Differentiation Question 106
Question: Let $ f(x)=\alpha (x)\beta (x)\gamma (x) $ for all real x, where $ \alpha (x),\beta (x) $ and $ \gamma (x) $ are differentiable functions of $ x. $ If $ f’(2)=18f(2),\alpha ‘(2)=3\alpha (2),\beta ‘(2)=-4\beta (2) $ and $ \gamma ‘(2)-k\gamma (2), $ then the value of k is
Options:
A) 14
B) 16
C) 19
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We have, $ f(x)=\alpha (x)\beta (x)\gamma (x), $ for all real x.
$ \Rightarrow f’(x)=\alpha ‘(x)\beta (x)\gamma (x)+\alpha (x)\beta ‘(x)\gamma (x)+\alpha (x)\beta (x)\gamma ‘(x) $
$ \Rightarrow f’(2)=\alpha ‘(2)\beta (2)\gamma (2)+\alpha (2)\beta ‘(2)\gamma (2)+\alpha (2)\beta (2)\gamma ‘(2) $
$ \Rightarrow 18f(2)=3\alpha (2)\beta (2)\gamma (2)-4\alpha (2)\beta ‘(2)\gamma (2)+k\alpha (2)\beta (2)\gamma (2) $
$ [\because f’(2)=18f(2),\alpha ‘(2)=3\alpha (2),\beta ‘(2) $
$ =-4\beta (2)and\gamma ‘(2)=k\gamma (2)] $
$ \Rightarrow 18f(2)=(-1+k)\alpha (2)\beta (2)\gamma (2)=(k-1)f(2) $
$ [\because f(2)=\alpha (2)\beta (2)\gamma (2)] $
$ \Rightarrow k-1=18\Rightarrow k=19. $
BETA
