The range of a∈R for which the
function f(x) = (4a - 3)(x + logₑ 5) + 2(a - 7) cot( x/2 ) sin²( x/2 ), x ≠ 2nπ, n∈N has critical points, is :

Solution

Maths sir · 5 steps

Spot that cot x/2 sin² x/2 = 1/2 sin x, then demand |cos x| ≤ 1.

1. Simplify first. cot x/2 sin² x/2 = cos x/2 sin x/2 = 1/2 sin x. So f(x) = (4a-3)(x + ln 5) + (a-7)sin x.
2. Set the derivative. f'(x) = (4a-3) + (a-7)cos x. A critical point needs f'(x) = 0. For a = 7, f'(x) = 25, which is never zero, so take a ≠ 7.

   cos x = (4a-3)/(7-a), -1 ≤ (4a-3)/(7-a) ≤ 1

3. Case a below 7. Here 7 - a > 0, so multiply through by it without flipping signs. a - 7 ≤ 4a - 3 gives a ≥ -4/3, and 4a - 3 ≤ 7 - a gives a ≤ 2.
4. Case a above 7. The numerator is positive and the denominator is negative, so the ratio is negative. Requiring it to be at least -1 gives a ≤ -4/3, which contradicts a > 7. No values of a work here.
5. So. Critical points exist exactly when -4/3 ≤ a ≤ 2, that is, a ∈ [-4/3, 2], option (c).

The idea to keep. When f'(x) = 0 reduces to cos x or sin x equal to some expression in a, critical points exist only if that expression lies in [-1, 1]. Check the denominator's sign before you cross-multiply.

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