Question

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

• A. 14
• B. 16
• C. 19
• D. None of these

Answer 1

The Correct Option is B

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 $\gamma'\left(2\right) = k\gamma \left(2\right)$] $\Rightarrow 18 f \left(2\right) = \left(-1+ k \right)\alpha\left( 2\right)\beta\left( 2\right) \gamma \left(2\right)$ $= \left(k -1\right) f \left( 2\right)$ $\left[\because f \left(2\right) = \alpha\left(2\right)\beta\left(2\right) \gamma \left(2\right)\right]$ $\Rightarrow\quad k -1 = 18$ $\therefore\quad k = 19$