Let f(x)=1/(7-sin 5 x) be a function defined on R. Then the range of the function f(x) is equal to :

Solution

Maths sir · 4 steps

Pin down the denominator first: 7 - sin 5x stays between 6 and 8, then flip it.

1. Range of sine. For every real x, -1 ≤ sin 5x ≤ 1. The 5 inside only changes how fast sine repeats, not how high or low it goes.
2. Build the denominator. Multiply by -1, which reverses the inequalities, then add 7. This gives 6 ≤ 7 - sin 5x ≤ 8, so the denominator is always positive and never zero.

   6 ≤ 7-sin 5x ≤ 8 ⇒ 1/8 ≤ 1/(7-sin 5x) ≤ 1/6

3. Take reciprocals. All three quantities are positive, so taking reciprocals reverses the order. The smallest denominator 6 gives the largest value 1/6, and the largest denominator 8 gives the smallest value 1/8. Both ends are reached, at sin 5x = 1 and sin 5x = -1.
4. So. The range of f is [1/8, 1/6], which is option (d).

The idea to keep. For 1/(a - sin θ), first find the range of the denominator, then take reciprocals and flip the order. This works because the denominator never changes sign.

Answer from the NTA final answer key. Solution by PrepWiser.