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JEE Mains · Maths · STD 11- 2. Relation and Function

ધારોકે \(f(x)=\frac{1}{7-\sin 5 x}\) એ \({R}\) પર વ્યાખ્યાયિત એક વિધેય છે. તો વિધેય \(f(x)\) નો વિસ્તાર ............. છે.

  1. A \(\left[\frac{1}{8}, \frac{1}{5}\right]\)
  2. B \(\left[\frac{1}{7}, \frac{1}{6}\right]\)
  3. C \(\left[\frac{1}{7}, \frac{1}{5}\right]\)
  4. D  \(\left[\frac{1}{8}, \frac{1}{6}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(D)  \(\left[\frac{1}{8}, \frac{1}{6}\right]\)

Step-by-step Solution

Detailed explanation

\( \sin 5 x \in[-1,1] \) \( -\sin 5 x \in[-1,1] \) \( 7-\sin 5 x \in[6,8] \) \( \frac{1}{7-\sin 5 x} \in\left[\frac{1}{8}, \frac{1}{6}\right]\)
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