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If $\frac{d}{dx}[ax^3+ ax^2 + ax + 1] = 9x^2 + 6x +3$, then $a$ is equal to |
1 2 3 4 |
3 |
The correct answer is Option (3) → 3 $\frac{d}{dx}[ax^{3}+ax^{2}+ax+1]=3ax^{2}+2ax+a$ $3ax^{2}+2ax+a=9x^{2}+6x+3$ $\Rightarrow\;3a=9,\;2a=6,\;a=3$ $a=3$ |
