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STD 12 - 6. Application of derivatives — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 6. Application of derivatives1 Mark
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
$f (x) =\frac{{t + 3x - {x^2}}}{{x - 4}}$ ;$f ' (x) = \frac{{(x - 4)(3 - 2x) - (t + 3x - {x^2})}}{{{{(x - 4)}^2}}}$

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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