Question text 2Marks Find the domain of [math: g(x)=ln⁡(x−1)6−x−x2]g(x)=\dfrac{\ln(x-1)}{\sqrt{6-x-x^2}}.Answer: [math: Dom(g)=(] ext{Dom}(g)=(Answer 1[input],Answer 2[input][math: )] Notes Report question issue Question 1 Notes

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Question text 2Marks Find the domain of [math: g(x)=ln⁡(x−1)6−x−x2]g(x)=\dfrac{\ln(x-1)}{\sqrt{6-x-x^2}}.Answer: [math: Dom(g)=(]	ext{Dom}(g)=(Answer 1[input],Answer 2[input][math: )] Notes Report question issue Question 1 Notes

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Question: Find the domain of g(x) = ln(x-1) / sqrt(6 - x - x^2). First, identify the constraints for the expression to be defined: - The natural logarithm ln(x-1) requires its argument to be strictly greater than 0, so x-1 > 0, which gives x > 1. - The square root sqrt(6 - x - x^2) requires the radicand to be strictly greater than 0 (since division by zero is undefined and we cannot take a square root of a negative number in the real numbers), so 6 - x - x^2 > 0. Solve the second inequality: 6 - x - x^2 > 0 is equivalent to -(x^2 + x - 6) > 0, or x^2 + x - 6 < 0. Factor the quadratic: x^2 + x - 6 = (x + 3)(x - 2). The roots are -3 and 2, and a positive-leading quadratic is negative between its roots. Therefore, -3 < x < 2. Combine the two domain restrictions by intersection: - From ln: x > 1 - From sqrt: -3 < x < 2 Intersection gives: 1 < x < 2. Thus the domain is (1, 2). Why the given interval (1, 3) is not correct: - The upper bound 3 would include values up to 3, but for x > 2 the inequality x^2 + x - 6 < 0 fails, meaning 6 - x - x^2 becomes non-positive and the square root would be undefined or zero inside the denominator (leading to division by zero at some point). In particular, at x = 2, the radicand is 6 - 2 - 4 = 0, which makes the denominator zero, not allowed. Therefore x cannot reach 2 or exceed it, so the interval cannot extend to 3. - Any x between 1 and 2 is allowed, but once x passes 2, the square root becomes imaginary or the denominator becomes undefined in the real-number context, violating the domain requirements. Summary: The domain is all real x such that 1 < x < 2, i.e., (1, 2). The proposed (1, 3) overstates the allowable x-values and includes x for which the expression is not defined.

Question text 2Marks Find the domain of [math: g(x)=ln⁡(x−1)6−x−x2]g(x)=\dfrac{\ln(x-1)}{\sqrt{6-x-x^2}}.Answer: [math: Dom(g)=(]\text{Dom}(g)=(Answer 1[input],Answer 2[input][math: )] Notes Report question issue Question 1 Notes

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