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Q. Let $f \left( x\right) = \alpha\left( x\right)\beta\left( x\right) \gamma \left( x\right)$ for all real x, where $\alpha\left(x\right), \beta\left(x\right)$ and $\gamma \left( x\right)$ are differentiable functions of x. If $f ' \left(2\right) = 18 f \left(2\right),\alpha' \left(2\right) = 3\alpha\left(2\right), \beta' \left(2\right) = -4\beta\left(2\right)$ and $\gamma'\left(2\right) = k\gamma \left(2\right)$ , then the value of k is

Limits and Derivatives

Solution:

We have, $f \left( x\right) = \alpha\left( x\right)\beta\left( x\right) \gamma \left( x\right)$ , for all real x
$\Rightarrow f' \left( x\right) = \alpha'\left( x\right)\beta\left( x\right) \gamma \left( x\right) + \alpha\left( x\right)\beta'\left( x\right) \gamma \left( x\right) +\alpha\left( x\right)\beta\left( x\right) \gamma'\left( x\right)$
$\Rightarrow f' \left(2\right) = \alpha' \left(2\right)\beta\left(2\right) \gamma \left(2\right) + \alpha\left(2\right)\beta'\left(2\right)g \left(2\right)+a\left( 2\right)\beta\left(2\right) \gamma'\left(2\right)$
$18 f \left(2\right) = 3\alpha\left( 2\right)\beta\left(2\right) \gamma \left(2\right) - 4\alpha\left(2\right)\beta\left( 2\right) \gamma \left(2\right)+ k \alpha\left(2\right)\beta\left( 2\right) \gamma \left(2\right)$
[$\because f'\left( 2\right) =18 f \left(2\right),\alpha'\left( 2\right) = 3\alpha \left(2\right) ,\beta'\left(2\right)= -4\beta\left(2\right)$ and $\gamma'\left(2\right) = k\gamma \left(2\right)$]
$\Rightarrow 18 f \left(2\right) = \left(-1+ k \right)\alpha\left( 2\right)\beta\left( 2\right) \gamma \left(2\right)$
$= \left(k -1\right) f \left( 2\right)$
$\left[\because f \left(2\right) = \alpha\left(2\right)\beta\left(2\right) \gamma \left(2\right)\right]$
$\Rightarrow \quad k -1 = 18$
$\therefore \quad k = 19$