Considering only the principal values of the inverse trigonometric functions, the domain of the function f(x)=cos ⁻¹((x²-4 x+2)/(x²+3)) is :

Solution

Maths sir · 4 steps

Trap the fraction between -1 and 1; since x²+3 is always positive, clear it and solve two inequalities.

1. Condition for arccos. cos⁻¹ t is defined only for -1 ≤ t ≤ 1. The denominator x²+3 is always positive, so you can multiply through without flipping any sign.

   -(x²+3) ≤ x²-4x+2 ≤ x²+3

2. Upper bound. x²-4x+2 ≤ x²+3 gives -4x ≤ 1. Dividing by -4 flips the sign, so x ≥ -1/4.
3. Lower bound. x²-4x+2 ≥ -x²-3 gives 2x²-4x+5 ≥ 0. Its discriminant is 16-40=-24<0 and the leading coefficient is positive, so this holds for every real x.
4. So. Only the upper bound restricts x. At x=-1/4 the fraction equals exactly 1 and cos⁻¹ 1=0 is defined, so the endpoint is included. The domain is [-1/4, ∞), option (b).

The idea to keep. For cos⁻¹ or sin⁻¹ of a fraction, first check the sign of the denominator. If it is always positive, cross-multiply both bounds and use the discriminant to settle any quadratic.

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