JEE Advanced Questions: Inequalities and Functional Equations

Transcript

0:00Introduction

Hello everyone! Here is a puzzle. You know only two things about a function f: its derivative is always more than two times the function, and its value at zero is one. With no formula for f, can you prove that f of x is bigger than e to the power two x, for every positive x? Sounds impossible, right? One clever multiplication will do it, in a JEE Advanced question. Let's begin.

0:28Recap: the integrating factor

First, a quick recap from linear equations. The linear form is d y by d x, plus P of x times y, equals Q of x. Okay? I F on screen means the integrating factor: e to the power integral P d x, with no constant inside. It turns the left side into one derivative: d by d x of y times the I F, equal to Q times the I F, right? Then we integrate both sides. The tall S is the integral sign: y times the I F equals the integral of Q times the I F, d x, plus capital C, okay? Now a quick one for you: if P is one over x, what is the integrating factor, for positive x? The integral of one over x is natural log of x, and e to the power natural log of x is just x. On screen is the general rule, and the double arrow is read as gives: P equals n over x gives an integrating factor of x to the power n. Got it? And one trap: add capital C first, then divide by the I F, so the term with capital C in y is capital C over the I F, not capital C alone. With me?

1:47The same trick for an inequality

Now, today's key idea. Suppose f prime minus k times f is greater than zero, with k a constant; the dash on f is a prime, the derivative. This is the linear form with P equal to minus k, so the integrating factor is e to the power minus k x. Okay? So let me ask you this: what is the derivative of e to the power minus k x, times f? By the product rule, it is e to the power minus k x times f prime, minus k times e to the power minus k x times f. Take e to the power minus k x common, and the bracket is exactly f prime minus k f. Beautiful, right? And e to the power anything is positive, so this derivative is greater than zero. Multiplying by a positive number keeps the inequality the same way. Clear? So, e to the power minus k x times f is increasing, and we never even found f. Got it?

2:48Example 1: f' greater than 2f with f(0) = 1, JEE Adv 2017

Now, Example one, our puzzle, from JEE Advanced twenty seventeen. The double-struck R means the real numbers, and the arrow says f maps real numbers to real numbers. The bracket zero to infinity, with the infinity sign, means all positive x. Options A and B say f is increasing or decreasing there. Option C says f of x is greater than e to the power two x, and option D says f prime of x is less than e to the power two x. Pause the video and try it first. Done? Okay, let's check. Many students say: f prime is greater than two f, so f prime is positive, so f is increasing, right? Oh no, careful here, because nobody told us that f is positive. So first, the trick with k equal to two: e to the power minus two x times f has derivative e to the power minus two x, times f prime minus two f, which is positive. So it is increasing. Okay? So for x greater than zero, it beats its value at zero, which is e to the power zero times f of zero, and that equals one. Multiply by e to the power two x: f of x is greater than e to the power two x, so option C is true. Right? Now back to the trap. f is bigger than e to the power two x, which is positive. So now, is f prime positive? Yes, exactly! Now f is positive, so f prime is greater than two f, which is greater than zero. So f is increasing: option A is true, and option B is false. Same answer, now with a real reason. Clear? For option D, f prime is greater than two f, which is greater than two times e to the power two x, so it beats e to the power two x. So option D is false, okay? So the answer is options A and C, together. Puzzle solved, and we never found f at all. Nice, isn't it?

4:57Example 2: f'(x) = 2 - f(x)/x so f = x + C/x, JEE Adv 2018

Now, Example two, from JEE Advanced twenty eighteen. Here f prime of x equals two, minus f of x over x, and f of one is not equal to one, shown by the equals sign with a slash. In the options, l i m means the limit, the arrow means tends to, and the little plus means from the positive side. Option A says the limit of f prime of one over x is one, and option B says the limit of x times f of one over x is two. Option C says the limit of x squared times f prime of x is zero, and option D says the modulus of f, the two vertical bars, is less than or equal to two on the interval zero to two. Pause the video and try it first. Done? Okay, let's check. Move f over x to the left. Which form do you see? Yes, exactly, the linear form! f prime, plus one over x times f, equals two. P is one over x, so the integrating factor is x, and the left side becomes the derivative of x times f, equal to two x. Right? Integrating both sides, x times f equals x squared, plus capital C, okay? Divide by x, remembering the trap: f equals x, plus capital C over x. Since f of one, which is one plus capital C, is not one, capital C is not zero. With me? For option A, differentiate f, then put one over x in place of x. What do you get? f prime of x is one, minus capital C over x squared. With one over x in place of x, it becomes one, minus capital C times x squared, which tends to one. So option A is true! For option B, f of one over x is one over x, plus capital C times x. Times x, that is one, plus capital C times x squared, which tends to one, not two, so option B is false, okay? Now for option C, x squared goes to zero, so x squared times f prime of x goes to zero too, right? No, no, wait, that is the trap, because f prime has capital C over x squared inside, which blows up. Multiplying out gives x squared, minus capital C, which tends to minus capital C, not zero. So option C is false, and that is why we needed capital C to be non zero. And finally, option D: near zero, capital C over x becomes as large as you like in size, so the modulus of f cannot stay below two, and option D is false. So the answer is option A alone. That's it!

7:49Quick revision: I.F., inequality trick, prove f positive

So, quick revision. For a linear equation, the integrating factor is e to the power integral P d x; add capital C before dividing by it. If f prime minus k f is greater than zero, e to the power minus k x times f is increasing. And prove f is positive first; only then does f prime greater than two f make f increasing. Now try the practice problem on this page. All the best!