A) 14
B) 16
C) 19
D) None of these
Correct Answer: C
Solution :
[c] We have, \[f(x)=\alpha (x)\beta (x)\gamma (x),\]for all real x. \[\Rightarrow f'(x)=\alpha '(x)\beta (x)\gamma (x)+\alpha (x)\beta '(x)\gamma (x)+\alpha (x)\beta (x)\gamma '(x)\]\[\Rightarrow f'(2)=\alpha '(2)\beta (2)\gamma (2)+\alpha (2)\beta '(2)\gamma (2)+\alpha (2)\beta (2)\gamma '(2)\]\[\Rightarrow 18f(2)=3\alpha (2)\beta (2)\gamma (2)-4\alpha (2)\beta '(2)\gamma (2)+k\alpha (2)\beta (2)\gamma (2)\]\[[\because f'(2)=18f(2),\alpha '(2)=3\alpha (2),\beta '(2)\] \[=-4\beta (2)and\gamma '(2)=k\gamma (2)]\] \[\Rightarrow 18f(2)=(-1+k)\alpha (2)\beta (2)\gamma (2)=(k-1)f(2)\] \[[\because f(2)=\alpha (2)\beta (2)\gamma (2)]\] \[\Rightarrow k-1=18\Rightarrow k=19.\]You need to login to perform this action.
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