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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
Given $f(x) = -\frac{{{x^3}}}{3} + x^2 sin 1.5 a - x sin a . sin 2a - 5 \,\,sin^{-1} (a^2 - 8a + 17)$ then :
  • A
    $f(x)$ is not defined at $x = sin 8$
  • B
    $f ‘ (sin 8) > 0$
  • C
    $f ‘ (x)$ is not defined at $x = sin 8$
  • ✓
    $f ‘ (sin 8) < 0$
✓

Answer

Correct option: D.
$f ‘ (sin 8) < 0$
d
$f(x) =  - \frac{{{x^3}}}{3} + {x^2}sin\,6 - xsin\,4.sin\,8 - 5\,si{n^{ - 1}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} ({(a - 4)^2} + 1)$

$f'(x) =  - {x^2} + 2x\,sin\,6 - sin\,4\,sin\,8$             $      (a = 4)$

$f'(sin\,8) =  - si{n^2}8 + 2\,sin\,6\,sin\,8 - sin\,4\,sin\,8$

$ = {\rm{ }}sin\,8\,\,\left[ {{\rm{ }} - {\rm{ }}sin\,8{\rm{ }} + {\rm{ }}2{\rm{ }}sin\,6{\rm{ }} - {\rm{ }}sin\,4} \right]$

$\; = {\rm{ }} - {\rm{ }}sin\,8{\rm{  }}\,\left[ {sin\,8{\rm{ }} + {\rm{ }}sin\,4{\rm{ }} - {\rm{ }}2\,sin\,6} \right]{\rm{ }} = {\rm{ }} - {\rm{ }}sin\,8{\rm{  }}\left[ {2sin\,6{\rm{ }}cos\,2{\rm{ }} - {\rm{ }}2\,sin\,6} \right]$

$ = {\rm{ }}2{\rm{ }}sin\,8{\rm{  }}sin\,6{\rm{ }}\left[ {{\rm{ }}1{\rm{ }} - {\rm{ }}cos\,2} \right]\;$

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