JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 15)

Let a function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as :

$$f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases}$$

where $$\mathrm{b} \in \mathbb{R}$$. If $$f$$ is continuous at $$x=4$$, then which of the following statements is NOT true?

$$f$$ is not differentiable at $$x=4$$

$$f^{\prime}(3)+f^{\prime}(5)=\frac{35}{4}$$

$$f$$ is increasing in $$\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)$$

$$f$$ has a local minima at $$x=\frac{1}{8}$$

$$\because$$ f(x) is continuous at x = 4

$$ \Rightarrow f({4^ - }) = f({4^ + })$$

$$ \Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt} $$

$$ = \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} } $$

$$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$$

$$ = 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$$

$$16 + 4b = 15$$

$$ \Rightarrow b = {{ - 1} \over 4}$$

$$ \Rightarrow f(x) = \left\{ {\matrix{ {\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr {{x^2} - {x \over 4}} & {x \le 4} \cr } } \right.$$

$$ \Rightarrow f'(x) = \left\{ {\matrix{ {5 - |x - 3|} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$$

$$ \Rightarrow f'(x) = \left\{ {\matrix{ {8 - x} & {x > 4} \cr {2x - {1 \over 4}} & {x \le 4} \cr } } \right.$$

$$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$$

$$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$$

$$f'(x) = 0 \Rightarrow x = {1 \over 8}$$ have local minima

$$\therefore$$ (C) is only incorrect option.