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Question 11 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

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Question 11 (15 marks) Use a SEPARATE writing booklet. (a) Find the inverse of the function $y = x^3 - 2$. (b) Use the substitution $u = x - 4$ to find \( \int \sq... show full transcript

Worked Solution & Example Answer:Question 11 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2016 - Paper 1

Step 1

Find the inverse of the function $y = x^3 - 2$

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Answer

To find the inverse, we start with the equation:

y=x32y = x^3 - 2

We interchange xx and yy:

x=y32x = y^3 - 2

Next, we solve for yy:

y3=x+2y^3 = x + 2

Taking the cube root:

y=x+23y = \sqrt[3]{x + 2}

Thus, the inverse function is:

f1(x)=x+23f^{-1}(x) = \sqrt[3]{x + 2}

Step 2

Use the substitution $u = x - 4$ to find \( \int \sqrt{x-4} \, dx \)

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Answer

Using the substitution:

Let u=x4u = x - 4, then:

x=u+4x = u + 4

Also, we have:

dx=dudx = du

The integral becomes:

x4dx=udu\int \sqrt{x-4} \, dx = \int \sqrt{u} \, du

Evaluating this:

udu=23u3/2+C=23(x4)3/2+C\int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (x - 4)^{3/2} + C

Step 3

Differentiate $3 \tan^{-1}(2x)$

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Answer

To differentiate 3tan1(2x)3 \tan^{-1}(2x), we use the chain rule. The derivative of ( \tan^{-1}(u) ) is ( \frac{1}{1 + u^2} ) and we substitute u=2xu = 2x:

ddx[3tan1(2x)]=311+(2x)22=61+4x2\frac{d}{dx}[3 \tan^{-1}(2x)] = 3 \cdot \frac{1}{1 + (2x)^2} \cdot 2 = \frac{6}{1 + 4x^2}

Step 4

Evaluate \( \lim_{x \to 0} \frac{2 \sin x \cos x}{3x} \)

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Answer

We use the fact that ( \lim_{x \to 0} \frac{\sin x}{x} = 1 ). Rewrite the limit:

limx02sinxcosx3x=limx02sinx3xcosx\lim_{x \to 0} \frac{2 \sin x \cos x}{3x} = \lim_{x \to 0} \frac{2 \sin x}{3x} \cdot \cos x

Evaluating:

=2311=23= \frac{2}{3} \cdot 1 \cdot 1 = \frac{2}{3}

Step 5

Solve \( \frac{3}{2x + 5} - x > 0 \)

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Answer

First, isolate the fraction:

32x+5>x    3>x(2x+5)\frac{3}{2x + 5} > x \implies 3 > x(2x + 5)

This leads to:

3>2x2+5x3 > 2x^2 + 5x

Rearranging gives:

2x2+5x+3<02x^2 + 5x + 3 < 0

We can use the quadratic formula to find the roots:

x=b±b24ac2a=5±5242322=5±14x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{-5 \pm 1}{4}

Thus, the roots are x=1x = -1 and x=3/2x = -3/2. Testing intervals:

The solution is:

3/2<x<1-3/2 < x < -1

Step 6

Find the probability that she hits the bullseye with exactly one of her first three throws

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Answer

Let P(B)=25P(B) = \frac{2}{5} and P(N)=35P(N) = \frac{3}{5}. The probability of hitting the bullseye exactly once in three attempts (BNN, NBN, NNB) is:

P(1B)=(31)P(B)P(N)2=325(35)2=325925=54125P(1B) = \binom{3}{1} P(B) P(N)^2 = 3 \cdot \frac{2}{5} \cdot \left( \frac{3}{5} \right)^2 = 3 \cdot \frac{2}{5} \cdot \frac{9}{25} = \frac{54}{125}

Step 7

Find the probability that she hits the bullseye with at least two of her first six throws

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Answer

Calculating the complementary event, we find the probabilities for 0 and 1 bullseyes:

  • Probability of hitting 0:

P(0B)=(35)6=72915625P(0B) = \left( \frac{3}{5} \right)^6 = \frac{729}{15625}

  • Probability of hitting 1:

P(1B)=(61)P(B)P(N)5=625(35)5=6252433125=291615625P(1B) = \binom{6}{1} \cdot P(B) \cdot P(N)^5 = 6 \cdot \frac{2}{5} \cdot \left( \frac{3}{5} \right)^5 = 6 \cdot \frac{2}{5} \cdot \frac{243}{3125} = \frac{2916}{15625}

Adding these:

P(0B)+P(1B)=729+291615625=364515625P(0B) + P(1B) = \frac{729 + 2916}{15625} = \frac{3645}{15625}

Thus the probability of at least 2 bullseyes is:

1364515625=11980156251 - \frac{3645}{15625} = \frac{11980}{15625}

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