Maths-
General
Easy

Question

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

  1. r equals 2 S i n space theta    
  2. r equals 2 C o s space theta    
  3. r equals 2 C o s space theta plus 2 S i n space theta    
  4. None    

The correct answer is: r equals 2 S i n space theta

Related Questions to study

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

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.

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

Maths-General

Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.

parallel
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
General
Maths-

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

The roots (also called as zeros, y = 0) of the quadratic equation a x squared space plus space b x space plus space c space equals space 0
 are given by x = fraction numerator negative b space plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction . The quantity b squared minus space 4 a c is called the discriminant of the equation and determines the nature of its roots.
If b squared minus space 4 a c  ≥ 0, the roots are real.
If b squared minus space 4 a c  = 0, the roots are real and equal.
If b squared minus space 4 a c  < 0, the roots are complex and conjugates of each other.

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

Maths-General

The roots (also called as zeros, y = 0) of the quadratic equation a x squared space plus space b x space plus space c space equals space 0
 are given by x = fraction numerator negative b space plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction . The quantity b squared minus space 4 a c is called the discriminant of the equation and determines the nature of its roots.
If b squared minus space 4 a c  ≥ 0, the roots are real.
If b squared minus space 4 a c  = 0, the roots are real and equal.
If b squared minus space 4 a c  < 0, the roots are complex and conjugates of each other.

General
Maths-

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

The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows
The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.
The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.
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.

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

Maths-General

The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows
The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.
The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.
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.

parallel
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 ?

Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.

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

Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.

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 :

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
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
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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
parallel
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) :

Alternatively, we can use the formula for the sum of numbers as
(n - 1)! cross times (sum of digits) cross times (11111 ............ntimes). We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

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

Alternatively, we can use the formula for the sum of numbers as
(n - 1)! cross times (sum of digits) cross times (11111 ............ntimes). We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.

General
Maths-

Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to :

We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to :

Maths-General

We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.

parallel

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