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Sat Apr 19, 2014 9:04 pm  

Ans: U s e space t h i s space f o r m u l a : 2 tan to the power of minus 1 end exponent x equals sin to the power of minus 1 end exponent open parentheses fraction numerator 2 x over denominator 1 plus x squared end fraction close parentheses

Sat Apr 19, 2014 6:08 pm  

Ans: assume any three sets A, B, C and prove it.

Sat Apr 19, 2014 6:05 pm  

Ans: A union B equals A intersection B  rightwards double arrow open curly brackets x : x element of A space o r space x element of B close curly brackets equals open curly brackets x : x element of A space a n d space x element of B close curly brackets  rightwards double arrow A equals B space  o r A union B equals A intersection B  rightwards double arrow open curly brackets x : x element of A space o r space x element of B close curly brackets equals open curly brackets x : x element of A space a n d space x element of B close curly brackets  rightwards double arrow A equals B space  o r  

Sat Apr 19, 2014 3:35 pm  

Ans: There are many good books of Physcis like books of Resnick Halliday, Lakhmir Singh-Manjit Kaur, Tata McGrawHill etc.There are also good books for particular classes and course.

Sat Apr 19, 2014 11:52 am  

Ans: a set which has elements x > 3, will also be greater than 2 . therefore, set is open parentheses 3 comma space infinity close parentheses

Sat Apr 19, 2014 8:44 am  

Ans: Consider three numbers 24, 36 and 60.   Our aim is to find the HCF of 24, 36 and 60.   First let us write the prime factorization of all the numbers.   24 equals 2 cross times 2 cross times 2 cross times 3 equals 2 cubed cross times 3 36 equals 2 cross times 2 cross times 3 cross times 3 equals 2 squared cross times 3 squared 60 equals 2 cross times 2 cross times 3 cross times 5 equals 2 squared cross times 3 to the power of 1 cross times 5 to the power of 1 T h u s comma space h i g h e s t space c o m m o n space f a c t o r space i s space t h e space p r o d u c t space o f space t h e space f a c t o r s space t h a t space a r e space c o m m o n space t o space a l l space t h e space n u m b e r s. S o comma space H C F space o f space 24 comma space 36 space a n d space 60 space equals 2 squared cross times 3 to the power of 1 equals 12

Fri Apr 18, 2014 11:44 pm  

Ans: Torque is a measure of how much a force acting on an object causes that object to rotate.  Mathematically, Torque is defined as  =  r  x  F  =  r F  sin() In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, '' being the angle between r and F.
Using the right hand rule, we can find the direction of the torque vector. If we put our fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector.

Fri Apr 18, 2014 10:19 pm  

Ans: A plane mirror is a mirror with a planar reflective surface. For light rays striking a plane mirror, the angle of reflection equals the angle of incidence The mage formed by a plane mirror is always at the same distance behind the mirror as the object is in front of the mirror and the image formed is upright, virtual, of the same size as the object and laterally inverted.

Fri Apr 18, 2014 10:00 pm  

Ans: It is not possible for something  to occupy space and not have mass or for something to have mass but not occupy any space and be called as 'matter'. It is against the fundamentals of physics as matter is something that occupies space and has mass. Also, anything that would have had mass but would occupy no space, would have infinite density.

Fri Apr 18, 2014 9:43 pm  

Ans: We can construct a simple circuit with a battery and a bulb to produce electricity. The requirements of this experiment is a battery, a bulb and two copper wires.  Procedure:
Remove the insulating material from the two ends of the copper wires. Connect one end of the wire to the negative end of the battery. Make sure that the wire touching the lower part of the battery. Connect the other end of the battery to the bulb using other wire. We can light the bulb by touching the two ends of the battery accordingly. We can use some optional materials to make this process easier. Use switches to on and off the bulb. The cell provides a direct current to the bulb. This current has a unique direction and it flows from the positive terminal of cell to its negative terminal through the bulb. The magnitude of the current given by the cell remains constant for a sufficiently long time. When the cell gets discharged, it stops giving current and becomes useless.

Fri Apr 18, 2014 8:19 pm  

Ans: please mention the R.H.S

Fri Apr 18, 2014 2:23 pm  

Ans: First find the determinant of the given square matrix.   That is we need to check whether A is singular (open vertical bar A close vertical bar equals 0) or non-singular (open vertical bar A close vertical bar not equal to 0).   If the determinant of the matrix is equal to zero, then the inverse of the matrix does not exist.   If the determinant of the matrix is not equal to zero, then the inverse of the matrix exist.

Fri Apr 18, 2014 9:56 am  

Ans: The characteristic of log356 is 3-1=2   Mantissa part is 0.5127   Thus log 356 = 2.5127

Fri Apr 18, 2014 9:42 am  

Ans: Dear bhuvisen

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.
We are happy that you are pleased and it is our duty to give answers.   Try to post many queries so that we will answer and clear your doubts.   Regards

Topperlearning Team.

Fri Apr 18, 2014 9:39 am  

Ans: java.lang.Exception: The input exceeds the limit of 3072 characters of length. N o w space l e t space u s space f a c t o r i z e space 2 x squared minus x minus 1 : 2 x squared minus x minus 1 equals 0 rightwards double arrow 2 x squared minus 2 x plus x minus 1 equals 0 rightwards double arrow 2 x open parentheses x minus 1 close parentheses plus 1 open parentheses x minus 1 close parentheses equals 0 rightwards double arrow open parentheses 2 x plus 1 close parentheses open parentheses x minus 1 close parentheses equals 0 rightwards double arrow 2 x plus 1 equals 0 space o r space x minus 1 equals 0 rightwards double arrow 2 x equals minus 1 space o r space x equals 1 rightwards double arrow x equals fraction numerator minus 1 over denominator 2 end fraction space o r space x equals 1 T h u s space t h e space z e r o s space o f space t h e space g i v e n space p o l y n o m i a l space a r e : 2 plus square root of 3 comma 2 minus square root of 3 comma space fraction numerator minus 1 over denominator 2 end fraction comma 1

Fri Apr 18, 2014 9:35 am  

Ans: java.lang.Exception: The input exceeds the limit of 3072 characters of length.

Fri Apr 18, 2014 1:12 am  

Ans: g i v e n space t h a t space x equals 3 plus 2 square root of 2 W e space n e e d space t o space f i n d space t h e space v a l u e space o f space square root of x minus fraction numerator 1 over denominator square root of x end fraction : square root of x minus fraction numerator 1 over denominator square root of x end fraction equals square root of 3 plus 2 square root of 2 end root minus fraction numerator 1 over denominator square root of 3 plus 2 square root of 2 end root end fraction equals fraction numerator square root of 3 plus 2 square root of 2 end root cross times square root of 3 plus 2 square root of 2 end root minus 1 over denominator square root of 3 plus 2 square root of 2 end root end fraction equals fraction numerator 3 plus 2 square root of 2 minus 1 over denominator square root of 3 plus 2 square root of 2 end root end fraction equals fraction numerator 2 plus 2 square root of 2 over denominator square root of 3 plus 2 square root of 2 end root end fraction cross times fraction numerator square root of 3 minus 2 square root of 2 end root over denominator square root of 3 minus 2 square root of 2 end root end fraction equals fraction numerator 2 plus 2 square root of 2 over denominator square root of 3 squared minus 2 squared cross times 2 end root end fraction cross times begin inline style square root of 3 minus 2 square root of 2 end root end style equals open parentheses begin inline style 2 plus 2 square root of 2 end style close parentheses open parentheses begin inline style square root of 3 minus 2 square root of 2 end root end style close parentheses

Fri Apr 18, 2014 1:07 am  

Ans: java.lang.Exception: The input exceeds the limit of 3072 characters of length.

Fri Apr 18, 2014 12:42 am  

Ans:   We know that Distance = Speed x Time Distance travelled with 10 km/h for 30 mins = 10 space k m divided by h r space cross times 1 half h r space equals space 5 space k m Distance travelled with x km/h for 30 mins = x space k m divided by h r space cross times 1 half h r space equals space x over 2 k m   The total distance to be travelled is 15 km. Hence, Distance with 10 km/hr speed + Distance with x km/hr = 15 km rightwards double arrow 5 plus x over 2 equals 15 rightwards double arrow x equals 20 space k m divided by h r

Thu Apr 17, 2014 8:00 pm  

Ans: just check the continuty of the function at x = 1 and -1.   or   Use definition of modulus function and try to get the graph of the given function.
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