Let the set of all values of p, for which f(x)=(p²-6 p+8)(sin ² 2 x-cos ² 2 x)+2(2-p) x+7 does not have any critical point, be the interval (a, b). Then 16 a b is equal to _________.

Solution

Maths sir · 5 steps

Turn the trig part into one cosine, then ask when f'(x)=0 can never be solved.

1. Simplify first. sin² 2x-cos² 2x=-cos 4x, and p²-6p+8=(p-2)(p-4). So f(x)=-(p-2)(p-4)cos 4x+2(2-p)x+7.
2. Differentiate. f is differentiable everywhere, so a critical point can only be a zero of f'. Since d/dx cos 4x=-4 sin 4x, we get f'(x)=4(p-2)(p-4)sin 4x-2(p-2).
3. Special case. At p=2, f'(x)=0 for every x, so every point is critical. For p≠ 2, divide by 2(p-2). Then f'(x)=0 becomes the equation below.

   sin 4x=1/2(p-4)

4. No solution. This has no solution when |1/2(p-4)|>1, that is |p-4|<1/2. At p=4 itself, f'(x)=-4, which is never zero, so p=4 is included. This gives p∈(7/2,9/2), and p=2 is already outside it.
5. So. a=7/2 and b=9/2, so 16ab=16×63/4= 252.

The common slip: 240. Taking the derivative of cos 4x as -2 sin 4x instead of -4 sin 4x gives |p-4|<1, the interval (3,5), and 240.

The idea to keep. No critical point means f'(x)=0 has no solution. Isolate sin or cos, then demand that the other side lies outside [-1,1]. Always check the parameter values where your division breaks down.

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