Practice Start Test

Limits, Continuity and Differentiability

JEE Mathematics · 196 questions · Page 11 of 20 · Click an option or "Show Solution" to reveal answer

Q101
Let a be an integer such that lim⁡x→718−[1−x][x−3a]\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}} exists, where [t] is greatest integer ≤\le t. Then a is equal to :
A −-6
B −-2
C 2
D 6
Correct Answer
Option A
Solution
lim⁡x→718−[1−x][x−3a]\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}

exist &

a∈Ia \in I

.

=lim⁡x→717−[−x][x]−3a= \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}

exist

RHL=lim⁡x→7+17−[−x][x]−3a=257−3aRHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}
[a≠73]\left[ {a \ne {7 \over 3}} \right]
LHL=lim⁡x→7−17−[−x][x]−3a=246−3aLHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}
[a≠2]\left[ {a \ne 2} \right]

For limit to exist

LHL=RHLLHL = RHL
257−3a=246−3a{{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}
⇒257−3a=82−a\Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}

∴\therefore

a=−6a = - 6
Q102
lim⁡x→12sin⁡(cos⁡−1x)−x1−tan⁡(cos⁡−1x)\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}} is equal to :
A 2\sqrt 2
B −2 - \sqrt 2
C 12{1 \over {\sqrt 2 }}
D −12 - {1 \over {\sqrt 2 }}
Correct Answer
Option D
Solution
lim⁡x→12sin⁡(cos⁡−1x)−x1−tan⁡(cos⁡−1x)\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}

Let

cos⁡−1x=t{\cos ^{ - 1}}x = t
⇒x=cos⁡t\Rightarrow x = \cos t

When

x→12x \to {1 \over {\sqrt 2 }}

, then

t→cos⁡−1(12)→π4t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}

∴\therefore

lim⁡t→π4sin⁡t−cos⁡t1−tan⁡(t)\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}
=lim⁡t→π4sin⁡t−cos⁡t1−sin⁡tcos⁡t= \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}
=lim⁡t→π4(sin⁡t−cos⁡t)(cos⁡t)(cos⁡t−sin⁡t)= \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}
=lim⁡t→π4−cos⁡t= \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t
=−lim⁡t→π4cos⁡t= - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t
=−12= - {1 \over {\sqrt 2 }}
Q103
Let f, g : R →\to R be two real valued functions defined as f(x) = \left\{ {\matrix{ { - |x + 3|} & , & {x 1 and k2 are real constants. If (gof) is differentiable at x = 0, then (gof) (-$$ 4) + (gof) (4) is equal to :
A 4(e4+1)4({e^4} + 1)
B 2(2e4+1)2(2{e^4} + 1)
C 4e44{e^4}
D 2(2e4−1)2(2{e^4} - 1)
Correct Answer
Option D
Solution

∵\because gof is differentiable at x = 0 So R.H.D = L.H.D

ddx(4ex+k2)=ddx((−∣x+3∣)2−k1∣x+3∣){d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)
⇒4=6−k1⇒k1=2\Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2

Also

f(f(0+))=g(f(0−))f(f({0^ + })) = g(f({0^ - }))
⇒4+k2=9−3k1⇒k2=−1\Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1

Now

g(f(−4))+g(f(4))g(f( - 4)) + g(f(4))
=g(−1)+g(e4)=(1−k1)+(4e4+k2)= g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})
=4e4−2= 4{e^4} - 2
=2(2e4−1)= 2(2{e^4} - 1)
Q104
lim⁡x→0cos⁡(sin⁡x)−cos⁡xx4\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} is equal to :
A 13{1 \over 3}
B 14{1 \over 4}
C 16{1 \over 6}
D 112{1 \over 12}
Correct Answer
Option C
Solution
lim⁡x→0cos⁡(sin⁡x)−cos⁡xx4=lim⁡x→02sin⁡(x+sin⁡x) . sin⁡(x−sin⁡x2)x4\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}
=lim⁡x→02 . ((x+sin⁡x2)(x−sin⁡x2)x4)= \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}} \right)
=lim⁡x→012 . ((x+x−x33!+x55!...)(x−x+x33!...)x4)= \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)
=lim⁡x→012 . (2−x23!+x45!...)(13!−x25!−1)= \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)
=16= {1 \over 6}
Q105
Let f(x) = min {1, 1 + x sin x}, 0 ≤\le x ≤\le 2π\pi . If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to
A (2, 0)
B (1, 0)
C (1, 1)
D (2, 1)
Correct Answer
Option B
Solution
f(x)=min⁡{1, 1+xsin⁡x}f(x) = \min \{ 1,\,1 + x\sin x\}

,

0≤x≤x0 \le x \le x
f(x) = \left\{ {\matrix{ {1,} & {0 \le x Now at

x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)

\therefore

f(x)iscontinuousin[0,2f(x) is continuous in [0, 2

\pi

]Now,atx=] Now, at x =

\pi

L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0

R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi

\therefore

f(x)isnotdifferentiableatx=f(x) is not differentiable at x =

\pi

\therefore$$ (m, n) = (1, 0)

Q106
lim⁡x→π2(tan⁡2x((2sin⁡2x+3sin⁡x+4)12−(sin⁡2x+6sin⁡x+2)12))\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{{\tan }^2}x\left( {{{(2{{\sin }^2}x + 3\sin x + 4)}^{{1 \over 2}}} - {{({{\sin }^2}x + 6\sin x + 2)}^{{1 \over 2}}}} \right)} \right) is equal to
A 112{1 \over {12}}
B −-118{1 \over {18}}
C −-112{1 \over {12}}
D 16{1 \over {6}}
Correct Answer
Option A
Solution
lim⁡x→π2tan⁡2x{2sin⁡2x+3sin⁡x+4−sin⁡2x+6sin⁡x+2}\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}
=lim⁡x→π2tan⁡2x(sin⁡2x−3sin⁡x+2)2sin⁡2x+3sin⁡x+4+sin⁡2x+6sin⁡x+2= \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}
=16lim⁡x→π2(1−sin⁡x)(2−sin⁡x)cos⁡2x . sin⁡2x= {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x
=16lim⁡x→π2(2−sin⁡x)sin⁡2x1+sin⁡x= {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}
=112= {1 \over {12}}
Q107
Let f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1f(x) = \left\{ \begin{array}{ll}{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \\ { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \end{array} \right.. Then the set of all values of b, for which f(x) has maximum value at x = 1, is :
A (−-6, −-2)
B (2, 6)
C [−6,−2)∪(2,6][ - 6, - 2) \cup (2,6]
D [−6,−2)∪(2,6]\left[ {-\sqrt 6 , - 2} \right) \cup \left( {2,\sqrt 6 } \right]
Correct Answer
Option C
Solution
f(x)={x3−x2+10x−7,x≤1−2x+log⁡2(b2−4),x>1f(x) = \left\{ \begin{array}{ll}{{x^3} - {x^2} + 10x - 7,} & {x \le 1} \\ { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \end{array} \right.

If

f(x)f(x)

has maximum value at

x=1x = 1

then

f(1+)≤f(1)f(1 + ) \le f(1)
−2+log⁡2(b2−4)≤1−1+10−7- 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7
log⁡2(b2−4)≤5{\log _2}({b^2} - 4) \le 5
0(i)0 (i)

{b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )

(ii)(ii)

{b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]

IntersectionofabovetwosetsIntersection of above two sets

b \in [ - 6, - 2) \cup (2,6]$$

Q108
Let f(x) be a polynomial function such that f(x)+f′(x)+f′′(x)=x5+64f(x) + f'(x) + f''(x) = {x^5} + 64. Then, the value of lim⁡x→1f(x)x−1\mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x - 1}} is equal to:
A −-15
B −-60
C 60
D 15
Correct Answer
Option A
Solution

Given,

f(x)+f′(x)+f′′(x)=x5+64f(x) + f'(x) + f''(x) = {x^5} + 64

.........(i) ⇒f(x)\Rightarrow f(x) is a polynomial in xx whose degree is 5. Let

f(x)=x5+ax4+bx3+cx2+dx+ef(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e
f′(x)=5x4+4ax3+3bx2+2cx+df'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d
f′′(x)=20x3+12ax2+6bx+2cf''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c

On substituting the value of f(x),f′(x)f(x), f^{\prime}(x) and f′′(x)f^{\prime \prime}(x) in Eq. (i), we get

x5+(a+5)x4+(b+4a+20)x3+(c+3b+12a)x2+(d+2c+6b)x+e+d+2c=x5+64{x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64

Now, equating the coefficient, we get

⇒a+5=0\Rightarrow a + 5 = 0
b+4a+20=0b + 4a + 20 = 0
c+3b+12a=0c + 3b + 12a = 0
d+2c+6b=0d + 2c + 6b = 0
e+d+2c=64e + d + 2c = 64

∴\therefore

a=−5, b=0, c=60, d=−120, e=64a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64

∴\therefore

f(x)=x5−5x4+60x2−120x+64f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64

Now,

lim⁡x→1x5−5x4+60x2−120x+64x−1\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}

is (

00{0 \over 0}

form) By L' Hospital rule

lim⁡x→15x4−20x3+120x−1201\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}
=−15= - 15
Q109
Let f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & {,\,|x| where [t] denotes greatest integer \let.Ifmisthenumberofpointswhere t. If m is the number of points where fisnotcontinuousandnisthenumberofpointswhere is not continuous and n is the number of points where f$$ is not differentiable, then the ordered pair (m, n) is :
A (3, 3)
B (2, 4)
C (2, 3)
D (3, 4)
Correct Answer
Option C
Solution
f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & , & {|x|

f(x) = \left\{ {\matrix{ {{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr {2x} & , & {x \in (0,1)} \cr 1 & , & {otherwise} \cr } } \right.

Itclearlyshowsthatf(x)isdiscontinuousAtx=It clearly shows that f(x) is discontinuous At x =

-

1,1alsonondifferentiableandat1, 1 also non differentiable and at

x = 0

,,

L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0

R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2

\therefore

f(x)isnotdifferentiableatx=0f(x) is not differentiable at x = 0

\therefore$$ m = 2, n = 3

Q110
If lim⁡n→∞(n2−n−1+nα+β)=0\mathop {\lim }\limits_{n \to \infty } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0, then 8(α+β)8(\alpha+\beta) is equal to :
A 4
B −-8
C −-4
D 8
Correct Answer
Option C
Solution
lim⁡n→α(n2−n−1+nα+β)=0\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0

[ This limit will be zero when α\alpha 0 then overall limit will be ∞\infty. ]

⇒lim⁡n→α(n2−n−1+nα+β)(n2−n−1−(nα+β))n2−n−1−(nα+β)=0\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0
⇒lim⁡n→α(n2−n−1)−(nα+β)2n2−n−1−(nα+β)=0\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0
⇒lim⁡n→αn2−n−1−n2α2−2nαβ−β2n2−n−1−(nα+β)=0\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0
⇒lim⁡n→αn2(1−α2)−n(1+2αβ)−(1+β2)n2−n−1−(nα+β)\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}

Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.

To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite (∞\infty).

∴\therefore Coefficient of n2 should be 0 in this case.

∴\therefore

1−α2=01 - {\alpha ^2} = 0
⇒α= ± 1\Rightarrow \alpha = \, \pm \,1

But α\alpha should be ∴\therefore

α= + 1\alpha = \, + \,1

not possible ∴\therefore

α=−1\alpha = - 1

.

⇒lim⁡n→α0−n(1+2αβ)−(1+β)n[1−1n−1n2−α−βn]=0\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{0 - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + \beta } \right)} \over {n\left[ {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}} \right]}} = 0

Divide numerator and denominator by n then we get,

⇒lim⁡n→α−(1+2αβ)−(1+β)n1−1n−1n2−α−βn=0\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0
⇒−(1+2αβ)−01−0−0−α−0=0\Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0
⇒−(1+2αβ)1−α=0\Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0
⇒−(1+2αβ)=0\Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0
⇒1+2αβ=0\Rightarrow 1 + 2\alpha \beta = 0
⇒2αβ=−1\Rightarrow 2\alpha \beta = - 1
⇒β=−12α=−12(−1)=12\Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}

∴\therefore

8(α+β)8\left( {\alpha + \beta } \right)
=8(−1+12)= 8\left( { - 1 + {1 \over 2}} \right)
=8×−12= 8 \times - {1 \over 2}
=−4= - 4
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