Tìm GTLN-GTNN của các hàm số: a, y=2cos(x+π/2)-3 b, y = √[1-sin(x^2)]-1 c, y = 4sin(√x)

Tìm GTLN-GTNN của các hàm số:
a, y=2cos(x+π/2)-3
b, y = √[1-sin(x^2)]-1
c, y = 4sin(√x)

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1 Answer

  1. Đáp án:

    a)

    $\min y=1$ khi $ x=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    Và $\max y=5$ khi $ x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    b)

    $\min y=-1$ khi $ x^2=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    $\max y=\sqrt2-1$ khi $ x^2=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    c)

    $\min y=-4$ khi $ \sqrt x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z, K>0)$

    $\max y=4$ khi $\sqrt x=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z, k\ge0)$

    Giải thích các bước giải:

    a)

    $y=2\cos\left({x+\dfrac{\pi}2}\right)-3$

    Do $-1\le\cos\left({x+\dfrac{\pi}2}\right)\le1$ $\forall x$

    $\Rightarrow -2\le2\cos\left({x+\dfrac{\pi}2}\right)\le2$

    $\Rightarrow-2+3\le2\cos\left({x+\dfrac{\pi}2}\right)+3\le2+3$

    hay $1\le y\le5$

    Vậy $\min y=1$ khi $\cos\left({x+\dfrac{\pi}2}\right)=-1\Leftrightarrow x=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    Và $\max y=5$ khi $\cos\left({x+\dfrac{\pi}2}\right)=1\Leftrightarrow x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$.

    b)

    $y=\sqrt{1-\sin x^2}-1$

    Do $-1\le\sin x^2\le1$ $\forall x$

    $\Rightarrow 1\ge-\sin x^2\ge-1$

    $\Rightarrow 1+1\ge1-\sin x^2\ge1-1$

    $\Rightarrow 0\le\sqrt{1-\sin x^2}\le\sqrt2$

    $\Rightarrow-1\le\sqrt{1-\sin x^2}-1\le\sqrt2-1$

    Hay $-1\le y\le\sqrt2-1$

    Vậy $\min y=-1$ khi $\sin x^2=-1\Leftrightarrow x^2=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$

    $\max y=\sqrt2-1$ khi $\sin x^2=1\Leftrightarrow x^2=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z)$.

    c) 

    $y=4\sin\sqrt x$

    Do $-1\le\sin\sqrt x\le1$ $\forall x\ge0$

    $\Leftrightarrow-4\sin\sqrt x\le4$

    Hay $-4\le y\le4$

    Vậy $\min y=-4$ khi $\sin\sqrt x=-1\Leftrightarrow \sqrt x=-\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z, K>0)$

    $\max y=4$ khi $\sin\sqrt x=1\Leftrightarrow \sqrt x=\dfrac{\pi}2+k2\pi$ $(k\in\mathbb Z, k\ge0)$.

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