Maths-
General
Easy

Question

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

  1. a parabola    
  2. an ellipse    
  3. a circle    
  4. a straight line    

The correct answer is: a circle

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Related Questions to study

General
maths-

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

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

maths-General
General
maths-

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

maths-General
General
Maths-

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
maths-

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
General
Maths-

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
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 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 
General
maths-

The foot of the perpendicular from the point left parenthesis 3 , 3 pi divided by 4 right parenthesis on the line r left parenthesis c o s space theta minus s i n space theta right parenthesis equals 6 square root of 2 is

The foot of the perpendicular from the point left parenthesis 3 , 3 pi divided by 4 right parenthesis on the line r left parenthesis c o s space theta minus s i n space theta right parenthesis equals 6 square root of 2 is

maths-General
General
maths-

The point of intersection of the lines 2 c o s space theta plus s i n space theta equals 1 over r comma c o s space theta plus s i n space theta equals 1 over r is

The point of intersection of the lines 2 c o s space theta plus s i n space theta equals 1 over r comma c o s space theta plus s i n space theta equals 1 over r is

maths-General
General
maths-

The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

The line passing through open parentheses negative 1 comma fraction numerator pi over denominator 2 end fraction close parentheses and perpendicular to square root of 3 s i n space theta plus 2 c o s space theta equals 4 over r is

maths-General
General
maths-

The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

The equation of the line passing through left parenthesis negative 1 comma pi divided by 6 right parenthesis comma left parenthesis 1 comma pi divided by 2 right parenthesis is

maths-General
General
maths-

The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

The length of the perpendicular from (-1, π/6) to the line r left parenthesis 3 s i n space theta plus square root of 3 c o s space theta right parenthesis equals 3 is

maths-General
General
Maths-

The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Complete step-by-step answer:

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits.
We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3!=6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.&lt;br&gt; &lt;/br&gt;
Now, we get a sum of digits in units place for all the numbers as 14×6=84.&lt;br&gt; &lt;/br&gt;
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.


So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is
(84×1000)+(84×100)+(84×10)+(84×1)Error converting from MathML to accessible text.
Sum =  84000+8400+840+84&lt;br&gt; &lt;/br&gt;
Sum = 93324
We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.
Error converting from MathML to accessible text.
Error converting from MathML to accessible text.

The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :

Maths-General
Complete step-by-step answer:

According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.

Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits.
We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3!=6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits. We get sum as 2+3+4+5=14.&lt;br&gt; &lt;/br&gt;
Now, we get a sum of digits in units place for all the numbers as 14×6=84.&lt;br&gt; &lt;/br&gt;
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.


So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is
(84×1000)+(84×100)+(84×10)+(84×1)Error converting from MathML to accessible text.
Sum =  84000+8400+840+84&lt;br&gt; &lt;/br&gt;
Sum = 93324
We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.
Error converting from MathML to accessible text.
Error converting from MathML to accessible text.