Q90 / 155 · WBJEE 2019 · Mathematics

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

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

Solution

Given that $f:[1,3] \rightarrow R$ be a continuous and differentiable in (1,3) and $f^{\prime}(x)=|f(x)|^{2}+4$ By applying $L M V T,$ there exist at least one point $c \in(1,3)$ such that $$ \begin{array}{c} \frac{f(3)-f(1)}{3-1}=f^{\prime}(c)\left[\because f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\right] \\ \Rightarrow \quad \frac{f(3)-f(1)}{2}=f^{\prime}(c) \\ \Rightarrow \quad \frac{f(3)-f(1)}{2}=\left[f|c|^{2}+4\right] \\ \Rightarrow f^{\prime}(c)=|f(c)|^{2}+4 \\ \Rightarrow \quad f(3)-f(1)=2 \cdot|f(c)|^{2}+8 \end{array} $$ $\Rightarrow \quad f(3)-f(1) \geq 8$

Hence, the correct answer is option: 'B'.

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