If $f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{a | x^{2} - 15 x + 56 |}{x - 8} & , x > 9 b & , x = 9 \frac{x - [x]}{x - 8} & , x < 9 \end{matrix}$ , where $[.]$ denotes greatest integer function and the function $f$ is continuous then

a = \frac{1}{2}, b = 1

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a = 0, b = 1

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a = \frac{- 1}{2}, b = 1

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a = \frac{- 1}{2}, b = - 1

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Solution

The correct option is A $a = \frac{1}{2}, b = 1$
The function is continuous,
$∴$ Left hand limit $f (9^{-}) =$ Right hand limit $f (9^{+}) = f (9)$
$f (9^{-}) = {lim}_{h \to 0} \frac{9 - h - [9 - h]}{9 - h - 8} = {lim}_{h \to 0} \frac{9 - h - 8}{9 - h - 8} = 1$
$f (9^{+}) = {lim}_{h \to 0} \frac{a | (9 + h)^{2} - 15 (9 + h) + 56 |}{9 + h - 8} = {lim}_{h \to 0} \frac{a | (h + 2) (h + 1) |}{1 - h} = 2 a$ , $h > 0$
$f (9) = b$
$∴ f (9^{-}) = f (9^{+}) = f (9)$
$\Rightarrow 1 = 2 a = b$
$\Rightarrow a = \frac{1}{2}, b = 1$

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