Finding the Range

Transcript

0:00Introduction

Hello everyone! Okay, one small challenge before we start. Take this function: x, over one plus x squared. Put x equals one, and you get one half. Put x equals two, you get two fifths. Put x equals ten, ten over a hundred and one, even smaller. Now my question. Is there any real number x that gives the output one? Or even point six? Hmm. You can try values all day and still not be sure, right? So today we will find the range, all the outputs of a function, with proper methods, no guessing. And look, in JEE, range questions hide inside functions, inequalities, everywhere. So, let's begin.

0:49Method 1: build up from known ranges

So, first, what exactly is the range? It is the set of outputs, f of x, that the function actually gives, as x runs over the domain. Not what we hope for, only what we actually get. Clear? Method one: build up from ranges you already know. Start with the simplest one, x squared. Tell me, can x squared ever be negative? Never, of course! Square any real number, and you get zero or more. So x squared is greater than or equal to zero. Add one to both sides, and x squared plus one is greater than or equal to one. Simple, isn't it? Now flip it: one over x squared plus one. The bottom is at least one, so the fraction is at most one. So the answer is zero to one, both ends included, right? Wait. Can a fraction with one on top ever become zero? Never! The top is one, so the fraction always stays above zero. So zero is strictly less than one over x squared plus one, which is less than or equal to one. Zero is not included, one is included. See, this is exactly where careless marks are lost. One more tool for reading a range: the graph. Sketch the function first. On an interval where f is increasing, the range simply runs from the value at the left end to the value at the right end, okay? Draw first, then read the range.

2:18Method 2: solve for x

Method two: solve for x. Put y equals f of x, and turn the equation around, so that x is written in terms of y. Why? Because a value y is in the range only if some x actually gives it, right? Then keep every y for which this x exists. But tell me, is that enough? No! That x must also lie in the domain. If the formula gives you a forbidden x, that y is not really obtained. Clear? Now a topper's trick, for one very common shape: a x plus b, over c x plus d. Here c is not zero, and a d is not equal to b c, okay? The domain is all real numbers, this outlined R on screen, except minus d over c, where the bottom becomes zero. And the range? All real numbers except a over c, the ratio of the two x coefficients. No algebra at all. Just one look, and you know the answer.

3:18Method 3: the discriminant method

Method three, the discriminant method. Put y equals f of x and cross multiply. Sometimes you get a quadratic in x, where the coefficients, capital A, B and C, depend on y, okay? We need a real x. So tell me, when does a quadratic equation have a real root? Yes, exactly! When the discriminant is not negative. So, B squared minus four A C, greater than or equal to zero. Solve this for y, and you have the range. Almost! Now, there are two traps here, and both are classic JEE traps. Trap one. The coefficient of x squared, capital A, depends on y. For the y that makes it zero, the equation is not quadratic at all; it becomes linear. So test that y separately, okay? Trap two: a common factor on top and bottom. Look at x squared minus one, over x minus one. It simplifies to x plus one. And x plus one takes every real value, so the range is all real numbers, right? No, no, careful! We cancelled x minus one, but x equal to one is still not allowed. And the output two comes only from x equal to one. So two is never produced: the range is all real numbers except two. Whenever you cancel, check what the cancelled point would have given. Got it?

4:48Example 1: range of (x-2)/(x-3) by solving for x

Alright, Example one. Find the range of x minus two, over x minus three. Use method two: solve for x. Pause the video, and try it first. Done? Okay, let's check. Put y equals x minus two, over x minus three. Cross multiply: x y minus three y equals x minus two. Bring the x terms together. x y minus x is x into y minus one, and that equals three y minus two. So x equals three y minus two, over y minus one. Now look at this formula carefully. Which value of y breaks it? y equals one, yes! The bottom becomes zero. So this x exists for every y, except one. Now the domain check. Could this x ever be three, the forbidden input? If it were, three y minus three would equal three y minus two. Minus three equals minus two? Impossible! So nothing is lost. So the range is all real numbers except one. And now our topper's trick: a over c is one over one, that is one. Same answer, in two seconds. Super quick!

6:04Example 2: JEE Main 2017 range of x/(1+x^2)

Now, remember our puzzle from the start? Let's finish it. Find the range of x over one plus x squared, for all real x. This one came in JEE Main twenty seventeen. Hint: method three. Pause the video and try it. Done? Okay, let's check. Put y equals x over one plus x squared, and cross multiply: y into one plus x squared equals x. Rearrange: y x squared, minus x, plus y, equals zero. So it is a quadratic in x, right? Wait, wait. Before you jump to the discriminant: what is the coefficient of x squared here? It is y itself! And y can be zero. Trap one, remember? So, first the case y equals zero. The equation becomes minus x equals zero, so x equals zero. That works, so zero is in the range. Now y not zero, so it is a genuine quadratic. Here B is minus one, so B squared is one; and four A C is four y squared. So we need one minus four y squared, greater than or equal to zero. That means y squared is at most one fourth. So y lies between minus one half and one half, both included. With me? So the range is minus one half to one half, closed. The extremes come at x equals one and x equals minus one, the highest and lowest dots on the graph. So, our puzzle: can this function ever give one? Never! Not even point six. The curve never goes above the dashed line at one half. And a lower line, at height point three, meets it twice: at x equals one third, and at x equals three.

7:59Quick revision: three methods and the two traps

So, quick revision. One: build up from ranges you know, or solve for x, and check that x lies in the domain. Two: if you get a quadratic in x, use the discriminant, then test the y that kills the x squared term, and any cancelled point. Three: for a x plus b over c x plus d, the range is all real numbers except a over c. Now go and try the practice problems on this page. Range questions will start feeling easy, I promise. All the best!