Mathematics · 2020
JEE Main · 7 January 2020, Shift 2 · Q59
The value of c in the Lagrange's mean value theorem for the function f(x)=x^3-4 x^2+8 x+11, when x ∈[0,1] is:
The value of $\displaystyle c$ in the Lagrange's mean value theorem for the function $\displaystyle f(x)=x^3-4 x^2+8 x+11$, when $\displaystyle x \in[0,1]$ is :
Official answer
(2)
$\displaystyle \frac{4-\sqrt{7}}{3}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.