Mixed Problems: Set 2 (Equations)

Transcript

0:00Introduction

Hello everyone! Let us start with a scary-looking equation. Sine of e to the power x equals five to the power x, plus five to the power minus x. How many real solutions does it have? Most students start taking logs, and get lost. Today we will settle it in one line. This is Set two of our mixed problems, on equations, where JEE Main often asks you to count the solutions, or add them. So, let's begin!

0:30Four methods to try first

So, in Set two, every question hands you an equation to solve, or asks you to count or add its solutions. Now tell me, before you write a single line, what should you decide first? The method, yes, exactly! There are four to try first. The first is to factorise, using sum-to-product or multiple-angle formulas, then set each factor equal to zero. Okay? Second, one ratio and its square, like cos squared x and cos x, make a quadratic. Solve it, and keep only the roots in the range of that ratio; cos x can never be two, right? Third, when sine x plus cos x comes with sine two x, put t equal to sine x plus or minus cos x; that is the plus-or-minus sign. Then sine x times cos x equals plus or minus, t squared minus one, over two; the small two up top means squared. And mod t, in vertical bars, is at most root two; the tick is the square root sign, and the sign with a line under it means less than or equal to. With me? And fourth, bounds. If one side is at most some number M, and the other is at least that same M, then both must equal M, exactly. The sign with a line under it, opening the other way, means greater than or equal to. Got it?

1:56Bounds settle our puzzle

Now, back to our puzzle: sine of e to the power x equals five to the power x plus five to the power minus x; the small letters up top are powers. Logs will not help. We use bounds. Okay? Look at the left side first. Whatever e to the power x is, the sine of it is at most one. Now the right side, five to the power x plus five to the power minus x. How small can it ever get? Yes, it can never go below two! Put t equal to five to the power x, always positive; the sign pointing right means greater than. Then the right side is t plus one over t, which is at least two for positive t. Clear? So the left side is at most one, and the right side is at least two; they can never meet. Zero real solutions, and we solved nothing at all! Right?

2:50Counting solutions in an interval

Now, counting solutions in an interval. Solve for the simplest ratio, then list the angles. In the left graph, the solid curve is sine x, and the dashed one is cos x. Sine x equals k, with mod k less than one, has exactly two solutions in each interval of length two pi, from a to a plus two pi, with one end point left out; pi is the Greek letter, and the sign pointing left means less than. Okay? On the right is tan x. The dashed vertical lines mark where tan x is undefined. It repeats every pi, and takes each value once in every period. So tan x equals k has one solution in each interval of length pi, again with one end point left out. Right? Next, the brackets. Square brackets mean a closed interval: the end points are included, so test them. Round brackets leave them out. For example, sine x equals zero has three solutions in the closed interval from zero to two pi: zero, pi and two pi. Clear? Now two warnings. Never cancel a factor that can be zero: divide by cos x, and every root where cos x is zero is lost. With me? And reject every value where a term of the original equation is undefined, like tan x at pi over two. We will need both warnings soon. Got it?

4:15Example 1: sum of roots of 2 cos squared x + cos x - 1 = 0

Now, Example one, a warm-up. Find the sum of the solutions of two cos squared x plus cos x minus one equals zero, in the closed interval from zero to two pi. Pause the video and try it first. Done? Okay, let's check. One ratio and its square, so it is a quadratic in cos x, our second method. It factorises as two cos x minus one, times cos x plus one, equals zero. Right? So cos x equals one half, or cos x equals minus one. Both lie between minus one and one, so we keep both. Okay? Now, cos x equals one half. Everyone knows pi over three. So that is the only solution here, right? Oh no, careful, this is where marks are lost! Cos x is positive in the fourth quadrant too, so we also get two pi minus pi over three, which is five pi over three. Two solutions, just as our counting rule said. Clear? Next, cos x equals minus one gives x equals pi. And the end points? Cos zero and cos two pi are both one, so they are not roots. Okay? So the sum is pi over three, plus five pi over three, plus pi. The first two make two pi, so the sum is three pi. Got it?

5:43Example 2: sin x + cos x - sin 2x = 1 by t = sin x + cos x

Now a harder one, Example two. Solve sine x plus cos x minus sine two x equals one. Hint: which of our four methods fits? Pause the video and try it first. Done? Okay, let's check. Sine x plus cos x with sine two x: the third method. Put t equal to sine x plus cos x. Squaring, and using sine squared plus cos squared equals one, t squared equals one plus sine two x. So sine two x equals t squared minus one. Okay? Now substitute: t minus, in brackets, t squared minus one, equals one. The ones cancel, and we get t minus t squared equals zero; the double arrow means implies. Right? Now, many students divide both sides by t, and write t equals one. Is that allowed? No, no, wait, that is exactly the trap! Since t can be zero, we never cancel it. We factorise: t times one minus t equals zero, so t is zero or one. Both lie within mod t at most root two, so both are allowed. Clear? Take t equals zero first: sine x plus cos x is zero. Here cos x is not zero, or sine x would be plus or minus one, and the sum could not be zero. So divide by cos x: tan x equals minus one, and x equals n pi minus pi over four. Here n belongs to the double-lined capital Z, the integers; the curved E sign means belongs to. With me? Now t equals one. Sine x plus cos x is root two times cos of x minus pi over four, so that equals one. Then cos of x minus pi over four is one over root two, which is cos of pi over four. So x minus pi over four equals two n pi plus or minus pi over four. Okay? Add pi over four: the plus sign gives two n pi plus pi over two, and the minus sign gives two n pi. So x equals n pi minus pi over four, two n pi, or two n pi plus pi over two. Cancel t, and you lose the whole first family! Got it?

8:06Quick revision: four methods, never cancel, check end points

So, quick revision. First, pick the method: factorise, a quadratic, t equals sine x plus or minus cos x, or bounds. Second, never cancel a factor that can be zero, and reject undefined values. Third, when you count, check the end points and every quadrant. Now go and try the practice problems on this page. All the best!