The range of the function,
f(x) = log _(√5)( 3 + cos ( 3π/4 + x ) + cos ( π/4 + x ) + cos ( π/4 - x ) - cos ( 3π/4 - x ) ) is :

Solution

Maths sir · 4 steps

Pair the cosines with matching angles, collapse them with sum-to-product, then push the range through the log.

1. Pair the cosines. Group cos(3π/4+x) - cos(3π/4-x), which gives -2 sin 3π/4 sin x = -√2 sin x. Then group cos(π/4+x) + cos(π/4-x), which gives 2 cos π/4 cos x = √2 cos x.

   √2(cos x - sin x) = 2 cos(x + π/4) ∈ [-2, 2]

2. Range of the argument. Add 3 to get the expression inside the log: 3 + 2 cos(x+π/4) takes every value in [1, 5]. It is always positive, so the log is defined for every x.
3. Apply the log. The base √5 is greater than 1, so the log is increasing. The ends map to log_(√5) 1 = 0 and log_(√5) 5 = 2, because (√5)² = 5.
4. So. The range of f is [0, 2], which is option (d).

The idea to keep. Find the range of the inside expression first, then pass both ends through the outer function. An increasing log keeps the order, so the endpoints map straight across.

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