Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 5 of 20 · Click an option or "Show Solution" to reveal answer

Q41
For each t ∈R \in R, let [t] be the greatest integer less than or equal to t. Then lim⁡x→0+x([1x]+[2x]+.....+[15x])\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)
A does not exist in R
B is equal to 0
C is equal to 15
D is equal to 120
Correct Answer
Option D
Solution

Given,

limx→0+  x ([1x]+[2x]+\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. +
 . . . . . +[15x])\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)

as we know that

1x=[1x]+{1x}{1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}
⇒    [1x]=1x−{1x}\Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}
=    lim⁡x→0+  x[1x−{1x}+22−{2x}+........15x−{15x}]= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]
=    lim⁡x→0+ [x.1x+x.2x+.....x.15x]= \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]
                − [x.{12}+..+x.{15x}]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]

We know

{1x}\left\{ {{1 \over x}} \right\}

is fractional part of

1x.{1 \over x}.

So, the range of

 {1x}\,\left\{ {{1 \over x}} \right\}

is

0≤{1x}<10 \le \left\{ {{1 \over x}} \right\} < 1

So,

limx→0+  x {1x}=0.\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.

(finite no)

=0=0

Similarly

lim⁡x→0+x.{2x}=0\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0
=    (1+2+...+15)−(0+0...)= \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)
=    15×162= \,\,\,\,{{15 \times 16} \over 2}
=    120= \,\,\,\,120
Q42
lim⁡x→0sin⁡2x2−1+cos⁡x\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }} equals:
A 2 \sqrt 2
B 222 \sqrt 2
C 4
D 424 \sqrt 2
Correct Answer
Option D
Solution
lim⁡x→0sin⁡2x2−1+cos⁡x\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}

=

lim⁡x→0(sin⁡2x2−1+cos⁡x)(2+1+cos⁡x2+1+cos⁡x)\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right)\left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)

=

lim⁡x→0(sin⁡2x1−cos⁡x)(2+1+cos⁡x)\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {1 - \cos x}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)

=

lim⁡x→0(sin⁡2x2sin⁡2(x2))(2+1+cos⁡x)\mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}} \right)\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)

=

lim⁡x→012(sin⁡xx)2x2(x2sin⁡x2)21x24(2+1+cos⁡x)\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)

=

12×4×22{1 \over 2} \times 4 \times 2\sqrt 2

=

424\sqrt 2
Q43
If f(x)=[x]−[x4]f(x) = [x] - \left[ {{x \over 4}} \right] ,x ∈ \in 4 , where [x] denotes the greatest integer function, then
A Both lim⁡x→4−f(x)\mathop {\lim }\limits_{x \to 4 - } f(x) and lim⁡x→4+f(x)\mathop {\lim }\limits_{x \to 4 + } f(x) exist but are not equal
B f is continuous at x = 4
C lim⁡x→4+f(x)\mathop {\lim }\limits_{x \to 4 + } f(x) exists but lim⁡x→4−f(x)\mathop {\lim }\limits_{x \to 4 - } f(x) does not exist
D lim⁡x→4−f(x)\mathop {\lim }\limits_{x \to 4 - } f(x) exists but lim⁡x→4+f(x)\mathop {\lim }\limits_{x \to 4 + } f(x) does not exist
Correct Answer
Option B
Solution
f(x)=[x]−[x4]f(x) = [x] - \left[ {{x \over 4}} \right]

Here check continuty at x = 4 LHL =

lim⁡x→4−f(x)\mathop {\lim }\limits_{x \to {4^ - }} f\left( x \right)

=

lim⁡x→4−[x]−[x4]\mathop {\lim }\limits_{x \to {4^ - }} \left[ x \right] - \left[ {{x \over 4}} \right]

=

lim⁡h→0[4−h]−[4−h4]\mathop {\lim }\limits_{h \to 0} \left[ {4 - h} \right] - \left[ {{{4 - h} \over 4}} \right]

= 3 - 0 = 3 h →\to 0 means h > 0 (very small positive value) So 4 - h = 3.something (means less than 4) ∴\therefore [4 - h] = [3.something] = 3 and

[4−h4]\left[ {{{4 - h} \over 4}} \right]

= [0.something] = 0 RHL =

lim⁡x→4+f(x)\mathop {\lim }\limits_{x \to {4^ + }} f\left( x \right)

=

lim⁡x→4+[x]−[4x]\mathop {\lim }\limits_{x \to {4^ + }} \left[ x \right] - \left[ {{4 \over x}} \right]

=

lim⁡h→0[4+h]−[4+h4]\mathop {\lim }\limits_{h \to 0} \left[ {4 + h} \right] - \left[ {{{4 + h} \over 4}} \right]

= 4 - 1 = 3 Here [4 + h] = [4.something] = 4 and

[4+h4]\left[ {{{4 + h} \over 4}} \right]

= [1.something] = 1 And f(4) =

[4]−[44]\left[ 4 \right] - \left[ {{4 \over 4}} \right]

= 4 - 1 = 3 As LHL = RHL = f(4) ∴\therefore f is continuous at x = 4.

Q44
lim⁡x→a(a+2x)13−(3x)13(3a+x)13−(4x)13\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}} (aa ≠ \ne 0) is equal to :
A (29)(23)13\left( {{2 \over 9}} \right){\left( {{2 \over 3}} \right)^{{1 \over 3}}}
B (23)(29)13\left( {{2 \over 3}} \right){\left( {{2 \over 9}} \right)^{{1 \over 3}}}
C (23)43{\left( {{2 \over 3}} \right)^{{4 \over 3}}}
D (29)43{\left( {{2 \over 9}} \right)^{{4 \over 3}}}
Correct Answer
Option B
Solution

L =

lim⁡x→a(a+2x)13−(3x)13(3a+x)13−(4x)13\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}
=lim⁡h→0(a+2(a+h))1/3−(3(a+h))1/3(3a+a+h)1/3−(4(a+h))1/3= \mathop {\lim }\limits_{h \to 0} {{{{(a + 2(a + h))}^{1/3}} - {{(3(a + h))}^{1/3}}} \over {{{(3a + a + h)}^{1/3}} - {{(4(a + h))}^{1/3}}}}

=

lim⁡h→0(3a)1/3(1+2h3a)1/3−(3a)1/3(1+ha)1/3(4a)1/3(1+h4a)1/3−(4a)1/3(1+ha)1/3\mathop {\lim }\limits_{h \to 0} {{{{(3a)}^{1/3}}{{\left( {1 + {{2h} \over {3a}}} \right)}^{1/3}} - {{(3a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}} \over {{{(4a)}^{1/3}}{{\left( {1 + {h \over {4a}}} \right)}^{1/3}} - {{(4a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}}}

=

lim⁡h→0(31/341/3)[(1+2h9a)−(1+h3a)(1+h12a)−(1+h3a)]\mathop {\lim }\limits_{h \to 0} \left( {{{{3^{1/3}}} \over {{4^{1/3}}}}} \right)\left[ {{{\left( {1 + {{2h} \over {9a}}} \right) - \left( {1 + {h \over {3a}}} \right)} \over {\left( {1 + {h \over {12a}}} \right) - \left( {1 + {h \over {3a}}} \right)}}} \right]
=(34)1/3(29−13)(112−13)=(34)1/3(8−123−12)= {\left( {{3 \over 4}} \right)^{1/3}}{{\left( {{2 \over 9} - {1 \over 3}} \right)} \over {\left( {{1 \over {12}} - {1 \over 3}} \right)}} = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{8 - 12} \over {3 - 12}}} \right)
=(34)1/3(−4−9)=41−1332−13=42/335/3= {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{ - 4} \over { - 9}}} \right) = {{{4^{1 - {1 \over 3}}}} \over {{3^{2 - {1 \over 3}}}}} = {{{4^{2/3}}} \over {{3^{5/3}}}}
=(8×2)1/3(27×9)1/3=23(29)1/3= {{{{(8 \times 2)}^{1/3}}} \over {{{(27 \times 9)}^{1/3}}}} = {2 \over 3}{\left( {{2 \over 9}} \right)^{1/3}}
Q45
If the function f(x)={a∣π−x∣+1,x≤5b∣x−π∣+3,x>5f(x) = \left\{ \begin{array}{ll}{a|\pi - x| + 1,x \le 5} \\ {b|x - \pi | + 3,x > 5} \end{array} \right. is continuous at x = 5, then the value of a – b is :-
A 2π−5{2 \over {\pi - 5 }}
B 25−π{2 \over {5 - \pi }}
C −2π+5{-2 \over {\pi + 5 }}
D 2π+5{2 \over {\pi + 5 }}
Correct Answer
Option B
Solution

As f(x) is continuous at x = 5 then

lim⁡x→5−f(x)=f(5)=lim⁡x→5+f(x)\mathop {\lim }\limits_{x \to {5^ - }} f\left( x \right) = f\left( 5 \right) = \mathop {\lim }\limits_{x \to {5^ + }} f\left( x \right)

⇒\Rightarrow

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

⇒\Rightarrow

lim⁡h→0(a∣π−(5−h)∣+1)=a∣π−5∣+1\mathop {\lim }\limits_{h \to 0} \left( {a\left| {\pi - \left( {5 - h} \right)} \right| + 1} \right) = a\left| {\pi - 5} \right| + 1

=

lim⁡h→0(b∣5+h−π∣+3)\mathop {\lim }\limits_{h \to 0} \left( {b\left| {5 + h - \pi } \right| + 3} \right)

⇒\Rightarrow

a∣π−5∣+1{a\left| {\pi - 5} \right| + 1}

=

a∣π−5∣+1a\left| {\pi - 5} \right| + 1

=

b∣5−π∣+3{b\left| {5 - \pi } \right| + 3}

Note : As π\pi = 3.14, then π\pi - 5 = 3.14 - 5 = -1.84 < 0 ∴\therefore

∣π−5∣{\left| {\pi - 5} \right|}

= - (π\pi - 5) and

∣5−π∣{\left| {5 - \pi } \right|}

= (5 - π\pi) ⇒\Rightarrow

−a(π−5)+1- a\left( {\pi - 5} \right) + 1

=

−a(π−5)+1- a\left( {\pi - 5} \right) + 1

=

b(5−π)+3{b\left( {5 - \pi } \right) + 3}

⇒\Rightarrow

−a(π−5)+1- a\left( {\pi - 5} \right) + 1

=

b(5−π)+3{b\left( {5 - \pi } \right) + 3}

⇒\Rightarrow

(a−b)(5−π)=2\left( {a - b} \right)\left( {5 - \pi } \right) = 2

⇒\Rightarrow

(a−b)=2(5−π)\left( {a - b} \right) = {2 \over {\left( {5 - \pi } \right)}}
Q46
Iff(x)={sin⁡(p+1)x+sin⁡xx,x<0q,x=0x+x2−xx3/2,x>0f(x) = \left\{ \begin{array}{ll}{{{\sin (p + 1)x + \sin x} \over x}} & {,x < 0} \\ q & {,x = 0} \\ {{{\sqrt {x + {x^2}} - \sqrt x } \over {{x^{{{3}/{2}}}}}}} & {,x > 0} \end{array} \right. is continuous at x = 0, then the ordered pair (p, q) is equal to
A (−32,−12)\left( { - {3 \over 2}, - {1 \over 2}} \right)
B (−12,32)\left( { - {1 \over 2},{3 \over 2}} \right)
C (−32,12)\left( { - {3 \over 2}, {1 \over 2}} \right)
D (52,12)\left( { {5 \over 2}, {1 \over 2}} \right)
Correct Answer
Option C
Solution
f(x)={sin⁡(p+1)x+sin⁡xxx<0qx=0x2+x−xx32x>0f(x) = \left\{ \begin{array}{ll}{{{\sin (p + 1)x + \sin x} \over x}} & {x < 0} \\ q & {x = 0} \\ {{{\sqrt {{x^2} + x} - \sqrt x } \over {{x^{{3 \over 2}}}}}} & {x > 0} \end{array} \right.

is continuous at x = 0 So f(0–) = f(0) = f (0+) ... (1)

f(0−)=lth→0f(0−h)f({0^ - }) = \mathop {lt}\limits_{h \to 0} f(0 - h)
⇒lth→0sin⁡(p+1)(−h)+sin⁡(−h)−h\Rightarrow \mathop {lt}\limits_{h \to 0} {{\sin (p + 1)( - h) + \sin ( - h)} \over { - h}}
⇒lth→0[−sin⁡(p+1)h−h+sinh⁡h]\Rightarrow \mathop {lt}\limits_{h \to 0} \left[ {{{ - \sin (p + 1)h} \over { - h}} + {{\sinh } \over h}} \right]
⇒lim⁡h→0sin⁡(p+1)hh(p+1)×(p+1)+lim⁡h→0sinh⁡h\Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sin (p + 1)h} \over {h(p + 1)}} \times (p + 1) + \mathop {\lim }\limits_{h \to 0} {{\sinh } \over h}

= (p + 1) + 1 = p + 2 ...... (2) Now

f(0+)=lim⁡h→0(0+h)=lim⁡h→0h2+h−hh3/2f({0^ + }) = \mathop {\lim }\limits_{h \to 0} (0 + h) = \mathop {\lim }\limits_{h \to 0} {{\sqrt {{h^2} + h} - \sqrt h } \over {{h^{3/2}}}}
⇒lim⁡h→0(h)12[h+1−1]h(h12)\Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(h)}^{{1 \over 2}}}\left[ {\sqrt {h + 1} - 1} \right]} \over {h\left( {{h^{{1 \over 2}}}} \right)}}
⇒lim⁡h→0h+1−1h×h+1+1h+1+1\Rightarrow \mathop {\lim }\limits_{h \to 0} {{\sqrt {h + 1} - 1} \over h} \times {{\sqrt {h + 1} + 1} \over {\sqrt {h + 1} + 1}}
⇒lim⁡h→0h+1−1h(h+1+1)\Rightarrow \mathop {\lim }\limits_{h \to 0} {{h + 1 - 1} \over {h\left( {\sqrt {h + 1} + 1} \right)}}
⇒lim⁡h→01h+1+1=11+1=12\Rightarrow \mathop {\lim }\limits_{h \to 0} {1 \over {\sqrt {h + 1} + 1}} = {1 \over {1 + 1}} = {1 \over 2}

..... (3) Now, from equation (1) f(0–) = f(0) = f(0+) p + 2 = q = 1/2 So,

q=12q = {1 \over 2}

and

p=12−2=−32p = {1 \over 2} - 2 = {{ - 3} \over 2}
(p,q)≡(−32,12)(p,q) \equiv \left( { - {3 \over 2},{1 \over 2}} \right)
Q47
If lim⁡x→1x2−ax+bx−1=5\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5, then a + b is equal to :
A 1
B - 4
C - 7
D 5
Correct Answer
Option C
Solution
lim⁡x→1x2−ax+bx−1=5\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5

⇒\Rightarrow

(1)2−a(1)+b=0{(1)^2} - a(1) + b = 0

⇒\Rightarrow

1−a+b=01 - a + b = 0

⇒\Rightarrow

a−b=1  ......(1)a - b = 1\,\,......(1)

Now 'L' hospital rule 2x - a = 5 ⇒\Rightarrow2 - a = 5 (∵\because x = 1) ⇒\Rightarrow a = - 3 By putting a = -3 in (1) ⇒\Rightarrow b = -4 ∴\therefore a + b = -7

Q48
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x ∈ \in R. If f(x) attains maximum value at α\alpha and g(x) attains minimum value at β\beta , then lim⁡x→−αβ(x−1)(x2−5x+6)x2−6x+8\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}} is equal to :
A 12{1 \over 2}
B −12-{1 \over 2}
C 32{3 \over 2}
D −32-{3 \over 2}
Correct Answer
Option A
Solution

From f(x) = 5 - | x - 2 | maximum value of f(x) is at x = 2 From g(x) = | x + 1 | minimum value of g(x) is at x = -1 ∴\therefore

αβ\alpha \beta

= - 2 ⇒\Rightarrow

lim⁡x→2(x−1)(x−2)(x−3)(x−2)(x−4)\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}

⇒\Rightarrow

(2−1)(2−3)(2−4)=12{{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}
Q49
If α\alpha and β\beta are the roots of the equation 375x2 – 25x – 2 = 0, then lim⁡n→∞∑r=1nαr+lim⁡n→∞∑r=1nβr\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\alpha ^r}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\beta ^r}} is equal to :
A 7116{7 \over {116}}
B 29348{{29} \over {348}}
C 112{1 \over {12}}
D 21346{{21} \over {346}}
Correct Answer
Option B
Solution

α\alpha and β\beta are the two root of 375x2 - 25x - 2 = 0 Both of the roots are lie in (-1, 1) hence sum of given series is finite

lim⁡n→∞(α1−α+β1−β)=α(1−β)+β(1−α)(1−α)(1−β)\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}
⇒(α+β)−2αβ1−(α+β)+αβ=25−2(−2)375−25−2=29348\Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}
Q50
lim⁡y→01+1+y4−2y4\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}
A exists and equals 122{1 \over {2\sqrt 2 }}
B exists and equals 142{1 \over {4\sqrt 2 }}
C exists and equals 122(1+2){1 \over {2\sqrt 2 (1 + \sqrt {2)} }}
D does not exists
Correct Answer
Option B
Solution
lim⁡y→01+1+y4−2y4\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}

If you put y = 0 at

1+1+y4−2y4{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}

it is in

00{0 \over 0}

form. So we can use L' Hospital's Rule. =

lim⁡y→0121+1+y4×(121+y4)×4y34y3\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}

=

lim⁡y→0121+1+y4×121+y4\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}

=

142{1 \over {4\sqrt 2 }}
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