Domain of a function

Q. 25 of IIT 2000 (Screening): What is the domain of the function y(x) which is given by the equation 2^x + 2^y = 2.

Sol: Given 2^x + 2^y = 2.

ð     2^y = 2 - 2^x  è 2^y = 2(1 - 2^x-1)  è y log 2 = log 2 + log (1 - 2^x-1)

ð     y = 1+ log (1 - 2^x-1)/(log 2)

ð     In order that the function is defined, the value of log (1 - 2^x-1) must exist.

ð     1 - 2^x-1 must be greater than zero.

ð     2^x-1 < 1. Taking log both the sides: (x-1) log 2 < 0. But the value of log 2 is a non-zero number.

ð     Hence, dividng both sides by log 2, we have (x-1) < 0

ð     x<1. So the domain is (-∞ < x < 1).

Reality shows by imaginary roots of a quadratic equation

1. A simple question appeared in IIT 2000 (Screening). For the equation 3x² + px + 3 = 0, p>0. If one of the roots is square of the other, then p is equal to what?

This question looks a bit complex, it is not. It is simply a matter of refinement of your grey cells (of brain) when you solve it.

You must know the relation between imaginary roots of the quadratic equation x² + x + 1 = 0. The roots of this equation are 1, ω, ω², where ω is: {-1+(√3 i)}/2.

Hence if we compare the given equation with x² + x + 1 = 0, we must have p = 3, in order to have the two roots out of a total of three, one being the square of the other (i.e. ω, ω²).

Hence p = 3.


2. If one root is square of the other root of the equation x² + px + q = 0, the relation between p and q is:

(a) p³ – (3p – 1)q + q² = 0       (b) p³ – (3p + 1)q + q² = 0

(c) p³ + (3p – 1)q + q² = 0       (d) p³ + (3p + 1)q + q² = 0

The question written above appeared in IIT (Screening) 2004.

This question looks tough, but banking upon the choices given in the question, it is very easy. Though the procedure I am going to employ, may not be feasible for all such type of questions. But we are more concerned (at IIT entrance exam) for opting for the most obvious choice in the shortest period and save our time for those questions which need a little more brain storming.

Here is the clue.

We know the relation between imaginary roots of the quadratic equation x² + x + 1 = 0. The roots of this equation are 1, ω, ω², where ω is: {-1+(√3 i)}/2.

Given the condition that one root is square of the other root of the equation, comparing the given equation x² + px + q = 0 with x² + x + 1 = 0, we must have p = 1 and q = 1, in order to have the two roots out of a total of three, one being the square of the other (i.e. ω, ω²). Now putting these values of p and q in the given four options, we find that option (a) satisfies these values.

Please note that if more than one option satisfies for the values of p and q, this method will not work, and you will have to solve it fully.

In the present scenario, option (a) is the correct answer.

3. This question regarding the application of cube root of unity appeared in IIT 2004 (Screening).

If ω (≠ 1) be a cube root of unity and (1 +  ω²)ⁿ  = (1 +  ω⁴)ⁿ , then the least positive value of n is:

(a) 2       (b) 3      (c) 5       (d) 6

We know the relation between imaginary roots of the quadratic equation x² + x + 1 = 0. The roots of this equation are 1, ω, ω², where ω is: {-1+(√3 i)}/2. And sum of these roots is 0.

(1 +  ω²)ⁿ  = (1 +  ω⁴)ⁿ  è (1 +  ω² + ω - ω)ⁿ  = {(1 +  ω²)² - 2ω²)}ⁿ 

è (1 +  ω² + ω - ω)ⁿ  = {(1 +  ω²+ ω - ω)² - 2ω²)}ⁿ 

è (0 - ω)ⁿ  = {(0 - ω)² - 2ω²)}ⁿ è (-ω)ⁿ  = {-ω²}ⁿ 

è Obviously ωⁿ  = ω²ⁿ  è Obviously ωⁿ  = 1

But  ω is the cube root of unity and given to be not equal to 1.

Hence n = 3.

Knowledge of expansion of many functions expands your horizon of understanding calculus

I came across a question (IIT - 2002), which read as:

What is the integer n for which limit x tends to 0 for [{cos x - 1}{cos x – e^x}/xⁿ] is a finite non-zero number? The options were: 1 or 2 or 3 or 4.
Without expansion, it is rather difficult to solve it. Even application of L'Hospital's Rule will not be a providential help. But if we expand the exponential and trigonometric functions, the question is easily solvable.