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Differential Equations

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Question
CBSEENMA12036165
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How many real solutions does the equation x7+ 14x5+ 16x3+ 30x – 560 = 0 have?

  • 7

  • 1

  • 3

  • 5

Solution

B.

1

x7+ 14x5+ 16x3+ 30x – 560 = 0
Let f(x) = x7+ 14x5+ 16x3+ 30x
⇒ f′(x) = 7x6+ 70x4+ 48x2+ 30 > 0 ∀ x.
∴ f (x) is an increasing function ∀ x.

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Question
CBSEENMA12036128
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If left parenthesis 2 space sin space straight x right parenthesis space dy over dx space plus space left parenthesis space straight y space plus 1 right parenthesis space cos space straight x space equals space 0 space and space straight y left parenthesis 0 right parenthesis space equals 1 space then space straight y open parentheses straight pi over 2 close parentheses is equal to

  • 4/3

  • 1/3

  • -2/3

  • -4/3

Solution

B.

1/3

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Question
CBSEENMA12036026
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If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

  • 3a2−26a+55=0

  • 3a2−32a+84=0

  • 3a2−34a+91=0

  • 3a2−23a+44=0

Solution

B.

3a2−32a+84=0

We know that, if x1, x2..... xn are n observations, then their standard deviation is given by

square root of 1 over straight n sum straight x subscript straight i superscript 2 space minus open parentheses fraction numerator space stack sum straight x subscript straight i with space below over denominator straight n end fraction close parentheses squared end root
we begin inline style space end style begin inline style have end style begin inline style comma end style begin inline style space end style begin inline style space end style begin inline style left parenthesis end style begin inline style 3 end style begin inline style. end style begin inline style 5 end style begin inline style right parenthesis squared end style begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator left parenthesis 2 squared space plus 3 squared space plus straight a squared plus 11 squared right parenthesis over denominator 4 end fraction begin inline style space end style begin inline style minus end style begin inline style space end style begin inline style open parentheses begin display style fraction numerator 2 plus 3 plus straight a plus 11 over denominator 4 end fraction end style close parentheses squared end style begin inline style space end style begin inline style space end style begin inline style space end style
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49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 4 straight a squared space plus 536 minus 256 minus straight a squared minus 32 straight a over denominator 16 end fraction
space begin inline style space end style begin inline style 49 end style begin inline style space end style begin inline style straight x end style begin inline style space end style begin inline style 4 end style begin inline style space end style begin inline style equals end style begin inline style space end style begin inline style 3 end style begin inline style straight a squared end style begin inline style minus end style begin inline style 32 end style begin inline style straight a end style begin inline style plus end style begin inline style 280 end style begin inline style space end style begin inline style 3 end style begin inline style straight a squared end style begin inline style minus end style begin inline style 32 end style begin inline style straight a end style begin inline style space end style begin inline style plus end style begin inline style space end style begin inline style 84 end style begin inline style space end style begin inline style equals end style begin inline style 0 end style

Question
CBSEENMA12036025
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Let bold a with hat on top space bold b with hat on top space and space bold c with hat on top space be three unit vectors such that straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x straight c with hat on top right parenthesis space equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top plus straight c with hat on top right parenthesis. space If space straight b with hat on top space is space not space parallel space to space stack straight c. with hat on top then the angle between bold a with bold hat on top bold space and space bold b with bold hat on top is

  • 3π/4

  • π/2

  • 2π/3

  • 5π/6

Solution

D.

5π/6

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Question
CBSEENMA12036129
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Let In ∫tan x dx,(n> 1) . I4 + I6 = a tan5x + bx5 + C, where C is a constant of integration, then the ordered pair (a, b) is equal to

  • open parentheses negative 1 fifth comma 0 close parentheses
  • open parentheses negative 1 fifth comma 1 close parentheses
  • open parentheses 1 fifth comma 0 close parentheses
  • open parentheses 1 fifth comma negative 1 close parentheses

Solution

C.

open parentheses 1 fifth comma 0 close parentheses

I4 +I6 = ∫(tan4x + tan6x) dx
= ∫tan4 x sec2 x dx
= 1/5 tan5 x + c
⇒ a = 1/5, b = 0

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