The value of k so that the function f(x) = l k(2x - x^2 ), ; ; ; … - Vidyadip

MCQ

The value of $k$ so that the function $f(x) = \left\{ \begin{array}{l}k(2x - {x^2}),\;\;\;{\rm{when\,}}\,x < 0\\\,\,\,\,\,\,\,\,\,\cos x,\,\,\,\,\,\,{\rm{when\,}}\,x \ge {\rm{0}}\end{array} \right.$ is continuous at $x = 0$, is

A
$1$
B
$2$
C
$4$
D
None of these

✓

Answer

$f(0 - ) = \mathop {\lim }\limits_{x \to 0 - } k(2x - {x^2}) = 0$; $f(0 + ) = \mathop {\lim }\limits_{x \to 0 + } \cos x = 1$
$\therefore f(0) = \cos x = 1$
Hence no value of $k$ can make $f(0 - ) = 1.$

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