Question

Let $f:[1,3]\to R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime }\left(x\right)=|f\left(x\right){|}^2+4$ for all $x\in (1,3)$. Then,

• A. $f\left(3\right)-f\left(1\right)=5$ is true
• B. $f\left(3\right)-f\left(1\right)=5$ is false
• C. $f\left(3\right)-f\left(1\right)=7$ is false
• D. $f\left(3\right)-f\left(1\right)<0$ only at one point of $(1,3)$

Answer 1

Answer: (B), (C)

The correct options are
B $f\left(3\right)-f\left(1\right)=5$ is false
C $f\left(3\right)-f\left(1\right)=7$ is false

Using LMVT
$\frac{f\left(3\right)-f\left(1\right)}{3-1}=f^{{}^{\prime }}\left(c\right)$ for atleast one $c\in (1,3)$

Using $f^{\prime }\left(c\right)=|f\left(c\right){|}^2+4$
$\Rightarrow f\left(3\right)-f\left(1\right)=2\left(f\right(c){)}^2+8$ for atleast one $c\in (1,3)$
$\Rightarrow f\left(3\right)-f\left(1\right)\ge 8$