Problem for the year

Calculus Level 2

$\lambda=\left( \frac{f(5102)-f(2015)}{f'(c)} \right) \left( \frac{f^2(2015)+f^2(5102)+f(2015)f(5102)}{f^2(c)} \right)$

Let $f:[2015,5102] \rightarrow [0,\infty)$ be any continuous and differentiable function. Find the value of $\lambda$ , such that there exists some $c\in [2015,5102]$ which satisfies the equation above.

1629 6174 3087 9261

1 solution

Vishwak Srinivasan
Jul 2, 2015

This problem is an extrapolation of Lagrange's Mean Value Theorem.

By LMVT, for a differentiable function $g(x)$ for $x \in (a,b)$ , there exists a $c \in (a,b)$ such that

$f'(c) = \dfrac{f(b)-f(a)}{b-a}$

So here we will take $f^3(x) = g(x)$

$g'(c) = \dfrac{g(b) -g(a)}{b-a}$

$3f^2(c)f'(c) = \dfrac{f^3{b} - f^3{a}}{b-a} = \dfrac{(f(b)-f(a))(f^2(b) + f^2(a) + f(b)f(a))}{b-a}$

Here, in this question, $b = 5102 \text { & } a = 2015$

$\lambda = 3 \times (b-a) = \dfrac{(f(b)-f(a))(f^2(b) + f^2(a) + f(b)f(a))}{(f^2(c))(f'(c))}$ $\Rightarrow \lambda = 3 \times 2997 = \boxed{ 9261 }$

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