Limits, Continuity and Differentiability Questions

MathematicsLimits, Continuity and Differentiability118 Questions Found

MathematicsContinuity at a Point and in an IntervalJEE Main 2011Easy

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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

(A)

p = 5/2, q = 1/2

(B)

p = -3/2, q = 1/2

(C)

p = 1/2, q = 3/2

(D)

p = -1/2, q = 3/2

Answer:B

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2011Moderate

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$lim_{x \to 2} \frac{1 - c o s { 2 ( x - 2 )}}{x - 2}$

(A)

equals

2

(B)

equals

- 2

(C)

equals

1/ 2

(D)

does not exist

Answer:D

MathematicsContinuity at a Point and in an IntervalJEE Main 2012Moderate

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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.

Answer:A

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2013Easy

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$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

(A)

1/4

(B)

1/2

(C)

(D)

Answer:D

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2014Easy

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$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

(A)

- π

(B)

π

(C)

π /2

(D)

Answer:B

MathematicsRelationship Between Continuity and DifferentiabilityJEE Main 2015Moderate

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If the function $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is

(A)

10/3

(B)

(C)

(D)

16/5

Answer:C

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2015Easy

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$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

(A)

(B)

1/2

(C)

(D)

Answer:A

MathematicsEvaluation of Limits & L'Hopital's RuleJEE Main 2016Easy

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Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

(A)

1/2

(B)

1/4

(C)

(D)

Answer:A

...11

.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; } limx→2​x−21−cos{2(x−2)}​​

.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; } limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function g(x)={kx+1​,mx+2,​0≤x≤33<x≤5​ is differentiable, then the value of k+m is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p=limx→0+​(1+tan2x​)2x1​ then logp is equal to

.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; } limx→2​x−21−cos{2(x−2)}​​

.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; } limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function g(x)={kx+1​,mx+2,​0≤x≤33<x≤5​ is differentiable, then the value of k+m is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p=limx→0+​(1+tan2x​)2x1​ then logp is equal to

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

$lim_{x \to 2} \frac{1 - c o s { 2 ( x - 2 )}}{x - 2}$

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

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

If the function $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

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

$lim_{x \to 2} \frac{1 - c o s { 2 ( x - 2 )}}{x - 2}$

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

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

If the function $g (x) = {k x + 1, m x + 2, 0 \leq x \leq 3 3 < x \leq 5$ is differentiable, then the value of $k + m$ is

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to