Write the maximum values of 12 -9 ^2x. - Vidyadip

Question

Write the maximum values of $12\sin\text{x}-9\sin^2\text{x}.$

✓

Answer

$12\sin\text{x}-9\sin^2\text{x}=3\sin\text{x}(4-3\sin\text{x})$
We know that $0\le\sin\text{x}\le1$
$\therefore$ Maximum value of $12\sin\text{x}-9\sin^2\text{x}$
$=12\Big(\frac23\Big)-9\Big(23\Big)^2$
$=\frac{24}{3}-\frac{36}{9}$
$=\frac{36}{9}$
$=4$
$\because$ for other than $\sin\text{x}=\frac{4}{3}$ it is less than 4 Maximum value of $12\sin\text{x}-9\sin^2\text{x}$
$=4$

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