(A) Let $f(x) = -x^2 + 6x + 12$. To find the extreme value, we find the derivative $f'(x) = -2x + 6$. Setting $f'(x) = 0$, we get $-2x + 6 = 0$, which

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(A) Let $f(x) = -x^2 + 6x + 12$. To find the extreme value, we find the derivative $f'(x) = -2x + 6$. Setting $f'(x) = 0$, we get $-2x + 6 = 0$, which implies $x = 3$. Thus, $\alpha = 3$. The extreme value $\beta$ is $f(\alpha) = f(3) = -(3)^2 + 6(3) + 12 = -9 + 18 + 12 = 21$. Since $\alpha = 3$, we can write $\beta = 21 = 7 \times 3 = 7 \alpha$. Therefore, $\beta = 7 \alpha$.

Class 12 Mathematics Applications of Derivatives Maxima and Minima

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If $6x - x^2 + 12$ attains its extreme value $\beta$ at $x = \alpha$, then $\beta =$

TS EAMCET 2022 All TS EAMCET

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$7 \alpha$
B
$5 \alpha$
C
$3 \alpha$
D
$\alpha$

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Applications of Derivatives

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If $6x - x^2 + 12$ attains its extreme value $\beta$ at $x = \alpha$, then $\beta =$

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