JEE Main 2022 — Mathematics Calculus

Given \displaystyle f(x)={\begin{matrix}{\int }_{0}^{x}(5-|t-3|)dt, & x>4 \\ {x}^{2}+bx, & x\leq 4\end{matrix}

Also given, $f(x)$f(x) is continuous at $x=4$x=4

So $\underset{x\rightarrow {4}^{-}}{\mathrm{lim}}f(x)=\underset{x\rightarrow {4}^{+}}{\mathrm{lim}}f(x)=f(4)$x→4−lim​f(x)=x→4+lim​f(x)=f(4)

So $\displaystyle 16+4b={\int }_{0}^{3}(2-t)dt+{\int }_{3}^{4}(8-t)dt$16+4b=∫03​(2−t)dt+∫34​(8−t)dt

$$\Rightarrow 16+4b={(2t-\frac{{t}^{2}}{2})}_{0}^{3}+{(8t-\frac{{t}^{2}}{2})}_{3}^{4}$$⇒16+4b=(2t−2t2​)03​+(8t−2t2​)34​

$$\Rightarrow 16+4b=15$$⇒16+4b=15

So $b=\frac{-1}{4}$b=4−1​

Now differentiating $f(x)$f(x) we get,

{f}^{'}(x)={\begin{matrix}(5-|x-3|), & x>4 \\ 2x+b, & x<4\end{matrix}

Now at $x=4$x=4

$$\mathrm{LHD}=2x+b=2\times 4-\frac{1}{4}=\frac{31}{4}$$LHD=2x+b=2×4−41​=431​

$$\mathrm{RHD}=5-|x-3|=4$$RHD=5−∣x−3∣=4

$$\mathrm{LHD}\neq \mathrm{RHD}$$LHD=RHD

So, option $(A)$(A) is true

And ${f}^{'}(3)+{f}^{'}(5)=\frac{23}{4}+3=\frac{35}{4}$f′(3)+f′(5)=423​+3=435​

So, option $(B)$(B) is true

Now $f(x)={x}^{2}-\frac{x}{4}$f(x)=x2−4x​ at $x\leq 4$x≤4

$${f}^{'}(x)=2x-\frac{1}{4}$$f′(x)=2x−41​

Thus the function is not increasing in the interval in $x\in (-\infty ,\frac{1}{8})$x∈(−∞,81​)

So, option $(C)$(C) is NOT TRUE.

And function $f(x)$f(x) is also have local minima at $x=\frac{1}{8}$x=81​ so option $(D)$(D) is also true.