Differential Equations

More Topic from Mathematics

Question 1

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.

Question 2

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

left parenthesis 2 space plus space sin space straight x right parenthesis dy over dx plus left parenthesis straight y space plus 1 right parenthesis space cos space straight x space equals 0
straight d over dx left parenthesis 2 space plus space sin space straight x right parenthesis space left parenthesis straight y plus 1 right parenthesis space equals 0
left parenthesis 2 space plus space sin space straight x right parenthesis space left parenthesis straight y plus 1 right parenthesis space equals space straight c
straight x space equals space 0 comma space straight y space equals 1
rightwards double arrow space straight c space equals 4
straight y space plus 1 space equals space fraction numerator 4 over denominator 2 space plus space sin space straight x end fraction
straight y open parentheses straight pi over 2 close parentheses space equals space 4 over 3 minus 1 space equals space 1 third
Question 3

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
space begin inline style rightwards double arrow end style begin inline style space end style 49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 4 space plus space 9 space plus straight a squared plus 121 over denominator 4 end fraction begin inline style minus end style begin inline style open parentheses begin display style fraction numerator 16 plus straight a 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 begin inline style space end style begin inline style space end style
begin inline style rightwards double arrow end style begin inline style space end style 49 over 9 begin inline style space end style begin inline style equals end style begin inline style space end style fraction numerator 134 space plus straight a squared over denominator 4 end fraction begin inline style minus end style fraction numerator 256 space plus straight a squared plus 32 straight a over denominator 16 end fraction begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style begin inline style space end style
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 4

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

Given comma space vertical line straight a with hat on top vertical line space equals space vertical line stack straight b vertical line with hat on top space equals vertical line straight c with hat on top vertical line space equals space 1
and space space straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x space straight c with hat on top space right parenthesis equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top space plus space straight c with hat on top space right parenthesis
Now comma space consider space straight a with hat on top space straight x space left parenthesis straight b with hat on top space straight x space straight c with hat on top space 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 space plus space straight c with hat on top space right parenthesis
rightwards double arrow space left parenthesis straight a with hat on top. straight c with hat on top right parenthesis straight b with hat on top space minus space left parenthesis straight a with hat on top. straight b with hat on top right parenthesis straight c with hat on top space equals space fraction numerator square root of 3 over denominator 2 end fraction space left parenthesis straight b with hat on top space plus fraction numerator square root of 3 over denominator 2 end fraction straight c with hat on top
On space comparing comma space we space get

straight a with hat on top. straight b with hat on top space equals space minus space fraction numerator square root of 3 over denominator 2 end fraction space rightwards double arrow space vertical line straight a with hat on top vertical line space vertical line straight b with hat on top space vertical line space cosθ space equals space minus fraction numerator square root of 3 over denominator 2 end fraction
rightwards double arrow space cos space straight theta space equals space minus fraction numerator square root of 3 over denominator 2 end fraction space vertical line because space vertical line straight a with hat on top vertical line space equals space vertical line straight b with hat on top vertical line space equals space 1
rightwards double arrow space cos space straight theta space equals space cos space open parentheses straight pi space minus straight pi over 6 close parentheses rightwards double arrow space straight theta space equals space fraction numerator 5 straight pi over denominator 6 end fraction space
Question 5

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