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If $6 sin^{-1}(x^2 -6x +12) = 2 \pi $, then the value of x, is |
1 2 3 does not exist |
does not exist |
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. |
