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 show that ( \frac{d}{dx}(x) = 1 ) using first principles, we start with the definition of the derivative:

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

Thus, we have shown that ( \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

We will use mathematical induction to prove this.

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} ). For n = k + 1:

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

Thus, by induction, ( \frac{d}{dx}(x^n) = nx^{n-1} ) holds 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}{h}}{x^2 + h(a + h)}\right]. $

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Given that the height of the billboard is ( h ) and the point ( P ) is positioned at a distance ( x ) from the building, we can express the angle ( \theta ) using the tangent of the angles.

Using the identity: $\tan \theta = \tan\left(\tan^{-1}\left(\frac{h}{x}\right) - \tan^{-1}\left(\frac{a}{h}\right)\right).$

Applying the identity: $\tan \theta = \frac{\frac{h}{x} - \frac{a}{h}}{1 + \frac{h}{x} \cdot \frac{a}{h}} = \frac{\frac{a}{h}}{x^2 + h(a + h)}.$

Thus, $\theta = \tan^{-1}\left[\frac{\frac{a}{h}}{x^2 + h(a + h)}\right].$

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 value of ( \theta ), we first differentiate: $\frac{d\theta}{dx} = \frac{d}{dx}\left\{\tan^{-1}\left[\frac{\frac{a}{h}}{x^2 + h(a + h)}\right]\right\} = 0.$

Using the chain rule and the quotient rule, we can solve for ( x ). Setting ( \frac{d\theta}{dx} = 0 ) gives us the critical points where we can evaluate whether ( \theta ) is maximized.

Step 5

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

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Answer

When ( P ) and ( T ) are different points, the angle subtended at point ( P ) will always be less than the angle at point ( T ) due to the geometrical properties of angles in a triangle. This implies that: $\Theta < \phi.$

When ( P ) coincides with ( T ), the angle ( \theta ) will reach its maximum value. This relationship can be established using the concept of angles subtended by chords in a circle.

Step 6

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

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Answer

Applying the properties of tangents and secants in circles, we know that: $TO^2 = OR \cdot OQ,$ where ( O ) is the center of the circle, and ( R, Q ) are points on the circle. By applying the appropriate identities, we can derive the distance of point ( T ) from the building to be ( d ), solving for ( d ) we will arrive at the required expression.

(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 \( \theta = \tan^{-1}\left[\frac{\frac{a}{h}}{x^2 + h(a + h)}\right]. \)

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 \( T \) 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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