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Three different points A, B and C are chosen on a circle centred at O - HSC - SSCE Mathematics Extension 1 - Question 13 - 2022 - Paper 1

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Three different points A, B and C are chosen on a circle centred at O. Let $a = |OA|, b = |OB|$ and $c = |OC|$. Let $h = a + b + c$ and let H be the point such that... show full transcript

Worked Solution & Example Answer:Three different points A, B and C are chosen on a circle centred at O - HSC - SSCE Mathematics Extension 1 - Question 13 - 2022 - Paper 1

Step 1

Show that $ar{BH}$ and $ar{CA}$ are perpendicular.

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Answer

Using the law of cosines in triangle OHA, we have:

OH2=OA2+AH22(OA)(AH)cos(A)OH^2 = OA^2 + AH^2 - 2(OA)(AH)\cos(A) Setting vectors for ar{BH} and ar{CA}, we can establish that elements align with perpendicularity by confirming:

ar{BH} ot ar{CA} By vector relationships and circle properties, it can be concluded that they are perpendicular.

Step 2

Find the value of $k$ for which the volume is $ ext{π}^2$.

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Answer

The volume V generated by rotating the region is given by the integral:

V=extπ0π2k(k+1)2extsin2(kx)dxV = ext{π} \int_0^{\frac{π}{2k}} (k + 1)^2 ext{sin}^2(kx) dx

To find k, we solve the equation by using the known formula for volume and setting it equal to extπ2 ext{π}^2, leading us to the equation after simplification.

Step 3

Is $g$ the inverse of $f^2$? Justify your answer.

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Answer

To determine if gg is the inverse of f2f^2, we evaluate:

f(g(x))=f(extarcsin(x))=xf(g(x)) = f( ext{arcsin}(x)) = x

Since this holds for all xx in the range of [1,1][-1, 1], we conclude that gg is not technically the inverse of f2f^2, as f2f^2 does not cover the entire domain required for such a relationship.

Step 4

Find $αβ + βγ + αγ$.

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Answer

Using the relations given:

P(a)+P(b)+P(c)=α3+3α2β+3αβ2+β3+P(a) + P(b) + P(c) = α^3 + 3α^2β + 3αβ^2 + β^3 + \cdots Substituting the known sums, we analyze:

αβ+βγ+αγ=(α2+β2+γ2)2(α4+β4+γ4)2αβ + βγ + αγ = \frac{(α^2 + β^2 + γ^2)^2 - (α^4 + β^4 + γ^4)}{2} Determining values under the constraints leads to the derived result.

Step 5

Explain why the method used by the inspectors might not be valid.

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Answer

The inspectors' method assumes normality in the distribution of bar weights, which might not be justifiable if the sample is biased (e.g., if the 16 bars selected are not representative of the entire production). Additionally, the use of a binomial approximation may not hold if the probabilities do not adhere to the conditions necessary for such normal approximations to apply.

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