CBSE 2023 · Region 1 · Set 1 · Q38 · 4 marks
The equation of the path traced by a roller-coaster is given by the polynomial $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{a}(\mathrm{x}+9)(\mathrm{x}+1)(\mathrm{x}-3)$. If the roller-coaster crosses y -axis at a point $\displaystyle (0,-1)$, answer the following:
(i)Find the value of 'a'.(ii)Find $\displaystyle \mathrm{f}^{\prime \prime}(\mathrm{x})$ at $\displaystyle \mathrm{x}=1$.
The equation of the path traced by a roller-coaster is given by the polynomial $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{a}(\mathrm{x}+9)(\mathrm{x}+1)(\mathrm{x}-3)$. If the roller-coaster crosses y -axis at a point $\displaystyle (0,-1)$, answer the following:
(i)
Find the value of 'a'.
(ii)
Find $\displaystyle \mathrm{f}^{\prime \prime}(\mathrm{x})$ at $\displaystyle \mathrm{x}=1$.
Marking-scheme solution
(i)
$\displaystyle -1 = a(-27) \implies a = \dfrac{1}{27}$
(ii)
$\displaystyle f(x) = \dfrac{1}{27}(x+9)(x+1)(x-3)$
$\displaystyle = \dfrac{1}{27}\left(x^{3} + 7x^{2} - 21x - 27\right)$
$\displaystyle f'(x) = \dfrac{1}{27}\left(3x^{2} + 14x - 21\right)$
$\displaystyle f''(x) = \dfrac{6x + 14}{27}$
$\displaystyle f''(1) = \dfrac{20}{27}$
Continuity and DifferentiabilitySecond Order DerivativeApplycase_studymedium
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.