m,$
$D > 0$
$64 - 4(12 + t) > 0$
$16 - 12 - t > 0$
==>$4 > t$ or $t < 4$
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