Maths
General
Easy
Question
If x is real, then maximum value of is
 1


 41
The correct answer is: 41
Related Questions to study
Maths
The value of 'c' of Lagrange's mean value theorem for is
The value of 'c' of Lagrange's mean value theorem for is
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Maths
The value of 'c' of Rolle's mean value theorem for is
The value of 'c' of Rolle's mean value theorem for is
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Maths
The value of 'c' of Rolle's theorem for – on [–1, 1] is
The value of 'c' of Rolle's theorem for – on [–1, 1] is
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Maths
For in [5, 7]
For in [5, 7]
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Maths
The value of 'c' in Lagrange's mean value theorem for in [0, 1] is
The value of 'c' in Lagrange's mean value theorem for in [0, 1] is
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Maths
The value of 'c' in Lagrange's mean value theorem for in [0, 2] is
The value of 'c' in Lagrange's mean value theorem for in [0, 2] is
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Maths
The equation represents
The equation represents
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Maths
The polar equation of the circle whose end points of the diameter are and is
The polar equation of the circle whose end points of the diameter are and is
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Maths
The radius of the circle is
The radius of the circle is
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Maths
The adjoining figure shows the graph of Then –
The adjoining figure shows the graph of Then –
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Maths
Graph of y = ax^{2} + 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 (a) is correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =
On further solving this, we get
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 (c) will 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.
Hence, the option (a) is correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =
On further solving this, we get
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 (c) will 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 = ax^{2} + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –
MathsGeneral
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 (a) is correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =
On further solving this, we get
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 (c) will 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.
Hence, the option (a) is correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be =
On further solving this, we get
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 (c) will 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.
Maths
For the quadratic polynomial f (x) = 4x^{2} – 8kx + k, the statements which hold good are
For the quadratic polynomial f (x) = 4x^{2} – 8kx + k, the statements which hold good are
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Maths
The graph of the quadratic polynomial y = ax^{2} + bx + c is as shown in the figure. Then :
Clearly, y = represent a parabola opening downwards. Therefore, a < 0
y = cuts negative y axis , Putting x = 0 in the given equation
y = c
y = c
c < 0
Thus, from the above graph c < 0.
The graph of the quadratic polynomial y = ax^{2} + bx + c is as shown in the figure. Then :
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Clearly, y = represent a parabola opening downwards. Therefore, a < 0
y = cuts negative y axis , Putting x = 0 in the given equation
y = c
y = c
c < 0
Thus, from the above graph c < 0.
Maths
The greatest possible number of points of intersections of 8 straight line and 4 circles is :
Complete stepbystep 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
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 = = 28
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 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 = 64
Hence, the total number of points of intersection is = 28 + 12 + 64 = 104
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
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 = = 28
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 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 = 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 :
MathsGeneral
Complete stepbystep 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
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 = = 28
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 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 = 64
Hence, the total number of points of intersection is = 28 + 12 + 64 = 104
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
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 = = 28
For calculating the points of intersection between two circles, we can use the formula which is mentioned in the hint as follows = = 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 = 64
Hence, the total number of points of intersection is = 28 + 12 + 64 = 104
Maths
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 stepbystep 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.
Number of ways to arrange the odd digits in 4 even places =
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 =
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places =
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
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.
X−X−X−X−X
Here, symbol ( − ) is for the even places and (X) is 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
Number of odd digits = 4
We have to arrange the odd digits in even places.
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 =
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places =
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 ?
MathsGeneral
Complete stepbystep 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.
Number of ways to arrange the odd digits in 4 even places =
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 =
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places =
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
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.
X−X−X−X−X
Here, symbol ( − ) is for the even places and (X) is 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
Number of odd digits = 4
We have to arrange the odd digits in even places.
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 =
On finding the value of the factorials, we get
Number of ways to arrange the even digits in 5 odd places =
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