# If x=1 is a critical point of a function $f(x) = (3x^2 + ax - 2-a)e^x$ then Option: 1  is a local maxima and  is a local minima of f. Option: 2  is a local maxima and   is a local minima of f. Option: 3  and   is a local minima of f. Option: 4  and   is a local maxima of f.

$\\f(x) = ( 3x^2 +ax-2-a)e^x \\ \\f'(x) = ( 6x +a)e^x +( 3x^2 +ax-2-a)e^x \\ \\f'(x)|_{at x =1} = ( 6 +a)e^1 +( 3 +a-2-a)e^1 = 0\\ a + 7 = 0 \\ a = -7 \\ \\f'(x) = ( 3x^2 -x-2)e^x \\ \text{for minima and maxima f'(x) = 0 } \\ \\f'(x) = ( 3x^2 -x-2)e^x = 0 \\ 3x^2 -x-2 = 0 \\ 3x^2 -3x+2x-2 = 0 \\ 3x(x-1)+2(x-1) = 0 \\ (x-1)(3x+2) = 0 \\ x = 1 \ or \ x = \frac{-2}{3}$

$\\f''(x) = ( 6x -1)e^x +( 3x^2 -x-2)e^x \\ \\f''(x) |_{at x = 1}= 5e > 0\\ \rightarrow local minima at x =1 \\ \\f''(x) |_{at x = -2/3}= -5e^{\frac{-2}{3}} < 0\\ \rightarrow local maxima at x =-2/3$

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