Let $f : R \to R$ be defined as $f(x) = \begin{cases} x^5 \sin\left(\frac{1}{x}\right) + 5x^2, & x 0 \end{cases}$. The value of $\lambda$ for which $f''(0)$ exists, is

### Analyzing the Setup The function is defined piecewise as: $$f(x) = \begin{cases} x^5 \sin\left(\frac{1}{x}\right) + 5x^2, & x 0 \end{cases}$$ Our objective is to determine the value of $\lambda$ such that the second derivative $f''(0)$ exists. ### The Foundation of Continuity Before evaluating derivatives, we must confirm continuity at $x=0$. As $x \to 0^-$, the term $x^5 \sin\left(\frac{1}{x}\right)$ approaches $0$ because $x^5$ vanishes while $\sin\left(\frac{1}{x}\right)$ remains bounded. Similarly, for $x \to 0^+$, $x^5 \cos\left(\frac{1}{x}\right)$ approaches $0$. Since both one-sided limits equal $f(0) = 0$, the function is continuous at the origin. ### The First Derivative Existence We apply the first principle of derivatives: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$. For the left-hand limit ($h 0$): $$\lim_{h \to 0^+} \frac{h^5 \cos\left(\frac{1}{h}\right) + \lambda h^2}{h} = \lim_{h \to 0^+} \left(h^4 \cos\left(\frac{1}{h}\right) + \lambda h\right) = 0$$ Since both limits are $0$, we conclude that $f'(0) = 0$. ### The Anatomy of the Derivative To find $f''(0)$, we first determine the general expression for $f'(x)$ in the neighborhood of $0$. For $x 0$: $$f'(x) = 5x^4 \cos\left(\frac{1}{x}\right) + x^3 \sin\left(\frac{1}{x}\right) + 2\lambda x$$ ### The Second Derivative Limit We now evaluate $f''(0) = \lim_{h \to 0} \frac{f'(h) - f'(0)}{h}$. Given $f'(0) = 0$, we divide the expressions by $h$. For the left-hand limit ($h 0$): $$\lim_{h \to 0^+} \left(5h^3 \cos\left(\frac{1}{h}\right) + h^2 \sin\left(\frac{1}{h}\right) + 2\lambda\right) = 2\lambda$$ ### Final Calculation For the second derivative $f''(0)$ to exist, the left-hand limit must equal the right-hand limit. Equating the two results: $$10 = 2\lambda$$ Solving for $\lambda$, we find: **$\lambda = 5$**

JEE Main 2020 - 6 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ x^{5} sin (\frac{1}{x}) + 5 x^{2}, 0, x^{5} cos (\frac{1}{x}) + λ x^{2}, x < 0 x = 0 x > 0$ . The value of $λ$ for which $f^{''} (0)$ exists, is

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. To determine if

f^{''} (0)

exists, you cannot simply differentiate the piecewise expressions and plug in

x = 0

. You must use the formal limit definition of the second derivative at the point

x = 0

, which involves the first derivative

f^{'} (x)

x = 0

Hint 2

Hint 2: The Arsenal. Recall that

f^{''} (0) = lim_{h \to 0} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h}

. You must first calculate

f^{'} (0)

using the first principle of derivatives to ensure it exists, and then evaluate the limit of the difference quotient from both the left (

h \to 0^{-}

) and the right (

h \to 0^{+}

Hint 3

Hint 3: The Execution. Start by finding

f^{'} (x)

for

x < 0

and

x > 0

using the product and chain rules. Then, set up the limit

lim_{h \to 0^{-}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h} = lim_{h \to 0^{+}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h}

and solve for

λ

by equating the two results.

Visualized Solution

Understanding the Piecewise Function

Objective: Find $λ$ such that $f^{''} (0)$ exists.
The function is defined differently for $x < 0$ , $x = 0$ , and $x > 0$ .
We need to analyze the behavior of the function and its derivatives at the boundary $x = 0$ .

Checking Continuity at $x = 0$

For $f (x)$ to be differentiable, it must first be continuous at $x = 0$ .
$lim_{x \to 0^{-}} (x^{5} sin (\frac{1}{x}) + 5 x^{2}) = 0 \cdot (bounded) + 0 = 0$
$lim_{x \to 0^{+}} (x^{5} cos (\frac{1}{x}) + λ x^{2}) = 0 \cdot (bounded) + 0 = 0$
Since $L H L = R H L = f (0) = 0$ , $f (x)$ is continuous.

Existence of $f^{'} (0)$

$f^{'} (0^{-}) = lim_{h \to 0^{-}} \frac{h ^{5} s i n ( \frac{1}{h} ) + 5 h ^{2} - 0}{h} = lim_{h \to 0^{-}} (h^{4} sin (\frac{1}{h}) + 5 h) = 0$
$f^{'} (0^{+}) = lim_{h \to 0^{+}} \frac{h ^{5} c o s ( \frac{1}{h} ) + λ h ^{2} - 0}{h} = lim_{h \to 0^{+}} (h^{4} cos (\frac{1}{h}) + λh) = 0$
Therefore, $f^{'} (0)$ exists and equals $0$ .

Differentiating $f (x)$ for $x < 0$

For $x < 0$ , $f (x) = x^{5} sin (\frac{1}{x}) + 5 x^{2}$
Apply the Product Rule and Chain Rule.
$f^{'} (x) = 5 x^{4} sin (\frac{1}{x}) + x^{5} cos (\frac{1}{x}) \cdot (- \frac{1}{x ^{2}}) + 10 x$
Simplified: $f^{'} (x) = 5 x^{4} sin (\frac{1}{x}) - x^{3} cos (\frac{1}{x}) + 10 x$

Differentiating $f (x)$ for $x > 0$

For $x > 0$ , $f (x) = x^{5} cos (\frac{1}{x}) + λ x^{2}$
Apply the Product Rule and Chain Rule.
$f^{'} (x) = 5 x^{4} cos (\frac{1}{x}) + x^{5} (- sin (\frac{1}{x})) \cdot (- \frac{1}{x ^{2}}) + 2 λ x$
Simplified: $f^{'} (x) = 5 x^{4} cos (\frac{1}{x}) + x^{3} sin (\frac{1}{x}) + 2 λ x$

Calculating Left Hand Second Derivative $f^{''} (0^{-})$

$f^{''} (0^{-}) = lim_{h \to 0^{-}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h}$
Substitute $f^{'} (h)$ for $h < 0$ and $f^{'} (0) = 0$ .
$lim_{h \to 0^{-}} \frac{5 h ^{4} s i n ( \frac{1}{h} ) - h ^{3} c o s ( \frac{1}{h} ) + 10 h - 0}{h}$
Divide each term by $h$ : $lim_{h \to 0^{-}} (5 h^{3} sin (\frac{1}{h}) - h^{2} cos (\frac{1}{h}) + 10)$
As $h \to 0$ , the first two terms vanish. Result: $10$ .

Calculating Right Hand Second Derivative $f^{''} (0^{+})$

$f^{''} (0^{+}) = lim_{h \to 0^{+}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h}$
Substitute $f^{'} (h)$ for $h > 0$ and $f^{'} (0) = 0$ .
$lim_{h \to 0^{+}} \frac{5 h ^{4} c o s ( \frac{1}{h} ) + h ^{3} s i n ( \frac{1}{h} ) + 2 λh - 0}{h}$
Divide each term by $h$ : $lim_{h \to 0^{+}} (5 h^{3} cos (\frac{1}{h}) + h^{2} sin (\frac{1}{h}) + 2 λ)$
As $h \to 0$ , the first two terms vanish. Result: $2 λ$ .

Equating and Solving for $λ$

For $f^{''} (0)$ to exist, the left-hand and right-hand second derivatives must be equal.
Condition: $f^{''} (0^{-}) = f^{''} (0^{+})$
Equation: $10 = 2 λ$
Solving for $λ$ : $λ = \frac{10}{2} = 5$
Final Answer: $λ = 5$

00:00 / 00:00

The Sigma Insight: Differentiability of a Function

Solution Diagram

Analyzing the Setup

The function is defined piecewise as:

f (x) = ⎩ ⎨ ⎧ x^{5} sin (\frac{1}{x}) + 5 x^{2}, 0, x^{5} cos (\frac{1}{x}) + λ x^{2}, x < 0 x = 0 x > 0

Our objective is to determine the value of

λ

such that the second derivative

f^{''} (0)

exists.

The Foundation of Continuity

Before evaluating derivatives, we must confirm continuity at

x = 0

x \to 0^{-}

, the term

x^{5} sin (\frac{1}{x})

approaches

0

because

x^{5}

vanishes while

sin (\frac{1}{x})

remains bounded. Similarly, for

x \to 0^{+}

x^{5} cos (\frac{1}{x})

approaches

0

Since both one-sided limits equal

f (0) = 0

, the function is continuous at the origin.

The First Derivative Existence

We apply the first principle of derivatives:

f^{'} (0) = lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h}

For the left-hand limit (

h < 0

h \to 0^{-} lim \frac{h ^{5} sin ( \frac{1}{h} ) + 5 h ^{2}}{h} = h \to 0^{-} lim (h^{4} sin (\frac{1}{h}) + 5 h) = 0

For the right-hand limit (

h > 0

h \to 0^{+} lim \frac{h ^{5} cos ( \frac{1}{h} ) + λ h ^{2}}{h} = h \to 0^{+} lim (h^{4} cos (\frac{1}{h}) + λh) = 0

Since both limits are

0

, we conclude that

f^{'} (0) = 0

The Anatomy of the Derivative

To find

f^{''} (0)

, we first determine the general expression for

f^{'} (x)

in the neighborhood of

0

For

x < 0

f^{'} (x) = 5 x^{4} sin (\frac{1}{x}) - x^{3} cos (\frac{1}{x}) + 10 x

For

x > 0

f^{'} (x) = 5 x^{4} cos (\frac{1}{x}) + x^{3} sin (\frac{1}{x}) + 2 λ x

The Second Derivative Limit

We now evaluate

f^{''} (0) = lim_{h \to 0} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h}

. Given

f^{'} (0) = 0

, we divide the expressions by

h

For the left-hand limit (

h < 0

h \to 0^{-} lim (5 h^{3} sin (\frac{1}{h}) - h^{2} cos (\frac{1}{h}) + 10) = 10

For the right-hand limit (

h > 0

h \to 0^{+} lim (5 h^{3} cos (\frac{1}{h}) + h^{2} sin (\frac{1}{h}) + 2 λ) = 2 λ

Final Calculation

For the second derivative

f^{''} (0)

to exist, the left-hand limit must equal the right-hand limit.

Equating the two results:

10 = 2 λ

Solving for

λ

, we find:

$λ = 5$

Common Traps & FAQs

Wait, why can't I just differentiate the piecewise function directly and plug in

x = 0

That is a classic trap! Differentiating piecewise functions directly at the junction point often fails because the derivative might not be defined there. The function

f (x)

is only defined by its pieces, but the derivative

f^{'} (x)

is a limit. You must use the limit definition of the derivative at

x = 0

to account for the behavior of the function *as it approaches* the junction, not just the value at the junction.

What happens to the

sin (1/ x)

and

cos (1/ x)

terms when

x \to 0

Excellent question. These terms oscillate rapidly, but they are bounded between

- 1

and

1

. When you multiply them by

x^{n}

(where

n > 0

), the Squeeze Theorem tells us that the product approaches

0

x \to 0

. This is why those terms 'vanish' in our limit calculations, leaving only the polynomial terms behind.

Do I really need to check continuity and the first derivative before finding the second derivative?

Yes, absolutely. Differentiability requires continuity, and the existence of the second derivative requires the first derivative to exist and be continuous (or at least defined) at that point. If

f^{'} (0)

didn't exist,

f^{''} (0)

couldn't possibly exist. Always build your foundation step-by-step!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f : R \to R$ be defined as $f (x) = ⎩ ⎨ ⎧ x^{5} sin (\frac{1}{x}) + 5 x^{2}, 0, x^{5} cos (\frac{1}{x}) + λ x^{2}, x < 0 x = 0 x > 0$ . The value of $λ$ for which $f^{''} (0)$ exists, is

Enter Numerical Value:

Visualized Solution

Understanding the Piecewise Function

Checking Continuity at $x = 0$

Existence of $f^{'} (0)$

Differentiating $f (x)$ for $x < 0$

Differentiating $f (x)$ for $x > 0$

Calculating Left Hand Second Derivative $f^{''} (0^{-})$

Calculating Right Hand Second Derivative $f^{''} (0^{+})$

Equating and Solving for $λ$

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Foundation of Continuity

The First Derivative Existence

The Anatomy of the Derivative

The Second Derivative Limit

Final Calculation

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Enter Numerical Value:

Visualized Solution

Understanding the Piecewise Function

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

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

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

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

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

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

Equating and Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

The Sigma Insight: Differentiability of a Function

Analyzing the Setup

The Foundation of Continuity

The First Derivative Existence

The Anatomy of the Derivative

The Second Derivative Limit

Final Calculation

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