If f(x) = acos(bx + c) + d,then the range of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $f(x) = a\cos(bx + c) + d$. We know that the range of the cosine function $\cos(	heta)$ is $[-1, 1]$. Therefore,$-1 \le \cos(b

Vedclass · Accepted Answer

$[d - a, d + a]$

Vedclass · Answer

(D) Given the function $f(x) = a\cos(bx + c) + d$. We know that the range of the cosine function $\cos(	heta)$ is $[-1, 1]$. Therefore,$-1 \le \cos(bx + c) \le 1$. Multiplying by $a$ (assuming $a > 0$),we get $-a \le a\cos(bx + c) \le a$. Adding $d$ to all parts,we get $d - a \le a\cos(bx + c) + d \le d + a$. Thus,the range of $f(x)$ is $[d - a, d + a]$.

If $f(x) = a\cos(bx + c) + d$,then the range of $f(x)$ is

Similar Questions

The range of $f(x) = \sec \left( \frac{\pi}{4} \cos^2 x \right)$ for $-\infty < x < \infty$ is

The domain of the function $f(x) = \sqrt{2 - x} - \frac{1}{\sqrt{9 - x^2}}$ is:

If $A$ and $B$ are two sets,then $A \cap (A \cup B)'$ is equal to

If $A = \{a, b\}$,$B = \{2, 3, 5, 6, 7\}$ and $C = \{5, 6, 7, 8, 9\}$,find $A \times (B \cap C)$.

If $U = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$,$A = \{3, 5, 7, 11\}$ and $B = \{7, 8, 9, 10, 11\}$,compute $(A - B)'$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module