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Mathematics Question on limits and derivatives

Let f(x)=α(x)β(x)γ(x)f \left( x\right) = \alpha\left( x\right)\beta\left( x\right) \gamma \left( x\right) for all real x, where α(x),β(x)\alpha\left(x\right), \beta\left(x\right) and γ(x)\gamma \left( x\right) are differentiable functions of x. If f(2)=18f(2),α(2)=3α(2),β(2)=4β(2)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 γ(2)=kγ(2)\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

16

Explanation

Solution

We have, f(x)=α(x)β(x)γ(x)f \left( x\right) = \alpha\left( x\right)\beta\left( x\right) \gamma \left( x\right) , for all real x f(x)=α(x)β(x)γ(x)+α(x)β(x)γ(x)+α(x)β(x)γ(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) f(2)=α(2)β(2)γ(2)+α(2)β(2)g(2)+a(2)β(2)γ(2)\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) 18f(2)=3α(2)β(2)γ(2)4α(2)β(2)γ(2)+kα(2)β(2)γ(2)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) [f(2)=18f(2),α(2)=3α(2),β(2)=4β(2)\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 γ(2)=kγ(2)\gamma'\left(2\right) = k\gamma \left(2\right)] 18f(2)=(1+k)α(2)β(2)γ(2)\Rightarrow 18 f \left(2\right) = \left(-1+ k \right)\alpha\left( 2\right)\beta\left( 2\right) \gamma \left(2\right) =(k1)f(2)= \left(k -1\right) f \left( 2\right) [f(2)=α(2)β(2)γ(2)]\left[\because f \left(2\right) = \alpha\left(2\right)\beta\left(2\right) \gamma \left(2\right)\right] k1=18\Rightarrow\quad k -1 = 18 k=19\therefore\quad k = 19