Let f(x) = x²/2, 0 ≤ x < 1; 2x² - 3x +...

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1983Moderate

Let $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$ . Discuss the continuity of $f, f^{'}$ and $f^{''}$ on $[0, 2]$ .

Answer:

f, f^{'}

continuous;

f^{''}

discontinuous

Visualized Solution (Hindi)

Visualizing the Piecewise Function

Given function: $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$
The critical point for continuity is $x = 1$ .
We need to examine $f (x)$ , $f^{'} (x)$ , and $f^{''} (x)$ at this point.

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

Left Hand Limit (LHL): $lim_{x \to 1^{-}} f (x) = lim_{x \to 1^{-}} \frac{x ^{2}}{2} = \frac{1}{2}$
Right Hand Limit (RHL): $lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} (2 x^{2} - 3 x + \frac{3}{2}) = 2 (1)^{2} - 3 (1) + \frac{3}{2} = \frac{1}{2}$
Since $L H L = R H L = f (1)$ , $f (x)$ is continuous at $x = 1$ .

Finding the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f^{'} (x)

Differentiating $f (x)$ piecewise:
For $0 \leq x < 1$ : $f^{'} (x) = \frac{d}{d x} (\frac{x ^{2}}{2}) = x$
For $1 < x \leq 2$ : $f^{'} (x) = \frac{d}{d x} (2 x^{2} - 3 x + \frac{3}{2}) = 4 x - 3$

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

Left Hand Derivative (LHD): $f^{'} (1^{-}) = 1$
Right Hand Derivative (RHD): $f^{'} (1^{+}) = 4 (1) - 3 = 1$
Since $L H D = R H D$ , $f^{'} (x)$ is continuous at $x = 1$ .

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

Differentiating $f^{'} (x)$ piecewise:
For $0 \leq x < 1$ : $f^{''} (x) = \frac{d}{d x} (x) = 1$
For $1 < x \leq 2$ : $f^{''} (x) = \frac{d}{d x} (4 x - 3) = 4$

Discontinuity 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

Left Hand Limit of $f^{''} (x)$ : $lim_{x \to 1^{-}} f^{''} (x) = 1$
Right Hand Limit of $f^{''} (x)$ : $lim_{x \to 1^{+}} f^{''} (x) = 4$
Since $1 \neq = 4$ , $f^{''} (x)$ is discontinuous at $x = 1$ .

Final Conclusion

Summary of Results:
1. $f (x)$ is continuous on $[0, 2]$ .
2. $f^{'} (x)$ is continuous on $[0, 2]$ .
3. $f^{''} (x)$ is discontinuous at $x = 1$ .
Key Takeaway: Continuity of lower-order derivatives does not guarantee continuity of higher-order ones.

00:00 / 00:00

Conceptually Similar Problems

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1985Moderate

View in:English Hindi

Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Answer:cont. on

(0, 2)

, diff. on

(0, 2) - {1}

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1994Easy

View in:English Hindi

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

* Multiple Correct Options

(A)

f^{'}

is differentiable

(B)

f

is differentiable

(C)

f^{'}

is continuous

(D)

f

is continuous

Answer:B, C, D

MathematicsLinear Differential EquationsJEE Advanced 2016Easy

View in:English Hindi

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

(A)

lim_{x \to 0^{+}} f^{'} (\frac{1}{x}) = 1

(B)

lim_{x \to 0^{+}} x f (\frac{1}{x}) = 2

(C)

lim_{x \to 0^{+}} x^{2} f^{''} (x) = 0

(D)

|f(x)| \leq 2 \text{ for all } x \in (0, 2)

Answer:A

MathematicsMean Value TheoremsJEE Advanced 2005Moderate

View in:English Hindi

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

(A)

f^{''} (x) = 2

for

\forall x \in (1, 3)

(B)

f^{''} (x) = f^{'} (x) = 5

for some

x \in (2, 3)

(C)

f^{''} (x) = 3

for

\forall x \in (2, 3)

(D)

f^{''} (x) = 2

for some

x \in (1, 3)

Answer:D

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1997Easy

View in:English Hindi

Determine the values of $x$ for which the following function fails to be continuous or differentiable: $f (x) = ⎩ ⎨ ⎧ 1 - x, (1 - x) (2 - x), 3 - x, x < 1 1 \leq x \leq 2 x > 2$ Justify your answer.

Answer:discontinuous at

x = 2

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1988Moderate

View in:English Hindi

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

* Multiple Correct Options

(A)

continuous at

x = 1

(B)

differentiable at

x = 1

(C)

continuous at

x = 3

(D)

differentiable at

x = 3

Answer:A, B, C

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 2011Moderate

View in:English Hindi

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

* Multiple Correct Options

(A)

f (x)

is continuous at

x = - \frac{π}{2}

(B)

f (x)

is not differentiable at

x = 0

(C)

f (x)

is differentiable at

x = 1

(D)

f (x)

is differentiable at

x = - \frac{3}{2}

Answer:A, B, C, D

MathematicsContinuity at a Point and in an IntervalJEE Advanced 1983Moderate

View in:English Hindi

Let $f (x) = {1 + x, 0 \leq x \leq 2 3 - x, 2 < x \leq 3$ . Determine the form of $g (x) = f (f (x))$ and hence find the points of discontinuity of $g$ , if any.

Answer:

x = 1, 2

MathematicsMaxima and MinimaJEE Advanced 2012Moderate

View in:English Hindi

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

* Multiple Correct Options

(A)

f

has a local maximum at

x = 2

(B)

f

is decreasing on

(2, 3)

(C)

there exists some

c \in (0, \infty)

, such that

f^{''} (c) = 0

(D)

f

has a local minimum at

x = 3

Answer:A, B, C, D

MathematicsDifferentiability of a FunctionJEE Advanced 2012Moderate

View in:English Hindi

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

(A)

differentiable both at

x = 0

and at

x = 2

(B)

differentiable at

x = 0

but not differentiable at

x = 2

(C)

not differentiable at

x = 0

but differentiable at

x = 2

(D)

differentiable neither at

x = 0

nor at

x = 2

Answer:B

Let $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$ . Discuss the continuity of $f, f^{'}$ and $f^{''}$ on $[0, 2]$ .

Visualized Solution (Hindi)

Visualizing the Piecewise Function

Continuity of $f (x)$ at $x = 1$

Finding the First Derivative $f^{'} (x)$

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

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

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

Final Conclusion

Conceptually Similar Problems

Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

Determine the values of $x$ for which the following function fails to be continuous or differentiable: $f (x) = ⎩ ⎨ ⎧ 1 - x, (1 - x) (2 - x), 3 - x, x < 1 1 \leq x \leq 2 x > 2$ Justify your answer.

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

Let $f (x) = {1 + x, 0 \leq x \leq 2 3 - x, 2 < x \leq 3$ . Determine the form of $g (x) = f (f (x))$ and hence find the points of discontinuity of $g$ , if any.

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

Let $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$ . Discuss the continuity of $f, f^{'}$ and $f^{''}$ on $[0, 2]$ .

Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

Determine the values of $x$ for which the following function fails to be continuous or differentiable: $f (x) = ⎩ ⎨ ⎧ 1 - x, (1 - x) (2 - x), 3 - x, x < 1 1 \leq x \leq 2 x > 2$ Justify your answer.

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

Let $f (x) = {1 + x, 0 \leq x \leq 2 3 - x, 2 < x \leq 3$ . Determine the form of $g (x) = f (f (x))$ and hence find the points of discontinuity of $g$ , if any.

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={2x2​,2x2−3x+23​,​0≤x<11≤x≤2​. Discuss the continuity of f,f′ and f′′ on [0,2].

Visualized Solution (Hindi)

Visualizing the Piecewise Function

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

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

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

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

Discontinuity 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

Final Conclusion

Conceptually Similar Problems

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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)=⎩⎨⎧​−x−2π​,−cosx,x−1,lnx,​x≤−2π​−2π​<x≤00<x≤1x>1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={1+x,0≤x≤23−x,2<x≤3​. Determine the form of g(x)=f(f(x)) and hence find the points of discontinuity of g, if any.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x2​et​(t−2)(t−3)dt for all x∈(0,∞), then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={x2cosxπ​0​x=0x=0​,x∈R then f is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={2x2​,2x2−3x+23​,​0≤x<11≤x≤2​. Discuss the continuity of f,f′ and f′′ on [0,2].

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x3−x2+x+1 and g(x)={max{f(t);0≤t≤x},3−x,​0≤x≤11<x≤2​. Discuss the continuity and differentiability of the function g(x) in the interval (0,2).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={0,x2,​x<0x≥0​ then for all x

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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) is a twice differentiable function and given that f(1)=1;f(2)=4,f(3)=9, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)={∣x−3∣,4x2​−23x​+413​,​x≥1x<1​ is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={1+x,0≤x≤23−x,2<x≤3​. Determine the form of g(x)=f(f(x)) and hence find the points of discontinuity of g, if any.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫0x2​et​(t−2)(t−3)dt for all x∈(0,∞), then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={x2cosxπ​0​x=0x=0​,x∈R then f is

Let $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$ . Discuss the continuity of $f, f^{'}$ and $f^{''}$ on $[0, 2]$ .

Continuity of $f (x)$ at $x = 1$

Finding the First Derivative $f^{'} (x)$

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

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

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

Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

Let $f (x) = {1 + x, 0 \leq x \leq 2 3 - x, 2 < x \leq 3$ . Determine the form of $g (x) = f (f (x))$ and hence find the points of discontinuity of $g$ , if any.

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is

Let $f (x) = {\frac{x ^{2}}{2}, 2 x^{2} - 3 x + \frac{3}{2}, 0 \leq x < 1 1 \leq x \leq 2$ . Discuss the continuity of $f, f^{'}$ and $f^{''}$ on $[0, 2]$ .

Let $f (x) = x^{3} - x^{2} + x + 1$ and $g (x) = {max {f (t); 0 \leq t \leq x}, 3 - x, 0 \leq x \leq 1 1 < x \leq 2$ . Discuss the continuity and differentiability of the function $g (x)$ in the interval $(0, 2)$ .

Let $f (x) = {0, x^{2}, x < 0 x \geq 0$ then for all $x$

Let $f : (0, \infty) \to R$ be a differentiable function such that $f^{'} (x) = 2 - \frac{f ( x )}{x}$ for all $x \in (0, \infty)$ and $f (1) \neq = 1$ . Then

If $f (x)$ is a twice differentiable function and given that $f (1) = 1; f (2) = 4, f (3) = 9$ , then

The function $f (x) = {∣ x - 3∣, \frac{x ^{2}}{4} - \frac{3 x}{2} + \frac{13}{4}, x \geq 1 x < 1$ is

If $f (x) = ⎩ ⎨ ⎧ - x - \frac{π}{2}, - cos x, x - 1, ln x, x \leq - \frac{π}{2} - \frac{π}{2} < x \leq 0 0 < x \leq 1 x > 1$

Let $f (x) = {1 + x, 0 \leq x \leq 2 3 - x, 2 < x \leq 3$ . Determine the form of $g (x) = f (f (x))$ and hence find the points of discontinuity of $g$ , if any.

If $f (x) = \int_{0}^{x^{2}} e^{t} (t - 2) (t - 3) d t$ for all $x \in (0, \infty)$ , then

Let $f (x) = {x^{2} cos \frac{π}{x} 0 x \neq = 0 x = 0, x \in R$ then $f$ is