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Question #05dd5

1 Answer

Ratnesh Bhosale

Mar 11, 2017

#a=3#

Explanation:

#f(x)= (ax^2-1)(3+2x)#

#f(x)= 3ax^2+2ax^3-3-2x#

#f(x)= 2ax^3+3ax^2 -2x-3#

By Differentiating,

#f'(x) = 6ax^2 +6ax-2#

Put #x=1#

#f'(1) = 12a-2#

#34 = 12a-2#

#12a = 36#

#a=3#

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