Let a function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as: $f(x)= \begin{cases}\int_{0}^{x}(5-|t-3|) dt, & x>4 \\ x^{2}+bx, & x \leq 4\end{cases}$ where $b \in \mathbb{R}$. If $f$ is continuous at $x=4$, then which of the following statements is NOT true?

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

(A) $f$ is not differentiable at $x=4$

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

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

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Continuity and Differentiability

Let a function

f : R \to R

be defined as:

f (x) = {\int_{0}^{x} (5 - ∣ t - 3∣) d t, x^{2} + b x, x > 4 x \leq 4

where

b \in R

. If

f

is continuous at

x = 4

, then which of the following statements is NOT true?

Answer: (C)

For continuity at

x = 4

lim_{x \to 4^{-}} f (x) = lim_{x \to 4^{+}} f (x)

16 + 4 b = \int_{0}^{3} (5 - (3 - t)) d t + \int_{3}^{4} (5 - (t - 3)) d t = \int_{0}^{3} (2 + t) d t + \int_{3}^{4} (8 - t) d t

Evaluating:

16 + 4 b = 15

, so

b = - \frac{1}{4}

.

For

x \leq 4

f (x) = x^{2} - \frac{x}{4}

, so

f^{'} (x) = 2 x - \frac{1}{4}

.

This is negative for

x < \frac{1}{8}

, so

f

is NOT increasing in

(- \infty, \frac{1}{8})

.

LHD at

x = 4

f^{'} (4^{-}) = 8 - \frac{1}{4} = \frac{31}{4}

RHD at

x = 4

f^{'} (4^{+}) = 5 - ∣4 - 3∣ = 4

Since LHD ≠ RHD,

f

is not differentiable at

x = 4

(option A is true).

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

(option B is true).

Option C is NOT TRUE.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Single correct option