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  • Maths Class 11

Limits and Derivatives, Class 11 Mathematics R.D Sharma Question Answer

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LIMITS

Page 29.11 Ex 29.1

Q1.

Answer :

limx→0 xx

Left hand limit:
limx→0- xx Let x=0-h, where h→0.⇒limh→0 0-h0-h=limh→0 -hh=-1

Right hand limit:
limx→0+ xxLet x=0+h, where h→0.limh→0 0+h0+h=limh→0 hh=1

Left hand limit ≠ Right hand limit
Thus,lim x→0 xx does not exist.

Q2.

Answer :

fx=2x+3,x≤2x+k,x>2Left hand limit:limx→2- fx=limx→2- 2x+3Let x=2-h, where h→0.limh→0 22-h+3=22-0+3=7Right hand limit:limx→2+ fx=limx→2+ x+kLet x=2+h, where h→0.limh→0 2+h+k=2+k

Now, limx→2 fx exists if the left hand limit is equal to the right hand limit.

⇒7 = 2 + k
k = 5

Q3.

Answer :

limx→0 1xLeft hand limit:limx→0- 1xLet x=0-h, where h→0limh→0 10-h=-∞Right hand limit:limx→0+ 1xLet x=0+h, where h→0limh→0 10+h=∞limx→0- 1x≠limx→0+ 1xThus, limx→0 1x does not exist.

Q4.

Answer :

limx→0 e-1xLeft hand limit:limx→0- e-1xLet x=0-h, where h→0.⇒limh→0 e-10-h=limh→0 e1h=limh→0 e1h When h→0, then 1h→∞.=e∞=∞Right hand limit:limx→0+ e-1xLet x=0+h, where h→0.limh→0 e-10+h=limh→0 1e1h=1∞=0 limx→0- e-1x≠limx→0+ e1xThus, limitx→0 e-1x does not exist.

Q5.

Answer :

limx→2- xxLet x=2-h, where h→0.limh→0 2-h2-h=21limx→2+ xxLet x=2+h, where h→0.limh→0 2+h2+h=22=1∴ limx→2- xx≠limx→2+xx

Q6.

Answer :

fx=3xx+2x,x≠0 0,x=0Left hand limit:limx→0- 3xx+2xLet x=0-h, where h→0.⇒limh→0 3-h-h+2-h=limh→0 -3hh-2h=limh→0 -3h-h=3Right hand limit:limx→0+ 3xx+2xLet x=0+h, where h→0.⇒limh→0 3hh+2h=limh→0 3hh+2h=1 limx→0- 3xx+2x≠limx→0+ 3xx+2xThus, limx→0 fx does not exist.

Q7.

Answer :

fx=x-x,x<24,x=23x-5,x>2LHL:limx→2- fx=limx→2- x-xLet x=2-h, where h→0.⇒limh→0 2-h-2-h=2-1=1RHL:limx→2+ fx=limx→2+ 3x-5Let x=2+h, where h→0.⇒limh→0 32+h-5=6-5=1Here, LHL=RHL=1∴limx→2fx=1

Q8.

Answer :

limx→0 sin 1xL.H.Llimx→0- fxlimx→0- sin 1xLet x=0-h where h→0=limh→0 sin -1h=-limh→0 sin 1h

limx→0- fx = – (An oscillating number that oscillates between –1 and 1)

Thus,limx→0-fx does not exist.Similarly, limx→0+fx does not exist.

Q9.

Answer :

limx→52 xLHL:limx→52- xLet x=52-h, where h→0.limh→0 52-h=2RHL:limx→52+ xLet x=52+h, where h→0.limh→0 52+h=2

We know:
limx→52x=limx→52-x=limx→52+x

∴limx→52x=2

Q10.

Answer :

limx→3+ xxLet x=3+h, where h→0.⇒limh→0 3+h3+h=33=1Also, limx→3- xxLet x=3-h, where h→0.limh→0 3-h3-h=32∴limx→3- xx≠limx→3+ xx

Q11.

Answer :

fx=x+1,x≥0x-1,x<0RHL:limx→0+ fx=limx→0 x+1Let x=0+h, where h→0.limh→0 0+h+1=1LHL:limx→0- fx=limx→0- x-1Let x=0-h, where h→0.limh→0 0-h-1=-1 LHL≠RHLThus,limx→0 fx does not exist.

Q12.

Answer :

limx→1+ 1x-1Let x=1+h, where h→0.limh→0 11+h-1=∞

Q13.

Answer :

limx→a+ xLet x=a+h, where h→0.limh→0 a+h=alimx→1- xLet x=1-h, where h→0.limh→0 1-h=0

Q14.

Answer :

(i)

limx→2 xLHL:limx→2- xLet x=2-h, where h→0.limh→0 2-h=1RHL:limx→2+ xLet x=2+h, where h→0.limh→0 2+h=2 LHL ≠RHLThus, lim x→2 x does not exist.
(ii)
limx→52 xLHL limx→52- xLet x=52-h, where h→0.limh→0 52-h=2RHL:limx→52+ xLet x=52+h, where h→0.⇒limh→0 52+h ∴limx→52 x=2=2∴limx→52 x=2

(iii)
limx→1 xLHL:limx→1- xLet x=1-h, where h→0.limh→0 1-h=0RHL:limx→1+ xLet x=1+h, where h→0.limh→0 1+h=1 LHL ≠RHLThus,lim x→1 x does not exist.

Q15.

Answer :

(i)
limx→2+ x-3×2-4Let x=2+h, where h→0.⇒limh→0 2+h-32+h2-22=limh→0 -1+h2+h-2 2+h+2=-∞

(ii)
limx→2- x-3×2-4Let x=2-h, where h→0.⇒limh→0 2-h-32-h2-22=limh→0 -h-12-h-2 2-h+2=limh→0 -h-1-h 4-h=limh→01+hh4-h=∞

(iii)
limx→0+ 13xLet x=0+h, where h→0.⇒limh→0 13h=∞

(iv)
limx→-8+ 2xx+8Let x=-8+h, where h→0.⇒limh→0 2-8+h-8+h+8=limh→0 -16+2hh=-∞

(v)
limx→0+ 2x15Let x=0+h, where h→0.⇒limh→0 2h15=∞

(vi)
limx→π2- tan xLet x=π2-h, where h→0.⇒limh→0 tanπ2-h=limh→0 cot h=∞

(vii)
limx→-π2+ sec xLet x=-π2+h, where h→0.limh→0 sec-π2+h=limh→0 sec π2-h sec-θ=sec θ=limh→0 cosec h=∞

(viii)
limx→0- x2-3x+2×3-2×2=limx→0- x2-2x-x+2x2x-1=limx→0- xx-2-1x-2×2 x-2=limx→0- x-1 x-2x2x-2Let x=0-h, where h→0.⇒limh→0 -h-1-h2=-∞

(ix)
limx→-2+ x2-12x+4Let x=-2+h, where h→0.limh→0 -2+h2-12-2+h+4=4-1-4+4=30=∞

(x)
limx→0- 2-cot xLet x=0-h, where h→0.limh→0 2-cot -h=limh→0 2+cot h=2+∞=∞

(xi)
limx→0- 1+cosec xLet x=0-h, where h→0.limh→0 1+cosec -h=limh→0 1-cosec h cosec -θ=-cosec θ=1-∞=-∞

Page 29.12 Ex 29.1

Q16.

Answer :

fx=x+1,x≥0x-1,x<0LHL:limx→0- fx=limx→0- x-1Let x=0-h, where h→0.⇒limh→0 0-h-1=-1RHL:limx→0+ fx=limx→0+ x+1Let x=0+h, where h→0.⇒limh→0 0+h+1=+1
LHL ≠ RHL
Thus, limx→0 fx does not exist.

Q17.

Answer :

limx→3 fx=?fx=4,x≥3x+1,x<3LHL:limx→3- fx=limx→3- x+1Let x=3-h, where h→0.limh→0 3-h+1=4RHL:limx→3+ fx=limx→3+ 4=4∴LHL=RHLlimx→3 fx=4

Q18.

Answer :

fx=2x+3,x≤03x+1,x>0LHL:limx→0- fx=limx→0- 2x+3Let x=0-h, where h→0.limh→0 20-h+3=limh→0 -2h+3=3RHL:limx→0+ fxlimx→0+ 3x+1Let x=0+h, where h→0.limh→0 30+h+1=limh→0 3h+3=3∴LHL=RHL∴limx→0 fx=3

(ii)
limx→1 fx=?LHL:limx→1- fx=limx→1- 3x+1Let x=1-h, where h→0.limh→0 31-h+1=6RHL:limx→1+ fx=limx→1+ 3x+1Let x=1+h, where h→0.limh→0 31+h+1=6∴LHL=RHLlimx→1 fx=6

Q19.

Answer :

limx→1 fx=?fx=x2-1,x≤1-x2-1,x>1LHL:limx→1- fx=limx→1- x2-1Let x=1-h, where h→0.limh→0 1-h2-1=0RHL:limx→1+ fxlimx→1+ -x2-1Let x=1+h, where h→0.limh→0 -1+h2-1=-2 LHL≠RHLThus, limx→1 fx does not exist.

Q20.

Answer :

fx=x-a1 x-a2 … x-anlimx→a1 fx=limx→a1 x-a1 x-a2 … x-an=a1-a1 a1-a2 … a1-an=0limx→a fx=limx→a x-a1 x-a2 … x-an=a-a1 a-a2 … a-an

Q21.

Answer :

fx=xx,x≠00,x=0LHL:limx→0- xxLet x=0-h, where h→0.⇒limh→0 0-h-h=limh→0 h-h=-1RHL:limx→0+ fx=limx→0+ xxLet x=0+h, where h→0.limh→0 0+h0+h=1LHL≠RHLThus, limx→0 fx does not exist.

Page 29.16 Ex 29.2

Q1.

Answer :

limx→1×2+1x+1=12+11+1=1

Q2.

Answer :

limx→02×2+3x+4×2+3x+2=2×0+3×0+40+3×0+2=42=2

Q3.

Answer :

limx→32x+3x+3=2×3+33+3=36=12

Q4.

Answer :

limx→1x+8x=1+81=3

Q5.

Answer :

limx→ax+ax+a=a+aa+a=2a2a=1a

Q6.

Answer :

limx→11+x-121+x2=1+1-121+12=12

Q7.

Answer :

limx→0x2/3-9x-27=0-90-27=13

Q8.

Answer :

limx→09=9

f(x) = 9 is a constant function.
Its value does not depend on x.

Q9.

Answer :

limx→23-x=3-2=1

Q10.

Answer :

limx→-14×2+2=4-12+2=4+2=6

Q11.

Answer :

limx→-1×3-3x+1x-1=-13-3-1+1-1-1=-1+3+1-2=-32

Q12.

Answer :

limx→03x+1x+3=3×0+10+3=13

Q13.

Answer :

limx→3×2-9x+2=32-93+2=9-95=0

Q14.

Answer :

limx→0ax+bcx+d=a×0+bc×0+d=bd

Page 29.21 Ex 29.3

Q1.

Answer :

limx→-52×2+9x-5x+5It is of the form 00.limx→-52×2+10x-x-5x+5=limx→-52xx+5-1x+5x+5=limx→-52x-1x+5x+5=limx→-52x-1=2-5-1=-11

Q2.

Answer :

limx→3×2-4x+3×2-2x-3It is of the form 00.limx→3×2-x-3x+3×2-3x+x-3=limx→3x-3x-1x+1x-3=limx→3x-1x+1=3-13+1=12

Q3.

Answer :

limx→3×4-81×2-9It is of the form 00.limx→3×22-92×2-9=limx→3×2-9×2+9×2-9=limx→3×2+9=32+9=18

Q4.

Answer :

limx→2×3-8×2-4It is of the form 00.limx→2x-2×2+2x+4x-2x+2 ∵A3-B3=A-BA2+AB+B2∵A2-B2=A-BA+B=22+2×2+42+2=124=3

Q5.

Answer :

limx→-1/28×3+12x+1It is of the form 00.limx→-1/22×3+12x+1=limx→-1/22x+12×2-2x×1+122x+1 ∵A3+B3=A+BA2-AB+B2=limx→-1/22×2-2x+1=2×-122-2×-12+1=1+1+1=3

Q6.

Answer :

limx→4×2-7x+12×2-3x-4It is of the form 00.limx→4×2-3x-4x+12×2-4x+x-4=limx→4xx-3-4x-3xx-4+1x-4=limx→4x-4x-3x-4x+1=4-34+1=15

Q7.

Answer :

limx→3×2-x-6×3-3×2+x-3It is of the form 00.limx→3×2-3x+2x-6x2x-3+1x-3=limx→3xx-3+2x-3×2+1x-3=limx→3x+2x-3×2+1x-3=3+232+1=510=12

Q8.

Answer :

Let p(x) = x3 + x2 + 4x + 12
p(–2) = 0
Thus, x = –2 is the root of p(x).
Now, x+2 is a factor of p(x).

x2 – x + 6x+2×3 + x2 + 4x + 12 x3 + 2×2 – – -x2 + 4x + 12 -x2 – 2x + + 6x + 12 6x + 12 – – 0

p(x) = x3 + x2 + 4x + 12
= (x + 2)(x2 – x + 6)

Let q(x) = x3 – 3x + 2
q-2 = -8 + 6 + 2
= 0
Thus, x = -2 is the root of q(x).
Now, x+2 is a factor of q(x).

x2-2x+1x+2×3-3x+2 x3+2×2 – – -2×2-3x+2 -2×2-4x + + x+2 x+2 – – ×××

q(x) = (x + 2)(x2 – 2x + 1)

⇒limx→-2×3+x2+4x+12×3-3x+2=limx→-2x+2×2-x+6x+2×2-2x+1=(–2)2–2+6-22-2-2+1=4+2+64+4+1=129=43

Q9.

Answer :

Let p(x) = x3 + 3×2 – 6x + 2
p(1) = 1 + 3 – 6 + 2
= 0
Now, x+2 is a factor of p(x).

x2+4x-2x-1×3+3×2-6x+2 x3-x2 – + -4×2-6x+2 -4×2-4x + + – 2x+2 – 2x+2 + – 0

p(x) = (x – 1)(x2 + 4x – 2)

q(x) = x3 + 3×2 – 3x + 2
q(1) = 1 + 3 – 3 – 1
= 0
Now, x+2 is a factor of p(x).

x2+4x+1 x-1×3+3×2-3x-1 x3-x2 – + 4×2-3x-1 4×2-4x – + x-1 x-1 – + 0

⇒limx→1×3+3×2-6x+2×3+3×2-3x-1=limx→1x-1×2+4x-2x-1×2+4x+1=(1)2+4×1-212+4×1+1=1+4-21+4+1=36=12

Q10.

Answer :

limx→5×3-125×2-7x+10It is of the form 00.limx→5×3-53×2-2x-5x+10=limx→5x-5×2+5x+52xx-2-5x-2 ∵A3-B3=A-BA2+AB+B2=limx→5x-5×2+5x+25x-2x-5=52+5×5+255-2=753=25

Q11.

Answer :

limx→2×2-2×2+2x-4It is of the form 00.limx→2×2-22×2+22x-2x-4=limx→2x-2x+2xx+22-2x+22=limx→2x-2x+2x-2x+22=2+22+22=2232=23

Q12.

Answer :

limx→3×2-3×2+33x-12It is of the form 00.limx→3×2-32×2+43x-3x-12=limx→3x-3x+3xx+43-3x+43=limx→3x-3x+3x-3x+43=3+33+43=25

Q13.

Answer :

limx→3×4-9×2+43x-15It is of the form 00.limx→3×22-32×2+53x-3x-15=limx→3×2-3×2+3xx+53-3x+53=limx→3×2-32×2+3x-3x+53=limx→3x-3x+3×2+3x-3x+53=3+33+33+53=23×663=2

Q14.

Answer :

limx→2xx-2-4×2-2x=limx→2xx-2-4xx-2=limx→2×2-4xx-2=limx→2x-2x+2xx-2=limx→2x+2x=2+22=2

Q15.

Answer :

limx→1×3-1-xx2+x-2×2+x-2×3-1=limx→1×3-1-x3-x2+2xx2+x-2×3-1=limx→1-x2+2x-1×2+x-2x-1×2+x+1=limx→1-x2-2x+1×2+x-2x-1×2+x+1=limx→1-x-12×2+x-2x-1×2+x+1=limx→1-x-1×2+2x-x-2×2+x+1=limx→1-x-1xx+2-1x+2×2+x+1=limx→1-x-1x-1x+2×2+x+1=-11+21+1+1=-19

Q16.

Answer :

limx→31x-3-2×2-4x+3=limx→31x-3-2×2-3x-x+3=limx→31x-3-2xx-3-1x-3=limx→31x-3-2x-1x-3=limx→3x-1-2x-3x-1=limx→31x-1=13-1=12

Q17.

Answer :

limx→21x-2-2×2-2x=limx→21x-2-2xx-2=limx→2x-2xx-2=limx→21x=12

Q18.

Answer :

limx→1/44x-12x-1It is of the form 00.limx→1/42×2-122x-1=limx→1/42x-12x+12x-1=214+1=2×12+1=2

Q19.

Answer :

limx→4×2-16x-2It is of the form 00.limx→4×2-42x-2=limx→4x-4x+4x-2=limx→4×2-22x+4x-2=limx→4x-2x+2x+4x-2=2+24+4=32

Q20.

Answer :

limx→0a+x2-a2xIt is of the form 00.limx→0a2+x2+2ax-a2x=limx→0xx+2ax=limx→0x+2a=0+2a=2a

Q21.

Answer :

limx→21x-2-4×3-2×2=limx→21x-2-4x2x-2=limx→2×2-4x2x-2=limx→2x-2x+2x2x-2=limx→2x+2×2=2+222=1

Q22.

Answer :

limx→31x-3-3×2-3x=limx→31x-3-3xx-3=limx→3x-3xx-3=limx→31x=13

Q23.

Answer :

limx→11x-1-2×2-1=limx→11x-1-2x-1x+1=limx→1x+1-2x-1x+1=limx→1x-1x-1x+1=limx→11x+1=11+1=12

Q24.

Answer :

limx→3×2-91x+3+1x-3=limx→3×2-9x-3+x+3x+3x-3=limx→3×2-92xx2-9=limx→32x=2×3=6

Q25.

Answer :

p(x) = x4 – 3x3 + 2
p(1) = 1 – 3 + 2
= 0
Now, x-1 is a factor of p(x).

x3-2×2-2x-2x-1×4-3×3+2 x4-x3 – + -2×3+2 -2×3+2×2 + – – 2×2+2 – 2×2+2x + – -2x+2 -2x+2 + – ___________ 0

q(x) = x3 – 5×2 + 3x + 1
q(1) = 1 – 5 + 3 + 1
= 0
Now, x-1 is a factor of q(x).

x2-4x-1x-1×3-5×2+3x+1 x3-x2 – + 4×2+3x+1 4×2+4x – – -x+1 -x+1 + – 0

⇒limx→1×4-3×3+2×3-5×2+3x+1=limx→1x-1×3-2×2-2x-2x-1×2-4x-1=(1)3-212-21-212-4×1-1=1-2-2-21-4-1=-5-4=54

Page 29.22 Ex 29.3

Q26.

Answer :

limx→2×3+3×2-9x-2×3-x-6

It is of the form 00.

Let p(x) = x3 + 3x2 – 9x – 2
p(2) = 8 + 12 – 18 – 2
= 0
Now, x-2 is a factor of p(x).

x2+5x+1x-2×3+3×2-9x-2 x3-2×2 – + 5×2-9x-2 5×2-10x – + x-2 x-2 – + 0

Let q(x) = x3 – x – 6
q(2) = 8 – 2 – 6
= 0
Now, x-2 is a factor of q(x).

x2+2x+3x-2×3-x-6 x3-2×2 – + 2×2-x-6 2×2-4x – + 3x-6 3x-6 – + ×××

⇒limx→2×3+3×2-9x-2×3-x-6=limx→2x-2×2+5x+1x-2×2+2x+3=limx→2×2+5x+1×2+2x+3=(2)2+5×2+122+2×2+3=4+10+14+4+3=1511

Q27.

Answer :

limx→11-x-1/31-x-2/3It is of the form 00.limx→11-x-1/312-x-1/32=limx→11-x-1/31-x-131+x-1/3=limx→111+x-1/3=11+1=12

Q28.

Answer :

limx→2×4-16x-2It is of the form 00.limx→2×22-42x-2=limx→2×2-4×2+4x-2=limx→2x-2x+2×2+4x-2=2+222+4=32

Q29.

Answer :

limx→5×2-9x+20×2-6x+5It is of the form 00.limx→5×2-4x-5x+20×2-x-5x+5=limx→5xx-4-5x-4xx-1-5x-1=limx→5x-5x-4x-1x-5=5-45-1=14

Q30.

Answer :

limx→-1×3+1x+1It is of the form 00.limx→-1×3+13x+1=limx→-1x+1×2-x+1x+1=-12–1+1=1+1+1=3

Q31.

Answer :

limx→21x-2-22x-3×3-3×2+2x=limx→21x-2-22x-3xx2-3x+2=limx→21x-2–22x-3xx2-2x-x+2=limx→21x-2-22x-3xxx-2-1x-2=limx→21x-2-22x-3xx-1x-2=limx→2xx-1-22x-3xx-1x+2=limx→2×2-x-4x+6xx-1x+2=limx→2×2-5x+6xx-1x+2=limx→2×2-2x-3x+6xx-1x-2=limx→2xx-2-3x-2xx-1x-2=limx→2x-3x-2xx-1x-2=2-322-1=-12

Q32.

Answer :

limx→1×2-1+x-1×2-1It is of the form 00.limx→1×2-1×2-1+x-1×2-1=limx→11+x-1x-1x+1=1+11+1=1+12=2+12

Q33.

Answer :

limx→1x-2×2-x-1×3-3×2+2x=limx→1x-2xx-1-1xx2-3x+2=limx→1x-2xx-1-1xx2-2x-x+2=limx→1x-2xx-1-1xxx-2-1x-2
=limx→1x-2xx-1-1xx-1x-2=limx→1x-22-1xx-1x-2

=limx→1x-1x-3xx-1x-2=1-311-2=-2-1=2

Page 29.26 Ex 29.4

Q1.

Answer :

limx→01+x+x2-1x

When x = 0, the expression 1+x+x2-1x takes the form 00.
Rationalising the numerator:

limx→01+x+x2-11+x+x2+1×1+x+x2+1=limx→01+x+x2-1×1+x+x2+1=limx→0x1+xx1+x+x2+1=1+01+0+0+1=12

Q2.

Answer :

limx→02xa+x-a-x

When x = 0, then the expression 2xa+x-a-x becomes 00.

Rationalising the denominator:

limx→02xa+x-a-x×a+x+a-xa+x+a-x

= limx→02xa+x+a-xa+x-a-x

= limx→02xa+x+a-x2x

= a+a
= 2a

Q3.

Answer :

limx→0a2+x2-ax2

On putting x = 0 in the expression a2+x2-a, it becomes 00.
Rationalising the numerator:

limx→0a2+x2-aa2+x2+ax2a2+x2+a

= limx→0a2+x2-a2x2a2+x2+a

= 1a2+a

= 12a

Q4.

Answer :

limx→01+x-1-x2x

It is of the form 00.

Rationalising the numerator:

limx→01+x-1-x1+x+1-x2x1+x+1-x

= limx→01+ x-1-x2x1+x+1-x

= limx→02x2x1+x+1-x

= 11+0+1-0
= 12

Q5.

Answer :

limx→23-x-12-x

It is of the form 00.
Rationalising the numerator:

limx→23-x-13-x+12-x3-x+1

=limx→23-x-12-x3-x+1

=limx→22-x2-x3-x+1

=13-2+1=11+1=12

Q6.

Answer :

limx→3x-3x-2-4-x

It is of form 00.

Rationalising the denominator:

limx→3x-3x-2+4-xx-2-4-xx-2+4-x

= limx→3x-3x-2+4-xx-2-4-x

= limx→3x-3x-2+4-x2x-6

= limx→3x-3x-2+4-x2x-3

= 3-2+4-32

= 1+12
= 22=1

Q7.

Answer :

limx→0x1+x-1-x

It is of the form 00.

Rationalising the denominator:

limx→0x1+x-1-x×1+x+1-x1+x+1-x

= limx→0x1+x+1-x1+x-1-x

= limx→0x1+x+1-x2x

= 1+12

= 22
= 1

Q8

Answer :

limx→15x-4-xx-1

It is of the form 00.
Rationalising the numerator:

limx→15x-4-xx-15x-4+x5x-4+x

= limx→15x-4-xx-15x-4+x

= limx→14x-1x-15x-4+x

= 45-4+1

= 42

= 2

Q9.

Answer :

limx→1x-1×2+3-2
It is of the form 00.
Rationalising the denominator:

limx→1x-1×2+3+2×2+3-2×2+3+2

= limx→1x-1×2+3+2×2+3-4

= limx→1x-1×2+3+2×2-1

= limx→1x-1×2+3+2x-1x+1

=1+3+21+1

= 42

= 2

Q10.

Answer :

limx→3x+3-16×2-9

It is of the form 00.
Rationalising the numerator:

limx→3x+3-6x+3+6×2-9x+3+6

= limx→3x+3-6x-3x+3x+3+6

= 16×26

= 1126

Q11.

Answer :

limx→15x-4-xx2-1

It is of the form 00.
Rationalising the numerator:

limx→15x-4-x5x-4+x5x-4+xx2-1

= limx→15x-4-x5x-4+xx-1x+1

= limx→14x-15x-4+xx-1x+1

= 45-4+11+1

= 42×2

= 1

Q12.

Answer :

limx→01+x-1x

It is of the form 00.
Rationalising the numerator:

limx→01+x-11+x+1×1+x+1

=limx→01+x-1×1+x+1

= 11+0+1

= 12

Q13.

Answer :

limx→2×2+1-5x-2

It is of the form 00.
Rationalising the numerator:

limx→2×2+1-5×2+1+5x-2×2+1+5

= limx→2×2+1-5x-2×2+1+5

= limx→2×2-4x-2×2+1+5

= limx→2x-2x+2x-2×2+1+5

= 425

= 25

Q14.

Answer :

limx→107+2x-5+2×2-102

= limx→107+2x-5+22x-10x+10

= limx→107+2x-5+2+252x-10x+10

= limx→107+2x-7+210x-10x+10

Rationalising the numerator:

limx→107+2x-7+210 7+2x+7+210x-10x+10 7+2x+7+210

= limx→107+2x-7+210x-10x+10 7+2x+7+210

= limx→102x-10x-10x+10 7+2x+7+210

= 210+10 27+210
=2210×27+210=12105+22

= 12105+2×5-25-2

=12105-252-22

Q15.

Answer :

limx→74-9+x1-8-x

It is of the form 00.
Rationalising the numerator and the denominator:

limx→74-9+x 1×4+9+x4+9+x×11-8-x×1+8-x1+8-x

= limx→716-9+x4+9+x×1+8-x1-8-x

= limx→7-1-7+x1+8-x4+9+x-7+x

= limx→7-1+8-x4+9+x

= -1+8-74+9+7

= -24+4

= -14

Q16.

Answer :

limx→0a+x-axa2+ax

It is of the form 00.
Rationalising the numerator:

limx→0a+x-a × a+x+axa2+ax × a+x+a

= limx→0a+x-axa2+ax a+x+a

= 1a2 a+a

= 12aa

Q17.

Answer :

limx→5x-56x-5-4x+5

It is of the form 00.
Rationalising the denominator:

limx→5x-5 6x-5+4x+56x-5-4x+5 6x-5+4x+5

= limx→5x-5 6x-5+4x+56x-5-4x+5

= limx→5x-5 6x-5+4x+52x-10

= limx→5x-56x-5+4x+52x-5

= 6×5-5+4×5+52

= 5+52

= 5

Q18.

Answer :

limx→15x-4-xx3-1

It is of the form 00.
Rationalising the numerator:

limx→15x-4-x 5x-4+xx3-1 5x-4+x

= limx→15x-4-xx-1×2+x+15x-4+x

= limx→14x-1x-1×2+x+15x-4+x

= 431+1

= 23

Q19.

Answer :

limx→21+4x-5+2xx-2

It is of the from 00.
Rationalising the numerator:

limx→21+4x-5+2x 1+4x+5+2xx-2 1+4x+5+2x

= limx→21+4x-5+2xx-21+4x+5+2x

= limx→22x-2x-21+4x+5+2x

= 21+4×2+5+2×2

= 23+3

= 13

Q20.

Answer :

limx→13+x-5-xx2-1

It is of the form 00.
Rationalising the numerator:

limx→13+x-5-x3+x+5-xx-1x+13+x+5-x

= limx→13+x-5-xx-1x+13+x+5-x

= limx→12x-2x-1x+1 3+x+5-x

= limx→12x-1x-1x+13+x+5-x

= 22 4+4

= 14

Q21.

Answer :

limx→01+x2-1-x2x

It is of the form 00.
Rationalising the numerator:

limx→01+x2-1-x21+x2+1-x21+x2+1-x2x

= limx→01+x2-1-x2x1+x2+1-x2

= limx→02x2x1+x2+1-x2

= 2×01+0+1-0

= 0

Q22.

Answer :

limx→01+x+x2-x+12×2

It is of the form 00.
Rationalising the numerator:

limx→01+x+x2-x+11+x+x2+x+11+x+x2+x+12×2

= limx→01+x+x2-x+11+x+x2+x+12×2

= limx→0x21+x+x2+x+12×2

= 11+0+0+0+1×2

= 12×12

= 14

Q23.

Answer :

limx→65+2x-3+2×2-6

= limx→65+2x-3+22×2-62

= limx→65+2x-3+2+26x-6x+6

= limx→65+2x-5+26x-6x+6

Rationalising the numerator:

limx→65+2x-5+265+2x+5+26x-6x+65+2x+5+26

= limx→65+2x-5+26x-6x+65+2x+5+26

= limx→62x-6x-6x+65+2x+5+26

= 26+65+26+5+26

= 1263+22

= 1263+2

= 1263+2×3-23-2

= 3-2263-2

= 3-226

Q24.

Answer :

limh→0x+h-xh

It is of the form 00.
Rationalising the numerator:
= limh→0x+h-xh×x+h+xx+h+x

= limh→0x+h-xhx+h+x

= 1x+x

= 12x

Q25.

Answer :

limx→02-x-2+xx

It is of the form 00.
Rationalising the numerator:

limx→02-x-2+x2-x+2+xx2-x+2+x

= limx→02-x-2+xx2-x+2+x

= limx→0-2xx2-x+2+x

= -222

= -12

Q27.

Answer :

limx→42-x4-x

= limx→42-x2-x2

= limx→42-x2-x2+x

= 12+4

= 12+2

= 14

Q29.

Answer :

limx→2x-2x-2

It is of the form 00.

⇒limx→2×2-22x-2

= limx→2x-2x+2x-2

= 2+2

= 22

Q30.

Answer :

limx→1×2-xx-1

It is of the form 00.

limx→1xxx-1x-1

= limx→1xx3/2-1×12-1

= limx→1xx3-13x-1 A3-B3=A-BA2+AB+B2

= limx→1xx-1x+x+1x-1

= 1 (1 + 1 + 1)

= 3

Q31.

Answer :

limx→12x-3x-13×2+3x-6

It is of the form 00.

⇒ limx→12x-3x-13×2+x-2

= limx→12x-3x-13×2+2x-x-2

= limx→12x-3x-13xx+2-1x+2

= limx→12x-3x-13x-1x+2

= limx→12x-3x-13×2-12x+2

= limx→12x-3x-13x+1x-1x+2

= -132×3

= -118

Q32.

Answer :

limx→0 1-x2-1+x1+x3-1+x

It is of the form 00.
Rationalising the numerator and the denominator:

limx→01+x2-1+x1×1+x2+1+x1+x2+1+x×11+x3-1+x×1+x3-1+x1+x3+1+x=limx→0 1+x2-1+x1+x3-1+x×1+x3+1+x1+x2+1+x=limx→0x2-xx3-x×1+x3+1+x1+x2+1+x=limx→0xx-1xx2-1 1+x3+1+x1+x2+1+x=limx→0 x-1 1+x3+1+xx-1 x+1 1+x2+1+x=1+0+1+00+1 1+0+1+0=22=1

Q33.

Answer :

limx→1 3+x-5-xx2-1

It is of the form 00.

Rationalising the numerator:

limx→1 3+x-5-x 3+x + 5-xx-1 x+1 3+x+5-x=limx→13+x-5-xx-1 x+1 3+x +5-x=limx→12x-1x-1 x+1 3+x +5-x=21+1 3+1+5-1=22×2+2=14

Q34.

Answer :

limx→0 1+3x-1-3xx

It is of the form 00.

Rationalising the numerator:

limx→01+3x-1-3x 1+3x+1-3xx 1+3x+1-3x=limx→0 1+3x-1-3xx1+3x+1-3x=limx→0 6xx 1+3x+1-3x=61+1=62=3

Page 29.30 Ex 29.5

Q1.

Answer :

limx→a x+252-a+252x-a=limx→a x+252-a+252x+2-a+2

Let y = x + 2 and b = a + 2.

When x → a, then x + 2 → a + 2.
⇒ y → b

limy→by52-b52y-b=52b52-1=52b32=52a+232

Q2.

Answer :

limx→a=x+232-a+232x-a=limx→a x+232-a+232x+2-a+2

Let y = x + 2 and b = a + 2.

When x → a and x + 2 → a + 2.
⇒y → b

limy→b y32-b32y-b=32 b32-1=32b12=32 a+212

Q3.

Answer :

limx→0 1+x6-11+x2-1=limx→0 1+x6-1x×x1+x2-1=limx→0 1+x6-161+x-1×1+x-11+x2-1

Let y = 1 + x

When x → 0, then 1 + x → 1.
⇒y → 1

limy→1y6-16y-1×y-1y2-12=6×16-12×12-1=62=3

Q4.

Answer :

limx→a x27-a27x-a=27 a27-1 ∵limx→a xn-anx-a=nan-1=27a-57

Q5.

Answer :

limx→a x57-a57x27-a27=limx→a x57-a57x-a×x-ax27-a27=57 a57-1×127 a27-1=52a-27a57=52a-27+57=52a37

Q6.

Answer :

limx→-12 8×3+12x+1=limx→-12 2×3–12x–1When x→-12, then 2x → –1.
Let y = 2x

limy→-1 y3–13y–1=3-13-1=3×-12=3

Q7.

Answer :

limx→27 x13+3 x13-3x-27=limx→27 x13+3 x13-3×133-33 x→27∴ x13→3Let y=x13limy→3y+3 y-3y3-33=3+33×33-1=63×9=29

Q8.

Answer :

limx→4 x3-64×2-16=limx→4 x3-64x-4×x-4×2-16=limx→4 x3-43x-4×x-4×2-42=343-1×1242-1=3×162×4=6

Q9.

Answer :

limx→3 xn-3nx-3=108

⇒ x(3)n-1 = 108
⇒ x(3)n-1 = 4 × 33

On comparing LHS and RHS, we observe that x is equal to 4.

21082543273933 1

Q10.

Answer :

limx→-1 x3+1x+1=limx→-1 x3–1x–1=limx→-1 x3–13x–1=3-13-1=3

Q11.

Answer :

limx→a x23-a23x34-a34=lima→a x23-a23x-a×x-ax34-a34=23a23-1×134a34-1=89 a-13a-14=89a-13+14=89a-112

Q12.

Answer :

limx→1 x15-1×10-1=limx→1 x15-115x-1×x-1×10-110=15115-110110-1=32

Page 29.31 Ex 29.5

Q13.

Answer :

limx→a x9-a9x-a=9⇒9a8=9⇒a8=1⇒a=±1

Q14.

Answer :

limx→a x5-a5x-a=405⇒5a4=405⇒a4=81⇒a2=9⇒a=±3

Q15.

Answer :

limx→a x9-a9x-a=limx→5 4+x⇒9a9+1=9⇒a8=1⇒a=±1

Q16.

Answer :

limx→a x3-a3x-a=limx→1 x4-14x-1⇒3a3-1=414-1⇒3a2=4⇒a2=43⇒a=±23

Page 29.35 Ex 29.6

Q1.

Answer :

limx→∞3x-1 4x-2x+8 x-1Dividing the numerator and the denominator by x2:limx→∞ 3x-1x 4x-2xx+8x x-1x=limx→∞ 3-1x 4-2×1+8x 1-1xWhen x →∞, then 1x→0.3×41=12

Q2.

Answer :

limx→∞ 3×3-4×2+6x-12×3+x2-5x+7Dividing the numerator and the denominator by x3:limx→∞ 3-4x+6×2-1×32+1x-5×2+7x3When x→∞, then 1x, 1×2, 1×3→0.⇒32

Page 29.36 Ex 29.6

Q3.

Answer :

limx→∞ 5×3-69+4x6Dividing the numerator and the denominator by x:limx→∞5×3-6×39+4x6x3=limx→∞ 5-6x39x6+4=54=52

Q4.

Answer :

limx→∞ x2+cx-x

It is of the form ∞ – ∞.

Rationalising the numerator:

limx→∞x2+cx-x x2+cx+xx2+cx+x=limx→∞x2+cx-x2x2+cx+xDividing the numerator and the denomiator by x:limx→∞ cxxx2+cx+xx=limx→∞ c1+cx+1When x→∞, then 1x→0.⇒c1+1=c2

Q5.

Answer :

limx→∞ x+1-x

It is of the form ∞–∞.

On rationalising, we get:
limx→∞ x+1-x×x+1+xx+1+x=limx→∞ x+1-xx+1+x=1∞=0

Q6.

Answer :

limx→∞ x2+7x-xRationalising the numerator:limx→∞ x2+7x-x x2+7x+xx2+7x+x=limx→∞ x2+7x-x2x2+7x+xDividing the numerator and the denominator by x:limx→∞7×2+7xx+1=limx→∞ 7×2+7xx+1When x→∞, then 1x→0.⇒71+1=72

Q7.

Answer :

limx→∞ x4x2+1-1Rationalising the denominator:limx→∞x4x2+1-1 4×2+1+14×2+1+1=limx→∞ x4x2+1+14×2+1-1=limx→∞4×2+1+14xDividing the numerator and the denominator by x:limx→∞ 4×2+1x+1×4=limx→∞4×2+1×2+1×4=limx→∞ 4+1×2+1×4 x→∞∴ 1x, 1×2→0=44=24=12

Q8.

Answer :

limn→∞ n21+2+3…..nIt is of the form ∞∞.⇒limn→∞ n2nn+12=limn→∞2nn+1Dividing the numerator and the denominator by n:limn→∞21+1n=2

Q9.

Answer :

limx→∞ 3x-1+4x-25x-1+6x-2=limx→∞ 3x+4x25x+6×2=limx→∞3x+45x+6Dividing the numerator and the denominator by x:limx→∞ 3+4×5+6x=35

Q10.

Answer :

limx→∞ x2+a2-x2+b2x2+c2-x2+d2Rationalising the numerator and the denominator:limx→∞x2+a2-x2+b2x2+c2-x2+d2×x2+c2+x2+d2x2+c2+x2+d2×x2+a2+x2+b2x2+a2+x2+b2=limx→∞x2+a2-x2+b2 x2+a2+x2+b2 x2+c2+x2+d2x2+c2-x2+d2 x2+c2+x2+d2 x2+a2+x2+b2=limx→∞ x2+a2-x2+b2x2+c2-x2+d2×x2+c2+x2+d2x2+a2+x2+b2=limx→∞ a2-b2c2-d2 x2+c2+x2+d2x2+a2+x2+b2Dividing the numerator and the denominator by x:limx→∞a2-b2c2-d2 1+c2x2+1+d2x21+1×2+1+b2x2As x→∞, 1x, 1×2→0=a2-b2c2-d2 1+11+1=a2-b2c2-d2

Q11.

Answer :

limx→∞ x2+x+1-x ≠limx→∞ x2+1-xLHS:limx→∞ x2+x+1-xRationalising the numerator:limx→∞ x2+x+1-x x2+x+1+xx2+x+1+x=limx→∞ x2+x+1-x2x2+x+1+x=limx→∞ x+1×2+x+1+xDividing the numerator and the denominator by x:limx→∞ 1+1xx2+x+1x+1=limx→∞1+1xx2+x+1×2+1=limx→∞ 1+1×1+1x+1×2+1When x→∞, then 1x→0.11+1=12RHS:limx→∞ x2+1-x from ∞-∞

Rationalising the numerator:

limx→∞ x2+1-x x2+1+xx2+1+x=limx→∞ x2+1-x2x2+1+x=1∞=0∴limx→∞ x2+x+1-x≠limx→∞ x2+1-x

Q12.

Answer :

limx→∞xx2+1-x2-1Rationalising the numerator:limx→∞x x2+1-x2-1 x2+1+x2-1×2+1+x2-1=limx→∞x x2+1-x2-1×2+1+x2-1=limx→∞x×2×2+1+x2-1Dividing the numerator and the denominator by x:limx→∞ 2×2+1×2+x2-1×2=limx→∞2×2+1×2+x2-1×2=limx→∞21+1×2+1-1×2=21+1=2

Q13.

Answer :

limx→∞x+1-xx+2Rationalising the numerator:limx→∞x+2 x+1-x x+1+xx+1+x⇒limx→∞x+2 x+1-xx+1+xDividing the numerator and the denominator by x:limx→∞ x+2xx+1+xx=limx→∞ 1+2×1+1x+1=11+1=12

Q14.

Answer :

=limn→∞12+22+… +n2n3=limn→∞ nn+1 2n+16n3=limn→∞ n+1 2n+16n2=limn→∞n+1n 2n+1n×16 =limn→∞ 1+1n 2+1n×16=26=13

Q15.

Answer :

⇒limn→∞1+2+3+…n-1n2⇒limn→∞nn-12 n2⇒limn→∞1-1n×12When n →∞, then 1n→0.=12

Q16.

Answer :

limn→∞ 13+23+… +n3n4=limn→∞nn+122n4=limn→∞n2n+124n4=limn→∞n2n+124n4=limn→∞ n+124n2=limn→∞ n+1n2×14=limn→∞ 1+1n2×14When n→∞, then 1n→0.⇒14

Q17.

Answer :

limn→∞ 13+23+… +n3n-14=limn→∞ nn+122n-14=limn→∞ n2n+124n-14

Dividing the numerator and the denominator by n4:

limn→∞ n2n+12n44n-14n4=limn→∞n+12n24n-1n4=limn→∞ 1+1n241-1n4When n →∞, then 1n→0.1+0241-04=14

Q18.

Answer :

limx→∞xx+1-x=limx→∞ xx+1-xx+1+xx+1+x=limx→∞ xx+1-xx+1+x

Dividing the numerator and the denominator by x:

limx→∞1x+1+xx=limx→∞11+1x+1As x→∞, 1x→0=11+0+1=12

Q19.

Answer :

limn→∞13+132+133+…13n=limn→∞ 131+13+132+…..13n-1=limn→∞ 13 1-13n1-13 a+ar+ar2+……arn-1=a1-rn1-r=limn→∞ 131-13n23=limn→∞12 1-13nAs n→∞, 3n→∞, 13n→0=121-0=12

Q20.

Answer :

limx→-∞4×2-7x+2x

Let x =-m
When n → – ∞, then m → ∞.

⇒limm→∞4m2+7m-2m =limm→∞ 4m2=7m-2m ×4m2+7m+2m4m2+7m+2m=limm→∞4m2+7m-2m24m2+7m+2m=limm→∞ 4m2+7m-4m24m2+7m+2m

Dividing the numerator and the denominator by m:

limm→∞ 74m2+7mm2+2mm=limm→∞74m2m2+7mm2+2=limm→∞ 74+7m+2As m →∞, 1m→0=74+2=74

Q21.

Answer :

limx→-∞x2-8x+x

Let x = –m

When x → –∞, then m → ∞.

limm→∞m2+8m-m=limm→∞ m2+8m-m m2+8m+mm2+8m+m=limm→∞ m2+8m-m2m2+8m+m

Dividing the numerator and the denominator by m:

limm→∞ 8m2+8m+1m=limm→∞8m2m2+8mm2+1=limm→∞81+8m+1As m→∞, 1m→0=81+0+1=4

Q22.

Answer :

limx→∞x4+7×3+46x+ax4+6

Dividing the numerator and the denominator by x4:

limx→∞ 1+7x+46×2+9×41+6x4As x→∞, 1x, 1×2, 1×3, 1×4→0=11

Q23.

Answer :

fx=ax2+bx2+1limx→0 fx=1⇒limx→0 ax2+bx2+1=1⇒a×0+ba+1=1⇒b=1Also, limx→∞ fx=1limx→∞ ax2+bx2+1=1

Dividing the numerator and the denominator by x2:

⇒limx→∞ a+bx21+1×2=1As x→∞, 1x, 1×2→0⇒a+01+0=1⇒a=1∴a=1, b=1⇒fx=x2+1×2+1=1f-2=1 Since fx is a constant function, its value does not depend on x.f2=1

Q24.

Answer :

limn→∞n+2!+n+1!n+2!-n+1!=limn→∞ n+2 n+1!+n+1!n+2 n+1!-n+1!=limn→∞ n+1!n+1!×n+2+1n+2-1=limn→∞ n+3n+1

Dividing the numerator and the denominator by n:

limn→∞ 1+3n1+1n When n→∞, then 1n→0.⇒11=1

Page 29.47 Ex 29.7

Q1.

Answer :

limx→0sin 3x5x

=15limx→0sin3x3x×3 ∵limx→0sin3x3x=1

= 15×1×3

= 35

Q2.

Answer :

We know that x°=π180x.

∴limx→0 sin x0x=limx→0 sin π180xx=limx→0 sin π180xπ180x×π180=π180×1=π180

Q3.

Answer :

limx→0x2sinx2

When x→0, then x2→0.

Let θ=x2

⇒ limθ→0θsinθ

= 1

Q4.

Answer :

limx→0sinx cosx3x

= 13limx→0sinxx×cosx

= 13×1×cos0

= 13×1

= 13

Q5.

Answer :

limx→03sinx-4sin3xx

= limx→0sin 3xx ∵sin3A=3sinA-4sin3A

= limx→0sin 3x3x×3

= 1 × 3

= 3

Q6.

Answer :

limx→0tan 8xsin 2x

⇒limx→0tan 8x8x×8xsin 2x2x×2x ∵limx→0tan 8x8x=1, limx→0sin 2x2x=1⇒4

Q7.

Answer :

limx→0tan mxtan nx

⇒limx→0tan mxmx×mxnx×nxtan nx⇒limx→0tan mxmx×mxtan nxnx×nx⇒mn ∵limx→0tan xx=1

Q8.

Answer :

limx→0sin 5xtan 3x

⇒limx→0limx→0sin 5x5x×5xtan 3x3x×3x⇒53 ∵limx→0sin xx=1, limx→0tan xx=1

Q9.

Answer :

limx→0sin xnxn

It is of the form 00.

Let y=xn⇒limx→0=limy→0⇒limy→0sin yy⇒1 ∵limx→0sin xx=1

Q10.

Answer :

limx→07x cos x-3 sin x4x+tan x

It is of the form 00.
Dividing the numerator and the denominator by x:

⇒limx→07cosx-3sin xx4+tan xx⇒7limx→0cos x – 3limx→0sin xx4+limx→0tan xx⇒7.1-3.14+1⇒45

Q11.

Answer :

limx→0cos ax-cos bxcos cx-cos dx

It is of the form 00.

⇒limx→0-2sinax+bx2sinax-bx2-2sincx+dxa2sincx-dx2= limx→0sinax+bx2ax+bx×ax+bx2×sinax-bx2ax-bx2×ax-bx2sincx+dx2cx+dx2×cx+dx2×sincx-dx2cx-dx2×cx-dx2=1×limx→0ax+bx2ax-bx2cx+dx2cx-dx2= limx→0x2 a+ba-bx2 c+dc-d= a2-b2c2-d2

Q12.

Answer :

limx→0tan2 3xx2

= limx→0tan 3x3x×tan 3x3x×9= 1 × 1 × 9 ∵limx→0tan xx=1= 9

Q13.

Answer :

limx→01-cos mxx2

= limx→02sin2 mx2x2 ∵1-cos A= 2sin2 A2= 2limx→0sin mx2mx2×sin mx2mx2×m2×m2 = 2×m2×m2 ∵limx→0sin xx=1= m22

Q14.

Answer :

limx→03 sin 2x+2x3x+2 tan 3x

= limx→03 sin 2xx+23+2 tan 3xx=3limx→0sin 2x2x×2 + 23+2limx→0 tan 3xx×3=6+23+6= 89

Q15.

Answer :

limx→0cos 3x-cos 7xx2
= limx→0-2sin3x+7x2sin3x-7x2x2 cos C-cos D=-2sinC+D2sinC-D2= limx→0-2sin 5x sin -2xx2= limx→02sin 5x sin 2xx2 ∵sin-θ = -sinθ= 2limx→0sin 5x5x×sin 2x2x×5×2= 2 × 5 × 2=20

Q16.

Answer :

limθ→0sin 3θtan 2θ

= limθ→0sin 3θ3θ×3θtan 2θ2θ×2θ=32

Q17.

Answer :

limx→0sin x2 1-cos x2x6
= limx→0sin x2×2sin2 x22x6 ∵ 1-cos A=2sin2A2= 2limx→0sin x2x2×sin x222×x22×sin x222×x22= 22×2= 12

Q18.

Answer :

limx→0sin2 4x2x4

= limx→0sin4x2x2×sin4x2x2= limx→0sin4x24x2×4×sin4x24x2×4= 4×4 ∵limx→0sin xx=1= 16

Q19.

Answer :

limx→0x cos x+2 sin xx2+tan x

Dividing the numerator and the denominator by x, we get:

limx→0cos x + 2sin xxx + tan xx= cos0+2×10+1 ∵limx→0sin xx=1, limx→0tan xx=1= 31

Q20.

Answer :

limx→02x-sin xtan x+ x

Dividing the numerator and the denominator by x, we get:

limx→02-sin xxtan xx+1= 2-11+1 ∵limx→0tanxx=1, limx→0sinxx=1= 12

Q21.

Answer :

limx→05 x cos x+3 sin x3x2+tan x

Dividing the numerator and the denominator by x:

limx→05cos x + 3sin xx3x + tan xx= 5cos0+33×0+1 ∵limx→0sinxx=1, limx→0tanxx=1= 5+30+1= 8

Q22.

Answer :

limx→0sin 3x -sin xsin x

= limx→02 cos3x+x2 sin3x-x2sinx ∵sin C-sin D = 2cosC+D2sinC-D2= limx→02 cos2x.sinxsinx= 2cos0= 2

Q23.

Answer :

limx→0sin 5x – sin 3xsin x

= limx→02cos5x+3x2sin5x-3x2sin x ∵sinC-sinD=2cosC+D2sinC-D2= limx→02cos4x. sin xsin x= 2 cos0= 2

Q24.

Answer :

limx→0cos 3x -cos 5xx2
= limx→0-2sin3x+5×2 sin3x-5x2x2 ∵cosC-cosD=-2sinC+D2sinC-D2= 2 limx→02sin 4x sin xx2 ∵sin-θ=-sinθ= 2limx→0sin 4x4x×4×sinxx= 2×4= 8.

Q25.

Answer :

limx→0tan 3x-2x3x-sin2x

Dividing the numerator and the denominator by x:

limx→0tan 3xx-23-sin2 xx= limx→0tan 3x3x×3 – 23-sinxx×sinx= 3-23-1×0 ∵limx→0tan xx=1, limx→0sinxx=1= 13

Q26.

Answer :

limx→0sin 2+x-sin 2-xx

⇒limx→02cos 2+x+2-x2 sin2+x-2+x2x⇒limx→02cos 2 sinxx⇒2 cos 2 ∵limx→0sinxx=1

Q27.

Answer :

limh→0a+h2 sin a+h-a2 sin ah

= limh→0a2 sin a+h + h2sin a+h+2ah sina+h – a2 sin ah= limh→0a2sina+h-sinah+h2sina+hh+2ah sina+hhDividing and multiplying the denominator by 2: limh→0a22cosa+h+a2 sina+h-a22×h2+hsina+h+2a sina+h= limh→0a2cos a+h+a2+h sin a+h+2a sin a+h= a2 cos a+0+2a sina= a2 cosa + 2a sina

Q28.

Answer :

limx→0tan x -sin xsin 3x – 3 sin x

= limx→0sinxcosx-sinx3sinx-4sin3x-3sinx= limx→0sinx1-cosxcosx×-4sin3x= limx→02sin2x2cosx×-4sin2x= limx→02sinx2×sinx2cosx×-4 sinx×sinx= limx→02sinx2x2×x2×sinx2x2×x2cosx×-4sinxx×x×sinxx×x= limx→02×x2×x2cosx×-4×x×x= -18 cos0= -18

Q29.

Answer :

limx→0sec 5x -sec 3xsec 3x-sec x

= limx→01cos 5x1cos 3x1cos 3x-1cos x= limx→0cos 3x – cos 5xcos 5x cos 3xcos x-cos 3xcos x cos 3x= limx→0cos3x-cos5xcosxcos 5xcosx-cos3x= limx→0-2sin3x+5x2sin3x-5×2×cosxcos5x-2sinx+3x2sinx-3×2 ∵cosC-cosD=-2sinC+D2sinC-D2= limx→0sin4x×sin-x×cosxcos5x×sin2x×sin-x= limx→0sin4x4x×4xsin2x2x×2x×cosxcos5x= 42cos0cos0= 2

Q30.

Answer :

limx→01-cos 2xcos 2x-cos 8x

= limx→02sin2 x2 sin2x+8x2sin8x-2×2 ∵cosC-cosD=2sinC+D2sinD-C2= limx→02sin x × sin x2sin 5x×sin 3x= limx→0sinx × sinxsin 5x × sin 3x ∵sin-θ=-sinθ= limx→0sinxx× x × sinxx × xsin 5x5x× 5x × sin 3x×3x3x= 115

Q31.

Answer :

limx→0 1-cos 2x+tan2 xx sin x=limx→0 2 sin2 x+tan2 xx sin x ∵1-cos 2A=2 sin2 ADividing numerator & denominator by x2:limx→0 2 sin2 xx2+tan2 xx2sin xx=212+121 ∵limx→0 sin2 xx2=1, limx→0 tan2 xx2=1=3

Q32.

Answer :

limx→0 sin a+x+sin a-x-2 sin ax sin x=limx→0 2 sin a+x+a-x2 cos a+x-a+x2-2 sin ax sin x ∵sin C+sin D=2 sin C+D2cos C-D2=limx→0 2 sin a cos x – 2 sin ax sin x=limx→0 2 sin a cos x-1x sin x=limx→0 2 sin a 1-2 sin2 x2-1x sin x=limx→0 2 sin a -2 sin2 x2x sin x=-4 sin alimx→0 1x sin xx2×sin x2x2×sinx2x2×14=-4 sin a×11×1×1×14=-sin a

Q33.

Answer :

limx→0 x2-tan 2xtan xDividing the numerator and the denominator by x:limx→0 x-tan 2xxtan xx=limx→0 x-tan 2x2x×2tan xx=0-211=-2

Q34.

Answer :

limx→0 2-1+cos xx2Rationalising the numerator:limx→0 2-1+cos x 2+1+cos xx2×2+1+cos x=limx→0 2-1+cos xx22+1+cos x=limx→0 1-cos xx2 2+1+cos x=limx→0 2 sin2 x2x2 2+1+cos x=2 limx→0 sin2 x24×x22×12+1+cos x=2×14×12+1+cos 0=242+2=142

Q35.

Answer :

limx→0 x tan x1-cos xDividing the numerator and the denominator by x2:limx→0 x tan xx21-cos xx2=limx→0 tan xx2 sin2 x2x2×x2×4=42=2

Q36.

Answer :

limx→0 x2+1-cos xx sin x=limx→0x2+2 sin2 x2x sin xDividing the numerator and the denominator by x2:limx→0 1+2 sin2 x2x2sin xx=limx→0 1+2 sin2 x2x22×4sin xx=1+241=321=32

Q37.

Answer :

limx→0 sin 2x cos 3x-cos xx3=limx→0 sin 2x×-2 sin3x+x2 sin3x-x2x3=-2 limx→0 sin 2x2x×sin 2x2x×sin xx×2×2=-8

Q38.

Answer :

limx→0 2 sin x°-sin 2x°x3=limx→0 2 sin πx180-sin 2πx180x3=limx→0 2 sin πx180-2 sin πx180×cosπx180x3=limx→0 2 sin πx180 1-cos πx180x3=limx→0 2 sin πx180×2 sin2 πx360x×x2=limx→0 4 sin πx180× sin2 πx360πx180×πx360×πx360×π180×π3602=4 limx→0 sin πx180πx180×sin πx360×sin πx360πx360×πx360×π3180×3602=4×1×1×1×π3180×360×360=π1803

Page 29.48 Ex 29.7

Q39.

Answer :

limx→0 x3 cot x1-cos x=limx→0 x3tan x 1-cos x=limx→0 x3tan x×2 sin2 x2=limx→0 xtan x×x22 sin2 x2=limx→0 xtan x×x24×42×sin2 x2=limx→0 xtan x×x2sin x22×42=1×1×42=2

Q40.

Answer :

limx→0 x tan x1-cos 2x=limx→0 x tan x2 sin2 xDividing the numerator and the denominator by x2:limx→0 x tan xx22sin2 xx2=limx→0 tan xx2sin xx2=12

Q41.

Answer :

limx→0 sin 3+x-sin 3-xx=limx→0 2 cos 3+x+3-x2 sin 3+x-3+x2x ∵sin C-sin D=2 cos C+D2 sin C-D2=limx→0 2 cos 3 sin xx=2 cos 3

Q42.

Answer :

limx→0 kx.cosec x=limx→0 x cosec kx⇒limx→0 kxsin x=limx→0 xsin kx⇒klimx→0 xsin x=limx→0 kxsin kx×1k⇒k=1k⇒k2=1⇒k=±1

Q43.

Answer :

limx→0 3 sin2 x-2 sin x23x2=limx→0 3 sin2 x3x2-2 sin x23x2=1-23=13

Q44.

Answer :
limx→0 1+sin x-1-sin xx=limx→0 1+sin x-1-sin x 1+sin x+1-sin xx1+sin x+1-sin x=limx→0 1+sin x-1-sin xx1+sin x+1-sin x=limx→0 2 sin xx1+ sin x+1-sin x=21+0+1-0=22=1

Q45.

Answer :

limx→0 1-cos 4xx2=limx→0 2 sin2 2xx2=limx→0 2sin 2xx×sin 2xx=limx→0 2×sin 2x2x×sin2x2x×4=2×1×1×4=8

Q46.

Answer :

limx→0 x cos x+sin xx2+tan xDividing the numerator and the denominator by x:limx→0 cos x+sin xxx+tan xx=cos 0+10+1=2

Q47.

Answer :

limx→0 1-cos 2×3 tan2 x=limx→0 2 sin2 x3 tan2 x=limx→0 23×sin2 xsin2 x×cos2 x=23 cos2 0=23

Q48.

Answer :

limθ→0 1-cos 4θ1-cos 6θ=limθ→0 2 sin2 2θ2 sin2 3θ ∵1-cos A=2 sin2 A2=limθ→0 sin2 2θ2θ2×2θ2sin2 3θ3θ2×3θ2=limθ→0 sin 2θ2θ2×3θsin 3θ2×49=49 ∵limX→0 sin xx=1

Q49.

Answer :

limx→0 ax+x cos xb sin xDividing the numerator and the denominator by x:limx→0 a+cos xbsin xx ∵limx→0 sin xx=1=a+cos 0 b=a+1b

Q50.

Answer :

limθ→0 sin 4θtan 3θ=limθ→0 sin 4θ4θ×4θtan 3θ3θ×3θ=43

Q51.

Answer :

limθ→0 1-cos θ2θ2=limθ→0 2 sin2 θ22θ2=limθ→0 sin2 θ2θ24×4=limθ→0 sin θ2θ22×14=14

Q52.

Answer :

limx→0 1-cos 5×1-cos 6x=limx→0 2 sin2 5×22 sin2 3x=limx→0 sin2 5x25x22×5x22sin2 3x3x2×3×2=limx→0 sin 5x25x22254x2sin 3x3x2×9×2=259×4=2536

Q53.

Answer :

limx→0 cosec x-cot xx=limx→0 1sin x-cos xsin xx=limx→0 1-cos xx sin x=limx→0 2 sin2 x2x sin x=limx→0 2 sin2 x2x22×x24x sin xx×x×x2=2×14=12

Q54.

Answer :

limx→0 sin 3x+7x4x+sin 2x=limx→0 sin 3x3x×3x + 7x4x + sin 2x2x×2x=limx→0 sin 3x3x×3+7×4+sin 2x2x×2x=3+74+2=106=53

Q55.

Answer :

limx→0 5x+4 sin 3×4 sin 2x+7x=limx→0 5x+4×sin 3x3x×3×4 sin 2x2x×2x+7x=limx→0 5+4 sin 3x×33xx4sin 2x2x×2+7x=5+4×34×2+7 ∵limx→0 sin 3x3x=1=1715

Q56.

Answer :

limx→0 3 sin x-sin 3xx3=limx→0 3 sin x-3 sin x-4 sin3 xx3=limx→0 4 sin3 xx3=limx→0 4sin xx3=4×1=4

Q57.

Answer :

limx→0 tan 2x-sin 2xx3=limx→0 sin 2xcos 2x-sin 2xx3=limx→0 sin 2x 1-cos 2xcoS 2x×x3=limx→0 sin 2x×2 sin2 xcos 2x×x3=limx→0 sin 2x2x×2cos 2x×2sin xx2=2×2cos 0=4

Q58.

Answer :

limx→0 sin ax+bxax+sin bx=limx→0 sin axax×ax+bxax+sin bxbx×bx=limx→0 sin axax×a+bxa+sin bxbx×bx=1×a+ba+b=1

Q59.

Answer :
limx→0 cosec x-cot x=limx→0 1sin x-cos xsin x=limx→0 1-cos xsin x                                 ∵1-cos A=2 sin2 A2=limx→0  2 sin2 x22 sin x2 cos x2=limx→0 tan x2=0

Q60.

Answer :

limx→0 cos 2x-1cos x-1=limx→0 1-2 sin2 x-11-2 sin2 x2-1=limx→0 sin2 xsin2 x2=limx→0 sin2 xx2×x2sin2 x2x24×x24=limx→0 sin xx2×1sin x2x22×4=4

Page 29.57 Ex 29.8

Q1.

Answer :

limx→π sin xπ-x=limh→0 sin π-hπ-π-h ∵limx→afx=limh→0fa-h=limh→0 sin hh ∵sin π-0=sin 0⇒1

Q2.

Answer :

limx→π2 sin 2xcos x sin 2x= 2 sin x cos x=limx→π2 2 sin x cos xcos x=limx→π2 2 sin x⇒2

Q3.

Answer :

limx→π2 cos2 x1- sin x=limx→π2 1-sin2 x1-sin x ∵ cos2 x=1-sin2 x a2-b2=a-b a+b=limx→π2 1-sin x 1+sin x1-sin x=limx→π2 1+sin x⇒1+1=2

Q4.

Answer :

limx→π2 1-sin xcos2 x=limx→π2 1-sin x1-sin2 x=limx→π2 1-sin x1-sin x 1+sin x=limx→π2 11+sin x⇒12

Q5.

Answer :

limx→a cos x-cos ax-a=limx→a -2 sin x+a2 sin x-a22x-a2 ∵cos A-cos B -2 sin A-B2 sin A+B2=limx→a -sin x+a2 ∵limθ→a sinθ-aθ-a=1=-sin 2a2⇒-sin a

Q6.

Answer :

limx→π4 1-tan xx-π4=limh→0 1-tan π4+hπ4+h-π4=limh→0 1-1+tan h1-tan hh ∵tan π4+0=1+tan θ1-tan θ=limh→0 1-tan h-1-tan h1-tan h h=limh→0 -2 tan hh1-tan h ∵limh→0 tan hh=1⇒-21-0=-2

Q7.

Answer :

limx→π21-sin xπ2-x2=limh→0 1-sin π2-hπ2-π2-h2=limh→0 1-cos hh2=limh→0 2 sin2 h24h24 ∵limh→0 sin hh=1=12 limh→0 sin h2h22⇒12

Q8.

Answer :

limx→π3 3-tan xπ-3x=limh→0 3-tan π3-hπ-3π3-h=limh→03 -tan π3-tan h1+tan π3 tan hπ-3π3-h=limh→0 3-3-tan h1+3 tan h3h=limh→0 3+3 tan h-3+tan h1+3 tan h 3h=limh→0 4 tan h3h 1+3tan h=43

Q9.

Answer :

limx→π2 π2-x sin x-2 cos xπ2-x+cot x=limh→0 π2-π2-h sin π2-h-2 cos π2-hπ2-π2-h+cot π2-h Plugging x=π2-h=limh→0 h cos h-2 sin hh+tan h Dividing the numerator and the denominator by h:limh→0 cos h- 2 sin hh1+tan hh⇒1-211+1=-12

Q10.

Answer :

limx→π4 cos x-sin xπ4-x cos x+sin xDividing the numerator and the denominator by2:limx→π412 cos x-12 sin xπ4-x cos x+sin x2 =limx→π4 2 sin π4 cos x-cos π4 sin xπ4-x cos x+sin x=limx→π4 2 sin π4-xπ4-x cos x+sin x=limx→π4 2sin x+cos x×limx→π4 sin π4-xπ4-x⇒212+12×1=1

Q11.

Answer :

limx→π2 2-sin x-1π2-x2=limh→0 2-sin π2-h-1π2-π2-h2=limh→0 2-cos h-1h2Dividing the numerator and the denominator by 2-cos h+1:limh→0 2-cos h-1 2-cos h+12-cos h+1 h2=limh→0 2-cos h-1h2 2-cos h+1=limh→0 1-cos hh22-cos h+1=limh→0 2 sin2 h24h24 2-cos h+1=12limh→0 sin h2h22×limh→0 1 2-cos h+1=12 2-1+1=14

Q12.

Answer :

limx→π4 2-cos x-sin xπ4-x2Dividing the numerator and the denominator by 2:limx→π41-12cos x+12sin x12 π4-x2=limx→π41-cos π4cos x+sin π4 sin x12 π4-x2=limx→π4 21-cos π4-xπ4-x2=limx→π4 2 2 sin2 π4-x2π4-x2 ∵1-cos θ=2 sin 2θ2=22limx→π4 sin2 π4-x24π4-x22=224=12

Q13.

Answer :

limx→π8 cot 4x-cos 4xπ-8×3=limh→0 cot 4π8-h-cos 4π8-hπ-8π8-h3=limh→0 tan 4h-sin 4h8 h3=limh→0 sin 4h-cos 4h sin 4h512 cos 4hh3=limh→0sin 4h 1-cos 4hcos 4h 512 h3=limh→0 tan 4h4h×2 sin2 2h32×4h2=116limh→0tan 4h4h×limh→0sin 2h2h2=116×1×1 = 116

Q14.

Answer :

limx→a cos x-cos ax-a=limx→a -2 sin x+a2 sin x-a2x-aDividing the numerator and the denominator by x+a:-2 limx→a sin x+a2sin x-a2 2x-a2x+a=-2limx→a sin x+a2×sin x-a22x-a2x+a⇒-2 sin 2a2 a+a2=-2a sin a

Q15.

Answer :

limx→π 5+cos x-2π-x2=limh→0 5+cos π-h-2π-π-h2=limh→0 5-cos h-2h2Rationalising the numerator, we get:limh→0 5-cos h-2 5-cos h+2h2 5-cos h+2=limh→0 5-cos h-4h2 5-cos h+2=limh→0 1-cos hh2 5-cos h+2=limh→0 2 sin2 h24h24 5-cos h+2=12 5-1+2=124=18

Q16.

Answer :

limx→a cos x-cos ax-a=limx→a -2 sin x+a2 sin x-a22x-a2 x+a=limx→a -sin x+a2 x+a ×sin x-a2x-a2=-1 sin a2a×1=-sin a2a

Q17.

Answer :

limx→a sin x-sin ax-a=limx→a 2 cos x+a2 sin x-a2x+a x-a=limx→a 2 cos x+a2x+a×sin x-a22x-a2=1a+a cos 2a2⇒12a cos a

Q18.

Answer :

limx→1 1-x2sin 2πx=limh→0 1-1-h2sin 2π1-h=limh→0 2h-h2- sin 2π h=limh→0 -2-hsin 2π hh=limh→0 h-22π sin 2π h2π h=0-22π×1=-1π

Q19.

Answer :

limx→π4 fx-fπ4x-π4=limh→0 fπ4+h-fπ4π4+h-π4It is given that fx=sin 2x.⇒limh→0 sin π2+2h-sin π2h=limh→0 cos 2h-1h=limh→0 -2sin2 hh×h h⇒limh→0 -2h=0

Q20.

Answer :

limx→1 1+cos π x1-x2=limh→0 1+cos π 1-h1-1-h2=limh→0 1-cos πhh2=limh→0 2 sin2 πh24π2π2h24=limh→0 sin2 πh22π2 πh22⇒π22

Q21.

Answer :

limx→1 1-x2sin πxlimh→0 1-1-h2sin π1-hlimh→0 2h-h2sin πh=limh→0 h2-hsin π h=limh→0 2-hπ×sin πhπh⇒2-0π=2π

Q22.

Answer :

limx→π4 1-sin 2×1+cos 4x=limh→0 1-sin 2π4-h1+cos 4π4-h=limh→0 1-cos 2h1-cos 4h=limh→0 2 sin2 h2 sin2 2h⇒limh→0 sin2 hh24 sin2 2h4h2⇒14

Q23.

Answer :

limx→π6 cot2 x-3cosec x-2=limx→π6 cosec2 x-1-3cosec x-2=limx→π6 cosec2 x-4cosec x-2=limx→π6 cosec x-2 cosec x+2cosec x-2=cosecπ6+2=4

Q24.

Answer :

limn→∞n sin π4ncos π4n=limn→∞ n sin π4n×limn→∞ cos π4n=limn→∞ n sin π4nπ4n×π4n×1=π4limn→∞ sin π4nπ4nLet y=π4nIf n→∞, then y→0.=π4limy→0sin yy=π4

Q25.

Answer :

limn→∞ 2n-1 sin a2n=limn→∞ 2n2×sin a2na2n× a2nLet y=a2nIf n→∞, then y→0.=limy→0 a2×sin yy⇒a2

Q26.

Answer :

limn→∞ sin a2nsin b2n=limn→∞ a2n sin a2na2n×b2n× sin b2nb2nLet: y=a2n z=b2nIf n→∞, then y→0 and z→0.=yzlimy→0sin yy×1limz→0sin zz=yz×1×11=a2n×1b2n×1 =ab

Q27.

Answer :

limx→-1 x2-x-2×2+x+sin x+1=limx→-1 x-2 x+1xx+1+sin x+1Let y=x+1If x→-1, then y→0.=limy→0 y-3yy-1y+sin yDividing the numerator and the denominator by y:=limy→0 y-3y-1 +sin yy=0-30-1+1⇒-30=∞

Q28.

Answer :

limx→2 x2-x-2×2-2x+sin x-2=limx→2 x-2 x+1xx-2+sin x-2Let y=x-2 x→2∴ y→0=limy→0 yy+3y+2y+sin yDividing the numerator and the denominator by y:=limy→0 y+3y+2+sin yy=32+1=33=1

Page 29.58 Ex 29.8

Q29.

Answer :

limx→1 1-x tanπx2=limh→0 1-1-h tan π2 1-h=limh→0 h tan π2-πh2=limh→0 h cot πh2=limh→0 htan πh2=limh→0 1tanπh2×π2πh2=1π2 ∵limh→0 tan hh=1=2π

Q30.

Answer :

limx→π4 1-tan x1-2 sin xIt is of 00 form.

Rationalising the denominator, we get:

limx→π4 1-tan x 1+2 sin x1-2 sin x 1+2 sin x=limx→π4 1-tan x 1+2 sin x1-2 sin2 x=limx→π4 1-sin xcos x 1+2 sin xcos 2x=limx→π4 cos x-sin x 1+2 sin xcos x cos 2x =limx→π4 cos x-sin x 1+2 sin xcos x·cos2 x-sin2 x=limx→π4 cos x-sin x 1+2 sin xcos x cos x-sin x cos x+sin x=1+2×1212 12+12=212×2=2

Q31.

Answer :

limx→π 2+cos x-1π-x2=limh→0 2+cos π-h-1π-π-h2=limh→0 2-cos h-1h2×2-cos h+12-cos h+1=limh→0 2-cos h-1h22-cos h+1=limh→0 1-cos hh2 2-cos h+1=limh→0 2 sin2 h24×h24× 2-cos h+1=1×122 2-1+1=14

Q32.

Answer :

limx→π4 2-cos x-sin x4x-π2=limh→0 2-cos π4+h+sin π4+h4π4+h-π2=limh→0 2-cos π4 cos h – sin π4 sin h + sin π4 cos h + cos π4 sin h4h2=limh→0 2-2cos h4h2=limh→0 2 1-cos h16 h2=limh→0 22 sin2 h264×h24=232×12=1162

Q33.

Answer :

limx→1 1-1xsin π x-1=limx→1 x-1x sin πx-1Let y=x-1 x→1∴ y→0=limy→0 yy+1 sin π y=limy→0 1πy+1×sin πyπy=1π0+1×1=1π

Q34.

Answer :

limx→11-1xsinπx-1=limx→1x-1xsinπx-1

Let y = x – 1
If x → 1, then y → 0.

=limy→0yy+1sinπy=limy→01πy+1 sin πyπy=1π0+1×1=1π

Q35.

Answer :

limx→π 1+cos xtan2 x=limh→0 1+cos π+htan2 π+h=limh→0 1-cos htan2 h=limh→0 2 sin2 h2tan2 h=limh→0 2 sin2 h24h24×tan2 hh2=12 limh→0sin h2h22limh→0tan hh2=12×1212=12

Q36.

Answer :

limx→π2 cot xπ2-x=limh→0 cot π2-hπ2-π2-h⇒limx→0 tan hh=1

Q37.

Answer :

limx→π4 cos x-sin xx-π4Rationalising the numerator:=limx→π4 cos x-sin xx-π4 cos x+sin xcos x+sin x=limx→π4 cos x-sin xx-π4 cos x+sin x=limh→0 cos π4+h-sin π4+hπ4+h-π4 cos π4+h+sin π4+h=limh→0 cos π4 cos h – sin π4 sin h – sin π4 cos h – cos π4 sin hhcos π4+h+sin π4+h=limh→0 -2 sin hhcos π4+h+sin π4+h⇒-2×121214=-1214=-2-14

Q38.

Answer :

limx→a a sin x-x sin aax2-xa2Let x=a+hWhen x→a, then h→0.=limh→0 a sin a+h-a+h sin aaa+h2-a+ha2=limh→0 a sin a+h-a sin a-h sin aaa2+h2+2ah-a3-a2h=limh→0 asin a+h-sin a-h sin aa3+ah2+2a2h-a3-a2h=limh→0 a×2 cos a+h+a2 sin a+h-a2-h sin aah2+a2h=limh→0 a×2 cos 2a+h2 sin h2-h sin aah h+a=limh→0 2a cos 2a+h2 sin h2aa+h×h2×2-h sin aaha+h=2a cos a × 1a2×2-sin aa2=a cos a-sin aa2

Q39.

Answer :

limx→π2 2-1+sin xcos2 xRationalising the numerator:=limx→π2 2-1+sin x 2+1+sin xcos2 x×2+1+sin x=limx→π2 2-1+sin xcos2 x·2+1+sin x=limx→π2 1-sin x1-sin x 1+sin x 2+1+sin x=11+1 2+2=12×22=142

Q40.

Answer :

limx→π2 tan 2xx-π2=limh→0 tan 2π2+hπ2+h-π2=limh→0 tan π+2hh=limh→0 2 tan 2h2h⇒2

Page 29.58 Ex 29.9

Q1.

Answer :

limx→π 1+cos xtan2 x 00form=limx→π 1+cos xsin2 x×cos2 x=limx→π 1+cos x1-cos2 x×cos2 x=limx→π 1+cos x1-cos x 1+cos x×cos2 x=limx→π cos2 x1-cos x=cos2 π1-c

Q2.

Answer :

limx→π4 cosec2 x-2cot x-1=limx→π4 1+cot2 x-2cot x-1=limx→π4 cot2 x-1cot x-1=limx→π4 cot x-1 cot x+1cot x-1=cot π4+1=2

Q3.

Answer :

limx→π6 cot2 x-3cosec x-2=limx→π6 cosec2 x-1-3cosec x-2=limx→π6cosec2 x-4cosec x-2=limx→π6 cosec x-2 cosec x+2cosec x-2=cosecπ6+2=2+2=4

Q4.

Answer :

limx→π4 2-cosec2 x1-cot x=limx→π4 2-1+cot2 x1-cot x=limx→π4 1-cot2 x1-cot x=limx→π4 1-cot x 1+cot x1-cot x=1+cot π4=1+1=2

Q5.

Answer :

limx→π 2+cos x-1π-x2Rationalising the numerator, we get:limx→π 2+cos x-1×2+cos x+1π-x2 2+cos x+1=limx→π 2+cos x-1π-x2 2+cos x+1=limx→π 1+cos xπ-x2 2+cos x+1

Let x = π – h
when x → π, then h → 0

=limh→0 1+cos π-hπ-π-h2 2+cos π-h+1=limh→0 1-cos hh2 2-cos h+1 ∵cos π-θ=-cos θ=limh→0 2 sin2 h24×h242-cos h+1=12limh→0 sin h2h22×12-cos h+1=12×1×12-cos 0+1=12×11+1=12×2=14

Q6.

Answer :

limx→3π3 1+cosec3 xcot2 x=limx→3π21+cosec x 12+cosec2 x-cosec xcosec2 x-1=limx→3π2 1+cosec x 1+cosec2 x-cosec xcosec x-1 cosec x+1=1+cosec2 3π2-cosec 3π2cosec 3π2-1=1+1+1-1-1=-32

Page 29.67 Ex 29.10

Q1.

Answer :

limx→0 5x-14+x-2

Rationalising the denominator, we get:

=limx→0 5x-1 4+x+24+x-2 4+x+2=limx→0 5x-1 4+x+24+x-4=limx→0 5x-1x 4+x+2 ∵limx→0 ax-1x=log a=log 5×4+0+2=4 l

Q2.

Answer :

limx→0 log 1+x3x-1

Dividing the numerator and the denominator by x:

limx→0 log 1+xx·3x-1x ∵limx→0 log 1+xx=1 limx→0 ax-1x=log a=1log 3

Q3.

Answer :

limx→0 ax+a-x-2×2=limx→0 ax22+a-x22-2ax2·a-x2x2=limx→0 ax2-a-x22x2=limx→0 ax2-1ax22x2=limx→0 ax-12×2×1ax22=limx→0 ax-1×2×1ax=log a2a0=log a2

Q4.

Answer :

limx→0 amx-1bnx-1=limx→0 amx-1mx×mxbnx-1nx×nx=loge aloge b×mn=mn log alog b

Q5.

Answer :

limx→0 ax+bx-2x=limx→0 ax-1+bx-1x=limx→0 ax-1x+limx→0 bx-1x=log a+log b=log ab

Q6.

Answer :

limx→0 9x-2.6x+4xx2=limx→0 3×2-2·3x·2x+2x2x2=limx→0 3x-2x2x2=limx→0 3x-2x2x2×2x2x2=limx→0 32x-1×2×22x=log 322×20=log 322

Q7.

Answer :

limx→0 8x-4x-2x+1×2=limx→0 2×3-2×2-2x+1×2=limx→0 2×2 2x-1-12x-1×2=limx→0 22x-1 2x-1×2=limx→0 222x-12x×2x-1x=2 log 2×log 2=log 22×log 2=log 4×log 2

Q8.

Answer :

limx→0 amx-bnxx=limx→0 amx-1-bnx+1x=limx→0 amx-1-bnx-1x=limx→0 amx-1mx×m-bnx-1nx×n=m log a-n log b=log am-log bn=log ambn

Q9.

Answer :

limx→0 an+bn+cn-3x=limx→0 an-1x+bn-1x+cn-1x=log a+log b+log c=log abc

Q10.

Answer :

limx→2 x-2loga x-1

Let x = 2 + h

x → 2
∴ h → 0

=limh→0 2+h-2log 2+h-1log a=log a limh→0 hlog 1+h=log a ×1

Q11.

Answer :

limx→0 5x+3x+2x-3x=limx→0 5x-1x+3x-1x+2x-1x=log 5+log 3+log 2=log 5×3×2=log 30

Q12.

Answer :

limx→∞ a1x-1xLet y=1x x→∞∴ y→0⇒limy→0 ay-1y=log a

Q13.

Answer :

limx→0 amx-bnxsin kx=limx→0 amx-1-bnx-1sin kxDividing the numerator and the denomiantor by x:=limx→0 mamx-1mx-nbnx-1nxk×sin kxkx=m log a-n log bk×1=1k log am-log bn=1k log ambn

Q14.

Answer :

limx→0 an+bn-cn-dnxlimx→0 an+bn-2-cn-dn+2x=limx→0 an-1+bn-1-cn-1-dn-1x=limx→0 an-1x+bn-1x-cn-1x-dn-1x=log a +log b-log c-log d=log a +log b-log c+log d=log ab-log cd=log abcd

Q15.

Answer :

limx→0 ex-1+sin xx=limx→0 ex-1x+sin xx=1+1=2

Q16.

Answer :

limx→0 sin 2xex-1

Dividing the numerator and the denominator by x:

=limx→0 sin 2xx×ex-1x=limx→0 sin 2x2x×2ex-1x=1×21=2

Q17.

Answer :

limx→0 esin x-1x=limx→0 esin x-1sin x×sin xx

x → 0
∴ sin x → 0

Let y=sin x

x → 0
∴ y → 0

⇒limy→0 ey-1y×limx→0 sin xx=1×1

Q18.

Answer :

limx→0 e2x-exsin 2x=limx→0 ex ex-1sin 2x=limx→0 ex ex-1x×2xsin 2x×12=e0×1×11×12=12

Q19.

Answer :

limx→a log x-log ax-a=limx→a log xaaxa-1=limx→a log 1+xa-1axa-1x→a∴ xa→1⇒xa-1→0Let y=xa-1x→a∴ y→0=limy→0 log 1+ya×y=1a×1=1a

Q20.

Answer :

limx→0 log a+x-log a-xx=limx→0 log a+xa-xx=limx→0 log 1+a+xa-x-1x=limx→0 log 1+a+x-a+xa-xx=limx→0 log 1+2xa-x2xa-x×a-x2x→0∴ 2xa-x→0Let y=2xa-x=limy→0 log 1+yy ×limx→0 1a-x2=1×2a

Q21.

Answer :

limx→0 log 2+x+log 0.5x=limx→0 log 2+x×0.5x=limx→0 log 1+x2x2×2=12×1=12

Q22.

Answer :

limx→0 log a+x-log ax=limx→0 log a+xax=limx→0 log 1+xaxa×a=1a×1=1a

Q23.

Answer :

limx→0 log 3+x-log 3-xx=limx→0 log 3+x3-xx=limx→0 log 1+3+x3-x-1x=limx→0 log 1+3+x-3+x3-xx=limx→0 log 1+2×3-xx=limx→0 log 1+2×3-x2x3-x×3-x2=1×23-0=23

Q24.

Answer :

limx→0 8x-2xx=limx→0 8x-1x-2x-1x=log 8-log 2=log 82=log 4

Q25.

Answer :

limx→0 x2x-11-cos x

Dividing the numerator and the denominator by x2:

=limx→0 2x-1x×x21-cos x=limx→0 2x-1x×x22 sin2 x2=limx→0 2x-1x×x24×42 sin2 x2=log 2×42×112=2 log 2=log 22=log 4

Q26.

Answer :

limx→0 1+x-1log 1+x

Rationalising the numerator:

=limx→0 1+x-1 1+x+1log 1+x 1+x+1=limx→0 1+x-1log 1+x 1+x+1=11×1+0+1=12

Q27.

Answer :

limx→0 log 1+x3sin3 x=limx→0 log 1+x3x3×x3sin3 x=1×113=1

Q28.

Answer :

limx→π2 acot x-acos xcot x-cos x=limx→π2 acos xacot x-cos x-1cot x-cos x x→π2∴ cot x-cos x→0⇒acos π2×log a=a0×log a=log a

Q29.

Answer :

limx→0 ex-11-cos xRationalising the denominator, we get:=limx→0 ex-11-cos x×1+cos x1+cos x=limx→0 ex-1 1+cos x1-cos2 x=limx→0 ex-1 1+cos xsin x

Dividing numerator and the denominator by x, we get:

=limx→0 ex-1x×1+cos xsin xxLeft hand limit:limx→0- ex-1x×1+cos xsin xx=limx→0- ex-1x×1+cos x-sin xx=-1×21=-2Right hand limit:limx→0+ ex-1x×1+cos xsin xx=limx→0+ ex-1x×1+cos xsin xx=1×21=2

Left hand limit ≠ Right hand limit
Thus, limit does not exist.

Q30.

Answer :

limx→5 ex-e5x-5=limx→5 e5 ex-5-1x-5=e5×1

Q31.

Answer :

limx→0 ex+2-e2x=limx→0 e2 ex-1x=e2×1=e2

Q32.

Answer :

limx→π2 ecos x-1cos xIf x→π2, then cos x→0.Let y=cos x=limy→0 ey-1y=1

Q33.

Answer :

limx→0 e3+x-sin x-e3x=limx→0 e3+x-e3x-sin xx=limx→0 e3 ex-1x-sin xx=e3×1-1=e3-1

Q34.

Answer :

limx→0 ex-x-12ex=1+x1!+x22!+x33!+…..∞=limx→0 1+x1!+x22!+x33!….∞-x-12=limx→0 x22!+x33!+….∞2=0

Q35.

Answer :

limx→0 e3x-e2xx=limx→0 e3x-1x-e2x-1x=limx→0 3e3x-13x-2e2x-12x=3×1-2×1=1

Q36.

Answer :

limx→0 etan x-1tan x

If x → 0, then tan x → 0.

Let y = tan x

=limy→0 ey-1y=1

Q37.

Answer :

limx→0 ebx-eaxx=limx→0 ebx-1x-eax-1x=limx→0 bebx-1bx-a×eax-1ax=b×1-a×1=b-a

Q38.

Answer :

limx→0 etan x-1x=limx→0 etan x-1tan x×tan xx=1×1

Q39.

Answer :

limx→0 ex-esin xx-sin x=limx→0 esin xexesin x-1x-sin x=limx→0 esin x ex-sin x-1x-sin x=esin 0=1

Q40.

Answer :

limx→0 32+x-9x=limx→0 32·3x-32x=32 limx→0 3x-1x=9 log 3=9 loge 3

Page 29.68 Ex 29.10

Q41.

Answer :

limx→0 ax-a-xx=limx→0 ax-1axx=limx→0 a2x-1ax·2x×2=limx→0 a2x-12x×2ax=limx→0 a2x-12x×2ax=log a×2a0=2 log a

Q42.

Answer :

limx→0 x ex-11-cos x

Dividing the numerator and the denominator by x2:

=limx→0 ex-1×1-cos xx2=limx→0 ex-1x×12 sin2 x24×x24=1×212=2

Q43.

Answer :

limx→π2 2-cos x-1xx-π2=limx→π2 2-sin π2-x-1xx-π2 ∵cos x=sin π2-x=limx→π2 2sin x-π2-1x-π2×x=limx→π2 2sin x-π2-1sin x-π2×sin x-π2x-π2×1x=loge 2 ×1×1π2=2πloge 2

Page 29.72 (Very Short Answers)

Q1.

Answer :

limx→0 1-cos 2xx=limx→0 2 sin2 xx=2 limx→0 sin2 xx=2 limx→0 sin xxLHL:=2 limx→0- sin xxLet x=0-h, where h→0.=2 limh→0 sin -h-h=2 limh→0 sin h-h=-2RHL:=2 limx→0+ sin xxLet x=0+h, where h→0.=2 limh→0 sin hh=2LHL ≠RHLThus, limx→0 1-cos 2xx does not exist.

Q2.

Answer :

limx→0- xIf x=0-h, then h→0.limh→0 0-h=-1

Q3.

Answer :

limx→0+ xLet x=0+h, where h→0.limh→0 0+h=0
Disclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.

Page 29.72 Ex. 29.11

Q1.

Answer :

limx→0 cos x+ sin x1xBy adding and subtracting 1, we get:=limx→0 1+cos x+sin x-11xUsing the theorem given below:If limx→afx=limx→agx=0 such that limx→afxgx exists, then limx→a1+fx1gx=elimx→afxgx.Here: fx=cos x+sin x-1 gx=x⇒elimx→0 cos x+sin x-1x⇒elimx→0 sin xx-1-cos xx⇒elimx→0 sin xx-2 sin2 x2x⇒elimx→0 sin xx-2 sinx2×sin x22×x2=e1-0=e1

Q2.

Answer :

limx→0 cos x1sin x=limx→0 1+cos x-11sin xUsing the theorem given below:If limx→afx=limx→agx=0 such that limx→afxgx exists, thenlimx→a1+fx1gx=elimx→afxgx.Here: fx=cos x-1 gx=sin x⇒elimx→0cos x-1sin x =elimx→0-2 sin2 x22 sin x2 cos x2=elimx→0 -tanx2=e0=1

Q3.

Answer :

limx→0 cos x+a sin bx1x=limx→0 1+cos x+a sin bx-11xUsing the theorem given below:If limx→afx=limx→agx=0 such that limx→afxgx exists, then limx→a1+fx1gx=elimx→afxgx.Here: fx=cos x+a sin bx-1 gx=x⇒elimx→0cos x+a sin bx-1x =elimx→0b×a sin bxbx-1-cos xx =elimx→0ab sin bxbx-2 sin2 x2x =eab

Q4.

Answer :

limx→0+ 1+tan2 x 12xUsing the theorem given below:If limx→afx=limx→agx=0 such that limx→afxgx exists, thenlimx→a1+fx1gx=elimx→afxgx.Here: fx=tan2 x gx=2x⇒elimx→0+tan2 x2x=elimx→0+tan xx×tan xx×12 =e1×1×12=e

Q5.

Answer :

limx→∞ 3x-43x+2x+13=limx→∞1+3x-43x+2-1x+13=limx→∞1+3x-4-3x-23x+2x+13=limx→∞ 1-63x+2x+13Using the theorem given below:If limx→afx=limx→agx=0 such that limx→afxgx exists, then limx→a1+fx1gx=elimx→afxgx.Here: fx=-63x+2 gx=2x⇒elimx→∞-63x+2×x+13 =elimx→∞-6x-69x+6 =elimx→∞-6-6×9+6x =e-23 (When x→∞, 1x→0)

Q6.

Answer :

limx→∞ x2+2x+32×2+x+53x-23x+2=limx→∞ 1+x2+2x+32×2+x+5-13x-23x+2=limx→∞ 1+x2+2x+3-2×2+x+52×2+x+53x-23x+2=limx→∞ 1+-x2+x-22×2+x+53x-23x+2=elimx→∞-x2+x-22×2+x+5×3x-23x+2 =elimx→∞-1+1x-2×22+1x+5×2×3-2×3+2x =e-12×1=1eDisclaimer: The solution given in the book is incorrect. However, the solution given here has been created according to the question given in the book.

Q7.

Answer :

limx→1×3+2×2+x+1×2+2x+31-cos x-1x-12=limx→1 1+x3+2×2+x+1×2+2x+3-11-cos x-1x-12=elimx→1×3+2×2+x+1-x2-2x-3×2+2x+3×1-cos x-1x-12 =elimx→1×3+x2-x-2×2+2x+3×2 sin2 x-124×x-124 =e1+1-1-21+2+3×12=e-112

Q8.

Answer :

limx→0 ex+e-x-2x21x2=limx→0 1+ex+e-x-2×2-11×2=e limx→0ex+e-x-2×2-1×1×2 ∵ex=1+x1!+x22!+x33!+……∝e-x=1-x1!+x22!-x33!+……∝⇒ ex+e-x=2+2×22!+2×44!+…..∝=elimx→02+2×22!+2×44!…∝-2×2-1×1×2 =elimx→02×22!+2×44!+……∝x4-1×2 =elimx→0 x2+x412+…..∝-x2x4=e112

Q9.

Answer :

limx→a sin xsin a1x-a=limx→a 1+sin xsin a-11x-a=limx→a 1+sin x-sin asin a1x-a=e limx→asin x-sin asin a×1x-a =e limx→a2 cos x+a2 sin x-a2sin a×x-a2×2 =ecos asin a=ecot a

Q10.

Answer :

limx→∞ 3×2+14×2-1×31+x=limx→∞ 1+3×2+14×2-1-1×31+x=limx→∞ 1+3×2+1-4×2+14×2-1×31+x=limx→∞ 1+2-x24x2-1×31+x=elimx→∞2-x24x2-1×x31+x =e limx→∞-x5+2x34x2-1 1+x Dividing Nr and Dr by x5:=e limx→∞-1+2x34x2-1×3 1+xx2 =e limx→∞-1+2x34x-1×3 1×2+1x =e-10=e-∞=0

Q11.

Answer :

limn→∞ 1+xnn=elimn→∞xn×n =ex

Page 29.73 (Very Short Answers)

Q4.

Answer :

limx→1- x-x x=1-h∴ h→0limh→0 1-h-1-h=1-0=1

Q5.

Answer :

limx→0- sin xx x=0-h∴ h→0=limh→0 sin 0-h0-h=sin -1-1=-sin 1-1=sin 1

Q6.

Answer :

limx→π sin xx-πLHL:limx→π- sin xx-πIf x=π-h, then h→0.=limh→0 sin π-hπ-h-π=limh→0 sin h-h=-1RHL:limx→π+ sin xx-πIf x=π+h, then h→0.=limh→0 sin π+hπ+h-π=limh→0 -sin hh=-1∴limx→π sin xx-π=-1

Q7.

Answer :

limx→∞ sin xxLet x=1yIf x→∞, then y→0.=limy→0 y·sin 1yLHL:Let y=0-hIf y→0, then h→0.=limh→0 0-h×sin 10-h0×The oscillating number between -1 and 1=0RHL:limy→0+ y·sin 1yLet y=0+hIf y→0, then h→0.=limh→0 h×sin 1h=0×The oscillating number between –1 and 1=0

Q8.

Answer :

limx→2 x-2x-2LHL:limx→2- x-2x-2Let x=2-hIf x→2, then h→0.=limh→0 2-h-22-h-2=-1RHL:limx→2+ x-2x-2Let x=2+hIf x→2, then h→0.limh→0 2+h-22+h-2=1LHL ≠ RHLThus,limx→2 x-2x-2 does not exist.

Q9.

Answer :

limx→0 sin x°x=limx→0 sin πx180π180x×π180=1×π180=π180

Q10.

Answer :

limx→0- sin xxLet x=0-hIf x→0, then h→0.limh→0 sin -h-hThe value does not exist.

Q11.

Answer :

limx→0 sin x1+x-1Rationalising the denominator, we get:=limx→0 sin x1+x-1×1+x+11+x+1=limx→0 sin x ×1+x+11+x-1=limx→0 sin xx 1+x+1=1×1+1=2

Q12.

Answer :

limx→-∞ 3x+9×2-xLet m=-xIf x→-∞, then m→∞.=limm→∞ -3m+9m2+mRationalising the numerator, we get:=limm→∞ -3m+9m2+m -3m-9m2+m-3m-9m2+m=limm→∞ –9m2+9m2+m-3m-9m2+m=limm→∞-m-3m-9m2+mDividing the numerator and the denominator by m, and applying limit, we get:=-1-3-9+1∞=16

Q13.

Answer :

limn→∞ n!+n+1!n+1!+n+2!=limn→∞ n!+n+1n!n+1!+n+2 n+1!=limn→∞ n! 1+n+1n+1! 1+n+2=limn→∞n! n+2n+1 n! n+3=limn→∞ n+2n+1 n+3Degree of the denominator is greater than that of the numerator.⇒0

Q14.

Answer :

limx→π2 2x-πcos xLHL:limx→π2- 2x-πcos xLet x=π2-hIf x→π2, then we have: h→0=limh→0 2π2-h-πcos π2-h=lim h→0 π-2h-πsin h=-2RHL:limx→π2+ 2x-πcos xLet x=π2+hIf x→π2, then we have:h→0=limh→0 2π2+h-πcos π2+h=limh→0 2h-sin h=-2⇒limx→π2 2x-πcos x=-2

Q15.

Answer :

limn→∞ 1+2+3…nn2=limn→∞ nn+12n2=limn→∞ n+12n=12limn→∞ 1+1n=12

Page 29.73 (Multiple Choice Questions)

Q1.

Answer :

(c) 1/3

limn→∞ 12+22+32…..n2n3=limn→∞ Σn2n3=limn→∞ nn+1 2n+16n3=limn→0 n+1 2n+16n2Dividing the numerator and the denominator by n2, we get:limn→∞ n+1n×2n+1n6=limn→∞ 1+1n 2+1n6⇒26=13

Q2.

Answer :

(d) 2
limx→0 sin 2xx=limx→0 2sin 2x2x=2×1=2

Q3.

Answer :

(b) 0
limx→0 fx=limx→0 x sin 1x, x≠0LHL at x=0:limx→0- fx=limh→0 f0-h⇒limh→0 -h sin 1-h=0RHL at x=0:limx→0+ fx=limh→0 f0+h⇒limh→0 h sin 1h=0limx→0- fx=limx→0+ fxHence, limx→0 fx exists and is equal to 0.

Page 29.74 (Multiple Choice Questions)

Q4.

Answer :

(a) 0

limx→0 1-cos 2xx=limx→0 2 sin2 xx=limx→0 2x×sin2 xx2⇒0

Q5.

Answer :

(a) 10/3

limx→0 1-cos 2x sin 5xx2 sin 3x=limx→0 2 sin2 x sin 5xx2 sin 3x=limx→0 2 sin2 xx2 sin 5x5x×5sin 3x3x3=10×12×11×3=103

Q6.

Answer :

(b) 1

limx→0 xtan x=limx→0 1tan xx=11=1

Q7.

Answer :

(b) -12

limn→∞ 11-n2+21-n2+……+n1-n2=limn→∞ 11-n2 1+2+3…..n=limn→∞ 11-n2 nn+12=limn→∞ 12nn+11-n 1+n=limn→∞ 12 n1-n=limn→∞ 12 11n-1=12-1=-12

Q8.

Answer :

(a)
limx→∞ sin xxLet x=1y x→∞∴ y→0=limy→0 sin 1y1y=limy→0 y sin 1y=limy→0 y×limy→0sin 1y=0×limy→0sin 1y=0

Q9.

Answer :

(d) π180

limx→0 sin x0x=limx→0 sin π180xx=limx→0 sin π180xπ180x×π180=π180×1=π180

Q10.

Answer :

(d) does not exist

limx→3 x-3x-3LHL at x=3:limx→3- x-3-x-3 ∵x-3=-x-3, when x<3⇒-1RHL at x=3:limx→3+ x-3x-3 ∵x-3=x-3, when x>3=1

LHL ≠ RHL
Therefore, limit does not exist.

Q11.

Answer :

(b) nan-1

limx→a xn-anx-a=limx→a+ xn-anx-a ∵fx exists, limx→a fx=limx→a+fx=limh→0 a+hn-ana+h-a=limh→0 an 1+han-1h=an limh→0 1+n·ha+nn-12!h2a2…+…-1=an limh→0 na+hh-12! ha2+…=an na=nan-1

Q12.

Answer :

(b) 1/2

limx→π4 2cos x-1cot x-1Rationalising the numerator, we get:=limx→π4 2 cos x-1cot x-1×2 cos x+12 cos x+1=limx→π4 2 cos2 x-1 cos x-sin x ×sin x2cos x+1=limx→π4 cos2 x-sin2 xcos x-sin x×sin x2cos x+1=limx→π4 cos x+sin x sin x2cos x+1=cos π4+sin π4 sin π42cos π4+1=12+12122.12+1=22×122=12
The correct answer is B.

Q13.

Answer :

(d) 1/2

limx→∞ x2-12x+1=limx→∞ 1-1×22+1xDividing the numerator and the denominator by x, we get:=1-02+0=12

Hence, the correct answer is d.

Q14.

Answer :

(d) 4/3

limh→0 23 sin π/6+h-cos π/6+h3 h3cos h-sin h=limh→0 232cos h+32 sin h-32cos h+sin h2h3 cos h-3 sin h=limh→0 22 sin hh×13 cos h-3sin h=limh→0 43 cos h-3 sin h=43

Q15.

Answer :

(d) −1/48

limh→01h 8+h3-12h=limh→01h18+h3-12=limh→01h2-8+h1/32×8+h3=limh→01h81/3-8+h1/32 8+h3 A3-B3=A-BA2+AB+B2 or A-B=A3-B3A2+AB+B2=limh→08-8+hh28+h34+28+h1/3+8+h2/3=-12×834+2×81/3+82/3=-12×24+4+4=-148

Page 29.75 (Multiple Choice Questions)

Q16.

Answer :

(b) 1/2

limn→∞ 11.3+13.5+15.7…12n+1 2n+3Here, Tn=12n-1 2n+1⇒Tn=A2n-1+B2n+1On equating A=12 and B=-12:Tn=122n-1-122n+1⇒T1=121-13⇒T2=1213-15⇒Tn-1=1212n-1-12n-1⇒Tn=1212n-1-12n+1⇒T1+T2+T3…Tn=121-12n+1⇒T1+T2+T3…Tn=122n2n+1⇒T1+T2+T3…Tn=n2n+1∴ limn→∞ 11.3+13.5+15.7…12n+1 2n+3=limn→∞ ∑n=1n12n-1 2n+1=limn→∞ n2n+1

=limn→∞ 12+1n Dividing Nr and Dr by n=12

Q17.

Answer :

(a) −π
limx→1 sin πxx-1=limh→0 sin π1+h1+h-1=limh→0 sin π+πhh=limh→0 -sin πhh=limh→0 -sin πhπhπ=-π

Q18.

Answer :

(b) 100

limx→1 x+x2+x3…xn-nx-1=5050⇒limx→1 x-1x-1+x2-1x-1+x3-1x-1…xn-1x-2=5050⇒1+2+3…n=5050 ∵xn-anx-a=nan-1⇒nn+12=5050⇒nn+1=10100⇒nn+1=100×101On comparing:n=100

Q19.

Answer :

(c) 2

limx→∞ 1+x4+1+x2x2=limx→∞ 1×4+1+1×2+1=2

Q20.

Answer :

(a) 12

limx→0 1+x-1x=limx→0 1+x-1 1+x+11+x+1 x=limx→0 1+x-1×1+x+1=limx→0 1×1+x+1=12

Q21.

Answer :

(c) 13

limx→π3 sin π3-x2 cos x-1=limh→0 sin π3-π3-h2 cos π3-h-1=limh→0 sin h2cos π3cos h+sin π3 sin h-1=limh→0 sin h212cos h+32 sin h-1=limh→0 sin hcos h+3 sin h-1=limh→0 sin h-2 sin2 h2+3 sin h Dividing Nr and Dr by h:=limh→0 sin hh-2×h4 sin2 h2h×h4+3 sin hh=13

Q22.

Answer :

(b) 2n-1×3n+14

limx→3 ∑r=1n xr-∑r=1n 3rx-3=limx→3 x1+x2+x3+……+xn-31+32+33…..3nx-3=limx→3 x-3x-3+x2-32x-3+x3-33x-3+….xn-3nx-3=1+2×3+3×32+…..+n3n-1 ∵xn-anx-a=nan-1

This series is an AGP of the form given below:

S = 1 + 2r + 3r2……..nrn–1
S=1-rn1-r2-nrn1-rr=3, a=1 d=1=1-3n+2n 3n4=3n 2n-1+14

Q23.

Answer :

(b) -12

limn→∞1-2+3-4+5-6+…2n-1-2nn2+1+n2-1=limn→∞1+3+5+…2n-1-2+4+6+…2nn2+1+n2-1=limn→∞n21+2n-1-n22+2nn2+1+n2-1=limn→∞n2-nn+1n2+1+n2-1=limn→∞-nn2+1+n2-1

Dividing the numerator and the denominator by n:

=limn→∞-11+1n2+1-1n2 =-11+1=-12

Q24.

Answer :

(b) 0

fx=xsin1x, x≠00, x=0

LHL:

limx→0-fx=limx→0-xsin1x

Let x = 0 – h, where h → 0.

=limh→0-h×sin-1h

= 0 × The oscillating number between –1 and 1
= 0
RHL
limx→0+fx

Let x = 0 + h, where h → 0.

=limh→0h×sin1h

= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0
∴limx→0fx=0

Q25.

Answer :

(b) 0

limn→∞n!n+1!+n!=limn→∞n!n+1×n!+n!=limn→∞n!n!n+1+1=limn→∞1n+2=0

Page 29.76 (Multiple Choice Questions)

Q26.

Answer :

(a)

limx→π4 42-cos x+sin x51-sin 2x=limx→π4 252-cos x+sin x2522-1+sin 2x=limx→π4 252-cos x+sin x2522-cos x+sin x2Let t=cos x+sin x2 x→π4∴ t=cos π4+sin π42→22=2=limt→2252-t522-t=52 232 ∵limx→axn-anx-a=nan-1=52

Q27.

Answer :

(a)∵limx→21+2+x-3x-2=limx→21+2+x-3x-2×1+2+x+31+2+x+3=limx→22+x-2x-21+2+x+3=limx→22+x-2x-21+2+x+3×2+x+22+x+2=limx→2x-2x-21+2+x+32+x+2=limx→211+2+x+32+x+2=11+2+2+32+2+2=123×4=183

Q28.

Answer :

(a) b

limx→∞ ax sin baxlimx→∞ bsin baxbaxLet bax=yx→∞ ∴ y→0limy→0 b sin yy=b

Q29.

Answer :

(c) 132
limx→0 8×8 1-cos x22-cos x24+cos x22cos x24=limx→0 8×8 1-cos x24-cos x221-cosx24=limx→0 8×8 1-cosx24 1-cosx22=limx→0 8×8 2 sin2 x28 2 sin2 x24=limx→0 4×8 sin2 x2864×x464 sin2 x2416x416=3264×16=132

Q30.

Answer :

(a)

limx→αtanax2+bx+cx-α2=limx→αtanax-αx-αax-α2×a=limx→αtanax-α2ax-α2×a=1×a limθ→0tanθθ=1=a

Q31.

Answer :

(d) -a

limx→0 a2-ax+x2-a2+ax+x2a+x-a-xMultiplying a2-ax+x2+a2+ax+x2 and a+x+a-x in Nr and Dr:limx→0 a2+x2-ax-a2-ax-x2 a+x+a-xa+x-a+x x2-ax+x2+a2+ax+x2limx→0 -2ax a+x+a-x2x a2+ax+x2+a2+ax+x2=-a2a2a=-a

Q32.

Answer :

(b) 2

limx→0 1-cos x+ 2 sin x-sin3 x-x2+3x4tan3 x-6 sin2 x+x-5x3Dividing Nr and Dr by x:limx→0 2sin2 x2x+2 sin xx-sin3 xx-x+3x3tan3 xx-6 sin2 xx+1-5x2limx→0 2×4 sin2 x2x24+2sin xx -x2 sin3 xx3-x+3x3x2 tan 3xx3-6x sin2 xx2+1-5×2=0+2-0-0+00-0+1-0=2

Q33.

Answer :

(c) 12

limθ→π2 1-sin θπ2-θcos θ=limh→0 1- cos hπ2-π2-h sin h=limh→0 2 sin2 h2h sin h=limh→0 2 sin2 h24h24sin hh=24=12

Q34.

Answer :

(d) 0

limx→π2 sec x-tan x=limh→0 sec π2-h-tan π2-h=limh→0 cosec h-cot h=limh→0 1-cos hsin h=limh→0 2 sin2 h2sin h=limh→0 2 sin2 h22 sin h2cos h2=limh→0 tan h2=0

Q35.

Answer :

(a) 0
limn→∞ n!n+1!-n!Dividing Nr and Dr by n!:limn→∞ 1n+1!n!-1=limn→∞ 1n+1n!n!-1=limn→∞ 1n+1-1=limn→∞=1n=0

Q36.

Answer :

(c) 1
limn→∞ n+2!+n+1!n+2!-n+1!Dividing Nr & Dr by n+1!:⇒limn→∞ n+2 n+1!n+1!+1n+2 n+1!n+1!-1=limn→∞ n+2+1n+2-1=limn→∞ n+3n+1=limn→∞ 1+3n1+1n=1

Page 29.77 (Multiple Choice Questions)

Q37.

Answer :

(b) 100

limx→∞ x+110+x+210+….+x+10010×10+1010Dividing Nr and Dr by x10:⇒limx→∞ 1+1×10+1+2×10+….+1+100×101+1010×10=1+1+1+…+100 times=100

Q38.

Answer :

(d) −1/2

limn→∞ 1+2+3+…..nn+2-n2=limn→∞ nn+12n+2-n2=limn→∞ n2 n+1-n-2n+2=limn→∞ n2-1n+2=limn→∞ -121+2n=-121+0=-12

DERIVATIVES

Page 30.3 Ex.30.1

Q1.

Answer :

We have:

f'(2) =lim h→0f(2+h)-f(2)h =limh→03(2+h)-3(2)h =limh→06+3h-6h =limh→03hh =3

Q2.

Answer :

We have:
f'(x)=limh→0f(10+h)-f(10)h =limh→0(10+h)2-2-(102-2)h =limh→0100+h2+20h-2-100+2h =limh→0h2+20hh =limh→0h(h+20)h =limh→0h+20 =0+20 =20

Q3.

Answer :

We have:
f'(100)=limh→0f(100+h)-f(100)h =limh→099(100+h)-99(100)h =limh→09900+99h-9900h =limh→099hh =limh→099 =99

Q4.

Answer :

We have:
f'(x)=limh→0f(1+h)-f(1)h =limh→01+h-1h =limh→01 =1

Q5.

Answer :

We have:
f'(x)=limh→0f(0+h)-f(0)h=limh→0f(h)-f(0)h=limh→0cosh-cos0h=limh→0cosh-1h=lim h→0 -2 sin2 h2h=limh→0 -2 sin2 h2h24 ×h4==limh→0 -1 ×h2=0

Q6.

Answer :

We have:
f'(x)=limh→0f(0+h)-f(0)h =limh→0f(h)-f(0)h =limh→0tanh-tan0h =limh→0tanhh = 1

Q7.

Answer :

i We have:f’π2=limh→0fπ2+h-fπ2h =limh→0sinπ2+h-sinπ2h =limh→0cos h-1h =lim h→0 -2 sin2 h2h =limh→0 -2 sin2 h2h24 ×h4 =limh→0-1 ×h2 =0
ii We have:f'(x)=limh→0f(1+h)-f(1)h =limh→01+h-1h =limh→01 =1

iii We have:f’π2=limh→0fπ2+h-fπ2h =limh→02cosπ2+h-cosπ2h =limh→0-2sin h-0h =-2limh→0sinhh =-2(1) =-2
iv We have:f’π2=limh→0fπ2+h-fπ2h =limh→0sin2π2+h-sin2π2h =limh→0sin(π+2h)-0h =limh→0-sin2hh×22 =limh→0 -sin 2h2h×2 =-2

Page 30.26 Ex.30.2

Q1.

Answer :

i) ddxf(x)=limh→0fx+h-fxh =limh→02x+h-2xh =limh→02x-2x-2hhx(x+h) =limh→0-2hhx(x+h) =limh→0-2x(x+h) =-2×2
ii) ddxf(x)=limh→0fx+h-fxh =limh→01x+h-1xh =limh→0x-x+hhxx+h×x+x+hx+x+h =limh→0x-x-hhxx+hx+x+h =limh→0-hhxx+hx+x+h =limh→0-1xx+hx+x+h =-1xxx+x =-1x× 2x =-12×32 =-12x-32
iii) ddxf(x)=limh→0fx+h-fxh =limh→01(x+h)3-1x3h =limh→0x3-(x+h)3h(x+h)3×3 =limh→0x3-x3-3x2h-3xh2-h3h(x+h)3×3 =limh→0-3x2h-3xh2-h3h(x+h)3×3 =limh→0h-3×2-3xh-h2h(x+h)3×3 =limh→0-3×2-3xh-h2(x+h)3×3 =-3x2x6 =-3×4 =-3x-4
iv) ddxf(x)=limh→0fx+h-fxh =limh→0(x+h)2+1x+h-x2+1xh =limh→0x2+2xh+h2+1x+h-x2+1xh =limh→0x3+2x2h+h2x+x-x3-x2h-x-hxh(x+h) =limh→0x2h+h2x-hx(x+h) =limh→0h(x2+hx-1)xh(x+h) =limh→0x2+hx-1x(x+h) =x2-1×2

v) ddxf(x)=limh→0fx+h-fxh =limh→0(x+h)2-1x+h-x2-1xh =limh→0x2+2xh+h2-1x+h-x2-1xh =limh→0x3+2x2h+h2x-x-x3-x2h+x+hxh(x+h) =limh→0x2h+h2x+hx(x+h) =limh→0h(x2+hx+1)xh(x+h) =limh→0x2+hx+1x(x+h) =x2+1×2

vi) ddxf(x)=limh→0fx+h-fxh =limh→0x+h+1x+h+2-x+1x+2h =limh→0x+h+1x+2-x+h+2x+1hx+h+2x+2 =limh→0x2+2x+hx+2h+x+2-x2-x-hx-h-2x-2hx+h+2x+2 =limh→0hhx+h+2x+2 =limh→01x+h+2x+2 =1x+0+2x+2 =1x+22

vii) ddxf(x)=limh→0fx+h-fxh =limh→0x+h+23x+h+5-x+23x+5h =limh→0x+h+23x+3h+5-x+23x+5h =limh→0x+h+23x+5-3x+3h+5x+2h3x+3h+53x+5 =limh→03×2+3xh+6x+5x+5h+10-3×2-3xh-5x-6x-6h-10h3x+3h+53x+5 =limh→0-hh3x+3h+53x+5 =limh→0-13x+3h+53x+5 =-13x+52

viii) ddxf(x)=limh→0fx+h-fxh =limh→0 k x+hn-kxnh =limx+h-x→0 k x+hn-xnx+h-xHere, we have: limx→axm-amx-a=m am-1 =k n xn-1

ix) ddxf(x)=limh→0fx+h-fxh =limh→0 13-x-h-13-xh =limh→0 3-x-3-x-hh3-x3-x-h =limh→0 3-x-3-x-hh3-x3-x-h× 3-x+3-x-h3-x+3-x-h =limh→0 3-x-3+x+hh3-x3-x-h3-x+3-x-h =limh→0 hh3-x3-x-h3-x+3-x-h =limh→0 13-x3-x-h3-x+3-x-h =13-x3-x-03-x+3-x-0 =13-x 23-x =123-x32

x) ddxf(x)=limh→0fx+h-fxh =limh→0 x+h2+x+h+3-x2+x+3h =limh→0x2+h2+2xh+x+h+3-x2-x-3h =limh→0h2+2xh+hh =limh→0 h(h+2x+1)h =limh→0 h+2x+1 =0+2x+1 =2x+1

xi) ddxf(x)=limh→0fx+h-fxh =limh→0 x+h+23-x+23h =limh→0x+h+2-x-2x+h+22+x+h+2x+2+x+22h =limh→0hx+h+22+x+h+2x+2+x+22h =limh→0 x+h+22+x+h+2x+2+x+22 =x+0+22+x+0+2x+2+x+22 =x+22+x+22+x+22 =3x+22

xii) ddxf(x)=limh→0fx+h-fxh =limh→0 x+h3+4x+h2+3x+h+2-x3+4×2+3x+2h =limh→0x3+3x2h+3xh2+h3+4×2+4h2+8xh+3x+3h+2-x3-4×2-3x-2h =limh→0 3x2h+3xh2+h3+4h2+8xh+3h+2h =limh→0 h3x2+3xh+h2+4h+8x+3h =limh→03×2+3xh+h2+4h+8x+3 =3×2+8x+3

xiii) x2+1x-5=x3-5×2+x-5ddxf(x)=limh→0fx+h-fxh =limh→0 x+h3-5x+h2+x+h-5-x3-5×2+x-5h =limh→0 x3+3x2h+3xh2+h3-5×2-5h2-10xh+x+h-5-x3+5×2-x+5h =limh→0 3x2h+3xh2+h3-5h2-10xh+hh =limh→0 h 3×2+3xh+h2-5h-10x+1h =limh→0 3×2+3xh+h2-5h-10x+1 =3×2-10x+1

xiv) ddxf(x)=limh→0fx+h-fxh =limh→0 2x+h2+1-2×2+1h =limh→02×2+2h2+4xh+1-2×2+1h =limh→02×2+2h2+4xh+1-2×2+1h×2×2+2h2+4xh+1+2×2+12×2+2h2+4xh+1+2×2+1 =limh→0 2×2+2h2+4xh+1-2×2-1h2x2+2h2+4xh+1+2×2+1 =limh→0 h2h+4xh2x2+2h2+4xh+1+2×2+1 =limh→0 2h+4x2x2+2h2+4xh+1+2×2+1 =4x2x2+1+2×2+1 =4x22x2+1 =2x2x2+1

xv) ddxf(x)=limh→0fx+h-fxh =limh→0 2x+h+3x+h-2-2x+3x-2h =limh→0 2x+2h+3x-2-x+h-22x+3hx+h-2x-2 =limh→0 2×2+2xh+3x-4x-4h-6-2×2-2xh+4x-3x-3h+6hx+h-2x-2 =limh→0 -7hhx+h-2x-2 =limh→0 -7x+h-2x-2 =-7x-2x-2 =-7x-22

Q2.

Answer :

i) ddxf(x)=limh→0fx+h-fxhddxex=limh→0e-(x+h)-e-xh =limh→0e-xe-h-e-xh =limh→0e-xe-h-1h =-e-xlimh→0e-h-1-h =-e-x1 =-e-x

ii) ddxf(x)=limh→0fx+h-fxhddxe3x=limh→0e3(x+h)-e3xh =limh→0e3xe3h-e3xh =limh→0e3xe3h-13h =3 e3xlimh→0e3h-13h =3 e3x1 =3 e3x

iii) ddxf(x)=limh→0fx+h-fxhddxeax+b=limh→0ea(x+h)+b-eax+bh =limh→0eax+beah-eax+bh =limh→0eax+beah-1h =a eax+blimh→0eah-1ah =a eax+b1 =a eax+b

iv) ddxf(x)=limh→0fx+h-fxhddxx ex=limh→0(x+h )e(x+h)-x exh =limh→0(x+h) exeh-x exh =limh→0x exeh+hexeh-x exh =limh→0x exeh-x exh+limh→0h exehh =limh→0x exeh-1h+limh→0exeh =xex1+exe0 =xex+ex

v) ddxf(x)=limh→0fx+h-fxhddxx2 ex=limh→0(x+h)2e(x+h)-x2exh =limh→0(x2+2xh+h2)exeh-x2exh =limh→0x2exeh+2xhexeh+h2exeh-x2exh =limh→0x2exeh-x2exh+limh→02 x h exehh+limh→0h2exehh =limh→0x2exeh-1h+limh→02 x exeh+limh→0 h exeh =x2ex1+2xex1+0 =x2ex+2xex =x2+2x ex

vi) ddxf(x)=limh→0fx+h-fxhddxex2+1=limh→0e(x+h)2+1-ex2+1h =limh→0ex2+h2+2xh+1-ex2+1h =limh→0ex2+1eh2+2xh-ex2+1h =limh→0ex2+1ehh+2x-1h×h+2xh+2x =ex2+1limh→0 ehh+2x-1hh+2x limh→0 h+2x =ex2+11 2x =2x ex2+1

vii) ddxf(x)=limh→0fx+h-fxhddxe2x=limh→0e2(x+h)-e2xh =2 limh→0e2x+2h-e2x2x+2h-2x =2 limh→0 e2xe2x+2h-2x-12x+2h2-2×2 =2 e2x limh→0 e2x+2h-2x-12x+2h-2x2x+2h+2x =2 e2x limh→0 e2x+2h-2x-12x+2h-2x limh→012x+2h+2x =2 e2x 1122x =e2x2x

viii) ddxf(x)=limh→0fx+h-fxhddxeax+b=limh→0eax+ah+b-eax+bh =a limh→0eax+ah+b-eax+bax+ah+b-ax+b =a limh→0eax+beax+ah+b-ax+b-1ax+ah+b2-ax+b2 =a eax+b limh→0 eax+ah+b-ax+b-1ax+ah+b-ax+bax+ah+b+ax+b =a eax+b limh→0 eax+ah+b-ax+b-1ax+ah+b-ax+b limh→01ax+ah+b+ax+b =a eax+b 112ax+b =a eax+b2ax+b

ix) ddxf(x)=limh→0fx+h-fxhddxax=limh→0ax+h-axh =limh→0axax+h-x-1x+h-x =axlimh→0ax+h-x-1x+h2-x2 =axlimh→0ax+h-x-1x+h-xx+h+x =axlimh→0ax+h-x-1x+h-x limh→01x+h+x =ax loge a 12x =12xax loge a

x) ddxf(x)=limh→0fx+h-fxhddx3x2=limh→03x+h2-3x2h =limh→0 3×2+2xh+h2-3x2h =limh→0 3×2 3×2+2xh+h2-x2-1h×h+2xh+2x =3×2 limh→0 3hh+2x-1hh+2xlimh→0h+2x =3×2 log 3 2x =2x 3×2 log 3

Q3.

Answer :

i ddxf(x)=limh→0fx+h-fxh =limh→0 x+h sinx+h – x sin xh =limh→0x+hsin x cos h + cos x sin h- x sin xh =limh→0 x sin x cos h + x cos x sin h +h sin x cos h + h cos x sin hh =limh→0 x sin x cos h – x sin x + x cos x sin h +h sin x cos h + h cos x sin h – x sin x h =x sin x limh→0 cos h -1h+x cos x limh→0 sin h h+sin xlimh→0 cos h+cos xlimh→0 sin h =x sin x limh→0 -2 sin2 h2h24×h4+ x cos x 1+ sin x 1+ cos x 0 = x sin x × -h2 + x cos x 1+ sin x 1+ cos x 0 =-2x sin x 120+ x cos x + sin x = x cos x + sin x

ii ddxf(x)=limh→0fx+h-fxh =limh→0 x+h cos x+h – x cos xh =limh→0x+hcos x cos h-sin x sin h- x cos xh =limh→0 x cos x cos h – x sin x sin h+h cos x cos h – h sin x sin h- x cos xh =limh→0 x cos x cos h – x cos x – x sin x sin h+h cos x cos h – h sin x sin hh =x cos x limh→0 cos h -1h-x sin x limh→0 sin h h+cos xlimh→0 cos h+sin xlimh→0 sin h =x cos x limh→0 -2 sin2 h2h24×h4-x sin x 1+cos x 1+ sin x 0 =xcosx lim h→0-h2-x sin x 1+cos x 1+ sin x 0 =-x cos x 0-x sin x+ cos x = -x sin x+ cos x

iii ddxf(x)=limh→0fx+h-fxh =limh→0 sin 2x+2h-3-sin 2x-3hWe know:sin C-sin D=2 cos C+D2 sinC-D2 =limh→0 2 cos 2x+2h-3+2x-32 sin 2x+2h-3+2x-32 h =limh→0 2 cos 4x+2h-62 sin h h =limh→0 2 cos 4x+2h-62 limh→0 sin hh =2 cos 4x-62 1 = 2 cos 2x-3

iv ddxf(x)=limh→0fx+h-fxh =limh→0 sin 2x+2h-sin 2xh×sin 2x+2h+sin 2xsin 2x+2h+sin 2x =limh→0sin 2x+2h- sin 2xh sin 2x+2h+sin 2xWe have:sin C-sin D= 2 cos C+D2 sin C-D2 =limh→02 cos 2x+2h+2×2 sin 2x+2h-2x2h sin 2x+2h+sin 2x =limh→02 cos 2x+h sin hh sin 2x+2h+sin 2x =limh→0 2 cos 2x+h limh→0 sin h hlimh→01sin 2x+2h+sin 2x = 2 cos 2x 1 1sin 2x+sin 2x =2 cos 2x2sin 2x = cos 2xsin 2x

v ddxf(x)=limh→0fx+h-fxh =limh→0 sin x+hx+h-sin xxh =limh→0 x sin x+h- x+h sin x h x x+h =limh→0 x sin x cos h + cos x sin h- x sin x – h sin xh x x+h =limh→0 x sin x cos h +x cos x sin h- x sin x – h sin xh x x+h =limh→0 x sin x cos h- x sin x +x cos x sin h – h sin xh x x+h =x sin xlimh→0cos h -1h+x cos x x limh→0 sin hhlimh→0 1x+h-sin xxlimh→01x+h =x sin x limh→0 -2 sin2 h2h+x cos x x limh→0 sin hhlimh→0 1x+h-sin xxlimh→01x+h =x sin x limh→0 -2 sin2 h2h24×h4+x cos x x limh→0 sin hhlimh→0 1x+h-sin xxlimh→01x+h =-x sin x ×limh→0h2+x cos x x limh→0 sin hhlimh→0 1x+h-sin xxlimh→01x+h =-x sin x 12 0+cos xx-sin xx2 =cos xx-sin xx2 =x cos x-sin xx2

vi ddxf(x)=limh→0fx+h-fxh =limh→0 cos x+hx+h-cos xxh =limh→0 x cos x+h- x+h cos x h x x+h =limh→0 x cos x cos h – sin x sin h- x cos x – h cos xh x x+h =limh→0 x cos x cos h -x sin x sin h- x cos x – h cos xh x x+h =limh→0 x cos x cos h – x cos x-x sin x sin h – h cos xh x x+h =xcos xlimh→0cos h -1h-xsin x x limh→0 sin hhlimh→0 1x+h-cos xxlimh→01x+h =x cos x limh→0 -2 sin2 h2h24×h4-xsin x x limh→0 sin hhlimh→0 1x+h-cos xxlimh→01x+h ∵limh→0 sin2 h2h24=limh→0sin h2h2×limh→0sin h2h2=1×1, i.e. 1 = -x cosxlimh→0 h2-xsin x x limh→0 sin hhlimh→0 1x+h-cos xxlimh→01x+h =-x cos x ×0-sin x 11x-cos x x1x =0-sin xx-cos xx2 =-sin xx-cos xx2 =-x sin x-cos xx2

vii ddxf(x)=limh→0fx+h-fxh =limh→0 x+h2 sin x+h-x2 sin xh =limh→0 x2+h2+2xhsin x cos h + cos x sin h-x2 sin xh =limh→0 x2 sin x cos h + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin h-x2 sin xh =limh→0x2 sin x cos h-x2 sin x + x2 cos x sin h + h2 sin x cos h +h2 cos x sin h+2xh sin x cos h +2xh cos x sin hh =x2 sin xlimh→0cos h -1h+x2 cos xlimh→0sin hh+sin x limh→0 h cos h+ cos x limh→0 h sin h +2x sin xlimh→0 cosh +2x cos x limh→0 sin h =x2 sin x limh→0 -2 sin2 h2h24×h4+x2 cos xlimh→0sin hh+sin x limh→0 h cos h+ cos x limh→0 h sin h +2x sin xlimh→0 cosh +2x cos x limh→0 sin h ∵limh→0 sin2 h2h24=limh→0sin h2h2×limh→0sin h2h2=1×1, i.e. 1 =-x2sinx× limh→0 h2+x2 cos xlimh→0sin hh+sin x limh→0 h cos h+ cos x limh→0 h sin h +2x sin xlimh→0 cosh +2x cos x limh→0 sin h =-x2 sin x ×0+x2 cos x 1+ sin x 0+ cos x 0+ 2x sin x 1+2x cos x 0 =0+x2 cos x+ 2x sin x =0+x2 cos x+ 2x sin x =x2 cos x+ 2x sin x

viii ddxf(x)=limh→0fx+h-fxh = limh→0 sin 3x+h+1-sin 3x+1h =limh→0 sin 3x+3h+1-sin 3x+1h×sin 3x+3h+1+sin 3x+1sin 3x+3h+1+sin 3x+1 =limh→0sin 3x+3h+1-sin 3x+1h sin 3x+3h+1+sin 3x+1We have: sin C-sin D= 2 cos C+D2 sin C-D2 =limh→02 cos 3x+3h+1+3x+12 sin 3x+3h+1-3x-12h sin 3x+3h+1+sin 3x+1 =limh→02 cos 6x+3h+22 sin 3h2h sin 3x+3h+1+sin 3x+1 =limh→0 2 cos 6x+3h+22 limh→0 sin 3h 2h×32×32×limh→01sin 3x+3h+1+sin 3x+1 = 2 cos 3x+1 ×32 ×1sin 3x+1+sin 3x+1 =3 cos 3x+12sin 3x+1

ix ddxf(x)=limh→0fx+h-fxh =limh→0 sin x+h+cos x+h-sin x-cos xh =limh→0sin x+h -sin xh+limh→0 cos x+h-cos xhWe have: sin C-sin D= 2 cos C+D2 sin C-D2And, cos C-cos D=-2 sin C+D2 sin C-D2 =limh→0 2 cos 2x+h2 sin h2h+limh→0 -2 sin 2x+h2 sin h2h =2 limh→0 cos 2x+h2 limh→0 sin h2h2×12 -2 limh→0 sin 2x+h2 limh→0 sin h2h2×12 =2 cos x× 12-2 sin x ×12 =cos x – sin x

Q4.

Answer :

i) ddxf(x)=limh→0fx+h-fxh =limh→0tan2 x+h-tan2 xh =limh→0 tan x+h+tan xtan x+h-tan xh =limh→0 sin x+hcos x+h+sin xcos xsin (x+h)cos (x+h)-sin xcos x h =limh→0 sin x+h cos x + cos x+h sin x sin x+h cos x – cos x+h sin x h cos2 x cos2 x+h =limh→0 sin 2x+hsin hh cos2 x cos2 x+h =1cos2 xlimh→0 sin 2x+h limh→0 sin hh limh→0 1cos2x+h =1cos2 x sin 2x 11cos2 x =1cos2 x 2 sin x cos x 1cos2 x =2× sin x cos x×1cos2 x =2 tan x sec2x

ii) ddxf(x)=limh→0fx+h-fxh =limh→0 tan 2x+2h+1-tan 2x+1h =limh→0sin 2x+2h+1cos 2x+2h+1-sin 2x+1cos 2x+1h =limh→0 sin 2x+2h+1 cos 2x+1-cos 2x+2h+1 sin 2x+1 h cos 2x+2h+1 cos 2x+1 =limh→0 sin 2x+2h+1-2x-1h cos 2x+2h+1 cos 2x+1 =1cos 2x+1limh→0 sin 2h2h×2 limh→01 cos 2x+2h+1 =1cos 2x+1 ×2×1cos 2x+1 =2cos22x+1 =2 sec2 2x+1

iii) ddxf(x)=limh→0fx+h-fxh =limh→0 tan 2x+2h-tan 2xh =limh→0sin 2x+2hcos 2x+2h-sin 2xcos 2xh =limh→0 sin 2x+2h cos 2x-cos 2x+2h sin 2x h cos 2x+2h cos 2x =limh→0 sin 2x+2h-2xh cos 2x+2h cos 2x =1cos 2xlimh→0 sin 2h 2h×2× limh→01 cos 2x+2h =1cos 2x×2×1cos 2x =2cos22x =2 sec2 2x

iv) ddxf(x)=limh→0fx+h-fxh =limh→0 tanx+h-tan xh×tanx+h+tan xtanx+h+tan x =limh→0tanx+h-tan xhtanx+h+tan x =limh→0sin x+hcos x+h-sin xcos xhtanx+h+tan x =limh→0 sin x+h cos x-cos(x+h) sin x htanx+h+tan x cos x+h cos x =limh→0 sin hhtanx+h+tan x cos x+h cos x =limh→0 sin hhlimh→0 1tanx+h+tan x cos x+h cos x =112 tan xcos2 x =sec2 x2 tan x

Q5.

Answer :

(i) Let f(x)= sin2x Thus, we have: f(x+h)=sin2x+hddxf(x)=limh→0fx+h-fxh=limh→0 sin 2x+2h-sin 2xhWe know:sin C- sin D=2 sin C-D2 cos C+D2=limh→0 2 sin2x+2h-2x cos2x+2h-2x h=limh→0 2×2 sin2x+2h-2×2 cos2x+2h+2×2 2h+2x-2x=limh→0 2×2 sin2x+2h-2×2 cos2x+2h-2×2 2x+2h-2x2x+2h+2x=limh→0 2×2 sin2x+2h-2×2 cos2x+2h-2×2 2× 2x+2h-2x22x+2h+2x= limh→0 sin2x+2h-2x22x+2h-2x2limh→0 2cos 2x+2h-2x22x+2h+2x =1× 2cos2x22x ∵limh→0sin2x+2h-2x22x+2h-2×2=1=cos2x2x
(ii) Let f(x) = cos x Thus, we have: f(x+h)=cos x+hddxfx=limh→0f(x+h)-f(x) h= limh→0cos x+h-cos xhWe know: cos C -cos D = -2sinC+D2 sinC-D2= limh→0 -2sin x+h+x2 sinx+h-x2h= limh→0 -2sin x+h+x2 sinx+h-x2x+h-x=limh→0 -2sin x+h+x2 sinx+h-x22×x+h+xx+h-x2=limh→0sinx+h-x2x+h-x2limh→0 -sinx+h+x2x+h+x =1×-sinx2x ∵limh→0sinx+h-x2x+h-x2=1 =-sinx2x

(iii) Let f(x) = tanxThus, we have:(x+h)=tanx+hddx(f(x))=lim h→0 f(x+h)-f(x) h=lim h→0 tanx+h-tanx h=lim h→0 sin x+h-xh cosx+h cos x ∵ tan A-tan B= sin(A-B)cos A cos B =lim h→0 sin x+h-xx+h-x cosx+h cos x =lim h→0 sin x+h-xx+h-xx+h-xcosx+h cos x =lim h→0sin x+h-xx+h-x.lim h→0 1x+h+xcosx+hcosx ∵ lim h→0sinx+h-xx+h-x=1=1×12xcosxcosx=12xsec2x
(iv) Let f(x)= tan x2 Thus, we have:f(x+h) = tan (x+h)2ddxf(x)=limh→0f(x+h)-f(x)h=limh→0 tan (x+h)2-tanx2h=limh→0sinx+h2-x2h cos x+h2cosx2 ∵ tan A-tan B=sin (A-B)cos A cos B=limh→0 sin(x2+h2+2hx-x2)hcosx+h2cosx2=limh→0sin(hh+2x)hh+2x cosx+h2cosx2×h+2x=limh→0 sin(hh+2x)(hh+2x)limh→0 h+2xcos(x+h)2cos x2 As limh→0 sin(hh+2x)(hh+2x)=1=1×2xcos2 x2=2x sec2x2

Q6.

Answer :

i ddxfx=limh→0fx+h-fxhddx-x=limh→0-x+h–xh =limh→0-x-h+xh =limh→0-hh =limh→0-1 =-1

ii -x-1=1-x ddxfx=limh→0fx+h-fxhddx1-x =limh→01-x+h-1-x h =limh→0-1x+h+1xh =limh→0-x+x+hh x x+h =limh→0hh x x+h =limh→01 x x+h =1x.x =1×2

iii ddxfx=limh→0fx+h-fxhddxsin x+1=limh→0sin x+h+1-sin x+1 hWe know:sin C-sin D=2 cos C+D2 sin C-D2 =limh→02 cos x+h+1+x+12 sin x+h+1-x-12h =limh→02 cos 2x+h+22 sin h2 h =2limh→0 cos 2x+h+22 limh→0 sin h2 h2×12 =2 cos x+1 ×12 =cos x+1

iv ddxfx=limh→0fx+h-fxhddxcos x-π8=limh→0cos x+h-π8-cos x-π8hWe know:cos C-cos D=-2 sin C+D2 sin C-D2 =limh→0-2 sin x+h-π8+x-π82 sin x+h-π8-x+π82h =limh→0-2 sin 2x+h-π42 sin h2 h =-2limh→0 sin 2x+h-π42 limh→0 sin h2 h2×12 =-2 sin x-π8 ×12 =-sin x-π8

Page 30.34 Ex.30.3

Q1.

Answer :

ddxx4-2 sin x+ 3 cos x=ddxx4-2ddx sin x+ 3ddxcos x=4×4-2 cos x – 3 sin x

Q2.

Answer :

ddx3x+x3+33=ddx3x+ddxx3+ddx33=3x log 3+3×2 + 0=3x log 3+3×2

Q3.

Answer :

ddxx33-2x+5×2=13ddxx3-2ddxx12+5ddxx-2=133×2-2.12.x-12+5-2x-3=x2-x-12-10x-3

Q4.

Answer :

ddxex log a+ea log x+ea log a=ddxex log a+ddxea log x+ddxea log a=ddxelog ax+ddxelog xa+ddxelog aa=ddxax+ddxxa+ddxaa=ax log a + axa-1+0 =ax log a + axa-1

Q5.

Answer :

ddx2x2+13x+2=ddx6x3+4×2+3x+2=6ddxx3+4ddxx2+3ddxx+ddx2=63×2+42x+31+0=18×2+8x+3

Q6.

Answer :

ddxlog3 x+ 3 loge x+2 tan x=ddxlog xlog 3 +3ddx loge x+2ddxtan x=1log 3.1x+3.1x+2 sec2 x=1x log 3+3x+2 se

Q7.

Answer :

ddxx+1xx+1x=ddxx+x-1×12+x-12=ddxx32+x12+x-12+x-32=ddxx32+ddxx12+ddxx-12+ddxx-32=32×12+12x-12-12x-32-32x-52

Q8.

Answer :

ddxx+1×3=ddxx3+3x21x+3x1x2+1×3=ddxx32+3ddxx12+3ddxx-12+ddxx-32=32×32-1+3.12×12-1+3-12x-12-1+-32x-32-1=32×12+32x-12-32x-32-32x-52

Q9.

Answer :

ddx2x2+3x+4x=ddx2x2x+ddx3xx+ddx4x=2ddxx+3ddx1+4ddxx-1=21+30+4-1x-2=2-4×2

Q10.

Answer :

ddxx3+1x-2×2=ddxx4-2×3+x-2×2=ddxx4x2-2ddxx3x2+ddxxx2-ddx2x2=ddxx2-2ddxx+ddxx-1-2ddxx-2=2x-2-1×2-2-2x-3=2x-2-1×2+4×3

Q11.

Answer :

ddxa cos x+b sin x+csinx=ddxa cos xsin x+ddxb sin xsin x+ddxcsin x=addxcot x+ddxb+cddxcosec x=-a cosec2x+0-c cosec x cot x=-a cosec2x-c cosec x cot x

Q12.

Answer :

ddx2 sec x+ 3 cot x-4 tan x=2ddxsec x+3ddxcot x-4ddxtan x=2 sec x tan x-3 cosec2x-4 sec2x

Q13.

Answer :

ddxa0 xn+ddxa1 xn-1+ddxa2 xn-2+…+an-1 x+ an=a0ddxxn+a1ddxxn-1+a2ddxxn-2+…+an-1ddxx+ddxan=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+….+an-11+0=na0 xn-1+n-1 a1 xn-2+n-2a2 xn-3+….+an-1

Q14.

Answer :

ddx1sin x+2x+3+4logx3=ddxcosec x+2x. 23+4log 3log x=ddxcosec x+23ddx2x+4log 3ddxlog x=-cosec x cot x+23.2x.log 2+4log 3.1x=-cosec x cot x+2x+3.log 2+4xlog 3

Q15.

Answer :

ddxx+52×2-1x=ddx2x3+10×2-x-5x=ddx2x3x+ddx10x2x-ddxxx-ddx5x=2ddxx2+10ddxx-ddx1-5ddxx-1=22x+101-0-5-1x-2=4x+10+5×2

Q16.

Answer :

ddxlog 1x+5xa-3ax+x23+6 x-34=ddxlog x-12+5ddxxa-3ddxax+ddxx23+6ddxx-34=ddx-12log x+5ddxxa-3ddxax+ddxx23+6ddxx-34=-12.1x+5axa-1-3ax log a+23x-13+6-34x-74=-12x+5axa-1-3ax log a+23x-13-92x-74

Q17.

Answer :

ddxcos x+a=ddxcos x cos a – sin x sin a=cos addxcos x-sin a ddxsin x=-cos a sin x-sin a cos x=-sin x cos a + cos x sin a=-sinx+a

Q18.

Answer :

ddxcosx-2sin x=ddxcos x cos 2+sin x sin 2sin x=ddxcos x cos 2sin x+ddxsin x sin 2sin x=cos 2ddxcot x+sin 2ddx1=cos 2 -cosec2x+sin 2 0=-cosec2x cos 2

Q19.

Answer :

dydx=ddxsin x2+cos x22 =ddxsin2 x2+cos2x2+2 sin x2cos x2 =ddx1+sinx =ddx1+ddxsin x =0+cos x =cos xdydx at x = π6 = cos π6 = 32

Q20.

Answer :

dydx=ddx2-3 cos xsin x =ddx2sin x-ddx3 cos xsin x =2ddxcosec x-3ddxcot x =-2 cosec x cot x+3 cosec2xdydx at x=π4=-2 cosec π4 cot π4+3 cosec2π4 =-221+322 =-22+6 =6-22

Q21.

Answer :

Slope of the tangent = f'(x) =ddx2x6+x4-1 =2ddxx6+ddxx4-ddx1 =12×5+4×3∴ Slope of the tangent at x=1:1215+413=12+4=16

Q22.

Answer :

y=xa+ax=1ax12+ax-12dydx=1a12x-12+a-12x-32LHS = 2xy dydx = 2x 1ax12+ax-121a12x-12+a-12x-32 = 2x12a-12x+12x-a2x2 = 2x12a-a2x2 = xa-ax = RHS Hence, proved.

Q23.

Answer :

Rate = f'(x) =ddxx4-2×3+3×2+x+5 =ddxx4-2ddxx3+3ddxx2+ddxx+ddx5 =4×3-23×2+32x+1+0 =4×3-6×2+6x+1

Q24.

Answer :

dydx=ddx2x93-57×7+6×3-x =23ddxx9-57ddxx7+6ddxx3-ddxx =239×8-577×6+63×2-1 =6×8-5×6+18×2-1dydx at x = 1: 618-516+1812-1=6-5+18-1=18

Q25.

Answer :

f’x=λddxx2+μddxx+ddx12f’x=2λx+μ 1Given: f’4=152λ4+μ=15 From 1⇒8λ+μ=15 2Also, given: f’2=112λ2+μ=11 From 14λ+μ=11 3Subtracting equation (3) from equation (2):4λ=4λ=1Substituting this in equation (3):41+μ=11μ=7∴ λ=1 and μ=7

Q26.

Answer :

f’x =ddxx100100+x9999+…+x22+x+1 =1100100 x99+19999 x98+…+122x+1+0 =x99+x98+…+x+1f’1=199+198+…+1+1 =99+1 =100f’0=0+0+…+0+1 =1RHS=100 f’0 =1001 =100 =f’1 =LHS∴ f’1=100 f’0

Page 30.39 Ex.30.4

Q1.

Answer :

Let u=x3; v= sin xThen, u’= 3×2; v’=cos xUsing the product rule:ddxuv=uv’+vu’ddxx3 sin x=x3 cos x+ sin x 3×2 =x2 x cos x+ 3 sin x

Q2.

Answer :

Let u= x3; v=exThen, u’=3×2; v’=exUsing the product rule:ddxuv=uv’+vu’ddxx3 ex=x3 ex + ex 3×2 =x2ex x+3

Q3.

Answer :

Let u=x2; v=ex; w=log xThen, u’=2x; v’=ex, w= 1xUsing the product rule:ddxuvw=u’vw++uv’w+uvw’ddxx2ex log x=2x ex log x+x2 ex log x+x2 ex 1x =2x ex log x+x2 ex log x+x ex =x ex 2 log x+ x log x+1

Q4.

Answer :

Let u= xn; v=tan xThen, u’=nxn-1; v’=sec2xUsing the product rule:ddxuv=uv’+vu’ddxxn tan x=xn sec2x+tan xnxn-1 =xn-1 x sec2x + n tan x

Q5.

Answer :

Let u=xn; v=loga x=log xlog aThen, u’=n xn-1; v’=1x log aUsing the product rule:ddxuv=uv’+vu’ddxxn loga x=xn . 1x log a+loga x n xn-1 =xn-11 log a+loga x n xn-1 =xn-1 1log a+n loga x

Q6.

Answer :

Let u=x3+x2+1; v= sin xThen,u’= 3×2+2x; v’=cos xBy product rule,ddxuv=uv’+vu’ddxx3+x2+1 sin x=x3+x2+1 cos x+3×2+2x sin x

Q7.

Answer :

Let u= sin x; v= cos xThen, u’= cos x; v’= – sin xUsing the product rule:ddxuv=uv’+vu’ddxsin x cos x=sin x -sin x+cos x . cos x = -sin2x+cos2x = cos 2x

Q8.

Answer :

2x cot xx=2x cot x x-12Let u=2x; v=cot x; w=x-12Then, u’=2x log 2; v’=-cosec2 x; w’=-12x-32Using the product rule:ddxuvw=u’vw+uv’w+uvw’ddx2x cot x x-12 =2x log 2.cot x.x-12+2x-cosec2 x x-12+2x cot x-12x-32 =2x log 2.cot x.1x+2x-cosec2 x1x+2x cot x-12xx =2xxlog 2 . cot x-cosec2x-cot x2x

Q9.

Answer :

Let u=x2; v=sin x; w=log xThen, u’=2x; v’=cos x; w’=1xUsing the product rule:ddxuvw=u’vw+uv’w+uvw’ddxx2 sin x log x=2x sin x log x+x2 cos x log x+x2 sin x.1x =2x sin x log x+x2 cos x log x+x sin x

Q10.

Answer :

ddxx5 ex + x6 log x=x5 ddxex+ex ddxx5+x6ddxlog x+ log x ddxx6=x5 ex +ex 5×4 +x6.1x+log x6x5=x5 ex +ex 5×4 +x5+log x6x5=x4 xex + 5ex +x+6x log x

Q11.

Answer :

u= x sin x+cos x; v= x cos x- sin xu’=x cos x+ sin x- sin x = x cos x ; v’=- x sin x+cos x- cos x=- x sin x Using the product rule:ddxuv=uv’+vu’ddxx sin x+cos xx cos x- sin x= x sin x+cos x- x sin x + x cos x- sin x x cos x =- x2 sin2 x- x cos x sin x +x2 cos2 x- x cos x sin x =x2 cos2x – sin2x- x2 sin x cos x =x2 cos 2x- xsin 2x = x x cos 2x- sin 2x

Q12.

Answer :

Let u=x sin x + cos x; v = ex + x2 log x Then, u’ = xddxsin x+sinx ddxx-sin x = x cos x +sinx – sinx = x cos x v’ = ex +x2 ddxlog x + log x ddxx2 = ex +x + 2x log x Using the product rule: ddxuv= u v ‘ + v u’ddxx sinx + cos x ex + x2 cos x=x sin x + cos x ex +x + 2x log x + ex + x2 log x x cos x

Q13.

Answer :

Let u=1-2 tan x; v = 5 + 4 sin x Then, u’ = -2 sec2x; v’ = 4 cos xUsing the product rule:ddxuv=u v’ + v u’ddx1-2 tan x 5 + 4 sin x =1-2 tan x 4 cos x+ 5 + 4 sin x -2 sec2x= 4 cos x -8 ×sinxcosxcosx-10sec2x-8× sinxcos2x=4 cos x – 8 sin x – 10sec2 x-8 sec x tan x =4cos x- 2 sin x-52 sec2 x-2 sec x tan x

Q14.

Answer :

Let u=1+x2; v= cos xThen ,u’=2x; v’ =-sinxUsing the product rule:ddxuv=uv’+vu’ddx1+x2cos x=1+x2-sinx+ cos x2x =-sinx – x2 sinx + 2x cos x

Q15.

Answer :

ddxsin2 x=2 sin x ddxsin x (Using the chain rule)= 2 sin x cos x= sin 2x

Q16.

Answer :

logx2 x=log xlog x2 (by change of base property) =log x 2 log x log x2=2 log x =12Now ddxlogx2 x =ddx12 =0 ∵12 is a constant

Q17.

Answer :

Let u=ex; v= log x; w=tan xThen , u’ = ex; v’= 1x×12x=12x; w’ = sec2 xUsing the product rule:ddxuvw=u’vw + uv’w + uvw’ =ex log xtan x+ex×12xtan x+exlog x sec2 x =ex log x12.tan x+tanx2x+log x12.sec2 x =ex 12 log x. tan x+tanx2x+ 12 log x. sec2 x =ex2log x.tan x+tan xx+log x.sec2x

Q18.

Answer :

Let u=x3; v=ex; w=cos xThen ,u’=3×2; v’=ex; w’=-sin xUsing the product rule:ddxuvw=u’vw+uv’w+uvw’ddxx3 ex cos x=3×2 ex cos x+x3 ex cos x+x3 ex -sin x =x2 ex 3 cos x+x cos x-x sin x

Q19.

Answer :

x2 cos π4sin x=x2 cos π4 cosec xLet u=x2; v=cos π4; w= cosec xThen, u’=2x; v’=0; w’=-cosec x cot xUsing the product rule:ddxuvw=u’vw+uv’w+uvw’ddxx2 cos π4 cosec x=2x cos π4cosec x+ x2 .0.cosec x+x2 cos π4-cosec x cot x = cos π42x cosec x -x2 cosec x cot x = cos π42xsin x -x2 cot xsin x

Q20.

Answer :

Let u=x4; v=5 sin x – 3 cos xThen, u’=4×3; v’= 5 cos x – 3 (-sinx) = 5 cos x + 3 sin x According to the product rule:ddxuv=u v’+v u’ddxx45 sin x – 3 cos x=x4 5 cos x + 3 sin x +5 sin x – 3 cos x 4×3 =x35x cos x + 3 x sin x + 20 sin x – 12 cos x =x33x+20 sin x +5x – 12 cos x =3 x4 sinx +20x3sin x+5x cos x -12 cos x

Q21.

Answer :

Let u=2×2-3; v=sin xThen,u’=4x; v’= cos xUsing the product rule:ddxuv=uv’+vu’ddx2x2-3 sin x =2×2-3 cos x+ 4x sin x

Q22.

Answer :

Let u=x5; v=3-6x-9Then, u’=5×4; v’=54x-10Using the product rule:ddxuv=uv’+vu’ddxx53-6x-9=x554x-10+3-6x-95×4 =54 x-5+15 x4-30x-5 =15 x4+24x-5

Q23.

Answer :

Let u=x-4; v=3-4x-5Then, u’=-4x-5; v’=20 x-6Using the product rule:ddxuv=uv’+vu’ddxx-4 3-4x-5=x-4 20 x-6+3-4x-5-4x-5 =20x-10-12x-5+16x-10 =-12x-5 +36x-10

Q24.

Answer :

Let u=x-3; v=5+3xThen, u=-3x-4; v’=3Using the product rule:ddxuv=uv’+vu’ddxx3 5+3x=x-3.3+5+3x -3x-4 =3x-3 – 15 x-4 – 9 x-3 =-15 x-4-6 x-3

Q25.

Answer :

Product rule (1st method):Let u=1+2 tan x; v=5+4 cos xThen, u’=2sec2x; v’=-4 sin xUsing the product rule:ddxuv=uv’+vu’ddx1+2 tan x5+4 cos x=1+2 tan x-4 sin x+5+4 cos x2sec2x =-4 sin x-8 tan x sin x+10 sec2x+8 sec x =-4 sin x+10 sec2x+8cosx- 8sin2xcos x = -4 sin x+10 sec2x+81-sin2xcos x =-4 sin x+10 sec2x+8cos2xcos x =-4 sin x+10 sec2x+8 cos x2nd method:1+2 tan x5+4 cos x=5+4 cos x+10 tan x+ 8 sin xNow, we have:ddx1+2 tan x5+4 cos x=ddx5+4 cos x+10 tan x+ 8 sin x =-4 sin x+10 sec2x+8 cos x Using both the methods, we get the same answer.

Q26.

Answer :

i Product rule (1st method):Let u=3×2+2; v=3×2+2Then, u’=6x; v’=6xUsing the product rule:ddxuv=uv’+vu’ddx3x2+23×2+2=3×2+26x+3×2+26x =18×3+12x+18×3+12x =36×3+24x 2nd method:ddx3x2+22=ddx9x4+12×2+4 =36×3+24xUsing both the methods, we get the same answer.

ii Product rule (1st method):Let u=x+2; v=x+3Then, u’=1; v’=1Using the product rule:ddxuv=uv’+vu’ddxx+2x+3=x+21+x+31 =x+2+x+3 =2x+52nd method:ddxx+2x+3=ddxx2+5x+6 =2x+5Using both the methods, we get the same answer.

iii Product rule (1st method):Let u=3 sec x-4 cosec x; v=-2 sin x+ 5 cos xThen, u’=3 sec x tan x+4 cosec x cot x; v’=- 2 cos x-5 sin xUsing the product rule:ddxuv=uv’+vu’ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=3 sec x-4 cosec x- 2 cos x-5 sin x+-2 sin x+ 5 cos x3 sec x tan x+4 cosec x cot x =-6+15 tan x+8 cot x+20 -6 tan2x-8 cot x-15 tan x+ 20 cot2x =-6+20 -6sec2x -1+ 20 cosec2x-1 =-6+20 -6sec2x +6+ 20 cosec2x-20 =-6 sec2x+20 cosec2x2nd method:ddx3 sec x-4 cosec x-2 sin x+ 5 cos x=ddx-6 sec x sin x +15 sec x cos x+8 cosec x sin x -20 cosec x cos x =ddx-6 sin xcos x +15cos xcos x+8 sin xsin x -20 cos xsin x =ddx-6 tan x+15 +8 -20 cot x =ddx-6tan x-20 cot x+23 =-6 sec2x+20 cosec2x

Using both the methods, we get the same answer.

Q27.

Answer :

(ax+b)(a+d)2Let u =ax+b, v= a+d2Then, u’=a, v’= 0Using the product rule:ddxuv=u v’ + v u’ddx(ax+b)(a+d)2=(ax+b)×0+ a+d2×a∴ ddx(ax+b)(a+d)2=aa+d2

Q28.

Answer :

ax+bn cx+dnLet u =ax+bn, v=cx+dnThen, u’=naax+bn-1, v’=nccx+dn-1Using the product rule:ddxuv=uv’+u’vddxax+bn cx+dn=ax+bn×nccx+dn-1+naax+bn-1×cx+dn =nax+bn-1cx+dn-1acx+cb+acx+ad =nax+bn-1cx+dn-12acx+cb+ad

Page 30.44 Ex.30.5

Q1.

Answer :

Let u=x2+1; v=x+1Then, u’=2x; v’=1Using the quotient rule:ddxuv=vu’-uv’v2ddxx2+1x+1=x+12x-x2+11(x+1)2 =2×2+2x-x2-1(x+1)2 =x2+2x-1(x+1)2

Q2.

Answer :

Let u=2x-1; v=x2+1;Then, u’=2; v’=2xUsing the quotient rule:ddxuv=vu’-uv’v2ddx2x-1×2+1=x2+12-2x-12x(x2+1)2 =2×2+2-4×2+2x(x2+1)2 =-2×2+2x+2(x2+1)2 =21+x-x2(x2+1)2

Q3.

Answer :

Let u=x+ex; v=1+logxThen, u’=1+ex; v’=1xUsing the quotient rule:ddxuv=vu’-uv’v2ddxx+ex1+logx=1+log x1+ex-x+ex1x(1+log x)2 =x+xex+x log x+x log x ex-x-exx(1+logx)2 =x log x+ x log x ex-ex+x exx(1+logx)2 =x log x 1+ex-ex1-xx(1+logx)2

Q4.

Answer :

Let u=ex-tan x; v=cot x-xnThen, u’=ex-sec2x; v’=-cosec2x-nxn-1Using the quotient rule:ddxuv=vu’-uv’v2ddxex-tan xcot x-xn=cot x-xnex-sec2x-ex-tan x-cosec2x-nxn-1cot x-xn2 =cot x-xnex-sec2x+ex-tan xcosec2x+nxn-1cot x-xn2

Q5.

Answer :

Let u=ax2+bx+c; v=px2+qx+rThen, u’=2ax+b; v’=2px+qUsing the quotient rule:ddxuv=vu’-uv’v2ddxax2+bx+cpx2+qx+r=px2+qx+r2ax+b-ax2+bx+c2px+qpx2+qx+r2 =2apx3+2aqx2+2arx+bpx2+bqx+br-2apx3-2bpx2-2pcx-aqx2-bqx-cqpx2+qx+r2 =aq-bpx2+2ar-xpx+br-cqpx2+qx+r2

Q6.

Answer :

Let u=x; v=1+tan xThen, u’=1; v’= sec2xUsing the quotient rule:ddxuv=vu’-uv’v2 =1+tan x×1-xsec2x1+tanx2 =1+tan x-xsec2x1+tanx2

Q7.

Answer :

ddx1ax2+bx+c=ddxax2+bx+c-1=-1ax2+bx+c-2ddxax2+bx+c (Using the chain rule)=-1ax2+bx+c-22ax+b=-2ax+bax2+bx+c2

Q8.

Answer :

Let u=ex; v=1+x2Then, u’=ex; v’=2xUsing the chain rule:ddxuv=vu’-uv’v2ddxex1+x2=1+x2ex-ex2x1+x22 =ex+x2ex-2xex1+x22 =ex1+x2-2×1+x22 =ex1-x21+x22

Q9.

Answer :

Let u=ex+sin x; v=1+ log xThen, u’=ex+cos x; v’= 1xUsing the quotient rule:ddxuv=vu’-uv’v2ddxex+sin x1+ log x=1+log xex+cos x-ex+sin x 1×1+ log x2 =x1+log xex+cos x-ex+sin xx1+ log x2

Q10.

Answer :

Let u=x tan x; v=sec x+tan xThen, u’=x sec2x+tan x; v’=sec x tan x + sec2xUsing the quotient rule:ddxuv=vu’-uv’v2ddxxtan x sec x +tan x =sec x+tan xx sec2x+tan x-x tan xsec x tan x + sec2xsec x+tan x2 =x sec3x+x sec2xtanx+secx tanx+tan2x-x sec x tan2x-x tanx sec2xsec x+tan x2 =sec x+tan xx sec2x+tan x-x tan x secxsec x+tan xsec x+tan x2 =x sec2x+tan x-x tanx secxsec x+ tan x =x sec xsec x -tan x+tan xsec x + tan x

Q11.

Answer :

Let u=x sinx; v=1+ cos xThen, u’ = x cos x + sinx; v’ = -sin xUsing the quotient rule:ddxuv=vu’-uv’v2ddxx sin x1+cos x=1+ cos xx cos x + sinx-x sinx-sin x1+ cos x2 =1+ cos xx cos x + sinx+x sin2 x1+ cos x2 =1+ cos xx cos x + sinx+x 1-cos2x1+ cos x2 =1+ cos xx cos x + sinx+x1+cos x1-cos x1+ cos x2 =1+ cos xx cos x + sinx+x- xcosx1+ cos x2 =1+ cos xx+sin x1+ cos x2

Q12.

Answer :

Let u=2xcot x; v=xThen, u’=-2xcosec2x+2xlog 2 cot x; v’=12xddxuv=vu’-uv’v2ddx2xcot xx=x-2xcosec2x+2xlog 2 cot x-2xcot x12xx2 =x-2xcosec2x+2xlog 2 cot x-2x-1cot xxx =2x-xcosec2x+x cot x log 2 -12cot xxx =2x-xcosec2x+x cot x log 2 -12cot xx32

Q13.

Answer :

Let u=sin x-x cos x; v=x sinx + cos xThen, u’=cosx+x sin x-cosx; v’=x cos x+sin x-sinx =x sin x =x cos xUsing the quotient rule:ddxuv=vu’-uv’v2ddxsin x-x cos xx sinx + cos x=x sinx + cos xx sinx -sin x-x cos xx cos xx sinx + cos x2 =x2sin2x+x cosx sinx -x cos x sin x+x2cos2xx sinx + cos x2 =x2sin2x+cos2xx sinx + cos x2 =x2x sinx + cos x2

Q14.

Answer :

Let u=x2-x+1; v= x2+x+1Then, u’=2x-1; v’=2x+1By quotient rule,ddxuv=vu’-uv’v2ddxx2-x+1×2+x+1=x2+x+12x-1-x2-x+12x+1×2+x+12 =2×3+2×2+2x-x2-x-1-2×3+2×2-2x-x2+x-1×2+x+12 =2×2-2×2+x+12 =2×2-1×2+x+12

Q15.

Answer :

Let u=a+x; v=a-xThen,u’=12x; v’=-12xUsing the quotient rule:ddxuv=vu’-uv’v2ddxa+xa-x=a-x12x-a+x-12xa-x2 =a-x+a+x2xa-x2 =2a2xa-x2 =axa-x2

Q16.

Answer :

Let us use the quotient rule here.
We have:
u = a + sin x and v =1 + a sin x
u’ = cos x and v’=a cos x

Using the quotient rule:ddxuv=vu’-uv’v2ddxa+sinx1+asinx=(1+asinx)(cosx)-(a+sinx)(acosx)(1+asin x)2 =cosx+asinx cosx -a2cosx-a sinx cosx(1+asin x)2 =cosx-a2cosx(1+asinx)2 =(1-a2)cosx(1+a sinx)2

Q17.

Answer :

Let u=10x; v=sin xThen,u’=10xlog 10; v’= cos xUsing the quotient rule:ddxuv=vu’-uv’v2ddx10xsin x=sin x 10xlog 10-10x cos xsin2x =sin x 10xlog 10sin2x-10x cos xsin2x =10xlog 10 cosec x -10xcosec x cot x =10x cosec xlog 10 – cot x

Q18.

Answer :

Let u=1+3x; v=1-3xThen,u’=3x log 3; v’=-3x log 3Using the quotient rule:ddxuv=vu’-uv’v2ddx1+3×1-3x=1-3x3x log 3-1+3x-3x log 31-3×2 =3x log 3-32x log 3+3x log 3+32x log 31-3×2 =2 . 3x log 31-3×2

Q19.

Answer :

Let u=3x; v=x+tan xThen, u’=3xlog 3; v’=1+sec2xBy quotient rule, we have:ddxuv=vu’-uv’v2ddx3xx+tan x=x+tan x3xlog 3-3x 1+sec2xx+tan x2 =3xx+tan x log 3- 1+sec2xx+tan x2

Q20.

Answer :

Let u=1+log x; v=1- log xThen, u’=1x; v’=-1xUsing the quotient rule:ddxuv=vu’-uv’v2ddx1+log x1- log x=1- log x1x-1+log x-1×1- log x2 =1-log x+1+log xx1- log x2 =2 x1- log x2

Q21.

Answer :

Let u=4x + 5 sin x; v=3x + 7 cos xThen, u’=4+5 cos x; v’=3-7 sin xUsing the quotient rule:ddxuv=vu’-uv’v2ddx4x + 5 sin x3x + 7 cos x=3x + 7 cos x4+5 cos x-4x + 5 sin x3-7 sin x3x + 7 cos x2 =12x+15 x cos x+28 cos x+35 cos2x-12x +28 x sin x-15 sin x +35 sin2x3x + 7 cos x2 =15 x cos x+ 28 x sin x+28 cos x15 sinx +35sin2x+cos2x3x + 7 cos x2 =15 x cos x+ 28 x sin x+28 cos x15 sinx +353x + 7 cos x2

Q22.

Answer :

Let u=x; v=1+tan xThen, u’=1; v’= sec2xUsing the quotient rule:ddxuv=vu’-uv’v2ddxx1+tan x=1+tan x1-xsec2x1+tan x2 =1+tan x-xsec2x1+tan x2

Q23.

Answer :

Let u=a+b sin x; v=c+d cos xThen, u’=b cos x; v’=-d sin xUsing the quotient rule:ddxuv=vu’-uv’v2ddxa+b sin xc+d cos x=c+d cos xb cos x-a+b sin x-d sin xc+d cos x2 =bc cos x+bd cos2x+ad sin x+bd sin2xc+d cos x2 =bc cos x+ad sin x+bd sin2x+cos2xc+d cos x2 =bc cos x+ad sin x+bd c+d cos x2

Q24.

Answer :

Let u=px2+qx+r; v=ax+bThen, u’=2px+q; v’=aUsing the quotient rule:ddxuv=vu’-uv’v2ddxpx2+qx+rax+b=ax+b2px+q-px2+qx+raax+b2 =2ap x2+aq x+2bp x+bq-ap x2-aq x-arax+b2 =ap x2+2bp x+bq -arax+b2

Q25.

Answer :

Let u=sec x-1; v=sec x+1Then,u’=sec x tan x; v’=sec x tan xUsing the quotient rule:ddxuv=vu’-uv’v2ddxsec x-1sec x+1=sec x+1sec x tan x-sec x-1sec x tan xsec x+12 =sec2x tan x+ sec x tan x-sec2x tan x+sec x tan xsec x+12 =2sec x tan xsec x+12

Q26.

Answer :

Let u=x5-cos x; v=sin xThen, u’= 5×4+sin x; v’=cos xUsing the quotient rule:ddxuv=vu’-uv’v2ddxx5-cos xsin x=sin x 5×4+sin x-x5-cos xcos xsin2x =-x5cos x + 5x4sin x+sin2x+cos2xsin2x =-x5cos x + 5x4sin x+1sin2x

Q27.

Answer :

Let u=x+cos x; v=tan xThen, u’=1-sin x; v’= sec2xUsing the quotient rule:ddxuv=vu’-uv’v2ddxx+cos xtan x=tan x1-sin x-x+cos xsec2xtan2x

Q28.

Answer :

Let u=x; v=sinnxThen,u’=1; v’=n sinn-1x.cos xUsing the quotient rule:ddxuv=vu’-uv’v2ddxxsinnx=sinnx.1-x n sinn-1 x.cos xsinnx2 =sinn-1xsinx-nx.cos xsin2n x =sinx-nx.cos xsin2n-n-1 x =sinx-nxcos xsinn+1 x

Q29.

Answer :

Let u=ax+b; v=px2+qx+rThen,u’=a; v’=2px+qUsing the quotient rule:ddxuv=vu’-uv’v2ddxax+bpx2+qx+r=px2+qx+ra-ax+b2px+qpx2+qx+r2 =ap x2+aq x+ar-2ap x2-2bp x-aq x-bqpx2+qx+r2 =-apx2-2bp x+ar-bqpx2+qx+r2

Q30.

Answer :

Let u=1; v=ax2+bx+cThen,u’=0; v’=2ax+bUsing the quotient rule:ddxuv=vu’-uv’v2ddx1ax2+bx+c=ax2+bx+c0-12ax+bax2+bx+c2 =-2ax+bax2+bx+c2

Page 30.45 (Very Short Answers)

Q1.

Answer :

Using the definition of derivative, we have:limx→cfx-fxx-c=f’c

Q2.

Answer :

limx→axfa-afxx-a=limx→axfa-afx-xfx+xfxx-a=limx→axfa-xfx+xfx-afxx-a=limx→a-xfx-fa+x-afxx-a=limx→a-xlimx→afx-fax-a+limx→ax-afxx-a=-a f’a+f(a)

Q3.

Answer :

Given: x<2∴ 2-x>0ddxx2-4x+4=ddx2-x2=ddx2-x (∵ 2-x>0)=0-1=-1

Q4.

Answer :

1+cos 2×2=2 cos2x2=cos2x=-cos x (∵ π2<x<π)ddx1+cos 2×2=ddx-cosx=–sinx=sinx

Q5.

Answer :

Case 1:x>0|x|=xThus, we have: ddxx|x|=ddxx.x=ddxx2=2x 1Case 2:x<0|x|=-xThus, we have: ddxx|x|=ddxx.-x=ddx-x2=-2x 2From (1) and (2), we have:ddxx|x|=2x, if x>0-2x, if x<0

Q6.

Answer :

Case 1:x>0x=xx+xx=x+xx=2x2ddxx+xx=ddx2x2=4x 1Case 2:x<0x=-xx+xx=x-xx=0ddxx+xx=ddx0=0 2From (1) and (2), we have:ddxx+xx=4x, if x>00, if x<0

Q7.

Answer :

fx=x+x-1Case 1: x<0 (∴ x-1<-1<0)x=-x; x-1=-x-1=-x+1fx=-x+-x+1=-2xf’x=-2Case 2: 0< x <1 (∴ x>0 and x-1<0)x=x; x-1=-x-1=1-xfx=x+1-x=1f’x=0Case 3: x>1 ∴ x>1>0 ⇒ x>0)x=x; x-1=x-1fx=x+x-1=2x-1f’x=2From case 1, case 2 and case 3, we have:f’x=-2, when x<0 0, when 0<x<12, when x>1

Q8.

Answer :

Let x = 2We know: 2>1 and 2<3∴ x>1 and x<3x-1=x-1 and x-3=-x-3=-x+3fx=x-1+x-3=x-1-x+3=2f’x=0

Q9.

Answer :

Case 1: x>0x=xfx=x2x=x2x=xf’x=1Case 2: x<0x=-xfx=x2x=x2-x=-xf’x=-1From case 1 and case 2, we have:f’x=1, if x>0-1, if x<0

Q10.

Answer :

Case 1: x>0x=x …1ddxlog x=log x =1x =1x (from (1))Case 2: x<0x=-x …2ddxlog x=log -x =1-x =1x (from (2))From case (1) and case(2), ddxlog x=1x

Q11.

Answer :

limx→1 fx-1x-1=limx→1 fx-1x-1×fx+1fx+1×x+1x+1=limx→1 fx-1x+1x-1fx+1=limx→1 fx-1x-1× limx→1x+1fx+1=f’1×1+1f1+1=2×21+1=2

Q12.

Answer :

Let x = -3We know:-3<-2Thus, we have: x<-2It gives x+2<0.∴ 2+x=x+2=-x+2=-x-2fx=3 2+x=3-x-2=-3x-6f’x=-3ddxx-ddx6=-3

Q13.

Answer :

The given series is a geometric series where a = 1 and r = x.
fx=1+x+x2+x3+…=11-xSum of the infinite series of a geometric series is a1-r.f’x=-1(1-x)2ddx(1-x)=-1(1-x)2(-1)=1(1-x)2

Page 30.46 (Very Short Answers)

Q14.

Answer :

f(x)=logx2x3 =logx3logx2 (Change of base property) =3 logx2 log x =32f’x=0 (Since 32 is a constant)

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