If the function f(x) = 2x^2 + 3x + 5 satisfie | vedclass

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(C) According to the Lagrange's Mean Value Theorem $(LMVT)$,there exists a point $c \in (1, a)$ such that $f'(c) = \frac{f(a) - f(1)}{a - 1}$.Given $c

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$5$

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(C) According to the Lagrange's Mean Value Theorem $(LMVT)$,there exists a point $c \in (1, a)$ such that $f'(c) = \frac{f(a) - f(1)}{a - 1}$.Given $c = 3$,we have $f'(3) = \frac{f(a) - f(1)}{a - 1}$.First,find the derivative $f'(x) = 4x + 3$. Thus,$f'(3) = 4(3) + 3 = 15$.Next,calculate $f(a) = 2a^2 + 3a + 5$ and $f(1) = 2(1)^2 + 3(1) + 5 = 10$.Substitute these into the equation: $15 = \frac{(2a^2 + 3a + 5) - 10}{a - 1}$.$15 = \frac{2a^2 + 3a - 5}{a - 1}$.Factor the numerator: $2a^2 + 3a - 5 = (2a + 5)(a - 1)$.Since $a 
eq 1$,we can simplify: $15 = 2a + 5$.$2a = 10$,which gives $a = 5$.

If the function $f(x) = 2x^2 + 3x + 5$ satisfies the Lagrange's Mean Value Theorem $(LMVT)$ at $x = 3$ on the closed interval $[1, a]$,then the value of $a$ is equal to:

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