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

MathematicsContinuity at a Point and in an IntervalJEE Advanced 1983Moderate

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

Visualized Solution (Hindi)

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

Given piecewise function: $f (x) = {1 + x, 3 - x, 0 \leq x \leq 2 2 < x \leq 3$
Interval 1: For $x \in [0, 2]$ , $f (x)$ increases from $1$ to $3$ .
Interval 2: For $x \in (2, 3]$ , $f (x)$ decreases from $1$ to $0$ .

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

To find $g (x) = f (f (x))$ , we substitute $f (x)$ into the definition of $f$ :
$g (x) = {1 + f (x), 3 - f (x), 0 \leq f (x) \leq 2 2 < f (x) \leq 3$
We must analyze the range of $f (x)$ for different domains of $x$ .

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0 \leq x \leq 2

For $x \in [0, 2]$ , $f (x) = 1 + x$ .
If $0 \leq 1 + x \leq 2 ⟹ - 1 \leq x \leq 1$ . Combined with $x \in [0, 2]$ , we get $x \in [0, 1]$ .
If $2 < 1 + x \leq 3 ⟹ 1 < x \leq 2$ .

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g (x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x \in [0, 1]

For $x \in [0, 1]$ , $f (x) \in [1, 2]$ .
$g (x) = 1 + f (x)$
$g (x) = 1 + (1 + x) = 2 + x$

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g (x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x \in (1, 2]

For $x \in (1, 2]$ , $f (x) \in (2, 3]$ .
$g (x) = 3 - f (x)$
$g (x) = 3 - (1 + x) = 2 - x$

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 < x \leq 3

For $x \in (2, 3]$ , $f (x) = 3 - x$ .
Since $0 \leq 3 - x < 1$ , which is $\leq 2$ :
$g (x) = 1 + f (x)$
$g (x) = 1 + (3 - x) = 4 - x$

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

At $x = 1$ :
LHL: $lim_{x \to 1^{-}} (2 + x) = 3$
RHL: $lim_{x \to 1^{+}} (2 - x) = 1$
Since $L H L \neq = R H L$ , $g (x)$ is discontinuous at $x = 1$ .

Checking Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x = 2

At $x = 2$ :
LHL: $lim_{x \to 2^{-}} (2 - x) = 0$
RHL: $lim_{x \to 2^{+}} (4 - x) = 2$
Since $L H L \neq = R H L$ , $g (x)$ is discontinuous at $x = 2$ .

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

Final form of $g (x)$ :
$g (x) = ⎩ ⎨ ⎧ 2 + x, 2 - x, 4 - x, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$
Points of discontinuity: $x = 1$ and $x = 2$ .

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 1997Easy

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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

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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

MathematicsClassification of FunctionsJEE Advanced 2001Easy

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Let $g (x) = 1 + x - [x]$ and $f (x) = ⎩ ⎨ ⎧ - 1, 0, 1, x < 0 x = 0 x > 0$ . Then for all $x$ , $f (g (x))$ is equal to

(A)

x

(B)

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(C)

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(D)

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Answer:B

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 2016Moderate

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Let $f : [- 1/2, 2] \to R$ and $g : [- 1/2, 2] \to R$ be functions defined by $f (x) = [x^{2} - 3]$ and $g (x) = ∣ x ∣ f (x) + ∣4 x - 7∣ f (x)$ , where $[y]$ denotes the greatest integer less than or equal to $y$ . Then

* Multiple Correct Options

(A)

f

is discontinuous exactly at three points in

[- 1/2, 2]

(B)

f

is discontinuous exactly at four points in

[- 1/2, 2]

(C)

g

is NOT differentiable exactly at four points in

(- 1/2, 2)

(D)

g

is NOT differentiable exactly at five points in

(- 1/2, 2)

Answer:B, C

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 1995Moderate

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The function $f (x) = max {(1 - x), (1 + x), 2}, x \in (- \infty, \infty)$ is

* Multiple Correct Options

(A)

continuous at all points

(B)

differentiable at all points

(C)

differentiable at all points except at

x = 1

and

x = - 1

(D)

continuous at all points except at

x = 1

and

x = - 1

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MathematicsComposite FunctionsJEE Advanced 2001Easy

View in:English Hindi

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

(A)

2

(B)

- 2

(C)

1

(D)

- 1

Answer:D

MathematicsDifferentiability of a FunctionJEE Advanced 1986Moderate

View in:English Hindi

Let $f (x)$ be defined in the interval $[- 2, 2]$ such that $f (x) = {- 1, x - 1, - 2 \leq x \leq 0 0 < x \leq 2$ and $g (x) = f (∣ x ∣) + ∣ f (x) ∣$ . Test the differentiability of $g (x)$ in $(- 2, 2)$ .

Answer:not differentiable at

x = 1

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Let $f (x) = {x + a ∣ x - 1∣ if x < 0 if x \geq 0$ and $g (x) = {x + 1 (x - 1)^{2} + b if x < 0 if x \geq 0$ where $a$ and $b$ are non-negative real numbers. Determine the composite function $g \circ f$ . If $(g \circ f) (x)$ is continuous for all real $x$ , determine the values of $a$ and $b$ . Further, for these values of $a$ and $b$ , is $g \circ f$ differentiable at $x = 0$ ? Justify your answer.

Answer:

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, differentiable at

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MathematicsClassification of FunctionsJEE Advanced 2014Moderate

View in:English Hindi

Let $f_{1} : R \to R, f_{2} : [0, \infty) \to R, f_{3} : R \to R$ and $f_{4} : R \to [0, \infty)$ be defined by $f_{1} (x) = {∣ x ∣ e^{x} if x < 0, if x \geq 0;$ $f_{2} (x) = x^{2}$ ; $f_{3} (x) = {sin x x if x < 0, if x \geq 0;$ and $f_{4} (x) = {f_{2} (f_{1} (x)) f_{2} (f_{1} (x)) - 1 if x < 0, if x \geq 0.$

List-I

(P)

f_{4}

(Q)

f_{3}

(R)

f_{2} \circ f_{1}

(S)

f_{2}

List-II

(1)

Onto but not one-one

(2)

Neither continuous nor one-one

(3)

Differentiable but not one-one

(4)

Continuous and one-one

Answer:P → 1, Q → 3, R → 2, S → 4

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.

Visualized Solution (Hindi)

Visualizing $f (x)$

Defining $g (x) = f (f (x))$

Case 1: $0 \leq x \leq 2$

Computing $g (x)$ for $x \in [0, 1]$

Computing $g (x)$ for $x \in (1, 2]$

Case 2: $2 < x \leq 3$

Checking Continuity at $x = 1$

Checking Continuity at $x = 2$

Final Form of $g (x)$

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 $(0, 2)$ .

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.

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The function $f (x) = max {(1 - x), (1 + x), 2}, x \in (- \infty, \infty)$ is

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List-I

List-II

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.

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 $(0, 2)$ .

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.

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 $g (x) = 1 + x - [x]$ and $f (x) = ⎩ ⎨ ⎧ - 1, 0, 1, x < 0 x = 0 x > 0$ . Then for all $x$ , $f (g (x))$ is equal to

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The function $f (x) = max {(1 - x), (1 + x), 2}, x \in (- \infty, \infty)$ is

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

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List-I

List-II

.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.

Visualized Solution (Hindi)

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

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

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤x≤2

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[0,1]

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(1,2]

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2<x≤3

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

Checking Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

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

Conceptually Similar Problems

.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)={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 g(x)=1+x−[x] and f(x)=⎩⎨⎧​−1,0,1,​x<0x=0x>0​. Then for all x, f(g(x)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=max{(1−x),(1+x),2},x∈(−∞,∞) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1αx​,x=−1. Then, for what value of α is f(f(x))=x?

List-I

List-II

.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; } 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)={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 g(x)=1+x−[x] and f(x)=⎩⎨⎧​−1,0,1,​x<0x=0x>0​. Then for all x, f(g(x)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=max{(1−x),(1+x),2},x∈(−∞,∞) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1αx​,x=−1. Then, for what value of α is f(f(x))=x?

List-I

List-II

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.

Visualizing $f (x)$

Defining $g (x) = f (f (x))$

Case 1: $0 \leq x \leq 2$

Computing $g (x)$ for $x \in [0, 1]$

Computing $g (x)$ for $x \in (1, 2]$

Case 2: $2 < x \leq 3$

Checking Continuity at $x = 1$

Checking Continuity at $x = 2$

Final Form of $g (x)$

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Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

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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 $[0, 2]$ .

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