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41.

limxαtanxcotx1x - α is equal to

  • 2csc2α

  • 12sin2α

  • - 2csc2α

  • None of these


D.

None of these

We have, limxαtanxcotα1x - α= lim xα1 + tanxcotα - 11x - α= elim xαtanxcotα - 1x - α × 1tanα= elim xαtanx - tanαx - α × 1tanα= esec2α/tanα = e2csc2α


42.

For each t ∈R, let [t] be the greatest integer less than or equal to t. Then

limx0+ x1x+2x+......+15x

  • does not exist (in R)

  • is equal to 0

  • is equal to 15

  • is equal to 120


D.

is equal to 120

limx0+ x1x+2x +.......+15x Now 1x - 1 < 1x1x2x-1<2x2x15x-1 <15x15xlimx0+ x   15x -15 < limx0+ x1x + 2x + ..... + 15xlimx0+ x   15 xlimx0+ x  15-5x < L limx0+ 15L 120


43. 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

  • 1/16


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
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44.

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

  • -7/9

  • -3/5

  • 1/3

  • 1/3


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

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