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NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives

Edited By Ravindra Pindel | Updated on Sep 12, 2022 01:56 PM IST

NCERT Exemplar Class 11 Maths Solutions Chapter 13 Limits and Derivatives is drafted by experts teachers of mathematics. The solutions are prescribed by NCERT and using the easiest or rather the most comprehensible methods. To provide reliable and authentic NCERT exemplar Class 11 Maths solutions chapter 13, we aligned to the guidelines of CBSE for the students. The lesson provides for practical application and usage of Limits and Derivatives in various functions of trigonometry, polynomials, and rational numbers.
Also, check - NCERT Class 11 Maths Solutions.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

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  1. NCERT Class 11 Exemplar Maths Solutions Chapter 13 - Question-Wise Solution
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  4. Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 13

NCERT Class 11 Exemplar Maths Solutions Chapter 13 - Question-Wise Solution

Question:1

Evaluate limx3x29x3

Answer:

Given limx3x29x3=limx3(x3)(x+3)x3=limx3x+3=6

Question:2

Evaluate limx1/24x212x1

Answer:

Given that limx1/24x212x1=limx1/2(2x1)(2x+1)2x1=limx1/22x+1=2

Question:3

Evaluate limh0x+hxh

Answer:

limh0x+hxh

=limh0x+hxh[x+h+x]×(x+h+x)  [Rationalizing the denominator] =limh0x+hxh[x+h+x] =limh01x+h+x
Put limit
=1x+x=12x

Question:4

Evaluate: limx0(x+2)13213x

Answer:

Given limx0(x+2)13213x   
Now put x=x-2 limits change from 0 to 2
=limy2y13213y2=13(2)131=13223 [usinglimxaxnanxa=nan1]

Question:5

Evaluate : limx1(1+x)61(1+x)21

Answer:

~Given that limx1(1+x)61(1+x)21=limx1((1+x)2)31(1+x)21
=limx1((1+x)21)[(1+x)4+(1+x)2+1](1+x)21=limx1(1+x)4+(1+x)2+1=24+22+1=21

Question:6

Evaluate:
limxa(2+x)52(2+a)52xa

Answer:

   ~~~~~~~~~~ Given that limxa(2+x)52(2+a)52xa

=limya+2(y)52(2+a)52y(a+2)=52(a+2)521=52(a+2)32 [usinglimxaxnanxa=nan1]

Question:7

Evaluate:
limx1x4xx1

Answer:

Given limx1x4xx1=limx1(x4x)(x+1)(x1)(x+1)

=limx1x4x+x4xxx1=limx1x(x41)+x(x31)x1=limx1(x1)(x(x3+x2+x+1)+x(x2+x+1))x1=limx1x(x3+x2+x+1)+x(x2+x+1)=14+13=7

Question:8

Evaluate:
limx2x243x2x+2

Answer:

Given limx2x243x2x+2=limx2x24(3x2x+2)3x2+x+23x2+x+2=limx2x24(3x2x2)(3x2+x+2)=limx2(x2)(x+2)2(x2)(3x2+x+2)=limx2(x+2)2(3x2+x+2)=42(322+2+2)=8

Question:9

Evaluate:
limx2x24x2+32x8

Answer:

=limx2(x22)(x2+2)x2+42x2x8=limx2(x+2)(x2)(x2+2)x(x+42)2(x+42)=limx2(x+2)(x2)(x2+2)(x+42)(x2)=limx2(x+2)(x2+2)x+42 Put limit =(2+2)(2+2)2+42=22×452=85

Question:10

Evaluate:
limx1(x72x5+1x33x2+2)

Answer:

Let us apply LH rule i.e. L.Hospita'srule to this question


  limx1(x72x5+1x33x2+2)limxa(f(x)g(x))=limxa(f(x)g(x))limx1(x72x5+1x33x2+2)=limx1(7x610x43x26x)=71036=1

Question:11

Evaluate:
limx0(1+x31x3x2)

Answer:

Given that limx0(1+x31x3x2)
=limx0(1+x31x3x21+x3+1x31+x3+1x3)=limx0(1+x31+x3x211+x3+1x3)=limx0(2x1+x3+1x3)=0

Question:12

Evaluate:
limx3(x3+27x5+243)

Answer:

=limx3x3+(3)3x+3x3+(3)3x+3 [Dividing the numerator and denominator by x+3]=limx3(x3+(3)3x+3)limx3(x3+(3)5x+3)[limxaf(x)g(x)=limxaf(x)limxag(x)]=limx3(x23x+9)limx3(x43x3+9x227x+81)=9+9+981+81+81+81+81=3×95×81=15×3=115

Question:13

Evaluate:
limx12(8x32x14x2+14x21)

Answer:


Given limx12(8x32x14x2+14x21)
=limx12((12x+1)((8x3)(2x+1)4x212x1))=limx12((12x+1)(12x2+2x42x1))


=limx12((12x+1)(12x2+8x6x42x1))=limx12((12x+1)((3x+2)(4x2)2x1) )=limx12((12x+1)2(3x+2) )=1212+12(312+2)=72

Question:14

Evaluate: Find ‘n’, if
limx2(xn2nx2)=80,nN

Answer:

We know that limx2(xn2nx2)=n(2)n1n(2)n1=80n=55(2)51=516=80

Question:15

Evaluate:
limxa(sin3xsin7x)

Answer:

Given limxa(sin3xsin7x)=sin3asin7a

Question:16

Evaluate:
limx0(sin22xsin24x)

Answer:

Given limx0(sin22xsin24x)=limx0((sin2x2x2x)2(sin4x4x4x)2)=limx0(416(sin2x2x)2(sin4x4x)2)=416=14

Question:17

Evaluate:
limx0(1cos2xx2)

Answer:

Given that
limx0(1cos2xx2)=limx0(2sin2xx2)=limx0(2(sinxx)2)=2

Question:18

Evaluate:
limx0(2sinxsin2xx3)

Answer:

Given limx0(2sinxsin2xx3)=limx0(2sinx(1cosx)x3)=limx0(2sinx(2sin2x2)x3)=limx0(2sinxx2(sinx2x2)214)=1

Question:19

Evaluate:
limx0(1cosmx1cosnx)

Answer:

Given that limx0(1cosmx1cosnx)=limx0(sin2mx2sin2nx2)=limx0(sin2mx2(mx2)2(mx2)2sin2nx2(nx2)2(nx2)2)=limx0(sin2mx2(mx2)2m2sin2nx2(nx2)2n2)=m2n2

Question:20

Evaluate:
limxπ31cos6x2(π3x)

Answer:

limxπ31cos6x2(π3x)   Herecos6x=12sin23x


=limxπ32sin23x2(π3x)=limxπ3|sin3x|(π3x3)=limxπ3|sin(π3x)|(π3x3)=limxπ33|sin(π3x)|π3x=31=3

Question:21

Evaluate:
limxπ4sinxcosxxπ4

Answer:


Given that limxπ4sinxcosxxπ4=limxπ42(sinxcosπ4cosxsinπ4)xπ4=limxπ42sin(xπ4)xπ4=2

Question:22

Evaluate:
limxπ63sinxcosxxπ6

Answer:

Given that limxπ63sinxcosxxπ6=limxπ62(32sinx12cosx)xπ6=limxπ62(cosπ6sinxsinπ6cosx)xπ6

=limxπ62sin(xπ6)xπ6=2

Question:23

Evaluate:
limx0sin2x+3x2x+tan3x

Answer:


Given that limx0sin2x+3x2x+tan3x=limx02xsin2x2x+3x2x+3xtan3x3x=limx0x(2sin2x2x+3)x(2+3tan3x3x)


=limx0(2sin2x2x+3)(2+3tan3x3x)=2+32+3=55=1

Question:24

Evaluate:
limxasinxsinaxa

Answer:

Given that limxasinxsinaxa=limxa(sinxsina)(x+a)(xa)(x+a)=limxa(sinxsina)(x+a)(xa)
=limxa(2cos(x+a2)sin(xa2))(x+a)(xa)=limxacos(x+a2)sin(xa2)xa2(x+a)=cos(a)1(2a)=2acosa

Question:25

Evaluate:
limxπ6cot2x3cosecx2

Answer:

Given that limxπ6cot2x3cosecx2=limxπ6cosec2x13cosecx2=limxπ6cosec2x13cosecx2=limxπ6(cosecx2)(cosecx+2)cosecx2

=limxπ6(cosecx+2)=2+2=4

Question:26

Evaluate:
limx0(21+cosx)sin2x

Answer:

Given that limx0(21+cosx)sin2x=limx0(21+cosx)(2+1+cosx)sin2x(2+1+cosx)=limx0(2(1+cosx))sin2x(2+1+cosx)=limx01cosxsin2x(2+1+cosx)
=limx01cosx(1cosx)(1+cosx)(2+1+cosx)=limx01(1+cosx)(2+1+cosx)=1(1+1)(22)=142

Question:27

Evaluate:
limx0(sinx2sin3x+sin5xx)

Answer:

.Given that limx0(sinx2sin3x+sin5xx)=limx0(sinxx2sin3x3x3+sin5x5x5)=123+5=0

Question:28

Evaluate: If
limx1(x41x1)=limxk(x3k3x2k2)

Answer:

Given that limx1(x41x1)=4(1)41=4limxk(x3k3x2k2)=limxk((xk)(x2+k2+xk)(xk)(x+k))=limxk(x2+k2+xkx+k)=3k2
3k2=4k=8/3

Question:29

Differentiate each of the functions w.r.t x in
x4+x3+x2+1x

Answer:

 Let y=x4+x3+x2+1xdydx=d(x3)dx+d(x2)dx+d(x)dx+d(x1)dx=3x2+2x+11x2=3x4+2x3+x21x2

Question:30

Differentiate each of the functions w.r. to x in
(x+1x)3

Answer:

Let y= (x+1x)3


dydx=ddx(x+1x)3=3(x+1x)2(11x2)=3(x2+2+1x2)(11x2)=3(x2+2+1x212x21x4)=3x2+33x23x4

Question:31

Differentiate each of the functions w.r. to x in
(3x+5)(1+tanx)

Answer:

Given that y=(3x+5)(1+tanx) 

Applying product rule of differentiation we get
dydx=(1+tanx)ddx(3x+5)+(3x+5)ddx(1+tanx)=3(1+tanx)+(3x+5)sec2x

Question:32

Differentiate each of the functions w.r. to x in
(secx1)(secx+1)

Answer:

y= (secx1)(secx+1)=sec2x1=tan2x 

Now applying the concept of chain rule


dydx=d(tan2x)dx=2tanxsec2x

Question:33

Differentiate each of the functions w.r. to x in
3x+45x27x+9

Answer:

Given that y=3x+45x27x+9

Applying division rule of differentiation that is


dydx=ddx(3x+45x27x+9)=(5x27x+9)ddx(3x+4)(3x+4)ddx(5x27x+9)(5x27x+9)2=(5x27x+9)(3)(3x+4)(10x7)(5x27x+9)2=5(3x2+8x11)(5x27x+9)2
=5(3x+11)(1x)(5x27x+9)2

Question:34

Differentiate each of the functions w.r. to x in
x5cosxsinx

Answer:

Given that y=x5cosxsinx=x5sinxcosxsinx

Applying division rule of differentiation that is

dydx=sinxddx(x5)x5ddx(sinx)sin2xddx(cotx)=5x4sinxx5cosxsin2x+cosec2x

Question:35

Differentiate each of the functions w.r. to x in
x2cosπ4sinx

Answer:

y=x2cosπ4sinx

Applying division rule of differentiation that is

dydx=cosπ4(sinxddx(x2)x2ddx(sinx))sin2x=12(2xsinxx2cosxsin2x)

Question:36
Differentiate each of the functions w.r. to x in

(ax\textsuperscript{2} + cotx) (p + q cosx)}} \\

Answer:

Given that y=(ax2+cotx)(p+qcosx)   

Applying division rule of differentiation that is


dydx=(p+qcosx)ddx(ax2+cotx)+(ax2+cotx)ddx(p+qcosx)
=(p+qcosx)(2axcosec2x)+(ax2+cotx)(qsinx)

Question:37

Differentiate each of the functions w.r. to x in
(a+bsinxc+dcosx)

Answer:

Given thaty=(a+bsinxc+dcosx)

Applying division rule of differentiation that is
dydt=(c+dcosx)ddx(a+bsinx)(a+bsinx)ddx(c+dcosx)(c+dcosx)2=bcosx(c+dcosx)(dsinx)(a+bsinx)(c+dcosx)2

=bccosx+bdcos2x+adsinx+bdsin2x(c+dcosx)2=bd+bccosx+adsinx(c+dcosx)2

Question:38

Differentiate each of the functions w.r. to x in

(sin x + cosx)^2

Answer:

Given that
y=(sinx+cosx)2=sin2x+cos2x+2sinxcosx=1+2sinxcosx=1+sin2x 
Applying the concept of chain rule
dydx=ddx(1+sin2x)=0+2cos2x=2cos2x=2(cos2xsin2x)

Question:39

Differentiate each of the functions w.r. to x in

(2x - 7)^{2} (3x + 5)^{3}

Answer:

This question will involve the concept of both chain rule and product rule

Given that y=(2x7)2(3x+5)3 

Applying product rule of differentiation
dydx=(3x+5)3ddx(2x7)2+(2x7)2ddx(3x+5)3=(3x+5)32(2x7)2+(2x7)23(3x+5)23=(2x7)(3x+5)2[4(3x+5)+9(2x7)]=(2x7)(3x+5)2(30x43)

Question:40

Differentiate each of the functions w.r. to x in
x\textsuperscript2sinx+cos2x

Answer:

This question will involve the concept of both chain rule and product rule
Giventhaty=x2sinx+cos2xdydx=ddx(x2sinx)+ddx(cos2x)=sinxddx(x2)+x2ddx(sinx)+ddx(cos2x)=2xsinx+x2cosx2sin2x

Question:41

Differentiate each of the functions w.r.to x in
Extra close brace or missing open brace

Answer:

The question involves the concept of chain rule
Giventhaty=sin3xcos3xy=18(2sinxcosx)3=18sin32xdydx=38sin22x(2cos2x)=34sin22xcos2x

Question:42

Differentiate each of the functions w.r. to x in
1ax2+bx+c

Answer:

The question involves the concept of chain rule
Giventhaty=1ax2+bx+c=(ax2+bx+c)1dydx=1(ax2+bx+c)2(2ax+b)=2ax+bax2+bx+c

Question:43

Differentiate each of the functions with respect to ‘x’Differentiate using first principle

\cos (x^{2} + 1)

Answer:

Let f(x)=cos(x2+1)..(i)f(x+Δx)=cos((x+Δx)2+1)(ii)~ Subtracting equation (i)fromequation(ii)f(x+Δx)f(x)Δx=cos((x+Δx)2+1)cos(x2+1)Δx
=limΔx0cos((x+Δx)2+1)cos(x2+1)Δx=limΔx02sin[(x+Δx)2+1x212]sin[(x+Δx)2+1+x2+12]Δx=limΔx02sin[2xΔx+(Δx)22]sin[2x2+2xΔx+22]Δx=limΔx0(2x+Δx2)( 2sin[2xΔx+(Δx)22]sin[2x2+2xΔx+22]Δx(2x+Δx)2)
=limΔx0(2x+Δx2)(2sin[2x2+2xΔx+22])( sin[2xΔx+(Δx)22]Δx(2x+Δx)2)=(2x+02)(2sin(x2+0+1))(1)=2xsin(x2+1)is the required answer

Question:44

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle ax+bcx+d

Answer:


Let f(x)=ax+bcx+d..(i)f(x+Δx)=a(x+Δx)+bc(x+Δx)+d(ii)Subtracting equation (i)from equation (ii)f(x+Δx)f(x)Δx=a(x+Δx)+bc(x+Δx)+dax+bcx+dΔx 


~ Taking the limitf(x)=limΔx0a(x+Δx)+bc(x+Δx)+dax+bcx+dΔx=limΔx0(cx+d)(ax+aΔx+b)(ax+b)(cx+cΔx+d)(cx+cΔx+d)(cx+d)Δx  =limΔx0(adbc)Δx(cx+cΔx+d)(cx+d)Δx

=limΔx0(adbc)(cx+cΔx+d)(cx+d)=adbc(cx+d)2 is the required answer

Question:45

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle X23

Answer:

f(x)=limΔx0(x+Δx)23x23Δx=limΔx0x23[(1+Δxx)231]Δx=limΔx0x23[(1+23Δxx+)1]Δx

Expanding by binomial theorem and rejecting the higher powers of ΔxasΔx0


=limΔx0x23(23Δxx)Δx=limΔx0(23x23x)=23x13

Question:46

Differentiate each of the functions with respect to ‘x’
Differentiate using first principle x cos x.

Answer:

Giventhaty=xcosx.(i)y+Δy=(x+Δx)cos(x+Δx).(ii) ~ Subtracting equation (i)fromequation(ii)y=limΔx0(((x+Δx)cos(x+Δx)xcosx)Δx)


=limΔx0(xcos(x+Δx)xcosxΔx)+limΔx0(Δxcos(x+Δx)Δx)=limΔx0(x(2sinx+Δxx2sinx+Δx+x2)Δx)+limΔx0cos(x+Δx)=limΔx0(2sin(x+Δx2)sin(Δx2)x)Δx+limΔx0cos(x+Δx)


=limΔx0xsin(x+Δx2)(sin(Δx2))Δx2+limΔx0cos(x+Δx)=xsinx+cosx

Question:47

Evaluate each of the following limits

limy0(x+y)sec(x+y)xsecxy

Answer:

Given\: \: that\: \: limy0(x+y)sec(x+y)xsecxy=limy0(xsec(x+y)xsecxy+ysec(x+y)y)=limy0(xy[cosxcos(x+y)cosxcos(x+y)]+sec(x+y))=limy0(xy[2sin(x+y2)sin(y2)cosxcos(x+y)]+sec(x+y))
=limy0(xsin(x+y2)cosxcos(x+y)[sin(y2)y2]+sec(x+y))=xsecxtanx+secx=secx(xtanx+1)

Question:48

Evaluate each of the following limits
limx0[sin(α+β)x+sin(αβ)x+sin2αx]cos2βxcos2αxx

Answer:


limx0[sin(α+β)x+sin(αβ)x+sin2αx]cos2βxcos2αxx=limx0[2sinαxcosβx+2sinαxcosαx][2sin(α+β)x sin(αβ)x]x=limx0[sinαx(cosβx+cosαx)][sin(α+β)x sin(αβ)x]x=limx0[sinαx(2cos(α+β)x2cos(αβ)x2)][sin(α+β)x sin(αβ)x]x
=limx0[sinαx][2sin(α+β)x2sin(αβ)x2]x=limx0[sinαxαx]αxx[2(sin(α+β)x2(α+β)x2)(sin(αβ)x2(αβ)x2)(α+β)x2(αβ)x2]=α2(α+β)2(αβ)2=2αα2β2

Question:49

Evaluate each of the following limits
limxπ4(tan3xtanxcos(x+π4))

Answer:


limxπ4(tan3xtanxcos(x+π4))=limxπ4(tanx)limxπ4(tan2x1cos(x+π4))=1limxπ4((1tanx)(1+tanx)cos(x+π4))=limxπ4(1+tanx)limxπ4((1tanx)cos(x+π4))=2limxπ4((cosxsinx)cos(x+π4)cosx)
=2limxπ4(2(cosπ4cosxsinπ4sinx)cos(x+π4)cosx)=2limxπ4(2(cos(x+π4))cos(x+π4)cosx)=2limxπ4(2cosx)=4

Question:50

Evaluate each of the following limits
limxπ(1sinx2cosx2(cosx4sinx4))

Answer:

limxπ(1sinx2cosx2(cosx4sinx4))=limxπ(1sinx2cosx2(cosx4sinx4))=limxπ(cos2x4+sin2x42sinx4cosx4(cos2x4sin2x4)(cosx4sinx4))
=limxπ((cosx4sinx4)2(cos2x4sin2x4)(cosx4sinx4))=limxπ((cosx4sinx4)(cos2x4sin2x4))=limxπ(1(cosx4+sinx4))=112+12=12

Question:51

Evaluate each of the following limits
Show that limx4|x4|x4
does not exist.

Answer:

limx4|x4|x4LHL=limx4x+4x4=1RHL=limx4+x4x4=1 LHLRHL

Hence, the limit doesnot exist

Question:52

Let f(x)=kcosxπ2x when xπ2and
f(x)=3 if limxπ2f(x)=f(π2) find the value of k.

Answer:


LHL=limxπ2(kcos(π2h)π2(π2h))=limh0(kcos(π2h)2h)=limh0(ksinh2h)=k2RHL=limxπ2+(kcos(π2+h)π2(π2+h))=limh0+(kcos(π2+h)2h)
=limh0(ksinh2h)=k2limxπ2f(x)=k2=3k=6

Question:53

Evaluate each of the following limits
Let f(x)=x+2x1cx2x>1 find ‘c’ if limx1f(x) exists.

Answer:

LHL=limx1(x+2)=1+2=1RHL=limx1+cx2=cLHL=RHLc=1

Question:54

Choose the correct answer out of 4 options given against each Question
limxπsinxxπ is

A. 1
B. 2
C. –1
D. –2

Answer:

limxπsinxxπ=limxπsin(πx)(πx)=1 limx0sinxx=1   πx0   xπ

Hence, the answer is option C

Question:55

Choose the correct answer out of 4 options given against each Question
limx0x2cosx1cosx  is

A. 2
B. 32
C. 32
D. 1

Answer:

limx0x2cosx1cosx=limx0x2cosx2sin2x2=limx0x2cosx2x24sin2x2x24=limx0cosx24sin2x2x24=2
Hence, the answer is option A

Question:56

Choose the correct answer out of 4 options given against each Question
limx0(1+x)n1x  is

A. n
B. 1
C. –n
D. 0

Answer:

limx0(1+x)n1x=limx0(1+x)n1n(1+x)1=n(1)n1=n
Hence, the answer is option A

Question:57

Choose the correct answer out of 4 options given against each Question
limx1xm1xn1 is

A. 1

B. mn
C. mn
D. m2n2

Answer:

limx1xm1xn1=limx1xm1x1xn1x1=m(1)m1n(1)n1=mn
Hence, the answer is option B

Question:58

Choose the correct answer out of 4 options given against each Question

limθ0(1cos4θ1cos6θ)is


A.49
B.12
C.12
D.1

Answer:

limθ0(1cos4θ1cos6θ)=limθ0(2sin22θ2sin23θ)=limθ0(sin2θsin3θ)2=limθ0(sin2θ2θ2θ(sin3θ3θ)3θ)2=49
Hence, the answer is option A

Question:59

Choose the correct answer out of 4 options given against each Question
limx0(cosecxcotxx) is

A. 12

B. 1
C. 12
D. –1

Answer:

limx0(cosecxcotxx)=limx0(1cosxxsinx)=limx0(2sin2x/22xsinx/2cosx/2)=limx0(sinx/2xcosx/2)=12
Hence, the answer is option C

Question:60

Choose the correct answer out of 4 options given against each Question
limx0(sinxx+11x) is

A. 2
B. 0
C. 1
D. –1

Answer:

limx0(sinxx+11x)=limx0(sinx(x+1+1x)(x+11x)(x+1+1x))=limx0(sinx(x+1+1x)(x+11+x))=limx0(sinx(x+1+1x)2x)=1
Hence, the answer is option C

Question:61

Choose the correct answer out of 4 options given against each Question

limxπ4(sec2x2tanx1) is
A. 3
B. 1
C. 0
D. 2

Answer:

limxπ4(sec2x2tanx1)=limxπ4(1+tan2x2tanx1)=limxπ4(tan2x1tanx1)=limxπ4(tanx+1)=2
Hence, the answer is option D

Question:62

Choose the correct answer out of 4 options given against each Question
limx1((x1)(2x3)2x2+x3) is

A. 110
B. 110
C. 1
D. None of these

Answer:

limx1((x1)(2x3)2x2+x3)=limx1((x1)(2x3)(2x+3)(x1))=limx1((x1)(2x3)(2x+3)(x1)(x+1))=limx1((2x3)(2x+3)(x+1))=110
Hence, the answer is option B

Question:63

Choose the correct answer out of 4 options given against each Question
If f(x)=sin[x][x],[x]00,[x]=0 where [.] denotes the greatest integer function, then limx0f(x) is equal to

A. 1
B. 0
C. –1
D. None of these

Answer:

LHL=limx0(sin[x][x])=limh0(sin[0h][0h])=limh0(sin[h][h])=limh0(sin(1)1)=sin1
RHL=limx0+(sin[x][x])=limh0+(sin[0+h][0+h])=limh0+(sin[h][h])=limh0+(sin(0)0)
Limit doesn’t exist
Hence, the answer is option D

Question:64

Choose the correct answer out of 4 options given against each Question

limx0(|sinx|x)  is

A. 1
B. –1
C. does not exist
D. None of these

Answer:

LHL=limx0(|sinx|x)=limh0(|sin(0h)|(0h))=limh0(sin(h)h)=1RHL=limx0+(|sinx|x)=limh0+(|sin(0+h)|(0+h))=limh0+(sin(h)h)=1
 So, limit doesnt existsHence, the answer is option C

Question:65

Choose the correct answer out of 4 options given against each Question
Let f(x)=x21,0<x<22x+3,2x<3 the quadratic equation whose roots are limx2f(x) and limx2f(x) is

A.x\textsuperscript26x+9=0B.x\textsuperscript27x+8=0C.x\textsuperscript214x+49=0D.x\textsuperscript210x+21=0

Answer:

LHL=limx2f(x)=limx2(x21)=3RHL=limx2+f(x)=limx2+(2x+3)=7The quadratic equation whose roots are 3 and 7 are (x3)(x7)=x210x+21
Hence, the answer is option D

Question:66

Choose the correct answer out of 4 options given against each Question
limx0(tan2xx)3xsinx  is

A. 2
B. 12
C. 12
D. 14

Answer:

limx0(tan2xx)3xsinx=limx0(tan2xx1)(3sinxx)=limx0(2 tan2x2x1)(3sinxx)=2131=12
Hence, the answer is option B

Question:67

Choose the correct answer out of 4 options given against each Question
Let f(x) = x – [x], R, then f12 is

A. 3/2
B. 1
C. 0
D. –1

Answer:

LHD=limh0(f(12h)f(12))h=limh0((12h)[(12h)](12)+[12])h=limh0hh=1RHD=limh0(f(12+h)f(12))h=limh0((12+h)[(12+h)](12)+[12])h=limh0hh=1
Hence, the answer is option B

Question:68

Choose the correct answer out of 4 options given against each Question
If Ify=x+1x,thendydx at x = 1 is

A. 1
B. 12
C. 12
D. 0

Answer:

y=x+1xdydx=12x1x2(dydx)x=1=121=12

Hence, the answer is option D

Question:69

Choose the correct answer out of 4 options given against each Question
If  f(x)=x42x then f’(1) is

A. 54
B. 45
C. 1
D. 0

Answer:

f(x)=x42xf(x)=12[x1(x4)12xx]=12[x+42x32]f(1)=12[1+42(1)]=54
Hence, the answer is option A

Question:70

Choose the correct answer out of 4 options given against each Question

If y=1+1x211x2 then dydx is

A.4x(x21)2
B.4xx21
C.1x24x
D.4xx21

Answer:

y=1+1x211x2=x2+1x21dydx=(x21)(2x)(x2+1)(2x)(x21)2=(2x)(x21x21)(x21)2=4x(x21)2
Hence, the answer is option A

Question:71

Choose the correct answer out of 4 options given against each Question

If y=sinx+cosxsinxcosx then dydxatx=0 is


A. –2
B. 0
C. 12
D. does not exist

Answer:

y=sinx+cosxsinxcosxdydx=(sinxcosx)(cosxsinx)(sinx+cosx)(sinx+cosx)(sinxcosx)2=(sin2x+cos2x2sinxcosx)(sin2x+cos2x+2sinxcosx)(sinxcosx)2=2(sinxcosx)2(dydx)x=0=2(1)2=2
Hence, the answer is option A

Question:72

Choose the correct answer out of 4 options given against each Question

If y=sin(x+9)cosx then dydx at x=0 is

A. cos 9
B. sin 9
C. 0
D. 1

Answer:

y=sin(x+9)cosxdydx=cosxcos(x+9)sin(x+9)(sinx)cos2x=cosxcos(x+9)+sin(x+9)sinxcos2x=cos(x+9x)cos2x=cos9cos2x(dydx)x=0=cos9(1)2=cos9
Hence, the answer is option A

Question:73

Choose the correct answer out of 4 options given against each Question

If f(x)=1+x+x22+x33++x100100 then f’(1) is equal to

A. 1/100
B. 100
C. does not exist
D. 0

Answer:

f(x)=1+x+x22+x33++x100100f(x)=0+1+2x2+3x23++100x99100f(x)=0+1+x+x2+x99f(1)=1+1+1++1  (100 times)=100
Hence, the answer is option B

Question:74

Choose the correct answer out of 4 options given against each Question

If f(x)=xnanxa for some constant ‘a’, then f’(a) is

A. 1
B. 0
C. does not exist
D. 1/2

Answer:

f(x)=xnanxaf(x)=(xa)(nxn1)(xnan)(1)(xa)2f(a)=(aa)(nan1)(anan)(1)(aa)2=00
Hence, the answer is option B

Question:75

Choose the correct answer out of 4 options given against each Question
If f(x)=x\textsuperscript100+x\textsuperscript99+ \ldots x+1,, then f’(1) is equal to

A. 5050
B. 5049
C. 5051
D. 50051

Answer:

f(x)=x100+x99++x+1f(x)=100x99+99x98++1+0f(1)=100+99+98++2+1=1001012=5050
Hence, the answer is option A

Question:76

Choose the correct answer out of 4 options given against each Question
If f(x)=1x+x\textsuperscript2x\textsuperscript3 \ldots x\textsuperscript99+x\textsuperscript100, then f’(1) is equal to

A. 150
B. –50
C. –150
D. 50

Answer:

f(x)=1x+x2x3+x99+x100f(x)=01+2x3x2+99x98+100x99f(1)=1+23+499+100=(21)+(43)+(65)+(10099)=1+1++1  (50 times)=50
Hence, the answer is option D

Question:77

Fill in the blanks
If f(x)=tanxxπ,limxπf(x)=

Answer:

limxπtanxxπ=limxπtan(πx)(πx)=limπx0tan(πx)(πx)=1

Question:78

Fill in the blanks
limx0sinmxcotx3=2 then m = ________

Answer:

limx0sinmxcotx3=2limx0mx(sinmxmx)(x3tanx3)(3x)=2limx0mx(3x)=23m=2m=23=233

Question:79

Fill in the blanks
If y=1+x1!+x22!+x33!+ then dydx= ........

Answer:

y=1+x1!+x22!+x33!+dydx=0+1+2x2!+3x23!+4x34!+dydx=1+x1!+x22!+x33!+dydx=y

Question:80

Fill in the blanks
limx3+x[x]=...............

Answer:

limx3+x[x]=limh0+3+h[3+h]=limh0+3+h3=33=1

More About NCERT Exemplar Solutions for Class 11 Maths Chapter 13

NCERT Exemplar Class 11 Maths chapter 13 solutions is a great way for students to learn and understand the concept of limits and derivatives through easier methods prescribed by the experts and develop their base for the same.

NCERT Exemplar Class 11 Maths solutions chapter 13 PDF download is useful for students to read offline. Use an online webpage to PDF tool for this. These solutions make learning more convenient and includes the detailed study of methods and guidelines of CBSE and NCERT.

The NCERT Exemplar solutions for Class 11 Maths chapter 13 will provide access to efficient and carefully drafted solutions to students to aid preparation and learning process for a better outcome.

Topics and Subtopics in Class 11 Maths NCERT Exemplar Solutions Chapter 13

  • 13.1 Introduction
  • 13.2 Intuitive Idea of Derivatives
  • 13.3 Limits
  • 13.3.1 Algebra of Limits
  • 13.3.2 Limits of polynomials and rational functions
  • 13.4 Limits of Trigonometric function
  • 13.5 Derivatives
  • 13.5.1 Algebra of Derivative of functions
  • 13.5.2 Derivatives of polynomials and trigonometric functions
  • 13.6 Miscellaneous Examples

What will the Students Learn in NCERT Exemplar Class 11 Maths Chapter 13 solutions?

  • NCERT Exemplar Class 11 Maths solutions chapter 13 covers the really important topic of limits and derivatives of any function which is a very important concept for mathematics as well as physics.

  • The students will learn about calculating the limits of any function which is important for calculus and mathematical analysis that are further used to define integrals, derivatives, and continuity of different functions.

NCERT Exemplar Class 11 Maths Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Maths Solutions Chapter 13

Some of the important topics for students to review are as follows:

  • The students will learn about the limits of different functions from NCERT exemplar solutions for Class 11 Maths chapter 13

  • The students will be able to define the limits and derivatives of different trigonometric, polynomial and rational number functions.

  • The NCERT exemplar Class 11 Maths solutions chapter 13 covers various solved examples along with solutions for better understanding and learning of different concepts.

  • The students should practice the application of different formulas provided in the chapter along with solutions and solved examples, take help from Class 11 Maths NCERT exemplar solutions chapter 13.

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Frequently Asked Questions (FAQs)

1. Where can we download the solutions to this chapter?

NCERT exemplar Class 11 Maths solutions chapter 13 pdf download can be accessed by using the webpage to PDF tool.

2. Is the chapter important from the perspective of Board and competitive exams?

Yes, the chapter has prominent weightage in both Board and competitive exams and helps to understand the basics terms and methods important from an examination perspective.

3. Which are the important topics that students must cover in this chapter?

The students must cover limits for trigonometric functions and also for polynomials. Derivatives also make a base for future learning in this NCERT exemplar Class 11 Maths chapter 13 solutions.

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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