Continuity at a Point and in an Interval

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JEE Main 2012

LEVELJEE Main

If $f : R \to R$ is a function defined by $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , where $[x]$ denotes the greatest integer function, then $f$ is

(A)

continuous for every real x.

(B)

discontinuous only at x = 0

(C)

discontinuous only at non-zero integral values of x.

(D)

continuous only at x = 0.

JEE Main 2011

LEVELJEE Main

The values of $p$ and $q$ for which the function $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x}, q, \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0 x = 0 x > 0$ is continuous for all $x$ in $R$ , are

Let $f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]$ . If $f (x)$ is continuous in $[0, \frac{π}{2}]$ , then $f (\frac{π}{4})$ is

$f$ is defined in $[- 5, 5]$ as $f (x) = x$ if $x$ is rational $= - x$ if $x$ is irrational. Then

(A)

f(x) is continuous at every x, except x = 0

(B)

f(x) is discontinuous at every x, except x = 0

(C)

f(x) is continuous everywhere

(D)

f(x) is discontinuous everywhere

JEE Advanced 2014

LEVELJEE Main

For every pair of continuous functions $f, g : [0, 1] \to R$ such that $max {f (x) : x \in [0, 1]} = max {g (x) : x \in [0, 1]}$ , the correct statement(s) is (are):

* Multiple Correct Options

(A)

(f (c))^{2} + 3 f (c) = (g (c))^{2} + 3 g (c)

for some

c \in [0, 1]

(B)

(f (c))^{2} + f (c) = (g (c))^{2} + 3 g (c)

for some

c \in [0, 1]

(C)

(f (c))^{2} + 3 f (c) = (g (c))^{2} + g (c)

for some

c \in [0, 1]

(D)

(f (c))^{2} = (g (c))^{2}

for some

c \in [0, 1]

JEE Advanced 2012

LEVELJEE Main

For every integer $n$ , let $a_{n}$ and $b_{n}$ be real numbers. Let function $f : R \to R$ be given by $f (x) = {a_{n} + sin π x, b_{n} + cos π x, for x \in [2 n, 2 n + 1] for x \in (2 n - 1, 2 n)$ for all integers $n$ . If $f$ is continuous, then which of the following hold(s) for all $n$ ?

* Multiple Correct Options

(A)

a_{n - 1} - b_{n - 1} = 0

(B)

a_{n} - b_{n} = 1

(C)

a_{n} - b_{n + 1} = 1

(D)

a_{n - 1} - b_{n} = - 1

JEE Advanced 2000

LEVELJEE Main

Let $f : R \to R$ be any function. Define $g : R \to R$ by $g (x) = ∣ f (x) ∣$ for all $x$ . Then $g$ is

(A)

onto if

f

is onto

(B)

one-one if

f

is one-one

(C)

continuous if

f

is continuous

(D)

differentiable if

f

is differentiable.

JEE Advanced 1999

LEVELJEE Main

The function $f (x) = [x]^{2} - [x^{2}]$ (where $[y]$ is the greatest integer less than or equal to $y$ ), is discontinuous at

(A)

all integers

(B)

all integers except 0 and 1

(C)

all integers except 0

(D)

all integers except 1

JEE Advanced 1995

LEVELJEE Main

The function $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , $[.]$ denotes the greatest integer function, is discontinuous at

(A)

All

x

(B)

All integer points

(C)

x

(D)

x which is not an integer

JEE Advanced 1993

LEVELJEE Main

Let $[.]$ denote the greatest integer function and $f (x) = [tan^{2} x]$ , then:

(A)

lim_{x \to 0} f (x)

does not exist

(B)

f (x)

is continuous at

x = 0

(C)

f (x)

is not differentiable at

x = 0

(D)

f^{'} (0) = 1

JEE Advanced 1991

LEVELJEE Main

The following functions are continuous on $(0, π)$ .

* Multiple Correct Options

(A)

tan x

(B)

\int_{0}^{x} t sin \frac{1}{t} d t

(C)

{1, 2 sin \frac{2 x}{9}, 0 < x \leq \frac{3 π}{4} \frac{3 π}{4} < x < π

(D)

{x sin x, \frac{π}{2} sin (π + x), 0 < x \leq \frac{π}{2} \frac{π}{2} < x < π

JEE Advanced 1989

LEVELJEE Main

If $f (x) = \frac{1}{2} x - 1$ , then on the interval $[0, π]$

(A)

tan [f (x)]

and

1/ f (x)

are both continuous

(B)

tan [f (x)]

and

1/ f (x)

are both discontinuous

(C)

tan [f (x)]

and

f^{- 1} (x)

are both continuous

(D)

tan [f (x)]

is continuous but

1/ f (x)

is not

JEE Advanced 1983

LEVELJEE Main

The function $f (x) = \frac{l n ( 1 + a x ) - l n ( 1 - b x )}{x}$ is not defined at $x = 0$ . The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ , is

(A)

a - b

(B)

a + b

(C)

ln a - ln b

Let $f (x)$ be a continuous function defined for $1 \leq x \leq 3$ . If $f (x)$ takes rational values for all $x$ and $f (2) = 10$ , then $f (1.5) = .........$

JEE Advanced 1990

LEVELJEE Main

Let $f (x) = ⎩ ⎨ ⎧ \frac{1 - c o s 4 x}{x ^{2}}, a, \frac{x}{16 + x - 4}, x < 0 x = 0 x > 0$ . Determine the value of $a$ , if possible, so that the function is continuous at $x = 0$ .

JEE Advanced 1981

LEVELJEE Main

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

JEE Advanced 1996

LEVELJEE Main

Let $f (x) = [x] sin (\frac{π}{[ x + 1 ]})$ , where $[∙]$ denotes the greatest integer function. The domain of $f$ is... and the points of discontinuity of $f$ in the domain are.....

JEE Advanced 1994

LEVELJEE Main

Let $f (x) = ⎩ ⎨ ⎧ {1 + ∣ sin x ∣}^{a /∣ s i n x ∣}, b, e^{t a n 2 x / t a n 3 x}, - π /6 < x < 0 x = 0 0 < x < π /6$ . Determine $a$ and $b$ such that $f (x)$ is continuous at $x = 0$ .

JEE Advanced 1989

LEVELJEE Main

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .

JEE Advanced 1983

LEVELJEE Advanced

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.

JEE Advanced 1981

LEVELJEE Main

Let $f (x + y) = f (x) + f (y)$ for all $x$ and $y$ . If the function $f (x)$ is continuous at $x = 0$ , then show that $f (x)$ is continuous at all $x$ .

JEE Main 2018 (15 April Evening)

LEVELJEE Main

Let $f (x) = {(x - 1)^{\frac{1}{2 - x}}, k, x > 1, x \neq = 2 x = 2$ . The value of k for which f is continuous at x = 2 is :-

(A)

e^{- 1}

(B)

(C)

e^{- 2}

(D)

JEE Main 2018 (16 April Shift 1)

LEVELJEE Advanced

If the function f defined as $f (x) = \frac{1}{x} - \frac{k - 1}{e ^{2 x} - 1}, x \neq = 0$ , is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :

(A)

(\frac{1}{3}, 2)

(B)

(3, 2)

(C)

(2, 1)

(D)

(3, 1)

JEE Main 2019 (9 January)

LEVELJEE Main

Let $f : R \to R$ be a function defined as : $f (x) = ⎩ ⎨ ⎧ 5, a + b x, b + 5 x, 30, if x \leq 1 if 1 < x < 3 if 3 \leq x < 5 if x \geq 5$ . Then, f is :

(A)

continuous if a = 5 and b = 5

(B)

continuous if a = -5 and b = 10

(C)

continuous if a = 0 and b = 5

(D)

not continuous for any values of a and b

JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

Let $f : [- 1, 3] \to R$ be defined as $f (x) = [x] + [x], - 1 \leq x < 1; x + [x], 1 \leq x < 2; x + [x], 2 \leq x \leq 3$ . Then, f is discontinuous at:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:R→R is a function defined by f(x)=[x]cos(22x−1​)π, where [x] denotes the greatest integer function, then f is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The values of p and q for which the function f(x)=⎩⎨⎧​xsin(p+1)x+sinx​,q,x3/2x+x2​−x​​,​x<0x=0x>0​ is continuous for all x in R, are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=4x−π1−tanx​,x=4π​,x∈[0,2π​]. If f(x) is continuous in [0,2π​], then f(4π​) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f is defined in [−5,5] as f(x)=x if x is rational =−x if x is irrational. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For every pair of continuous functions f,g:[0,1]→R such that max{f(x):x∈[0,1]}=max{g(x):x∈[0,1]}, the correct statement(s) is (are):

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be any function. Define g:R→R by g(x)=∣f(x)∣ for all x. Then g is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=[x]2−[x2] (where [y] is the greatest integer less than or equal to y), is discontinuous at

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=[x]cos(22x−1​)π, [.] denotes the greatest integer function, is discontinuous at

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [.] denote the greatest integer function and f(x)=[tan2x], then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The following functions are continuous on (0,π).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=21​x−1, then on the interval [0,π]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=xln(1+ax)−ln(1−bx)​ is not defined at x=0. The value which should be assigned to f at x=0 so that it is continuous at x=0, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a continuous function defined for 1≤x≤3. If f(x) takes rational values for all x and f(2)=10, then f(1.5)=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=⎩⎨⎧​x21−cos4x​,a,16+x​​−4x​​,​x<0x=0x>0​. Determine the value of a, if possible, so that the function is continuous at x=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−2)2(x3+x2−16x+20)​k​if x=2if x=2​. If f(x) is continuous for all x, then k=.........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=[x]sin([x+1]π​), where [∙] denotes the greatest integer function. The domain of f is... and the points of discontinuity of f in the domain are.....

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=⎩⎨⎧​{1+∣sinx∣}a/∣sinx∣,b,etan2x/tan3x,​−π/6<x<0x=00<x<π/6​. Determine a and b such that f(x) is continuous at x=0.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the values of a and b so that the function f(x)=⎩⎨⎧​x+a2​sinx,2xcotx+b,acos2x−bsinx,​0≤x<π/4π/4≤x≤π/2π/2<x≤π​ is continuous for 0≤x≤π.

.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+y)=f(x)+f(y) for all x and y. If the function f(x) is continuous at x=0, then show that f(x) is continuous at all x.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={(x−1)2−x1​,k,​x>1,x=2x=2​. The value of k for which f is continuous at x = 2 is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f defined as f(x)=x1​−e2x−1k−1​,x=0, is continuous at x=0, then the ordered pair (k,f(0)) 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 defined as : f(x)=⎩⎨⎧​5,a+bx,b+5x,30,​if x≤1if 1<x<3if 3≤x<5if x≥5​. Then, f is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[−1,3]→R be defined as f(x)=[x]+[x],−1≤x<1;x+[x],1≤x<2;x+[x],2≤x≤3. Then, f is discontinuous at:

If $f : R \to R$ is a function defined by $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , where $[x]$ denotes the greatest integer function, then $f$ is

The values of $p$ and $q$ for which the function $f (x) = ⎩ ⎨ ⎧ \frac{s i n ( p + 1 ) x + s i n x}{x}, q, \frac{x + x ^{2} - x}{x ^{3/2}}, x < 0 x = 0 x > 0$ is continuous for all $x$ in $R$ , are

Let $f (x) = \frac{1 - t a n x}{4 x - π}, x \neq = \frac{π}{4}, x \in [0, \frac{π}{2}]$ . If $f (x)$ is continuous in $[0, \frac{π}{2}]$ , then $f (\frac{π}{4})$ is

$f$ is defined in $[- 5, 5]$ as $f (x) = x$ if $x$ is rational $= - x$ if $x$ is irrational. Then

For every pair of continuous functions $f, g : [0, 1] \to R$ such that $max {f (x) : x \in [0, 1]} = max {g (x) : x \in [0, 1]}$ , the correct statement(s) is (are):

Let $f : R \to R$ be any function. Define $g : R \to R$ by $g (x) = ∣ f (x) ∣$ for all $x$ . Then $g$ is

The function $f (x) = [x]^{2} - [x^{2}]$ (where $[y]$ is the greatest integer less than or equal to $y$ ), is discontinuous at

The function $f (x) = [x] cos (\frac{2 x - 1}{2}) π$ , $[.]$ denotes the greatest integer function, is discontinuous at

Let $[.]$ denote the greatest integer function and $f (x) = [tan^{2} x]$ , then:

The following functions are continuous on $(0, π)$ .

If $f (x) = \frac{1}{2} x - 1$ , then on the interval $[0, π]$

The function $f (x) = \frac{l n ( 1 + a x ) - l n ( 1 - b x )}{x}$ is not defined at $x = 0$ . The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ , is

Let $f (x)$ be a continuous function defined for $1 \leq x \leq 3$ . If $f (x)$ takes rational values for all $x$ and $f (2) = 10$ , then $f (1.5) = .........$

Let $f (x) = ⎩ ⎨ ⎧ \frac{1 - c o s 4 x}{x ^{2}}, a, \frac{x}{16 + x - 4}, x < 0 x = 0 x > 0$ . Determine the value of $a$ , if possible, so that the function is continuous at $x = 0$ .

Let $f (x) = {\frac{( x ^{3} + x ^{2} - 16 x + 20 )}{( x - 2 ) ^{2}} k if x \neq = 2 if x = 2$ . If $f (x)$ is continuous for all $x$ , then $k = .........$

Let $f (x) = [x] sin (\frac{π}{[ x + 1 ]})$ , where $[∙]$ denotes the greatest integer function. The domain of $f$ is... and the points of discontinuity of $f$ in the domain are.....

Let $f (x) = ⎩ ⎨ ⎧ {1 + ∣ sin x ∣}^{a /∣ s i n x ∣}, b, e^{t a n 2 x / t a n 3 x}, - π /6 < x < 0 x = 0 0 < x < π /6$ . Determine $a$ and $b$ such that $f (x)$ is continuous at $x = 0$ .

Find the values of $a$ and $b$ so that the function $f (x) = ⎩ ⎨ ⎧ x + a 2 sin x, 2 x cot x + b, a cos 2 x - b sin x, 0 \leq x < π /4 π /4 \leq x \leq π /2 π /2 < x \leq π$ is continuous for $0 \leq x \leq π$ .

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 $f (x + y) = f (x) + f (y)$ for all $x$ and $y$ . If the function $f (x)$ is continuous at $x = 0$ , then show that $f (x)$ is continuous at all $x$ .

Let $f (x) = {(x - 1)^{\frac{1}{2 - x}}, k, x > 1, x \neq = 2 x = 2$ . The value of k for which f is continuous at x = 2 is :-

If the function f defined as $f (x) = \frac{1}{x} - \frac{k - 1}{e ^{2 x} - 1}, x \neq = 0$ , is continuous at $x = 0$ , then the ordered pair $(k, f (0))$ is equal to :

Let $f : R \to R$ be a function defined as : $f (x) = ⎩ ⎨ ⎧ 5, a + b x, b + 5 x, 30, if x \leq 1 if 1 < x < 3 if 3 \leq x < 5 if x \geq 5$ . Then, f is :

Let $f : [- 1, 3] \to R$ be defined as $f (x) = [x] + [x], - 1 \leq x < 1; x + [x], 1 \leq x < 2; x + [x], 2 \leq x \leq 3$ . Then, f is discontinuous at: