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Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 2 of 20 · Click an option or "Show Solution" to reveal answer

Q11
lim⁡x→π2[1−tan⁡(x2)][1−sin⁡x][1+tan⁡(x2)][π−2x]3\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}} is
A ∞\infty
B 18{1 \over 8}
C 0
D 132{1 \over 32}
Correct Answer
Option D
Solution
lim⁡x→π2tan⁡(π4−x2).(1−sin⁡x)(π−2x)3\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {{\pi \over 4} - {x \over 2}} \right).\left( {1 - \sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}

Let

x=π2+x;  y→0x = {\pi \over 2} + x;\,\,y \to 0

==

lim⁡y→0tan⁡(−y2).(1−cos⁡ y)(−2y)3\mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( {1 - \cos \,y} \right)} \over {{{\left( { - 2y} \right)}^3}}}

==

lim⁡y→0−tan⁡y22sin⁡2y2(−8).y38.8\mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}
=lim⁡y→0132tan⁡y2(y2).[sin⁡ y/2y/2]2=132= \mathop {\lim }\limits_{y \to 0} {1 \over {32}}{{\tan {y \over 2}} \over {\left( {{y \over 2}} \right)}}.{\left[ {{{\sin \,y/2} \over {y/2}}} \right]^2} = {1 \over {32}}
Q12
lim⁡x→0sin⁡(πcos⁡2x)x2\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}} is equal to :
A −π - \pi
B π \pi
C π2{\pi \over 2}
D 1
Correct Answer
Option B
Solution

Consider

lim⁡x→0sin⁡(πcos⁡2x)x2\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}
=lim⁡x→0sin⁡[π(1−sin⁡2x)]x2= \mathop {\lim }\limits_{x \to 0} {{\sin \left[ {\pi \left( {1 - {{\sin }^2}x} \right)} \right]} \over {{x^2}}}
=lim⁡x→0sin⁡(π−πsin⁡2x)x2= \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi - \pi {{\sin }^2}x} \right)} \over {{x^2}}}
[ \left[ \, \right.

As

sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta
 ]\left. \, \right]
=lim⁡x→0sin⁡(πsin⁡2x)πsin⁡2x×πsin⁡2xx2= \mathop {\lim }\limits_{x \to 0} \sin {{\left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}} \times {{\pi {{\sin }^2}x} \over {{x^2}}}
=lim⁡x→01×π(sin⁡xx)2=π= \mathop {\lim }\limits_{x \to 0} 1 \times \pi {\left( {{{\sin x} \over x}} \right)^2} = \pi
Q13
The value of lim⁡x→0(x1−sin⁡x8−1+sin⁡x8)\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\sqrt[8]{1 - \sin x} - \sqrt[8]{1 + \sin x} }}} \right) is equal to :
A 0
B 4
C −-4
D −-1
Correct Answer
Option C
Solution
lim⁡x→0(x1−sin⁡x8−1+sin⁡x8)\mathop {\lim }\limits_{x \to 0} \left( {{x \over {\sqrt[8]{1 - \sin x} - \sqrt[8]{1 + \sin x} }}} \right)

=

lim⁡x→0x[(1−sin⁡x)18−(1+sin⁡x)18]×[(1−sin⁡x)18+(1+sin⁡x)18(1−sin⁡x)18+(1+sin⁡x)18]\mathop {\lim }\limits_{x \to 0} {x \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}}}} \right]

=

lim⁡x→0x[(1−sin⁡x)18+(1+sin⁡x)18][(1−sin⁡x)14−(1+sin⁡x)14]×[(1−sin⁡x)14+(1+sin⁡x)14(1−sin⁡x)14+(1+sin⁡x)14]\mathop {\lim }\limits_{x \to 0} {{x\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 8}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 8}}}} \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 4}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 4}}}}}} \right]

=

lim⁡x→0x[2][2][(1−sin⁡x)12−(1+sin⁡x)12]×[(1−sin⁡x)12+(1+sin⁡x)12(1−sin⁡x)12+(1+sin⁡x)12]\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]} \over {\left[ {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} - {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \right]}} \times \left[ {{{{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}} \over {{{\left( {1 - \sin x} \right)}^{{1 \over 2}}} + {{\left( {1 + \sin x} \right)}^{{1 \over 2}}}}}} \right]

=

lim⁡x→0x[2][2][2][(1−sin⁡x)−(1+sin⁡x)]\mathop {\lim }\limits_{x \to 0} {{x\left[ 2 \right]\left[ 2 \right]\left[ 2 \right]} \over {\left[ {\left( {1 - \sin x} \right) - \left( {1 + \sin x} \right)} \right]}}
=lim⁡x→0(−12)(2)(2)(2)= \mathop {\lim }\limits_{x \to 0} \left( { - {1 \over 2}} \right)(2)(2)(2)

= -4 ∵\because

{lim⁡x→0sin⁡xx=1}\left\{ {\mathop {\lim }\limits_{x \to 0} {{\sin x} \over x} = 1} \right\}
Q14
If f(1)=1,f′(1)=2,f\left( 1 \right) = 1,{f'}\left( 1 \right) = 2, then lim⁡x→1f(x)−1x−1\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}} is
A 22
B 44
C 11
D 12{1 \over 2}
Correct Answer
Option A
Solution
lim⁡x→1f(x)−1x−1  (00)\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1 } \over {\sqrt x - 1}}\,\,\left( {{0 \over 0}} \right)

form using

L′L'

Hospital's rule

=lim⁡x→112f(x)f′(x)1/2x= \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}
=f′(1)f(1)=21=2.= {{f'\left( 1 \right)} \over {\sqrt {f\left( 1 \right)} }} = {2 \over 1} = 2.
Q15
lim⁡x→0log⁡xn−[x][x]\mathop {\lim }\limits_{x \to 0} {{\log {x^n} - \left[ x \right]} \over {\left[ x \right]}}, n∈Nn \in N, ( [x] denotes the greatest integer less than or equal to x )
A has value −1 -1
B has value 00
C has value 11
D does not exist
Correct Answer
Option D
Solution

Since

lim⁡x→0[x]\mathop {\lim }\limits_{x \to 0} \left[ x \right]

does not exist, hence the required limit does not exist.

Q16
lim⁡x→01−cos⁡2x2x\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}} is
A 11
B −1-1
C zero
D does not exist
Correct Answer
Option D
Solution
lim⁡1−cos⁡ 2x2x⇒lim⁡1−(1−2 sin⁡2 x)2x\lim {{\sqrt {1 - \cos \,2x} } \over {\sqrt 2 x}} \Rightarrow \lim {{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\,x} \right)} } \over {\sqrt 2 x}}
lim⁡x→02 sin⁡2 x2x⇒lim⁡x→0∣sin⁡x∣x\mathop {\lim }\limits_{x \to 0} {{\sqrt {2\,{{\sin }^2}\,x} } \over {\sqrt {2x} }} \Rightarrow \mathop {\lim }\limits_{x \to 0} {{\left| {\sin x} \right|} \over x}

The limit of above does not exist as

LHS=−1≠RHL=1LHS = - 1 \ne RHL = 1
Q17
lim⁡x→∞(x2+5x+3x2+x+2)x\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}
A e4{e^4}
B e2{e^2}
C e3{e^3}
D 11
Correct Answer
Option A
Solution
lim⁡x→∞(x2+5x+3x2+x+2)x\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}
=lim⁡x→∞(1+4x+1x2+x+2)x= \mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}
=lim⁡x→∞[(1+4x+1x2+x+2)x2+x+24x+1](4x+1)xx2+x+2= \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{{{x^2} + x + 2} \over {4x + 1}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}

=

lim⁡x→∞[(1+4x+1x2+x+2)14x+1x2+x+2](4x+1)xx2+x+2\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{1 \over {{{4x + 1} \over {{x^2} + x + 2}}}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}
=elim⁡x→∞4x2+xx2+x+2= {e^{\mathop {\lim }\limits_{x \to \infty } {{4{x^2} + x} \over {{x^2} + x + 2}}}}

[ As

lim⁡x→∞(1+λx)1x=eλ\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }

]

=elim⁡x→∞4+1x1+1x+2x2=e4= {e^{\mathop {\lim }\limits_{x \to \infty } {{4 + {1 \over x}} \over {1 + {1 \over x} + {2 \over {{x^2}}}}}}} = {e^4}
Q18
f(x) and g(x) are two differentiable functions on [0, 2] such that f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9 then f(x) - g(x) at x = 32{3 \over 2} is
A 0
B 2
C 10
D -5
Correct Answer
Option D
Solution

To find the value of

f(x)−g(x)f(x) - g(x)

at

x=32x = \frac{3}{2}

, we need to use the given conditions and properties of differentiable functions. First, we are told that:

f′′(x)−g′′(x)=0f''(x) - g''(x) = 0

This implies that:

f′′(x)=g′′(x)f''(x) = g''(x)

Since the second derivatives of both functions are equal, their difference,

f′(x)−g′(x)f'(x) - g'(x)

, must be a linear function. Let’s denote it as:

f′(x)−g′(x)=kf'(x) - g'(x) = k

We'll find the constant

kk

using the initial conditions of the derivatives:

f′(1)=2f'(1) = 2
g′(1)=4g'(1) = 4

Thus,

f′(1)−g′(1)=2−4=−2f'(1) - g'(1) = 2 - 4 = -2

Therefore,

f′(x)−g′(x)=−2f'(x) - g'(x) = -2

Integrating the above result, we get:

f(x)−g(x)=−2x+Cf(x) - g(x) = -2x + C

To determine the constant

CC

, we use the values of the functions at

x=2x = 2

:

f(2)=3f(2) = 3
g(2)=9g(2) = 9

Thus,

f(2)−g(2)=3−9=−6f(2) - g(2) = 3 - 9 = -6

Therefore,

−2⋅2+C=−6-2 \cdot 2 + C = -6
−4+C=−6-4 + C = -6
C=−2C = -2

So the expression for

f(x)−g(x)f(x) - g(x)

is:

f(x)−g(x)=−2x−2f(x) - g(x) = -2x - 2

We need to find

f(32)−g(32)f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)

:

f(32)−g(32)=−2⋅32−2f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2
=−3−2= -3 - 2
=−5= -5

Thus, the value of

f(x)−g(x)f(x) - g(x)

at

x=32x = \frac{3}{2}

is -5.

Q19
If f(x)={xe−(1∣x∣+1x),x≠00,x=0f(x) = \left\{ \begin{array}{ll}{x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}} & {,x \ne 0} \\ 0 & {,x = 0} \end{array} \right. then f(x)f(x) is
A discontinuous everywhere
B continuous as well as differentiable for all x
C continuous for all x but not differentiable at x = 0
D neither differentiable nor continuous at x = 0
Correct Answer
Option C
Solution
f(0)=0;  f(x)=xe−(1∣x∣+1x)f\left( 0 \right) = 0;\,\,f\left( x \right) = x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}
R.H.L.  R.H.L.\,\,
   lim⁡h→0(0+h)e−2/h\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - 2/h}}
= lim⁡h→0he2/h=0= \,\mathop {\lim }\limits_{h \to 0} {h \over {{e^{2/h}}}} = 0
L.H.L.L.H.L.
   lim⁡h→0(0−h)e−(1h−1h)=0\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} = 0

therefore,

f(x)f(x)

is continuous,

R.H.D=R.H.D=
   lim⁡h→0(0+h)e−(1h+1h)−0h=0\,\,\,\mathop {\lim }\limits_{h \to 0} {{\left( {0 + h} \right){e^{ - \left( {{1 \over h} + {1 \over h}} \right)}} - 0} \over h} = 0
L.H.D.L.H.D.
   =lim⁡h→0(0−h)e−(1h−1h)−0−h=1\,\,\, = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} - 0} \over { - h}} = 1

therefore,

L.H.D.≠R.H.D.L.H.D. \ne R.H.D.
f(x)f(x)

is not differentiable at

x=0.x=0.
Q20
Let f(x)=1−tan⁡x4x−πf(x) = {{1 - \tan x} \over {4x - \pi }}, x≠π4x \ne {\pi \over 4}, x∈[0,π2]x \in \left[ {0,{\pi \over 2}} \right]. If f(x)f(x) is continuous in [0,π2]\left[ {0,{\pi \over 2}} \right], then f(π4)f\left( {{\pi \over 4}} \right) is
A −1-1
B 12{1 \over 2}
C −12-{1 \over 2}
D 11
Correct Answer
Option C
Solution
f(x)=1−tan⁡x4x−πf\left( x \right) = {{1 - \tan x} \over {4x - \pi }}

is continuous in

[0,π2]\left[ {0,{\pi \over 2}} \right]

∴\therefore

f(π4)=lim⁡x→π4f(x)=lim⁡x→π4 +f(x)=lim⁡h→0f(π4+h)f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi \over 4} + h} \right)
=lim⁡h→01−tan⁡(π4+h)4(π4+h)−π,h>0=lim⁡h→01−1+tan⁡ h1−tan⁡ h4h= \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi \over 4} + h} \right)} \over {4\left( {{\pi \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}
=lim⁡h→0 −21−tan⁡ h.tan⁡ h4h=−24=−12= \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} = - {1 \over 2}
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