Question

Let $f\left(x\right)$ satisfy the requirements of Lagrange's mean value theorem in $[0,1]$, $f\left(0\right)=0$ and $f^{\prime }\left(x\right)\le 1-x,\forall xϵ(0,1)$, then

• A. $f\left(x\right)\ge x$
• B. $|f\left(x\right)|\ge 1$
• C. $f\left(x\right)\le x(1-x)$
• D. $f\left(x\right)\le \frac{1}{4}$

Answer 1

The correct option is C $f\left(x\right)\le x(1-x)$

Let $x\in$ (0,1).

Since, Lagrange's theorem holds for $f\left(x\right)$ in $[0,1]$. So,Lagrange's theorem also holds for $f\left(x\right)$ in $[0,x]$.

So, there exists $c\in$ (0,x) such that

$f^{\prime }\left(c\right)=\frac{f\left(x\right)-f\left(0\right)}{x}$

$f^{\prime }\left(c\right)=\frac{f\left(x\right)}{x}$

Since, $f^{\prime }\left(c\right)\le (1-x)$

$\Rightarrow \frac{f\left(x\right)}{x}\le (1-x)$

$\Rightarrow f\left(x\right)\le x(1-x)$