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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
If the function $f$ defined on $\left( {\frac{\pi }{6},\frac{\pi }{3}} \right)$ by $f\,(x)\, = \,\left\{ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 2 \,\cos \,x - \,1}}{{\cot \,x\, - \,1}}\,,\,x\, \ne \,\frac{\pi }{4}}\\
{k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \frac{\pi }{4}}
\end{array}} \right.$ is continuous, then $k$ is equal to
  • A
    $1$
  • B
    $2$
  • ✓
    $\frac {1}{2}$
  • D
    $\frac {1}{\sqrt 2}$
✓

Answer

Correct option: C.
$\frac {1}{2}$
c
$\therefore \,$ function should be continuous at $x = \frac{\pi }{4}$

$\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = f\left( {\frac{\pi }{4}} \right)$

$ \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\sqrt 2 \cos x - 1}}{{\cot x - 1}} = k$

$ \Rightarrow \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{ - \sqrt 2 \sin x}}{{ - \cos e{c^2}x}} = k$    (Using $L$ Hospital Rule)

$\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \sqrt 2 {\sin ^3}x = k$

$ \Rightarrow k = \sqrt 2 {\left( {\frac{1}{{\sqrt 2 }}} \right)^3} = \frac{1}{2}$

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