Chapter Summary: Mixed Solved Problems

Transcript

0:00Introduction

Hello everyone! First, one challenge before we start. Take x squared, minus two times k minus one, times x, plus k plus five, equals zero. Here k is an integer from minus ten to ten, so twenty one values. For how many of them does this equation have at least one positive root? Will you solve twenty one equations, one by one? Oh no, that would eat up your whole exam! Today, our chapter summary: three tools, and two solved problems. And look, with the right tool, this JEE question takes just a few lines. So, let's begin.

0:40Tool 1: read the conditions off the parabola

So, tool number one, from location of roots. Let f of x be a x squared, plus b x, plus c, and capital D, b squared minus four a c, is the discriminant. On the right, three standard pictures: both roots greater than k, k between the roots, and both roots in an interval. The red dashed lines are x equals k, or x equals k one and k two. The method in one line: draw the parabola, and read off the conditions. Okay? Look at the middle picture. The parabola opens upward, so a is positive, and the dashed line x equals k sits between the two roots. The picture writes one condition: a f of k less than zero. Why does that put k between the roots? Yes, exactly! Here a is positive, so f of k is negative: at x equals k, the curve is below the x axis. And an upward parabola goes below the axis only between its roots. In general, a and f of k have opposite signs. And this one condition also guarantees real, distinct roots. Very powerful. Now the third picture: both roots inside k one, k two. The little one and two are subscripts; round brackets mean the end points are not included. First, real roots: D greater than or equal to zero; that sign with a line below means greater than or equal to. Then, a f of k one and a f of k two, both greater than zero. And the vertex must sit between the dashed lines: k one, less than minus b over two a, less than k two. Minus b over two a is the x coordinate of the vertex. So, four conditions in all. Keep that in mind, okay?

2:29Tool 2: signs of the roots

Tool number two: the signs of the roots. Call the roots alpha and beta, the two Greek letters on screen. You know this already: the sum, alpha plus beta, is minus b over a, and the product, alpha beta, is c over a. Right? Now suppose both roots are positive. What can you say about their sum, and their product? Both positive, exactly! Sum greater than zero, and product greater than zero. But first, the roots must be real, so D greater than or equal to zero as well. Three conditions, okay? And roots of opposite signs? Just one condition: the product is negative, that is, c over a less than zero. No separate D check for this one. Nice and short.

3:17Tool 3, topper's trick: count the opposite

Now tool number three, and this is the topper's trick. Suppose a question says: at least one positive root. That can happen in many ways, right? So toppers don't count it directly. They count the opposite. So, with real roots, what is the opposite? Most students say: no positive root means both roots are negative. Sounds right, doesn't it? No, no, careful! What about a root equal to zero? Zero is not positive. So the true opposite is: both roots less than or equal to zero. This sign, with a line below, means less than or equal to. Remember this; it decides the answer in example two. In terms of the sum and the product: sum less than or equal to zero, and product greater than or equal to zero. Got it? So the count we want is: the real-root cases, minus the cases with both roots at most zero. Two easy counts, instead of one messy one. That's it!

4:16Example 1: both roots of x^2 - 2ax + a^2 - 1 in (0, 5)

Now, Example one. Find all a for which both roots of x squared, minus two a x, plus a squared, minus one, equals zero, lie in the open interval from zero to five. Pause the video, and try it first. Done? Okay, let's check. Tool one says four conditions, right? That works, but it is long. First, look at x squared, minus two a x, plus a squared. What does it remind you of? Yes, x minus a, the whole squared! So the equation is the same as: x minus a, the whole squared, equals one. The double headed arrow means the two forms are equivalent. Clear? So x minus a is plus or minus one; this sign means plus or minus. The roots are a minus one, and a plus one, and a minus one is the smaller. Clean roots, no formula needed! Now, both roots between zero and five. Which two conditions are enough? Just two, that's right! The smaller root above zero: a minus one greater than zero. The larger root below five: a plus one less than five. Then both roots sit inside automatically. This double arrow, pointing one way, means implies: a greater than one, and a less than four. So, one less than a, less than four. Look, when the roots come out clean, use them directly; keep tool one for when they don't. Got it?

5:46Example 2: integers k giving at least one positive root

Now a harder one, JEE pattern: our challenge from the start. k is an integer in minus ten to ten; this little symbol means belongs to, and square brackets include both ends. For how many such k does x squared, minus two times k minus one, times x, plus k plus five, equals zero, have at least one positive root? Pause the video and try it. Hint: tool three. Done? Okay, let's check. Step one, real roots. Here b is minus two times k minus one, so we use D by four, that is, D divided by four; its sign is the sign of D, okay? It is k minus one, the whole squared, minus k plus five. Simplify and factorise: k minus four, times k plus one, greater than or equal to zero. The product is at least zero, so the double arrow, implies: k less than or equal to minus one, or k greater than or equal to four. With me? Now tool three: count the opposite, both roots less than or equal to zero. Find the sum and the product, and write the two conditions. Yes! The sum, minus b over a, is two times k minus one: at most zero, so k is at most one. The product, c over a, is k plus five: at least zero, so k is at least minus five. Together, minus five less than or equal to k, less than or equal to one. And look, at k equals minus five the product is zero: one root is zero, not positive. Our trap from tool three! But this must hold together with real roots. Common part with step one: k from minus five to minus one. For these values of k, there is no positive root. So, remove this block from the real-root cases. The double headed arrow says: positive root if and only if k is less than minus five, or at least four. Last step. How many integers from minus ten to ten satisfy this? k equals minus ten, and so on up to minus six; the three dots mean and so on. That is five values. Then four up to ten: seven values. Total, twelve! Our challenge is solved, without solving twenty one equations. Smart, isn't it?

8:18Quick revision: D, af(k), vertex and signs of roots

So, quick revision. Location of roots: if the roots come out clean, use them directly; otherwise D, a f of k, and the vertex. Signs of the roots: D greater than or equal to zero, then the sum and the product. And for at least one positive root: count the opposite, both roots less than or equal to zero, because zero is not positive. That is our chapter summary! Now go and try the practice problems on this page. All the best!