*Tìm GTLN, GTNN của hàm số
a. y = sin2x.cos2x
b. y = 2sin^2x-5cos2x+2
c. y = cos^2x+2cos2x
d. y = cosx-sinx
e. y = cos5x+sin5x
f. y= sin3x-cos3x
Mình đâng cần gấp ạ, ai giỏi toán chỉ mình với
*Tìm GTLN, GTNN của hàm số a. y = sin2x.cos2x b. y = 2sin^2x-5cos2x+2 c. y = cos^2x+2cos2x d. y = cosx-sinx e. y = cos5x+sin5x f. y= sin3x-cos3x Mình
By Julia
Đáp án:
$\begin{array}{l}
a)y = \sin 2x.\cos 2x = \dfrac{1}{2}.\sin 4x\\
\Rightarrow – \dfrac{1}{2} \le y \le \dfrac{1}{2}\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – \dfrac{1}{2}\\
Khi:\sin 4x = – 1 \Rightarrow x = \dfrac{{ – \pi }}{8} + \dfrac{{k\pi }}{2}\\
GTLN:y = \dfrac{1}{2}\\
Khi:\sin 4x = 1 \Rightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{2}
\end{array} \right.\\
b)y = 2{\sin ^2}x – 5\cos 2x + 2\\
= 2{\sin ^2}x – 5\left( {1 – 2{{\sin }^2}x} \right) + 2\\
= 12{\sin ^2}x – 3\\
Do:0 \le {\sin ^2}x \le 1\\
\Rightarrow – 3 \le 12{\sin ^2}x \le 9\\
\Rightarrow – 3 \le y \le 9\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – 3\\
GTLN:y = 9
\end{array} \right.\\
c)y = {\cos ^2}x + 2\cos 2x\\
= {\cos ^2}x + 2\left( {2{{\cos }^2}x – 1} \right)\\
= 5{\cos ^2}x – 2\\
\Rightarrow – 2 \le y \le 3\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – 2\\
GTLN:y = 3
\end{array} \right.\\
d)y = \cos x – \sin x\\
= – \sqrt 2 .sin\left( {x – \dfrac{\pi }{4}} \right)\\
\Rightarrow – \sqrt 2 \le y \le \sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – \sqrt 2 \\
GTLN:y = \sqrt 2
\end{array} \right.\\
e)y = \cos 5x + \sin 5x\\
= \sqrt 2 \sin \left( {5x + \dfrac{\pi }{4}} \right)\\
\Rightarrow – \sqrt 2 \le y \le \sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – \sqrt 2 \\
GTLN:y = \sqrt 2
\end{array} \right.\\
f)y = \sin 3x – \cos 3x\\
= \sqrt 2 \sin \left( {3x – \dfrac{\pi }{4}} \right)\\
\Rightarrow – \sqrt 2 \le y \le \sqrt 2 \\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = – \sqrt 2 \\
GTLN:y = \sqrt 2
\end{array} \right.
\end{array}$