(B) For a function $f(t)$ to satisfy Rolle's theorem on $[a, b]$,it must satisfy $f(a) = f(b)$.Given $f(t) = t^3 - 6t^2 + pt + q$ on $[1, 3]$,we have

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(B) For a function $f(t)$ to satisfy Rolle's theorem on $[a, b]$,it must satisfy $f(a) = f(b)$.Given $f(t) = t^3 - 6t^2 + pt + q$ on $[1, 3]$,we have $f(1) = f(3)$.$f(1) = 1^3 - 6(1)^2 + p(1) + q = 1 - 6 + p + q = p + q - 5$.$f(3) = 3^3 - 6(3)^2 + p(3) + q = 27 - 54 + 3p + q = 3p + q - 27$.Equating $f(1) = f(3)$:$p + q - 5 = 3p + q - 27$.$2p = 22 \Rightarrow p = 11$.Since $q$ cancels out from both sides,$q$ can be any real number $(q \in R)$.Thus,$p = 11$ and $q \in R$.

Class 12 Mathematics Continuity and Differentiation Rolle’s theorem, Lagrange's mean value theorem

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Find the value of $p$ and $q$ if the function $f(t) = t^3 - 6t^2 + pt + q$ defined on $[1, 3]$ satisfies Rolle's theorem for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}}$.

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A
$p \in R, q = 11$
B
$p = 11, q \in R$
C
$p \in R, q \in R$
D
$p = 11, q = 11$

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Rolle’s theorem, Lagrange's mean value theorem

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Find the value of $p$ and $q$ if the function $f(t) = t^3 - 6t^2 + pt + q$ defined on $[1, 3]$ satisfies Rolle's theorem for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}}$.

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