Limits and Derivatives

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Question 1

limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style cos end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction space is space equal space to
  • π/2

  • 1

  • π

Solution

D.

π

limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style cos end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction
space equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style straight pi end style begin display style left parenthesis end style begin display style 1 end style begin display style minus end style begin display style begin display style sin end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction space
equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style minus end style begin display style begin display style πsin end style squared end style begin display style straight x end style begin display style right parenthesis end style over denominator straight x squared end fraction
equals limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style left parenthesis end style begin display style straight pi end style begin display style space end style begin display style begin display style sin end style squared end style begin display style space end style begin display style straight pi end style begin display style right parenthesis end style over denominator straight x squared end fraction space space space space space left square bracket because space sin space left parenthesis straight pi space minus space straight theta space equals space sin space straight theta right square bracket
equals space limit as straight x space rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style begin display style πsin end style squared end style begin display style straight x end style over denominator straight pi space sin squared straight x end fraction space straight x space left parenthesis straight pi right parenthesis open parentheses fraction numerator begin display style sin squared end style begin display style straight x end style over denominator straight x squared end fraction close parentheses space equals space straight pi space open square brackets because space limit as straight theta rightwards arrow 0 of space fraction numerator sin begin display style space end style begin display style straight theta end style over denominator straight theta end fraction space equals 1 space close square brackets
Question 2

limit as straight x space rightwards arrow 0 of fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis left parenthesis 3 plus cos space straight x right parenthesis over denominator straight x space tan space 4 straight x end fraction space is space equal space to space
  • -1/4

  • 1/2

  • 1

  • 2

Solution

D.

2

Let space l space equals space space limit as straight x rightwards arrow 0 of space fraction numerator left parenthesis 1 minus cos space 2 straight x right parenthesis over denominator straight x squared end fraction fraction numerator left parenthesis 3 plus cos space straight x right parenthesis over denominator 1 end fraction. fraction numerator straight x over denominator tan space 4 straight x end fraction
equals space limit as straight x rightwards arrow 0 of space fraction numerator 2 space sin squared straight x over denominator straight x squared end fraction. fraction numerator 3 plus cosx over denominator 1 end fraction. fraction numerator straight x over denominator tan space 4 straight x end fraction
equals space 2 space limit as straight x rightwards arrow 0 of open parentheses fraction numerator sin space straight x over denominator straight x end fraction close parentheses squared. space limit as straight x rightwards arrow 0 of left parenthesis 3 plus space cos space straight x right parenthesis. space limit as straight x rightwards arrow 0 of fraction numerator 4 straight x over denominator space 4 space tan space 4 straight x end fraction
2.4.1 fourth space equals space 2
Question 3

stack lim space with straight x space rightwards arrow straight pi over 2 below space fraction numerator cot space straight x space minus cos space straight x over denominator left parenthesis straight pi minus 2 straight x right parenthesis cubed end fraction space equals
  • 1/4

  • 1/24

  • 1/16

  • 1/8

Solution

C.

1/16

limit as straight x rightwards arrow straight pi over 2 of space fraction numerator cot space straight x space left parenthesis 1 minus sin space straight x right parenthesis over denominator negative 8 space open parentheses straight x space minus begin display style straight pi over 2 end style close parentheses cubed end fraction
space equals space limit as straight x rightwards arrow fraction numerator begin display style straight pi end style over denominator begin display style 2 end style end fraction of space fraction numerator tan begin display style space end style begin display style open parentheses straight pi over 2 minus straight x close parentheses end style over denominator 8 open parentheses begin display style straight pi over 2 end style minus straight x close parentheses end fraction space fraction numerator open parentheses 1 minus space cos space open parentheses begin display style straight pi over 2 end style minus straight x close parentheses close parentheses over denominator open parentheses begin display style straight pi over 2 minus straight x end style close parentheses end fraction
space equals space 1 over 8.1.1 half space equals space 1 over 16
Question 4

If limit as straight x space rightwards arrow infinity of space open parentheses 1 plus straight a over straight x plus straight b over straight x squared close parentheses to the power of 2 straight x end exponent space equals space straight e squared then the values of a and b, are

  • straight a element of space straight R with equals below space straight b element of space straight R with equals below
  • straight a space equals 1 comma space straight b element of space straight R with equals below
  • straight a element of space straight R with equals below space comma space straight b space equals 2
  • a = 1 and b = 2

Solution

D.

a = 1 and b = 2

limit as straight x space rightwards arrow infinity of space open parentheses 1 plus straight a over straight x plus straight b over straight x squared close parentheses to the power of 2 straight x end exponent
space equals space limit as straight x space rightwards arrow infinity of space open parentheses 1 plus fraction numerator begin display style straight a end style over denominator begin display style straight x end style end fraction plus fraction numerator begin display style straight b end style over denominator begin display style straight x squared end style end fraction close parentheses to the power of open parentheses fraction numerator 1 over denominator begin display style straight a over straight x plus straight b over straight x squared end style end fraction close parentheses space straight x space 2 straight x space straight x space open parentheses straight a over straight x plus straight b over straight x squared close parentheses end exponent
space equals space straight e to the power of 2 straight a end exponent
rightwards double arrow space straight a space equals 1
straight b element of straight R
Question 5

If 5(tan2x – cos2x) = 2cos 2x + 9, then the value of cos4x is

  • -7/9

  • -3/5

  • 1/3

  • 2/9

Solution

A.

-7/9

5 space open square brackets fraction numerator 1 minus straight t over denominator straight t end fraction minus straight t close square brackets space equals space 2 left parenthesis 2 straight t minus 1 right parenthesis space plus space 9
left curly bracket space Let space cos squared space straight x space equals space straight t right curly bracket

⇒5(1 – t – t2) = t(4t + 7)
⇒ 9t2 + 12t – 5 = 0
⇒ 9t2 + 15t – 3t – 5 = 0
⇒ (3t – 1) (3t + 5) = 0
⇒ t = t/3 as t≠-5/3.
cos2x = 2(1/3)-1 = -1/3
cos space 4 straight x space equals space 2 space open parentheses negative 1 third close parentheses squared minus 1 space equals space minus 7 over 9