CBSE Board Examination 2023 · Section E · 4 marks
Q38: Case study: a roller-coaster path f ( x ) = a ( x + 9 ) ( x + 1 ) ( x − 3 ) f(x) = a(x + 9)(x + 1)(x - 3) f ( x ) = a ( x + 9 ) ( x + 1 ) ( x − 3 )
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The question
The equation of the path traced by a roller-coaster is given by the polynomial f ( x ) = a ( x + 9 ) ( x + 1 ) ( x − 3 ) f(x) = a(x + 9)(x + 1)(x - 3) f ( x ) = a ( x + 9 ) ( x + 1 ) ( x − 3 ) ( 0 , − 1 ) (0, -1) ( 0 , − 1 )
(i) Find the value of ‘a’. (2 marks)
(ii) Find f ′ ′ ( x ) f''(x) f ′′ ( x ) x = 1 x = 1 x = 1 (2 marks)
Exam answer Show all explanations
(i) The point ( 0 , − 1 ) (0, -1) ( 0 , − 1 ) f ( 0 ) = − 1 f(0) = -1 f ( 0 ) = − 1
Why?
A point lies on a curve exactly when its coordinates satisfy the equation. The roller-coaster crosses the y y y ( 0 , − 1 ) (0, -1) ( 0 , − 1 ) x = 0 x = 0 x = 0 − 1 -1 − 1
− 1 = a ( 0 + 9 ) ( 0 + 1 ) ( 0 − 3 ) ⇒ − 1 = a ( − 27 ) \begin{aligned} -1 &= a(0 + 9)(0 + 1)(0 - 3) \\ \Rightarrow\ -1 &= a(-27) \end{aligned} − 1 ⇒ − 1 = a ( 0 + 9 ) ( 0 + 1 ) ( 0 − 3 ) = a ( − 27 ) Why?
Multiply the three brackets:
( 0 + 9 ) ( 0 + 1 ) ( 0 − 3 ) = 9 × 1 × ( − 3 ) = − 27 \begin{aligned} &(0 + 9)(0 + 1)(0 - 3) \\ &= 9 \times 1 \times (-3) = -27 \end{aligned} ( 0 + 9 ) ( 0 + 1 ) ( 0 − 3 ) = 9 × 1 × ( − 3 ) = − 27 Then divide both sides by − 27 -27 − 27
a = − 1 − 27 = 1 27 a = \frac{-1}{-27} = \frac{1}{27} a = − 27 − 1 = 27 1 ⇒ a = 1 27 \Rightarrow\ a = \frac{1}{27} ⇒ a = 27 1
(ii)
f ( x ) = 1 27 ( x + 9 ) ( x + 1 ) ( x − 3 ) = 1 27 ( x 3 + 7 x 2 − 21 x − 27 ) \begin{aligned} f(x) &= \frac{1}{27}(x + 9)(x + 1)(x - 3) \\ &= \frac{1}{27}\left(x^3 + 7x^2 - 21x - 27\right) \end{aligned} f ( x ) = 27 1 ( x + 9 ) ( x + 1 ) ( x − 3 ) = 27 1 ( x 3 + 7 x 2 − 21 x − 27 ) Why?
Expand before differentiating. First ( x + 9 ) ( x + 1 ) = x 2 + 10 x + 9 (x + 9)(x + 1) = x^2 + 10x + 9 ( x + 9 ) ( x + 1 ) = x 2 + 10 x + 9
( x + 9 ) ( x + 1 ) = x ⋅ x + x ⋅ 1 + 9 ⋅ x + 9 ⋅ 1 = x 2 + x + 9 x + 9 = x 2 + 10 x + 9 \begin{aligned} &(x + 9)(x + 1) \\ &= x \cdot x + x \cdot 1 + 9 \cdot x + 9 \cdot 1 \\ &= x^2 + x + 9x + 9 \\ &= x^2 + 10x + 9 \end{aligned} ( x + 9 ) ( x + 1 ) = x ⋅ x + x ⋅ 1 + 9 ⋅ x + 9 ⋅ 1 = x 2 + x + 9 x + 9 = x 2 + 10 x + 9 Then
( x 2 + 10 x + 9 ) ( x − 3 ) = x 3 − 3 x 2 + 10 x 2 − 30 x + 9 x − 27 \begin{aligned} &(x^2 + 10x + 9)(x - 3) \\ &= x^3 - 3x^2 + 10x^2 - 30x \\ &\quad + 9x - 27 \end{aligned} ( x 2 + 10 x + 9 ) ( x − 3 ) = x 3 − 3 x 2 + 10 x 2 − 30 x + 9 x − 27 Collect like terms: − 3 x 2 + 10 x 2 = 7 x 2 -3x^2 + 10x^2 = 7x^2 − 3 x 2 + 10 x 2 = 7 x 2 − 30 x + 9 x = − 21 x -30x + 9x = -21x − 30 x + 9 x = − 21 x
f ′ ( x ) = 1 27 ( 3 x 2 + 14 x − 21 ) \begin{aligned} f'(x) &= \frac{1}{27}\left(3x^2 + 14x - 21\right) \end{aligned} f ′ ( x ) = 27 1 ( 3 x 2 + 14 x − 21 ) Why?
Power rule: d d x ( x n ) = n x n − 1 \frac{d}{dx}\left(x^n\right) = nx^{n - 1} d x d ( x n ) = n x n − 1 x 3 x^3 x 3 3 x 2 3x^2 3 x 2 7 x 2 7x^2 7 x 2 14 x 14x 14 x − 21 x -21x − 21 x − 21 -21 − 21 − 27 -27 − 27
The constant factor 1 27 \frac{1}{27} 27 1
d d x ( x 3 ) = 3 x 2 d d x ( 7 x 2 ) = 7 ( 2 x ) = 14 x d d x ( − 21 x ) = − 21 d d x ( − 27 ) = 0 \begin{aligned} \frac{d}{dx}(x^3) &= 3x^2 \\ \frac{d}{dx}(7x^2) &= 7(2x) = 14x \\ \frac{d}{dx}(-21x) &= -21 \\ \frac{d}{dx}(-27) &= 0 \end{aligned} d x d ( x 3 ) d x d ( 7 x 2 ) d x d ( − 21 x ) d x d ( − 27 ) = 3 x 2 = 7 ( 2 x ) = 14 x = − 21 = 0 f ′ ′ ( x ) = 6 x + 14 27 \begin{aligned} f''(x) &= \frac{6x + 14}{27} \end{aligned} f ′′ ( x ) = 27 6 x + 14 Why?
Differentiate once more: 3 x 2 3x^2 3 x 2 6 x 6x 6 x 14 x 14x 14 x − 21 -21 − 21
d d x ( 3 x 2 ) = 3 ( 2 x ) = 6 x f ′ ′ ( x ) = 1 27 ( 6 x + 14 ) = 6 x + 14 27 \begin{aligned} \frac{d}{dx}(3x^2) &= 3(2x) = 6x \\[4pt] f''(x) &= \frac{1}{27}(6x + 14) \\ &= \frac{6x + 14}{27} \end{aligned} d x d ( 3 x 2 ) f ′′ ( x ) = 3 ( 2 x ) = 6 x = 27 1 ( 6 x + 14 ) = 27 6 x + 14 f ′ ′ ( 1 ) = 6 + 14 27 f''(1) = \frac{6 + 14}{27} f ′′ ( 1 ) = 27 6 + 14 Why?
Put x = 1 x = 1 x = 1 6 ( 1 ) + 14 = 20 6(1) + 14 = 20 6 ( 1 ) + 14 = 20
∴ f ′ ′ ( 1 ) = 20 27 \therefore\ f''(1) = \dfrac{20}{27} ∴ f ′′ ( 1 ) = 27 20
Where this comes from Chapter 5, Continuity and Differentiability : Second order derivative of a cubic.
Question type: (ii) Second derivative
How to spot it: click to see (ii) Type 9. Second derivative: differentiate twice, then prove
Spot it: d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y y ′ ′ y'' y ′′ d 2 y d x 2 + y = 0 \frac{d^2y}{dx^2} + y = 0 d x 2 d 2 y + y = 0
Here: f ′ ′ ( x ) f''(x) f ′′ ( x ) 1 27 ( x 3 + 7 x 2 − 21 x − 27 ) \frac{1}{27}(x^3 + 7x^2 - 21x - 27) 27 1 ( x 3 + 7 x 2 − 21 x − 27 ) 6 x + 14 27 \frac{6x + 14}{27} 27 6 x + 14 x = 1 x = 1 x = 1
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