Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 3 of 20 · Click an option or "Show Solution" to reveal answer

Q21
Suppose f(x)f(x) is differentiable at x = 1 and lim⁡h→01hf(1+h)=5\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5, then f′(1)f'\left( 1 \right) equals
A 3
B 4
C 5
D 6
Correct Answer
Option C
Solution
f′(1)=lim⁡h→0f(1+h)−f(1)h;f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h};

As function is differentiable so it is continuous as it is given that

lim⁡h→0f(1+h)h=5\mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5

and hence

f(1)=0f(1)=0

Hence

f′(1)f'(1)
=lim⁡h→0f(1+h)h=5= \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5
Q22
If ff is a real valued differentiable function satisfying ∣f(x)−f(y)∣\left| {f\left( x \right) - f\left( y \right)} \right| ≤(x−y)2 \le {\left( {x - y} \right)^2}, x,yx, y ∈R \in R and f(0)f(0) = 0, then f(1)f(1) equals
A -1
B 0
C 2
D 1
Correct Answer
Option B
Solution
f′(x)=lim⁡h→0f(x+h)−f(x)hf'\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {x + h} \right) - f\left( x \right)} \over h}
         ∣f′(x)∣=lim⁡h→0∣f(x+h)−f(x)h∣≤lim⁡h→0∣(h)2h∣\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|
⇒∣f′(x)∣≤0⇒f′(x)=0\Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0
⇒f(x)=\Rightarrow f\left( x \right) =

constant As

f(0)=0⇒f(1)=0f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0
Q23
The set of points where f(x)=x1+∣x∣f\left( x \right) = {x \over {1 + \left| x \right|}} is differentiable is
A (−∞,0)∪(0,∞)\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)
B (−∞,1)∪(−1,∞)\left( { - \infty ,1} \right) \cup \left( { - 1,\infty } \right)
C (−∞,∞)\left( { - \infty ,\infty } \right)
D (0,∞)\left( {0,\infty } \right)
Correct Answer
Option C
Solution
f(x)={x1−x,x<0x1+x,x≥0f\left( x \right) = \left\{ \begin{array}{ll}{{x \over {1 - x}},} & {x < 0} \\ {{x \over {1 + x}},} & {x \ge 0} \end{array} \right.
⇒f′(x)={x(1−x)2,x<0x(1+x)2x≥0\Rightarrow f'\left( x \right) = \left\{ \begin{array}{ll}{{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \\ {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \end{array} \right.

∴\therefore

f′(x)f'\left( x \right)

exist at everywhere.

Q24
Let f:R→Rf:R \to R be a function defined by f(x)=min⁡{x+1,∣x∣+1}f(x) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}, then which of the following is true?
A f(x)f(x) is differentiale everywhere
B f(x)f(x) is not differentiable at x = 0
C f(x)>1f(x) > 1 for all x∈Rx \in R
D f(x)f(x) is not differentiable at x = 1
Correct Answer
Option A
Solution
f(x)=min⁡{x+1,∣x∣+1}f\left( x \right) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}
⇒f(x)=x+1 ∀ x∈R\Rightarrow f\left( x \right) = x + 1\,\forall \,x \in R

Hence,

f(x)f(x)

is differentiable everywhere for all

x∈R.x \in R.
Q25
The function f:R/{0}→Rf:R/\left\{ 0 \right\} \to R given by f(x)=1x−2e2x−1f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}} can be made continuous at xx = 0 by defining ff(0) as
A 0
B 1
C 2
D −1-1
Correct Answer
Option B
Solution

Given,

f(x)=1x−2e2x−1f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}
⇒f(0)=lim⁡x→01x−2e2x−1\Rightarrow f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {1 \over x} - {2 \over {{e^{2x}} - 1}}
=lim⁡x→0(e2x−1)−2xx(e2x−1)= \mathop {\lim }\limits_{x \to 0} {{\left( {{e^{2x}} - 1} \right) - 2x} \over {x\left( {{e^{2x}} - 1} \right)}}
[ \left[ \, \right.
00{0 \over 0}

form

 ]\left. \, \right]

∴\therefore using,

LL

'Hospital rule

f(0)=lim⁡x→04e2x2(xe2x2+e2x.1)+e2x.2f\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {2\left( {x{e^{2x}}2 + {e^{2x}}.1} \right) + {e^{2x}}.2}}
=lim⁡x→04e2x4xe2x+2e2x+2e2x  = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,
[  \left[ {\,\,} \right.
00{0 \over 0}

form

  ]\left. {\,\,} \right]
=lim⁡x→04e2x4(xe2x+e2x)=4.e04(0+e0)=1= \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4\left( {x{e^{2x}} + {e^{2x}}} \right)}} = {{4.{e^0}} \over {4\left( {0 + {e^0}} \right)}} = 1
Q26
lim⁡x→0 (1−cos⁡2x)22x tan⁡x −xtan⁡2x\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}} is :
A −- 2
B −- 12{1 \over 2}
C 12{1 \over 2}
D 2
Correct Answer
Option A
Solution
lim⁡x→0(1−cos⁡2x)22xtan⁡x−xtan⁡2x\mathop {\lim }\limits_{x \to 0} {{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\tan x - x\tan 2x}}
=lim⁡x→0(2sin⁡2x)22x(x+x33+2x515+....)−x(2x+23x33+2.25x515+...)= \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{x^5}} \over {15}}+...} \right)}}
=lim⁡x→04(x−x33!+x55!−....)4x4(23−83)+x6(415−6415)+....= \mathop {\lim }\limits_{x \to 0} {{4{{\left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}^4}} \over {{x^4}\left( {{2 \over 3} - {8 \over 3}} \right) + {x^6}\left( {{4 \over {15}} - {{64} \over {15}}} \right) + ....}}
=lim⁡x→04(1−x23!+345!−....)4−2+x2(−6015)+....= \mathop {\lim }\limits_{x \to 0} {{4{{\left( {1 - {{{x^2}} \over {3!}} + {{{3^4}} \over {5!}} - ....} \right)}^4}} \over { - 2 + {x^2}\left( { - {{60} \over {15}}} \right) + ....}}

(By dividing numerator and denominator by x4)

=4(1−0)4−2+0= {{4{{\left( {1 - 0} \right)}^4}} \over { - 2 + 0}}
=4−2= {4 \over { - 2}}

== −- 2

Q27
Let f(x)={(x−1)sin⁡1x−1if x≠10if x=1f\left( x \right) = \left\{ \begin{array}{ll}{\left( {x - 1} \right)\sin {1 \over {x - 1}}} & {if\,x \ne 1} \\ 0 & {if\,x = 1} \end{array} \right. Then which one of the following is true?
A ff is neither differentiable at x = 0 nor at x = 1
B ff is differentiable at x = 0 and at x = 1
C ff is differentiable at x = 0 but not at x = 1
D ff is differentiable at x = 1 but not at x = 0
Correct Answer
Option C
Solution

We have

f(x)={(x−1)sin⁡(1x−1),if  x≠10                                if  x=1f\left( x \right) = \left\{ \begin{array}{ll}{\left( {x - 1} \right)\sin \left( {{1 \over {x - 1}}} \right),} & {if\,\,x \ne 1} \\ {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {if\,\,x = 1} \end{array} \right.
Rf′(1)=lim⁡h→0f(1+h)−f(1)hRf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}
=lim⁡h→0h sin⁡1h−0h=lim⁡h→0  sin⁡1h=a= \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a

finite number Let this finite number be

ll
LL
f′(1)f'(1)
=lim⁡h→0f(1−h)−f(1)−h= \mathop {\lim }\limits_{h \to 0} {{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}
=lim⁡h→0−h sin⁡(1−h)−h=lim⁡h→0  sin⁡(1−h)= \mathop {\lim }\limits_{h \to 0} {{ - h\,\sin \left( {{1 \over { - h}}} \right)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {{1 \over { - h}}} \right)
=−lim⁡h→0sin⁡(1h)=−( = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,

a finite number

 )\,)
=−l=-l

Thus

Rf′(1)≠Lf′(1)Rf'\left( 1 \right) \ne Lf'\left( 1 \right)

∴\therefore

ff

is not differentiable at

x=1x=1

Also,

f′(0)=sin⁡1(x−1)−x−1(x−1)2cos⁡(1x−1)]x=0f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}
=−sin⁡1+cos⁡ 1= - \sin 1 + \cos \,1

∴\therefore

ff

is differentiable at

x=0x=0
Q28
Let f:R→Rf:R \to R be a positive increasing function with lim⁡x→∞f(3x)f(x)=1\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1. Then lim⁡x→∞f(2x)f(x)=\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} =
A 23{2 \over 3}
B 32{3 \over 2}
C 3
D 1
Correct Answer
Option D
Solution
f(x)f(x)

is a positive increasing function. ∴\therefore

0<f(x)<f(2x)<f(3x)0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)
⇒0<1<f(2x)f(x)<f(3x)f(x)\Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}
⇒lim⁡x→∞1≤lim⁡x→∞f(2x)f(x)≤lim⁡x→∞f(3x)f(x)\Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}

By Sandwich Theorem.

⇒lim⁡x→∞f(2x)f(x)=1\Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1
Q29
lim⁡x→2(1−cos⁡{2(x−2)}x−2)\mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)
A Equals 2\sqrt 2
B Equals −2-\sqrt 2
C Equals 12{1 \over {\sqrt 2 }}
D does not exist
Correct Answer
Option D
Solution
lim⁡x→21−cos⁡{2(x−2)}x−2\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}
=lim⁡x→22∣sin⁡(x−2)∣x−2= \mathop {\lim }\limits_{x \to 2} {{\sqrt 2 \left| {\sin \left( {x - 2} \right)} \right|} \over {x - 2}}
L.H.L.=lim⁡x→2−2sin⁡(x−2)(x−2)=−2L.H.L. = \mathop {\lim }\limits_{x \to {2^ - }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = - \sqrt 2
R.H.L.=lim⁡x→2+2sin⁡(x−2)(x−2)=2R.H.L. = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {x - 2} \right)} \over {\left( {x - 2} \right)}} = \sqrt 2

Thus

L.H.L.≠R.H.L.L.H.L. \ne R.H.L.

Hence,

lim⁡x→21−cos⁡{2(x−2)}x−2  \mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}\,\,

does not exist.

Q30
If f:R→Rf:R \to R is a function defined by f(x)=[x]cos⁡(2x−12)πf\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi , where [x] denotes the greatest integer function, then ff is
A continuous for every real xx
B discontinuous only at x=0x=0
C discontinuous only at non-zero integral values of xx
D continuous only at x=0x=0
Correct Answer
Option A
Solution

Let

f(x)=[x]cos⁡(2x−12)f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)

Doubtful points are

x=n,n∈Ix = n,n \in I
L.H.LL.H.L
=lim⁡x→n−[x]cos⁡(2x−12)π= \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi
=(n−1)cos⁡(2n−12)π=0= \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0

( As

[x]\left[ x \right]

is the greatest integer function )

R.H.L.=lim⁡x→n+ [x]cos⁡(2x−12)π=ncos⁡(2n−12)π=0R.H.L. = \mathop {\lim }\limits_{x \to {n^ + }} \,\left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi = n\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0

Now, value of the function at

x=nx = n

is

f(n)=0f(n)=0

Since,

L.H.L.L.H.L.
=R.H.L.=f(n)= R.H.L. = f\left( n \right)

∴\therefore

f(x)=[x]cos⁡(2x−12)f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)

is continuous for every real

x.x.
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