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fraction numerator 1 over denominator x to the power of 4 plus 1 end fraction equals

  1. fraction numerator x plus square root of 2 over denominator 2 square root of 2 open parentheses x squared plus square root of 2 x minus 1 close parentheses end fraction plus fraction numerator square root of 2 minus x over denominator 2 square root of 2 open parentheses x squared plus square root of 2 x minus 1 close parentheses end fraction
  2. fraction numerator x plus square root of 2 over denominator 2 square root of 2 open parentheses x squared plus square root of 2 x plus 1 close parentheses end fraction plus fraction numerator square root of 2 minus x over denominator 2 square root of 2 open parentheses x squared minus square root of 2 x plus 1 close parentheses end fraction
  3. fraction numerator x plus square root of 2 over denominator 2 square root of 2 open parentheses x squared plus square root of 2 x minus 1 close parentheses end fraction plus fraction numerator square root of 2 minus x over denominator 2 square root of 2 open parentheses x squared minus square root of 2 x plus 1 close parentheses end fraction
  4. fraction numerator x plus square root of 2 over denominator 2 square root of 2 open parentheses x squared minus square root of 2 x plus 1 close parentheses end fraction plus fraction numerator square root of 2 minus x over denominator 2 square root of 2 open parentheses x squared minus square root of 2 x plus 1 close parentheses end fraction

The correct answer is: fraction numerator x plus square root of 2 over denominator 2 square root of 2 open parentheses x squared plus square root of 2 x plus 1 close parentheses end fraction plus fraction numerator square root of 2 minus x over denominator 2 square root of 2 open parentheses x squared minus square root of 2 x plus 1 close parentheses end fraction

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If fraction numerator x squared plus 5 x plus 1 over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis end fraction equals fraction numerator A over denominator x plus 1 end fraction plus fraction numerator B over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis end fraction then B=

If fraction numerator x squared plus 5 x plus 1 over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis end fraction equals fraction numerator A over denominator x plus 1 end fraction plus fraction numerator B over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis end fraction plus fraction numerator C over denominator left parenthesis x plus 1 right parenthesis left parenthesis x plus 2 right parenthesis left parenthesis x plus 3 right parenthesis end fraction then B=

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S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
fraction numerator table row 12 end table over denominator table row 4 end table table row 4 end table table row 4 end table end fraction×fraction numerator table row 3 end table over denominator table row 3 end table end fraction = fraction numerator table row 12 end table over denominator left parenthesis table row 4 end table right parenthesis to the power of 3 end exponent end fraction

The set S : = { 1, 2, 3 .........12} is to be partitioned into three sets A, B, C of equal size. Thus A unionunion C = S, A intersection B = B  C = A intersection C = ϕ. The number of ways to partition S is -

maths-General
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
fraction numerator table row 12 end table over denominator table row 4 end table table row 4 end table table row 4 end table end fraction×fraction numerator table row 3 end table over denominator table row 3 end table end fraction = fraction numerator table row 12 end table over denominator left parenthesis table row 4 end table right parenthesis to the power of 3 end exponent end fraction
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Reason (R): If f left parenthesis x right parenthesis is divided by x minus a then the remainder is f(a)

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A particle begins at the origin and moves successively in the following manner as shown, 1 unit to the right, 1/2 unit up, 1/4 unit to the right, 1/8 unit down, 1/16 unit to the right etc. The length of each move is half the length of the previous move and movement continues in the ‘zigzag’ manner indefinitely. The co-ordinates of the point to which the ‘zigzag’ converges is –

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General
Maths-

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

 Detailed Solution
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 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 space cross times 1 space equals space 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 space cross times 2 space equals space 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 space cross times C presuperscript 8 subscript 1 cross times 2 space equals space 64

 Hence, the total number of points of intersection is = 28 + 64 + 12 = 104

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

Maths-General
 Detailed Solution
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 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 space cross times 1 space equals space 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 space cross times 2 space equals space 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 space cross times C presuperscript 8 subscript 1 cross times 2 space equals space 64

 Hence, the total number of points of intersection is = 28 + 64 + 12 = 104
General
Maths-

Assertion (A):Iffraction numerator 1 over denominator left parenthesis x minus 2 right parenthesis open parentheses x squared plus 1 close parentheses end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator x squared plus 1 end fraction ,then Aequals 1 fifth comma B equals 1 fifth comma C equals negative 2 over 5
Reason (R) : fraction numerator 1 over denominator left parenthesis x minus a right parenthesis open parentheses x squared plus b close parentheses end fraction equals fraction numerator 1 over denominator a squared plus b end fraction open square brackets fraction numerator 1 over denominator x minus a end fraction minus fraction numerator x plus a over denominator x squared plus b end fraction close square brackets

Assertion (A):Iffraction numerator 1 over denominator left parenthesis x minus 2 right parenthesis open parentheses x squared plus 1 close parentheses end fraction equals fraction numerator A over denominator x minus 2 end fraction plus fraction numerator B x plus C over denominator x squared plus 1 end fraction ,then Aequals 1 fifth comma B equals 1 fifth comma C equals negative 2 over 5
Reason (R) : fraction numerator 1 over denominator left parenthesis x minus a right parenthesis open parentheses x squared plus b close parentheses end fraction equals fraction numerator 1 over denominator a squared plus b end fraction open square brackets fraction numerator 1 over denominator x minus a end fraction minus fraction numerator x plus a over denominator x squared plus b end fraction close square brackets

Maths-General
General
physics-

A force–time graph for a linear motion is shown in figure where the segments are circular. The linear momentum gained between zero and  is

As the area above the time axis is numerically equal to area below the time axis therefore net momentum gained by body will be zero because momentum is a vector quantity

A force–time graph for a linear motion is shown in figure where the segments are circular. The linear momentum gained between zero and  is

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As the area above the time axis is numerically equal to area below the time axis therefore net momentum gained by body will be zero because momentum is a vector quantity

General
Maths-

How many different nine digit numbers can be formed from the number 2,2,3,3,5,5,8,8,8 by rearranging its digits so that the odd digits occupy even position ?

 Detailed Solution
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.
O space e space O space e space O space e space O space e space O
Here, e is for the even places and O 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
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 cross times 2 factorial end fraction = 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 space equals space 10

Total number of 9 digits number = 6 cross times 10 space equals space 60


 
 
 

How many different nine digit numbers can be formed from the number 2,2,3,3,5,5,8,8,8 by rearranging its digits so that the odd digits occupy even position ?

Maths-General
 Detailed Solution
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.
O space e space O space e space O space e space O space e space O
Here, e is for the even places and O 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
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 cross times 2 factorial end fraction = 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 space equals space 10

Total number of 9 digits number = 6 cross times 10 space equals space 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 :

There are total 20 matches and the outcome can either be win, lose or tie.
We have to find the number of ways in which exactly 10 predictions are correct which can be shown by C presuperscript 20 subscript 10 ways in which his prediction is correct.
And in the remaining 10 matches, he makes wrong predictions i.e. out of 3 outcomes (win, lose, tie) he can pick 2 outcomes out of 3 , which can be done in 2 to the power of 10 ways.
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 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
There are total 20 matches and the outcome can either be win, lose or tie.
We have to find the number of ways in which exactly 10 predictions are correct which can be shown by C presuperscript 20 subscript 10 ways in which his prediction is correct.
And in the remaining 10 matches, he makes wrong predictions i.e. out of 3 outcomes (win, lose, tie) he can pick 2 outcomes out of 3 , which can be done in 2 to the power of 10 ways.
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 2 to the power of 10
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) :

Detailed Solution
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.
W e space g e t space s u m space a s space 2 space plus space 3 space plus space 4 space plus space 5 space equals space 14
N o w comma space w e space g e t space a space s u m space o f space d i g i t s space i n space u n i t s space p l a c e space f o r space a l l space t h e space n u m b e r s space a s space 14 cross times 6 equals space 84.
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.
S o comma space w e space g e t space t h e space s u m space o f space a l l space t h e space n u m b e r s space t h a t space c a n space b e space f o r m e d space w i t h space t h e space d i g i t s space 2 comma 3 comma 4 comma 5 space t a k e n space a l l space a t space a space t i m e space i s space
left parenthesis 84 cross times 1000 right parenthesis plus left parenthesis 84 cross times 100 right parenthesis plus left parenthesis 84 cross times 10 right parenthesis plus left parenthesis 84 cross times 1 right parenthesis space equals space 93324
Thus, 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) : 93324

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
Detailed Solution
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.
W e space g e t space s u m space a s space 2 space plus space 3 space plus space 4 space plus space 5 space equals space 14
N o w comma space w e space g e t space a space s u m space o f space d i g i t s space i n space u n i t s space p l a c e space f o r space a l l space t h e space n u m b e r s space a s space 14 cross times 6 equals space 84.
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.
S o comma space w e space g e t space t h e space s u m space o f space a l l space t h e space n u m b e r s space t h a t space c a n space b e space f o r m e d space w i t h space t h e space d i g i t s space 2 comma 3 comma 4 comma 5 space t a k e n space a l l space a t space a space t i m e space i s space
left parenthesis 84 cross times 1000 right parenthesis plus left parenthesis 84 cross times 100 right parenthesis plus left parenthesis 84 cross times 10 right parenthesis plus left parenthesis 84 cross times 1 right parenthesis space equals space 93324
Thus, 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) : 93324
General
Maths-

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

Detailed Solution
In this question, we have been asked to find the total number of divisors of 480 which are of the form  4n + 2, n greater or equal than 0 
In order to solve this question, we should know that the number of the divisor of any number
x equals a to the power of m space end exponent b to the power of n c to the power of p.... space space where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..
We know that 480 can be expressed as 480 space equals space 2 to the power of 5.3.5
S o comma space a c c o r d i n g space t o space t h e space f o r m u l a comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space o f space 480 space a r e
space left parenthesis 5 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 6 cross times 2 cross times 2 equals 24 space.
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. 
S o comma space t h e space t o t a l space n u m b e r space o f space o d d space d i v i s o r s space t h a t space a r e space p o s s i b l e space a r e space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 2 cross times 2 equals 4 space comma space a c c o r d i n g space t o space t h e space p r o p e r t y.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that, 480 space equals space 2 to the power of 5.3.5 
space S o comma space t h e space n u m b e r space o f space d i v i s o r s space t h a t space a r e space m u l t i p l e s space o f space 4 space a r e space left parenthesis 3 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 4 cross times 2 cross times 2 space space equals space 16
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
S o comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space w h i c h space a r e space e v e n space b u t space n o t space d i v i s i b l e space b y space 2 space c a n space b e space g i v e n space b y space 20 space – space 16 space equals space 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to 4.

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

Maths-General
Detailed Solution
In this question, we have been asked to find the total number of divisors of 480 which are of the form  4n + 2, n greater or equal than 0 
In order to solve this question, we should know that the number of the divisor of any number
x equals a to the power of m space end exponent b to the power of n c to the power of p.... space space where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..
We know that 480 can be expressed as 480 space equals space 2 to the power of 5.3.5
S o comma space a c c o r d i n g space t o space t h e space f o r m u l a comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space o f space 480 space a r e
space left parenthesis 5 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 6 cross times 2 cross times 2 equals 24 space.
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution. 
S o comma space t h e space t o t a l space n u m b e r space o f space o d d space d i v i s o r s space t h a t space a r e space p o s s i b l e space a r e space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 2 cross times 2 equals 4 space comma space a c c o r d i n g space t o space t h e space p r o p e r t y.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that, 480 space equals space 2 to the power of 5.3.5 
space S o comma space t h e space n u m b e r space o f space d i v i s o r s space t h a t space a r e space m u l t i p l e s space o f space 4 space a r e space left parenthesis 3 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 4 cross times 2 cross times 2 space space equals space 16
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
S o comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space w h i c h space a r e space e v e n space b u t space n o t space d i v i s i b l e space b y space 2 space c a n space b e space g i v e n space b y space 20 space – space 16 space equals space 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to 4.