JEE Advanced 2019

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f : R \to R$ be given by $f (x) = ⎩ ⎨ ⎧ x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1, x^{2} - x + 1, \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}, (x - 2) lo g_{e} (x - 2) - x + \frac{10}{3}, x < 0; 0 \leq x < 1; 1 \leq x < 3; x \geq 3$ . Then which of the following options is/are correct?

Q: Wait, why do we say $f$ is onto? I thought piecewise functions usually have gaps?

That is a fantastic intuition, but remember that 'piecewise' does not automatically mean 'discontinuous.' We must check the limits at the boundaries. As $x \to -\infty$, the term $x^5$ dominates, driving the function to $-\infty$. As $x \to \infty$, the term $(x-2)\log_e(x-2)$ dominates, driving the function to $\infty$. Since the function is continuous at every transition point, it must cover every value between $-\infty$ and $\infty$. Thus, it is onto.

Q: Why is $f'(x)$ not differentiable at $x=1$? I calculated the derivative and it seemed fine.

You are likely looking at $f'(x)$ being continuous, which is true! We found $f'(1^-) = 1$ and $f'(1^+) = 1$. However, differentiability of $f'(x)$ requires the *derivative of the derivative* (the second derivative) to be continuous. Since $f''(1^-) = 2$ and $f''(1^+) = -4$, the slope of the derivative function jumps abruptly. That jump is exactly why $f'(x)$ is not differentiable there.

Q: How can $f'(x)$ have a local maximum if it's not differentiable at that point?

Great question! A function does not need to be differentiable to have a local maximum. It only needs to be continuous. At $x=1$, the function $f'(x)$ is continuous (it reaches the value 1 from both sides). Because the slope of $f'(x)$ is positive on the left ($f''(x) > 0$) and negative on the right ($f''(x) < 0$), the graph of $f'(x)$ must peak at $x=1$. Think of it like a mountain peak—it's sharp, but it's still the highest point in that neighborhood.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the function as a journey through different terrains. The transition points at

x = 0

x = 1

, and

x = 3

are where the 'physics' of the function changes. Don't just look at the equations; look at the continuity. Does the function break, or does it just change its 'rule' of motion? The key to the 'onto' property lies in the behavior at the limits,

lim_{x \to - \infty} f (x)

and

lim_{x \to \infty} f (x)

Hint 2

Hint 2: The Arsenal. To analyze the derivative

f^{'} (x)

, you must treat it as a new function. Remember, for

f^{'} (x)

to have a local maximum, its derivative (which is

f^{''} (x)

) must change sign from positive to negative. You need to calculate the second derivative

f^{''} (x)

on both sides of the transition point

x = 1

Hint 3

Hint 3: The Execution. Start by calculating the left-hand derivative

f^{'} (1^{-})

using the piece

x^{2} - x + 1

and the right-hand derivative

f^{'} (1^{+})

using the piece

\frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3}

. Then, calculate

f^{''} (1^{-})

and

f^{''} (1^{+})

. If

f^{''} (1^{-}) > 0

and

f^{''} (1^{+}) < 0

, you have found your local maximum.

Visualized Solution

Analyzing the Piecewise Function

We are given a piecewise function $f : R \to R$ .
We need to check its range, monotonicity, and properties of its derivative $f^{'} (x)$ .
Let's break down the problem by analyzing each interval separately.

Derivative for $x < 0$

For $x < 0$ : $f (x) = x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1$
Differentiating with respect to $x$ :
$f^{'} (x) = 5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x + 3$
Notice the coefficients: $5 (x^{4} + 4 x^{3} + 6 x^{2} + 4 x) + 3$
Completing the binomial expansion: $f^{'} (x) = 5 (x + 1)^{4} - 2$

Checking Monotonicity

Option 3 claims $f (x)$ is increasing on $(- \infty, 0)$ .
For $f (x)$ to be increasing, we need $f^{'} (x) \geq 0$ for all $x < 0$ .
Let's test a value, say $x = - 1$ :
$f^{'} (- 1) = 5 (- 1 + 1)^{4} - 2 = - 2$
Since $f^{'} (- 1) < 0$ , $f (x)$ is not increasing on $(- \infty, 0)$ .

Checking Range of $f (x)$

Option 2 claims $f$ is onto (Range is $R$ ).
$f (x)$ is a continuous function across its domain.
As $x \to - \infty$ , $f (x) \to - \infty$ .
As $x \to \infty$ , $f (x) \to \infty$ .
Since it spans from $- \infty$ to $\infty$ continuously, Range = $R$ .

Derivatives for $x > 0$

For $0 \leq x < 1$ : $f (x) = x^{2} - x + 1 ⟹ f^{'} (x) = 2 x - 1$
For $1 \leq x < 3$ : $f (x) = \frac{2}{3} x^{3} - 4 x^{2} + 7 x - \frac{8}{3} ⟹ f^{'} (x) = 2 x^{2} - 8 x + 7$
For $x \geq 3$ : $f (x) = (x - 2) lo g_{e} (x - 2) - x + \frac{10}{3}$
$f^{'} (x) = 1 \cdot lo g_{e} (x - 2) + (x - 2) \frac{1}{x - 2} - 1 = lo g_{e} (x - 2)$

Continuity of $f^{'} (x)$ at $x = 1$

Let's evaluate $f^{'} (x)$ exactly at $x = 1$ .
Left-hand derivative: $f^{'} (1^{-}) = 2 (1) - 1 = 1$
Right-hand derivative: $f^{'} (1^{+}) = 2 (1)^{2} - 8 (1) + 7 = 1$
Since $f^{'} (1^{-}) = f^{'} (1^{+})$ , the derivative $f^{'} (x)$ is continuous at $x = 1$ .
Let's visualize the graph of $y = f^{'} (x)$ around $x = 1$ .

Second Derivative $f^{''} (x)$

To check differentiability of $f^{'} (x)$ and local extrema, we need $f^{''} (x)$ .
For $0 < x < 1$ : $f^{'} (x) = 2 x - 1 ⟹ f^{''} (x) = 2$
For $1 < x < 3$ : $f^{'} (x) = 2 x^{2} - 8 x + 7 ⟹ f^{''} (x) = 4 x - 8$

Differentiability of $f^{'} (x)$ at $x = 1$

Let's evaluate $f^{''} (x)$ at $x = 1$ .
Left-hand second derivative: $f^{''} (1^{-}) = 2$
Right-hand second derivative: $f^{''} (1^{+}) = 4 (1) - 8 = - 4$
$f^{''} (1^{-}) \neq = f^{''} (1^{+})$
Therefore, $f^{'} (x)$ is not differentiable at $x = 1$ . (Option 4 is correct).

Local Extrema of $f^{'} (x)$

We found $f^{''} (1^{-}) = 2 > 0$ and $f^{''} (1^{+}) = - 4 < 0$ .
This means the slope of $f^{'} (x)$ changes from positive to negative at $x = 1$ .
$f^{'} (x)$ is increasing before $x = 1$ and decreasing after $x = 1$ .
Hence, $f^{'} (x)$ has a local maximum at $x = 1$ . (Option 1 is correct).
Correct Options: 1, 2, and 4.

00:00 / 00:00

The Sigma Insight: Higher Order Derivatives

Solution Diagram

Analyzing the First Region

For the region

x < 0

, the function is defined as

f (x) = x^{5} + 5 x^{4} + 10 x^{3} + 10 x^{2} + 3 x + 1

. Recognizing the binomial coefficients from Pascal's Triangle, we can simplify the expression:

f (x) = (x + 1)^{5} - 2 x

Differentiating this expression with respect to

x

yields:

f^{'} (x) = 5 (x + 1)^{4} - 2

To test the claim that

f (x)

is increasing on

(- \infty, 0)

, we examine the sign of

f^{'} (x)

. At

x = - 1

, we find:

f^{'} (- 1) = 5 (0)^{4} - 2 = - 2

Since the derivative is negative at this point, the function is decreasing rather than increasing. Thus, the initial claim is false.

The Range Odyssey

A function is onto (surjective) if its range covers the entire set of real numbers. Given that

f (x)

is continuous at all transition points, we examine the end behavior of the function.

x \to - \infty

, the leading term

x^{5}

dominates, causing the function to approach

- \infty

. As

x \to \infty

, the term

(x - 2) ln (x - 2)

dominates, pushing the function toward

\infty

Because the function is continuous and spans from

- \infty

\infty

, by the Intermediate Value Theorem, it must take on every real value. Therefore, $f$ is onto.

The Mystery of the Sharp Corner

We now investigate the differentiability of

f^{'} (x)

at the junction

x = 1

. First, we verify the continuity of

f^{'} (x)

at this point.

For

0 \leq x < 1

, the derivative is

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

, which gives

f^{'} (1) = 1

. For

1 \leq x < 3

, the derivative is

f^{'} (x) = 2 x^{2} - 8 x + 7

, which yields:

f^{'} (1) = 2 (1)^{2} - 8 (1) + 7 = 1

Since the left-hand and right-hand limits of

f^{'} (x)

are equal,

f^{'} (x)

is continuous at

x = 1

. However, to check for differentiability, we must examine the second derivative

f^{''} (x)

For

x < 1

f^{''} (x) = 2

. For

x > 1

f^{''} (x) = 4 x - 8

. Evaluating these at

x = 1

f^{''} (1^{-}) = 2, f^{''} (1^{+}) = 4 (1) - 8 = - 4

Because $2 eq -4$, the derivative

f^{'} (x)

possesses a sharp corner at

x = 1

. Consequently, $f^{'} (x)$ is not differentiable at $x = 1$ .

The Peak of the Derivative

We have established that

f^{'} (x)

is continuous at

x = 1

. Furthermore, the slope of

f^{'} (x)

, represented by

f^{''} (x)

, is positive (

2

) just before

x = 1

and negative (

- 4

) just after

x = 1

In calculus, if a continuous function's slope changes from positive to negative, the function must attain a local maximum. The graph of

f^{'} (x)

climbs to a value of

1

x = 1

and immediately turns downward.

Therefore,

f^{'} (x)

indeed possesses a local maximum at $x = 1$ .

Common Traps & FAQs

Wait, why do we say

f

is onto? I thought piecewise functions usually have gaps?

That is a fantastic intuition, but remember that 'piecewise' does not automatically mean 'discontinuous.' We must check the limits at the boundaries. As

x \to - \infty

, the term

x^{5}

dominates, driving the function to

- \infty

. As

x \to \infty

, the term

(x - 2) lo g_{e} (x - 2)

dominates, driving the function to

\infty

. Since the function is continuous at every transition point, it must cover every value between

- \infty

and

\infty

. Thus, it is onto.

Why is

f^{'} (x)

not differentiable at

x = 1

? I calculated the derivative and it seemed fine.

You are likely looking at

f^{'} (x)

being continuous, which is true! We found

f^{'} (1^{-}) = 1

and

f^{'} (1^{+}) = 1

. However, differentiability of

f^{'} (x)

requires the *derivative of the derivative* (the second derivative) to be continuous. Since

f^{''} (1^{-}) = 2

and

f^{''} (1^{+}) = - 4

, the slope of the derivative function jumps abruptly. That jump is exactly why

f^{'} (x)

is not differentiable there.

How can

f^{'} (x)

have a local maximum if it's not differentiable at that point?

Great question! A function does not need to be differentiable to have a local maximum. It only needs to be continuous. At

x = 1

, the function

f^{'} (x)

is continuous (it reaches the value 1 from both sides). Because the slope of

f^{'} (x)

is positive on the left (

f^{''} (x) > 0

) and negative on the right (

f^{''} (x) < 0

), the graph of

f^{'} (x)

must peak at

x = 1

. Think of it like a mountain peak—it's sharp, but it's still the highest point in that neighborhood.

Select Answer:

Visualized Solution

Analyzing the Piecewise Function

Derivative for $x < 0$

Checking Monotonicity

Checking Range of $f (x)$

Derivatives for $x > 0$

Continuity of $f^{'} (x)$ at $x = 1$

Second Derivative $f^{''} (x)$

Differentiability of $f^{'} (x)$ at $x = 1$

Local Extrema of $f^{'} (x)$

The Sigma Insight: Higher Order Derivatives

Analyzing the First Region

The Range Odyssey

The Mystery of the Sharp Corner

The Peak of the Derivative

Similar Questions

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

Let $f (x)$ be a polynomial function such that $f (x) + f^{'} (x) + f^{''} (x) = x^{5} + 64$ . Then, the value of $lim_{x \to 1} \frac{f ( x )}{x - 1}$ is

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

Let $S$ be the set of all twice differentiable functions $f$ from $R$ to $R$ such that $\frac{d ^{2} f}{d x ^{2}} (x) > 0$ for all $x \in (- 1, 1)$ . For $f \in S$ , let $X_{f}$ be the number of points $x \in (- 1, 1)$ for which $f (x) = x$ . Then which of the following statements is(are) true?

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f^{'} (10)$ is equal to

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

Let $f : R \to R$ be a twice differentiable function such that $(sin x cos y) (f (2 x + 2 y) - f (2 x - 2 y)) = (cos x sin y) (f (2 x + 2 y) + f (2 x - 2 y))$ for all $x, y \in R$ . If $f^{'} (0) = 1/2$ , then the value of $24 f^{''} (5 π /3)$ is:

If $f (x) = {x^{3} sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ then

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then

Select Answer:

Visualized Solution

Analyzing the Piecewise Function

Derivative for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Checking Monotonicity

Checking Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Derivatives for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0

Continuity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Differentiability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x) at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Local Extrema of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

The Sigma Insight: Higher Order Derivatives

Analyzing the First Region

The Range Odyssey

The Mystery of the Sharp Corner

The Peak of the Derivative

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } There exist a function f(x), satisfying f(0)=1,f′(0)=−1,f(x)>0 for all x, and

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial function such that f(x)+f′(x)+f′′(x)=x5+64. Then, the value of limx→1​x−1f(x)​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3+x2f′(1)+2xf′′(2)+f′′′(3),x∈R. Then the value of f′(5) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=log(f(x)) where f(x) is twice differentiable positive function on (0,∞) such that f(x+1)=xf(x). Then, for N=1,2,3,……… g′′(N+21​)−g′′(21​)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R. Then f′(10) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a function such that f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3),x∈R. Then f(2) equal :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={x3sin(x1​),0,​x=0x=0​ then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=x3−x2f′(1)+xf′′(2)−f′′′(3),x∈R, then

Derivative for $x < 0$

Checking Range of $f (x)$

Derivatives for $x > 0$

Continuity of $f^{'} (x)$ at $x = 1$

Second Derivative $f^{''} (x)$

Differentiability of $f^{'} (x)$ at $x = 1$

Local Extrema of $f^{'} (x)$

There exist a function $f (x)$ , satisfying $f (0) = 1, f^{'} (0) = - 1, f (x) > 0$ for all $x$ , and

Let $f (x)$ be a polynomial function such that $f (x) + f^{'} (x) + f^{''} (x) = x^{5} + 64$ . Then, the value of $lim_{x \to 1} \frac{f ( x )}{x - 1}$ is

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3), x \in R$ . Then the value of $f^{'} (5)$ is:

Let $g (x) = lo g (f (x))$ where $f (x)$ is twice differentiable positive function on $(0, \infty)$ such that $f (x + 1) = x f (x)$ . Then, for $N = 1, 2, 3, \dots\dots\dots$ $g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =$

Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f^{'} (10)$ is equal to

Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equal :

If $f (x) = {x^{3} sin (\frac{1}{x}), 0, x \neq = 0 x = 0$ then

If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3), x \in R$ , then