For the function f(x) = x / (1 +...

MathematicsDifferentiability of a FunctionJEE Advanced 1983Moderate

For the function $f (x) = {\frac{x}{1 + e ^{1/ x}}, 0, x \neq = 0 x = 0$ , the derivative from the right, $f^{'} (0^{+}) = .........$ , and the derivative from the left, $f^{'} (0^{-}) = .........$

Answer:0, 1

Visualized Solution (Hindi)

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

Function: $f (x) = {\frac{x}{1 + e ^{1/ x}}, 0, x \neq = 0 x = 0$
Goal: Find $f^{'} (0^{+})$ and $f^{'} (0^{-})$ .
Check if the function is differentiable at $x = 0$ .

Right-Hand Derivative Formula

Definition: $f^{'} (0^{+}) = lim_{h \to 0^{+}} \frac{f ( 0 + h ) - f ( 0 )}{h}$
This represents the slope of the tangent from the right side.

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f^{'} (0^{+})

Substitute $f (h) = \frac{h}{1 + e ^{1/ h}}$ and $f (0) = 0$ :
f'(0^+) = \lim_{h \to 0^+} \frac{\frac{h}{1 + e^{1/h}} - 0}{h}

Simplifying the RHD Expression

Cancel $h$ from the numerator and denominator:
f'(0^+) = \lim_{h \to 0^+} \frac{1}{1 + e^{1/h}}

Evaluating the RHD Limit

As $h \to 0^{+}$ , $\frac{1}{h} \to \infty$ .
Therefore, $e^{1/ h} \to \infty$ .
f'(0^+) = \frac{1}{1 + \infty} = 0

Left-Hand Derivative Formula

Definition: $f^{'} (0^{-}) = lim_{h \to 0^{+}} \frac{f ( 0 - h ) - f ( 0 )}{- h}$
This represents the slope of the tangent from the left side.

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

Substitute $f (- h) = \frac{- h}{1 + e ^{- 1/ h}}$ and $f (0) = 0$ :
f'(0^-) = \lim_{h \to 0^+} \frac{\frac{-h}{1 + e^{-1/h}} - 0}{-h}

Simplifying the LHD Expression

Cancel $- h$ from the numerator and denominator:
f'(0^-) = \lim_{h \to 0^+} \frac{1}{1 + e^{-1/h}}

Evaluating the LHD Limit

As $h \to 0^{+}$ , $- \frac{1}{h} \to - \infty$ .
Therefore, $e^{- 1/ h} \to 0$ .
f'(0^-) = \frac{1}{1 + 0} = 1

Final Conclusion

$f^{'} (0^{+}) = 0$
$f^{'} (0^{-}) = 1$
Since $f^{'} (0^{+}) \neq = f^{'} (0^{-})$ , the function is not differentiable at $x = 0$ .
Final Answer: $0, 1$

00:00 / 00:00

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Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

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Let

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f

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(D)

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Question 2:

If the function $e^{- x} f (x)$ assumes its minimum in the interval $[0, 1]$ at $x = 1/4$ , which of the following is true?

(A)

f^{'} (x) < f (x), 1/4 < x < 3/4

(B)

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(C)

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(D)

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Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

* Multiple Correct Options

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For the function $f (x) = {\frac{x}{1 + e ^{1/ x}}, 0, x \neq = 0 x = 0$ , the derivative from the right, $f^{'} (0^{+}) = .........$ , and the derivative from the left, $f^{'} (0^{-}) = .........$

Visualized Solution (Hindi)

Visualizing the Function $f (x)$

Right-Hand Derivative Formula

Substitution for $f^{'} (0^{+})$

Simplifying the RHD Expression

Evaluating the RHD Limit

Left-Hand Derivative Formula

Substitution for $f^{'} (0^{-})$

Simplifying the LHD Expression

Evaluating the LHD Limit

Final Conclusion

Conceptually Similar Problems

Let $f : R \to R$ be such that $f (1) = 3$ and $f^{'} (1) = 6$ . Then $lim_{x \to 0} (\frac{f ( 1 + x )}{f ( 1 )})^{1/ x}$ equals

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The function $f (x) = \frac{l n ( π + x )}{l n ( e + x )}$ is

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Comprehension Passage

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Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y), \forall x, y \in R$ . If $f (x)$ is differentiable at $x = 0$ , then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the function f(x)={1+e1/xx​,0,​x=0x=0​, the derivative from the right, f′(0+)=........., and the derivative from the left, f′(0−)=.........

Visualized Solution (Hindi)

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

Right-Hand Derivative Formula

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

Simplifying the RHD Expression

Evaluating the RHD Limit

Left-Hand Derivative Formula

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

Simplifying the LHD Expression

Evaluating the LHD Limit

Final Conclusion

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

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Visualizing the Function $f (x)$

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Substitution for $f^{'} (0^{-})$

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The function $f (x) = \frac{l n ( π + x )}{l n ( e + x )}$ is

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Let $f : (0, \infty) \to R$ be given by $f (x) = \int_{1/ x}^{x} e^{- (t + 1/ t)} \frac{d t}{t}$ . Then

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