Let f: R rightarrow R by given by f(x) = begin cases x 5 + 5x 4 + 10x 3 + 10x 2 + 3x + 1, & x < 0; \\ x 2 - x + 1, & 0 le x < 1; \\ 2/3x 3 - 4x 2 + 7x - 8/3, & 1 le x < 3; \\ (x-2)log e(x-2) - x + 10/3, & x ge 3. end cases Then which of the following options is/are correct?

f prime has a local maximum at x =1

f prime is NOT differentiable at x =1

Mathematics · Application of Derivatives

JEE Advanced 2019 — Paper 1 — Question 38

Let $f: R \rightarrow R$ by given by $f(x) = \begin{cases} x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1, & x < 0; \\ x^2 - x + 1, & 0 \le x < 1; \\ \frac{2}{3}x^3 - 4x^2 + 7x - \frac{8}{3}, & 1 \le x < 3; \\ (x-2)\log_e(x-2) - x + \frac{10}{3}, & x \ge 3. \end{cases}$

Then which of the following options is/are correct?

Option A:
$f^{\prime}$ has a local maximum at $\mathrm{x}=1$
Correct
Option B:
$\quad \mathrm{f}^{\prime}$ is NOT differentiable at $\mathrm{x}=1$
Correct
Option C:
f is onto
Correct
Option D:
$f$ is increasing on $(-\infty, 0)$

Answer: A, B, C

Step-by-step solution

x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1, & x < 0 \rightarrow (-\infty, 1) \\ x^2 - x + 1, & 0 \le x < 1 \rightarrow [\frac{3}{4}, 1) \\ \frac{2}{3}x^3 - 4x^2 + 7x - \frac{8}{3}, & 1 \le x < 3 \rightarrow [\frac{1}{3}, 1] \\ (x-2)\ln(x-2) - x + \frac{10}{3}, & x \ge 3 \rightarrow [\frac{1}{3}, \infty) \end{cases}$$

\text{Range will contain set } (-\infty, \infty)

$$f'(x) = \begin{cases} 5(x^4 + 4x^3 + 6x^2 + 4x + 1) - 2, & x < 0 \\ 2x - 1, & 0 \le x < 1 \\ 2x^2 - 8x + 7, & 1 \le x < 3 \\ \ln(x-2), & x \ge 3 \end{cases}$$ (A) $f'(1^-) > f'(1^+)$ & $f'(1^+) > f'(1^-)$ so $ f'(x)$ has local max. at $x = 1$ (B) $L.H.D. = 2 $ are $R.H.D. = -2, f $ is not differentiable at $x = 1$ (C) $f$ is containing $ (-\infty, \infty),$ so $f $ is onto. (D) $f'(x) = 5(x+1)^4 - 2$ is changing sign in $(-\infty, 0),$ so $f $ is not increasing

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives
Topic: Monotonicity