JEE Advanced 2019 · Paper 1 · Mathematics Q10

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MathematicsMultiple correct+4 / -1Question 46 of 54

Let $f:\mathbb R\to\mathbb R$ be given by

$f(x) = \begin{cases}x^5 + 5x^4 + 10x^3 + 10x^2 + 3x + 1, & x < 0;\\ x^2 - x + 1, & 0 \le x < 1;\\ \frac23x^3 - 4x^2 + 7x - \frac83, & 1 \le x < 3;\\ (x-2)\log_e(x-2) - x + \frac{10}3, & x \ge 3.\end{cases}$

Then which of the following options is/are correct?

f

is increasing on

(-\infty, 0)

f'

has a local maximum at

x = 1

f

is onto

f'

is NOT differentiable at

x = 1

Answer and worked solution

Official answer: B, C, D

(A) For $x<0$ , $f=(x+1)^5-2x$ and $f'=5(x+1)^4-2$ . This is negative near $x=-1$ , so $f$ is not increasing there. False.

Continuity. The pieces join up: $f(0)=1$ , $f(1)=1$ and $f(3)=\frac13$ from both sides.

Near $x=1$ .

On $[0,1)$ : $f'=2x-1$ and $f''=2$ .
On $[1,3)$ : $f'=2x^2-8x+7$ and $f''=4x-8$ .

Both give $f'(1)=1$ , so $f'$ is continuous. But $f''$ is $+2$ on the left and $-4$ on the right.

(B) $f'$ rises up to $x=1$ and then falls, so it has a local maximum there. True.

(D) The one-sided derivatives of $f'$ at 1 differ, so $f'$ is not differentiable at $x=1$ . True.

(C) $f\to-\infty$ as $x\to-\infty$ and $f\to\infty$ as $x\to\infty$ , and $f$ is continuous, so it is onto. True.

(B), (C), (D)

Topics:Differentiability Monotonicity Types of functions

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JEE Advanced 2019 · Paper 1, Mathematics question 10