 Hi and welcome to the session. Let us discuss the following question. Question says, choose the correct answer. For all real values of x, minimum value of 1 minus x plus x square upon 1 plus x plus x square is 0, 1, 3 or 1 upon 3. We have to choose the correct answer from A, B, C and D. First of all, let us understand that if we are given a function f defined on open interval i, then let us assume that function f is continuous at a critical point C in interval i. Then if f dash x is less than 0 for x less than C and if f dash x is greater than 0 for x greater than C, then C is a point of local minima. Minimum value is given by f c. This is the key idea to solve the given question. Clearly we can see if f dash x changes its sign from negative to positive as x increases through C, then C is a point of local minima. Now let us start with the solution. Let function f is given by f x is equal to 1 minus x plus x square upon 1 plus x plus x square. Now differentiating both sides with respect to x, we get f dash x is equal to 1 plus x plus x square multiplied by 0 minus 1 plus 2x minus 1 minus x plus x square multiplied by 0 plus 1 plus 2x upon 1 plus x plus x square whole square. Here we have applied quotient rule to find the derivative of this term. Now this is equal to 1 plus x plus x square multiplied by 2x minus 1 minus 1 minus x plus x square multiplied by 1 plus 2x upon 1 plus x plus x square whole square. Now we get f dash x is equal to 2x plus 2x square plus 2x cube minus 1 minus x minus x square minus. Here we have multiplied these two brackets and now we will multiply these two brackets. We get 1 plus 2x minus x minus 2x square plus x square plus 2x cube upon 1 plus x plus x square whole square. Now this is equal to 2x cube plus x square plus x minus 1. We know 2x minus x is equal to x and 2x square minus x square is equal to x square. So we can write this bracket as 2x cube plus x square plus x minus 1. Now we can write this bracket as we know 2x minus x is equal to x and minus 2x square plus x square is equal to minus x square. So we can write 2x cube minus x square plus x plus 1 upon 1 plus x plus x square whole square. Now this is equal to 2x cube plus x square plus x minus 1 minus 2x cube plus x square minus x minus 1 upon 1 plus x plus x square whole square. Here we have multiplied this negative sign with all the terms of this bracket. Now we know 2x cube and 2x cube will cancel each other. And x and minus x will cancel each other. We know we get 2x square minus 2x square plus x square is equal to 2x square and minus 1 minus 1 is equal to minus 2. So we get 2x square minus 2 upon 1 plus x plus x square whole square. Let us find out all the points at which f dash x is equal to 0. Now we will put f dash x is equal to 0. Now this implies 2x square minus 2 upon 1 plus x plus x square whole square is equal to 0. Now multiplying both sides by square of 1 plus x plus x square we get 2x square minus 2 is equal to 0. Now taking 2 common on left hand side we get 2 multiplied by x square minus 1 is equal to 0. Now dividing both sides by 2 we get x square minus 1 is equal to 0. Now this can be written as x minus 1 x plus 1 is equal to 0 below a square minus b square is equal to a plus b multiplied by a minus b. Here we can write it as x square minus 1 square and x square minus 1 square is equal to x minus 1 multiplied by x plus 1. Now this implies x is equal to 1 or x is equal to minus 1. Now let us check if x is equal to 1 is a point of local minima we know f dash x is equal to 2x square minus 2 upon 1 plus x plus x square whole square. Now this is equal to 2 multiplied by x minus 1 multiplied by x plus 1 upon 1 plus x plus x square whole square. Now clearly we can see for x less than 1 f dash x is also less than 0 and for x greater than 1 f dash x is greater than 0 using key idea we get x is equal to 1 is a point of local minima. Clearly we can see f dash x is changing with sign from negative to positive as in x increases through whole. So x is equal to 1 is a point of local minima. Now let us find out the minimum value. Minimum value is given by f1 f1 is equal to 1 minus 1 plus 1 square upon 1 plus 1 plus 1 square. Now simplifying we get 1 upon 3 so we get minimum value is equal to 1 upon 3. We can now discuss about the point x is equal to minus 1 for x less than minus 1 f dash x is greater than 0 and x greater than minus 1 f dash x is less than 0. Now clearly we can see f dash x changes its sign from positive to negative as x increases through minus 1. So x is equal to minus 1 is a point of local maxima. So our required minimum value is f1 is equal to 1 upon 3. So our required correct answer is d. This completes the session. Hope you understood the session. Take care and have a nice day.