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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If the function $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{\sqrt {2  + \cos \,x} - 1}}{{\left( {\pi  - {x^2}} \right)}},}&{x \ne \pi } \\ 
  {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x = \pi } 
\end{array}} \right.$ is continuous at $x\, =\pi $ , then $k$ equals
  • A
    $0$
  • B
    $\frac{1}{2}$
  • C
    $2$
  • ✓
    $0.25$
✓

Answer

Correct option: D.
$0.25$
d
Since $f\left( x \right) = \frac{{\sqrt {2 + \cos x }- 1 }}{{{{\left( {\pi  - x} \right)}^2}}}$ is

Continuos at $x = \pi $

$\therefore L.H.L = R.H.L = f\left( \pi  \right)$

Let $\left( {\pi  - x} \right) = \theta ,\theta  \to 0$ when $x \to \pi $

$\therefore \mathop {\lim }\limits_{\theta  \to 0} \frac{{\sqrt {2 + \cos \theta  - 1} }}{{{\theta ^2}}}$

$ = \mathop {\lim }\limits_{\theta  \to 0} \frac{{\left( {2 + \cos \theta } \right) - 1}}{{{\theta ^2}}} \times \frac{1}{{\sqrt {2 + \cos \theta }  + 1}}$

$ = \mathop {\lim }\limits_{\theta  \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}}.\frac{1}{2}\,\,$           ($\because $ $\cos 0 = 1$)

$ = \mathop {\lim }\limits_{\theta  \to 0} \frac{{2{{\sin }^2}\theta /2}}{{{\theta ^2}}}$

$ = \frac{2}{2}\mathop {\lim }\limits_{\theta  \to 0} \frac{{{{\sin }^2}\theta /2}}{{\frac{{{\theta ^2}}}{4}.4}} = \frac{1}{4}$

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