JEE Main 2022 — Definite Integration Question with Solution

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

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

$$\Rightarrow 16+4b=\underset{0}{\overset{4}{\int }}(5-|t-3|)dt$$

$$=\underset{0}{\overset{3}{\int }}(2+t)dt+\underset{3}{\overset{4}{\int }}(8-t)dt$$

$$={2t+\frac{t^2}{2})}_0^3+{8t-\frac{t^2}{3}]}_3^4$$

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

$$16+4b=15$$

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

\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.

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

$$f^{\prime }(3)+f^{\prime }(5)=6-\frac{1}{4}=\frac{35}{4}$$

$$f^{\prime }(x)=0\Rightarrow x=\frac{1}{8}$$ have local minima

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