12 let k(x)=7 - 3sec(2x). which of the following gives the range of k? a) (-∞,-1 ∪ 1,∞) b) (-∞,-3 ∪ 3,∞) c)…

Explanation:

Step1: Recall range of secant function

The range of $y = \sec(t)$ is $(-\infty,- 1]\cup[1,\infty)$. Here $t = 2x$, so the range of $\sec(2x)$ is $(-\infty,-1]\cup[1,\infty)$.

Step2: Find range of $-3\sec(2x)$

Multiply the range of $\sec(2x)$ by - 3. When we multiply an inequality by a negative number, the direction of the inequality signs changes. So the range of $y=-3\sec(2x)$ is $(-\infty,3]\cup[- 3,\infty)$.

Step3: Find range of $k(x)=7 - 3\sec(2x)$

Add 7 to the range of $-3\sec(2x)$. For the lower - bound: if $y=-3\sec(2x)\leq3$, then $k(x)=7 - 3\sec(2x)\leq7 + 3=10$. For the upper - bound: if $y=-3\sec(2x)\geq - 3$, then $k(x)=7 - 3\sec(2x)\geq7-3 = 4$. So the range of $k(x)$ is $(-\infty,4]\cup[10,\infty)$.

Answer:

C. $(-\infty,4]\cup[10,\infty)$