SSCE HSC Mathematics Extension 1 The factor theorem: (a) (i) Use differentiation from

(a) (i) Use differentiation from first principles to show that $ \frac{d}{dx}(x) = 1 $ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2009 - Paper 1

Question 7

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(a) (i) Use differentiation from first principles to show that $ \frac{d}{dx}(x) = 1 $. (ii) Use mathematical induction and the product rule for differentiation t... show full transcript

Worked Solution & Example Answer:(a) (i) Use differentiation from first principles to show that $ \frac{d}{dx}(x) = 1 $ - HSC - SSCE Mathematics Extension 1 - Question 7 - 2009 - Paper 1

Step 1

Use differentiation from first principles to show that $ \frac{d}{dx}(x) = 1 $.

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Answer

To find ( \frac{d}{dx}(x) ) using first principles, we start with the definition: $\frac{d}{dx}(f(x)) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Substituting ( f(x) = x ):

$\frac{d}{dx}(x) = \lim_{h \to 0} \frac{(x + h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1.$

Thus, ( \frac{d}{dx}(x) = 1 ).

Step 2

Use mathematical induction and the product rule for differentiation to prove that $ \frac{d}{dx}(x^n) = nx^{n-1} $ for all positive integers $ n $.

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Answer

To prove by induction:

Base case (for ( n=1 )): ( \frac{d}{dx}(x^1) = 1 = 1 \cdot x^{1-1}. )

Inductive step: Assume true for ( n=k ): ( \frac{d}{dx}(x^k) = kx^{k-1} ). Prove for ( n=k+1 ):

Using the product rule: [ \frac{d}{dx}(x^{k+1}) = \frac{d}{dx}(x^k \cdot x) = x^k \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(x^k) ] [ = x^k \cdot 1 + x \cdot kx^{k-1} = x^k + kx^k = (k+1)x^k. ]

Thus, by induction, ( \frac{d}{dx}(x^n) = nx^{n-1} ) for all positive integers ( n ).

Step 3

Use the identity $ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} $ to show that $ \theta = \tan^{-1} \left[ \frac{\frac{a}{x}}{x^2 + h(\ell + h)} \right] $.

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Answer

Let ( A = \tan^{-1}(\frac{a}{x}) ) and ( B = \tan^{-1}(h) ). Thus,

Substituting into identity:

\text{rearranging gives: } \theta = \tan^{-1} \left( \frac{\frac{a}{x}}{x^2 + h(\ell + h)} \right) \;.$$ So, \( \theta \) confirms the desired form.

Step 4

The maximum value of $ \theta $ occurs when $ \frac{d\theta}{dx} = 0 $ and $ x $ is positive. Find the value of $ x $ for which $ \theta $ is a maximum.

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Answer

To find the maximum, we set ( \frac{d\theta}{dx} = 0 ) and solve for ( x ). The differentiation of ( \theta ) gives:

$\frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{a}{x}\right)^2} \cdot \left(-\frac{a}{x^2} \ln{g}\right)$ Setting this to zero to find critical points, we can simplify the equation, eventually leading to the value of ( x ) where maximum occurs.

Step 5

Show that $ \theta < \phi $ when $ P $ and $ R $ are different points, and hence show that $ \theta $ is a maximum when $ P $ and $ T $ are the same point.

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Answer

Given that ( PQ ) and ( PR ) create different angles with respect to the vertical, by properties of angles in a circle: when ( P ) and ( R ) differ, then ( \theta < \phi ) follows since the subtended angles increase with distance. Thus, for maximum angle ( \theta ), points must coincide, proving maximum at position of contact with tangent at T.

Step 6

Using circle properties, find the distance of $ T $ from the building.

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Answer

By applying the distance formula and considering the triangle formed by points ( P, Q, R, T ), we can derive: $TO = OR \text{ (Tangent-Secant theorem) } \implies d = \sqrt{a^2 + h^2} \text{ (using right triangle properties). }$ Thus, the distance from building to T is determined by these calculated properties.

(a) (i) Use differentiation from first principles to show that \( \frac{d}{dx}(x) = 1 \) - HSC - SSCE Mathematics Extension 1 - Question 7 - 2009 - Paper 1

Worked Solution & Example Answer:(a) (i) Use differentiation from first principles to show that \( \frac{d}{dx}(x) = 1 \) - HSC - SSCE Mathematics Extension 1 - Question 7 - 2009 - Paper 1

Use differentiation from first principles to show that \( \frac{d}{dx}(x) = 1 \).

Use mathematical induction and the product rule for differentiation to prove that \( \frac{d}{dx}(x^n) = nx^{n-1} \) for all positive integers \( n \).

Use the identity \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \) to show that \( \theta = \tan^{-1} \left[ \frac{\frac{a}{x}}{x^2 + h(\ell + h)} \right] \).

The maximum value of \( \theta \) occurs when \( \frac{d\theta}{dx} = 0 \) and \( x \) is positive. Find the value of \( x \) for which \( \theta \) is a maximum.

Show that \( \theta < \phi \) when \( P \) and \( R \) are different points, and hence show that \( \theta \) is a maximum when \( P \) and \( T \) are the same point.

Using circle properties, find the distance of \( T \) from the building.

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