Let f:ℝ→ℝ be a function defined by f(x) = log _(√m){ √2 (sin x - cos x) + m - 2}, for some m, such that the range of f is [0, 2]. Then the value of m is _________

Solution

Maths sir · 4 steps

Squeeze the argument into one sine, then make the two ends of its range give 0 and 2.

1. One sine. sin x - cos x = √2 sin(x - π/4), so √2(sin x - cos x) = 2 sin(x - π/4), which runs over [-2, 2].
2. Range of argument. The argument g(x) = 2 sin(x - π/4) + m - 2 therefore takes every value in [m-4, m]. The domain is all of ℝ, so g must stay positive, which needs m - 4 > 0.
3. Top end. With m > 4 the base √m is greater than 1, so the log is increasing. The largest value is log_(√m) m = 2. This matches the top of [0, 2] for every valid m, so it does not fix m.

   log_(√m)(m-4) = 0 ⇒ m - 4 = 1

4. So. The bottom end decides it: m = 5. Then g runs over [1, 5] and f runs over [log_(√5) 1, log_(√5) 5] = [0, 2]. The answer is (c).

The idea to keep. For log of a trig expression, first find the exact range of the inside. Then match each end separately, and make sure the inside stays positive for every x.

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