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The value of 'c' of Lagrange's mean value theorem for f space left parenthesis x right parenthesis equals x squared minus 3 x minus 2 text  for  end text x element of left square bracket negative 1 , 2 right square bracket is

  1. 1/2    
  2. 0    
  3. 1    
  4. LMVT is not applicable    

The correct answer is: 1/2

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The value of 'c' of Rolle's mean value theorem for f space left parenthesis x right parenthesis equals vertical line x vertical line equals i n space left square bracket negative 1 , 1 right square bracket is

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The value of 'c' of Rolle's theorem for f space left parenthesis x right parenthesis equals l o g space open parentheses x squared plus 2 close parenthesesl o g subscript 3 end subscript on [–1, 1] is

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For f space left parenthesis x right parenthesis equals 4 minus left parenthesis 6 minus x right parenthesis to the power of 2 divided by 3 end exponent in [5, 7]

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The value of 'c' in Lagrange's mean value theorem for f space left parenthesis x right parenthesis equals x cubed minus 2 x squared minus x plus 4 in [0, 1] is

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The value of 'c' in Lagrange's mean value theorem for f space left parenthesis x right parenthesis equals x space left parenthesis x minus 2 right parenthesis squared in [0, 2] is

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The equation r equals a c o s space theta plus b s i n space theta represents

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The polar equation of the circle whose end points of the diameter are open parentheses square root of 2 comma fraction numerator pi over denominator 4 end fraction close parentheses and open parentheses square root of 2 comma fraction numerator 3 pi over denominator 4 end fraction close parentheses is

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The radius of the circle r equals 8 s i n space theta plus 6 c o s space theta is

The radius of the circle r equals 8 s i n space theta plus 6 c o s space theta is

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The adjoining figure shows the graph of y equals a x to the power of 2 end exponent plus b x plus c Then –

The adjoining figure shows the graph of y equals a x to the power of 2 end exponent plus b x plus c Then –

Maths-General
General
Maths-

Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and
Hence, the option (ais correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be = fraction numerator b squared over denominator 2 a end fraction space greater than space 0
On further solving this, we get
b space less than space 0 
Therefore, the option (b) is also correct.
Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and
Hence, the option (cwill also be correct
On checking all the options, and we can see all options are correct and
Therefore, we conclude that all the options available are correct.

Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –

Maths-General
As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and
Hence, the option (ais correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be = fraction numerator b squared over denominator 2 a end fraction space greater than space 0
On further solving this, we get
b space less than space 0 
Therefore, the option (b) is also correct.
Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and
Hence, the option (cwill also be correct
On checking all the options, and we can see all options are correct and
Therefore, we conclude that all the options available are correct.
General
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For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are

For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are

maths-General
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The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :


Clearly, y = a x squared space plus space b x space plus space c
 represent a parabola opening  downwards. Therefore, a < 0
 y = a x squared space plus space b x space plus space c
 cuts negative y- axis , Putting x = 0 in the given equation
rightwards double arrow-y = c
rightwards double arrowy = -c
rightwards double arrowc < 0
Thus, from the above graph c < 0.

The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :

Maths-General

Clearly, y = a x squared space plus space b x space plus space c
 represent a parabola opening  downwards. Therefore, a < 0
 y = a x squared space plus space b x space plus space c
 cuts negative y- axis , Putting x = 0 in the given equation
rightwards double arrow-y = c
rightwards double arrowy = -c
rightwards double arrowc < 0
Thus, from the above graph c < 0.
General
Maths-

The greatest possible number of points of intersections of 8 straight line and 4 circles is :

 Complete step-by-step answer:
The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
For selecting r objects from n objects can be done by using the formula as follows
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
 As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 cross times 1 = 28

 For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 cross times 2 = 12
For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows  C presuperscript 4 subscript 1 cross times C presuperscript 8 subscript 1 cross times 2 space= 64
Hence, the total number of points of intersection is = 28 + 12  + 64 = 104

The greatest possible number of points of intersections of 8 straight line and 4 circles is :

Maths-General
 Complete step-by-step answer:
The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
For selecting r objects from n objects can be done by using the formula as follows
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction
 As mentioned in the question, we have to find the total number of intersection points.

For calculating the points of intersection between two lines, we can use the formula which is mentioned in the hint as follows = C presuperscript 8 subscript 2 cross times 1 = 28

 For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = C presuperscript 4 subscript 2 cross times 2 = 12
For calculating the points of intersection between one line and one circle, we can use the formula which is mentioned in the hint as follows  C presuperscript 4 subscript 1 cross times C presuperscript 8 subscript 1 cross times 2 space= 64
Hence, the total number of points of intersection is = 28 + 12  + 64 = 104
General
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How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

Complete step-by-step answer:
Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
XXXXX
Here, symbol ( is for the even places and (Xis for the odd places of the digit number.
The digits which are even are 2, 2, 8, 8 and 8.
Number of even digits  5
The digits which are odd are 3, 3, 5 and 5.
Number of odd digits 4
We have to arrange the odd digits in even places.
 
Number of ways to arrange the odd digits in 4 even places = fraction numerator 4 factorial over denominator 2 factorial space cross times 2 factorial end fraction
On finding the value of the factorials, we get
Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.
Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 factorial over denominator 2 factorial cross times 3 factorial end fraction
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 space cross times 4 cross times 3 cross times 2 cross times 1 over denominator 2 cross times 1 space cross times 3 cross times 2 cross times 1 end fraction

On further simplification, we get
Number of ways to arrange the even digits in 5 odd places 
=10
Total number of 9 digits number 6×10 60
Hence, the required number of 9 digit numbers 60
 
 
 
 

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

Maths-General
Complete step-by-step answer:
Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.
XXXXX
Here, symbol ( is for the even places and (Xis for the odd places of the digit number.
The digits which are even are 2, 2, 8, 8 and 8.
Number of even digits  5
The digits which are odd are 3, 3, 5 and 5.
Number of odd digits 4
We have to arrange the odd digits in even places.
 
Number of ways to arrange the odd digits in 4 even places = fraction numerator 4 factorial over denominator 2 factorial space cross times 2 factorial end fraction
On finding the value of the factorials, we get
Number of ways to arrange the odd digits in 4 even places = 6

Now, we have to arrange the even digits in odd places.
Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 factorial over denominator 2 factorial cross times 3 factorial end fraction
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places = fraction numerator 5 space cross times 4 cross times 3 cross times 2 cross times 1 over denominator 2 cross times 1 space cross times 3 cross times 2 cross times 1 end fraction

On further simplification, we get
Number of ways to arrange the even digits in 5 odd places 
=10
Total number of 9 digits number 6×10 60
Hence, the required number of 9 digit numbers 60
 
 
 
 
General
Maths-

A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

Matches whose prediction are correct can be selected in C presuperscript 20 subscript 10 ways.
Since each match can result either in a win, loss or tie for the home team.
Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)
and there are 10 matches for which he predicted wrong.
Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s
C presuperscript 20 subscript 10 cross times space 2 to the power of 10
Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to C presuperscript 20 subscript 10 cross times space 2 to the power of 10 

A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :

Maths-General
Matches whose prediction are correct can be selected in C presuperscript 20 subscript 10 ways.
Since each match can result either in a win, loss or tie for the home team.
Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)
and there are 10 matches for which he predicted wrong.
Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s
C presuperscript 20 subscript 10 cross times space 2 to the power of 10
Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to C presuperscript 20 subscript 10 cross times space 2 to the power of 10