If $6 sin^{-1}(x^2 -6x +12) = 2 \pi $, t

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

If $6 sin^{-1}(x^2 -6x +12) = 2 \pi $, then the value of x, is

Options:

does not exist

Correct Answer:

does not exist

Explanation:

We hav,

$x^2 - 6x + 12 = (x -3)^2 + 3 ≥ 3 $ for all x

∴ $sin^{-1}(x^2 -6x + 12)$ does not exist.

Thus, there is no value of x satisfying the given equation.