Physics-
General
Easy

Question

Frequency ranges for micro waves are :

  1. 3 cross times 10 to the power of 9 end exponent text  to  end text 3 cross times 10 to the power of 4 end exponent H z    
  2. 3 cross times 10 to the power of 13 end exponent text  to  end text 3 cross times 10 to the power of 9 end exponent H z    
  3. 3 cross times 10 to the power of 14 end exponent text  to  end text 3 cross times 10 to the power of 9 end exponent H z    
  4. 3 cross times 10 to the power of 11 end exponent text  to  end text 3 cross times 10 to the power of 9 end exponent H z    

The correct answer is: 3 cross times 10 to the power of 13 end exponent text  to  end text 3 cross times 10 to the power of 9 end exponent H z

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A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2 divided by pi revolutions per s e c o n d around the vertical axis through the fixed end as shown in figure, then tension in the string is

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T equals M omega to the power of 2 end exponent L
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T sin invisible function application theta equals M omega to the power of 2 end exponent R(i)
T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta(ii)
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T equals M omega to the power of 2 end exponent L
equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
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cos invisible function application left parenthesis alpha minus beta right parenthesis equals 1 text  and  end text cos invisible function application left parenthesis alpha plus beta right parenthesis equals 1 over c where alpha comma beta element of left square bracket negative pi comma pi right square bracket Pairs alpha comma beta which satisfy both the equations is/are


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The number of integral values of k for which the equation 7 cos invisible function application x plus 5 sin invisible function application x equals 2 k plus 1 has a solution is

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The number of distinct real roots of open vertical bar table attributes columnalign center center center columnspacing 1em end attributes row cell sin invisible function application x end cell cell cos invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell sin invisible function application x end cell cell cos invisible function application x end cell row cell cos invisible function application x end cell cell cos invisible function application x end cell cell sin invisible function application x end cell end table close vertical bar equals 0 in the interval negative pi over 4 less or equal than x less or equal than pi over 4 is


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If 0 less or equal than x less than 2 pi, then the number of real values of x, which satisfy the equationcos invisible function application x plus cos invisible function application 2 x plus cos invisible function application 3 x plus cos invisible function application 4 x equals 0 is :

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The possible values of theta element of left parenthesis 0 comma pi right parenthesis such thatsin invisible function application theta plus sin invisible function application 4 theta plus sin invisible function application 7 theta equals 0  are :

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The number of values of x in the interval [0, 3straight pi] satisfying the equation ,2 sin squared invisible function application x plus 5 sin invisible function application x minus 3 equals 0 is



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Over left square bracket negative pi comma pi right square bracket the solution of 2 sin squared invisible function application open parentheses x plus pi over 4 close parentheses plus square root of 3 cos invisible function application 2 x greater or equal than 0 is



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Let f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction The sum of all the solutions of f (x) = 0 in [0, 100p] is

table row cell table row blank cell f left parenthesis x right parenthesis equals fraction numerator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent over denominator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses plus 1 minus c o t invisible function application x plus c o t invisible function application x end fraction end cell row blank cell text  defined for  end text R minus n pi comma n element of I end cell row blank cell f left parenthesis x right parenthesis equals fraction numerator left parenthesis c o t invisible function application x right parenthesis to the power of 4 end exponent over denominator 1 plus c o t to the power of 2 end exponent invisible function application x end fraction end cell row blank cell f left parenthesis x right parenthesis equals 0 rightwards double arrow c o t invisible function application x equals 0 rightwards double arrow x equals left parenthesis 2 n minus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell row blank cell therefore x equals fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction comma fraction numerator 5 pi over denominator 2 end fraction comma horizontal ellipsis horizontal ellipsis comma fraction numerator 199 pi over denominator 2 end fraction end cell row blank cell s u m equals fraction numerator pi over denominator 2 end fraction left square bracket 1 plus 3 plus 5 plus horizontal ellipsis horizontal ellipsis plus 199 right square bracket end cell end table end cell row cell text end text text ( end text text 100 end text text end text text s end text text o end text text l end text text u end text text t end text text i end text text o end text text n end text text s end text text ) end text text end text end cell row cell equals fraction numerator pi over denominator 2 end fraction times fraction numerator 100 over denominator 2 end fraction times 200 equals 5000 pi end cell end table

Let f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction The sum of all the solutions of f (x) = 0 in [0, 100p] is

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table row cell table row blank cell f left parenthesis x right parenthesis equals fraction numerator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent over denominator open parentheses c o s e c to the power of 2 end exponent invisible function application x minus 1 close parentheses plus 1 minus c o t invisible function application x plus c o t invisible function application x end fraction end cell row blank cell text  defined for  end text R minus n pi comma n element of I end cell row blank cell f left parenthesis x right parenthesis equals fraction numerator left parenthesis c o t invisible function application x right parenthesis to the power of 4 end exponent over denominator 1 plus c o t to the power of 2 end exponent invisible function application x end fraction end cell row blank cell f left parenthesis x right parenthesis equals 0 rightwards double arrow c o t invisible function application x equals 0 rightwards double arrow x equals left parenthesis 2 n minus 1 right parenthesis fraction numerator pi over denominator 2 end fraction end cell row blank cell therefore x equals fraction numerator pi over denominator 2 end fraction comma fraction numerator 3 pi over denominator 2 end fraction comma fraction numerator 5 pi over denominator 2 end fraction comma horizontal ellipsis horizontal ellipsis comma fraction numerator 199 pi over denominator 2 end fraction end cell row blank cell s u m equals fraction numerator pi over denominator 2 end fraction left square bracket 1 plus 3 plus 5 plus horizontal ellipsis horizontal ellipsis plus 199 right square bracket end cell end table end cell row cell text end text text ( end text text 100 end text text end text text s end text text o end text text l end text text u end text text t end text text i end text text o end text text n end text text s end text text ) end text text end text end cell row cell equals fraction numerator pi over denominator 2 end fraction times fraction numerator 100 over denominator 2 end fraction times 200 equals 5000 pi end cell end table
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The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n element of I

We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x
x equals fraction numerator n pi over denominator 3 end fraction comma n element of I

The general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x · tan 2x tan 3x is
where n element of I

maths-General
We know that
tan x · tan 2x · tan 3x = tan 3x – tan 2x – tan x
hence tan x + tan 2x + tan 3x = tan 3x – tan 2x – tan x
Þ tan x + tan 2x = 0
\ tan 2x = tan (– x)
2x = np – x
x equals fraction numerator n pi over denominator 3 end fraction comma n element of I
General
maths-

The sum of all the solution of cotθ= sin 2θ(θ ≠ nstraight pi, n integer), 0 ,less or equal than theta less or equal than pi is

For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0
table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table
Then the sum of roots is 3p/2

The sum of all the solution of cotθ= sin 2θ(θ ≠ nstraight pi, n integer), 0 ,less or equal than theta less or equal than pi is

maths-General
For the given relation cos q = (2 sin q cos q) sin q, sin q¹ 0
table row cell text end text text o end text text r end text text end text s i n invisible function application theta equals plus-or-minus fraction numerator 1 over denominator square root of 2 end fraction text end text text o end text text r end text text end text c o s invisible function application theta equals 0 end cell row cell text end text text o end text text r end text text end text theta equals fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction comma fraction numerator pi over denominator 2 end fraction end cell end table
Then the sum of roots is 3p/2
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