If f : R → R and is defined by f(x) = \(\frac{1}{2-cos 3x}\) for ea

1.	If f : R → R and is defined by f(x) = \(\frac{1}{2-cos 3x}\) for each x ∈ R, then the range of f is(a) ( \(\frac{1}{3}\), 1) (b) [ \(\frac{1}{3}\), 1] (c) (1, 2) (d) [1, 2]
Answer» Answer : (b) [ \(\frac{1}{3}\),1] f(x) = \(\frac{1}{2-cos\,3x}\) – 1 ≤ cos 3x ≤ 1 ⇒ – 1 ≤ – cos 3x ≤ 1 (Multiplying all terms by – 1) ⇒ 2 – 1 ≤ 2 – cos 3x ≤ 3 (Adding 2 to each term) ⇒ 1 ≤ 2 – cos 3x ≤ 3 ⇒ \(1\geq \frac{1}{2-3\,cos\,3x}\geq \frac{1}{3}\) Range of f is [ \(\frac{1}{3}\),1]

If f : R → R and is defined by f(x) = \(\frac{1}{2-cos 3x}\) for each x ∈ R, then the range of f is(a) ( \(\frac{1}{3}\), 1) (b) [ \(\frac{1}{3}\), 1] (c) (1, 2) (d) [1, 2]

Answer»

Answer : (b) [ \(\frac{1}{3}\),1]

f(x) = \(\frac{1}{2-cos\,3x}\)

– 1 ≤ cos 3x ≤ 1

⇒ – 1 ≤ – cos 3x ≤ 1 (Multiplying all terms by – 1)

⇒ 2 – 1 ≤ 2 – cos 3x ≤ 3 (Adding 2 to each term)

⇒ 1 ≤ 2 – cos 3x ≤ 3

⇒ \(1\geq \frac{1}{2-3\,cos\,3x}\geq \frac{1}{3}\)

Range of f is [ \(\frac{1}{3}\),1]