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Limits And Derivatives

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Question
CBSEENMA11015471
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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

  • π

Solution

D.

π

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Question
CBSEENMA11015492
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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

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Question
CBSEENMA11015546
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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

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Question
CBSEENMA11015678
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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

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Question
CBSEENMA11015547
Wired Faculty App

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

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⇒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

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